{"id":1181,"date":"2021-02-15T21:47:08","date_gmt":"2021-02-15T21:47:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/?post_type=chapter&p=1181"},"modified":"2021-02-25T21:00:47","modified_gmt":"2021-02-25T21:00:47","slug":"exercises-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/chapter\/exercises-2\/","title":{"raw":"Exercises","rendered":"Exercises"},"content":{"raw":"Part I<\/strong><\/span>\r\n
    \r\n \t
  1. What is the total mass (amu) of carbon in each of the following molecules?\r\n
      \r\n \t
    1. CH4 \u00a0<\/sub><\/li>\r\n \t
    2. CHCl3 \u00a0<\/sub><\/li>\r\n \t
    3. C12<\/sub>H10<\/sub>O6 \u00a0<\/sub><\/li>\r\n \t
    4. CH3<\/sub>CH2<\/sub>CH2<\/sub>CH2<\/sub>CH3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
    5. What is the total mass of hydrogen in each of the molecules?\r\n
        \r\n \t
      1. CH4 \u00a0<\/sub><\/li>\r\n \t
      2. CHCl3 \u00a0<\/sub><\/li>\r\n \t
      3. C12<\/sub>H10<\/sub>O6 \u00a0\u00a0<\/sub><\/li>\r\n \t
      4. CH3<\/sub>CH2<\/sub>CH2<\/sub>CH2<\/sub>CH3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
      5. Calculate the molecular or formula mass of each of the following:\r\n
          \r\n \t
        1. P4 \u00a0\u00a0<\/sub><\/li>\r\n \t
        2. H2<\/sub>O<\/li>\r\n \t
        3. Ca(NO3<\/sub>)2 \u00a0\u00a0<\/sub><\/li>\r\n \t
        4. CH3<\/sub>CO2<\/sub>H (acetic acid)<\/li>\r\n \t
        5. C12<\/sub>H22<\/sub>O11<\/sub> (sucrose, cane sugar).<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        6. Determine the molecular mass of the following compounds:\r\n
            \r\n \t
          1. \"A<\/li>\r\n \t
          2. \"A<\/li>\r\n \t
          3. \"A<\/li>\r\n \t
          4. \"A<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
          5. Determine the molecular mass of the following compounds:\r\n
              \r\n \t
            1. \"A<\/li>\r\n \t
            2. \"A<\/li>\r\n \t
            3. \"A<\/li>\r\n \t
            4. \"A<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
            5. Which molecule has a molecular mass of 28.05 amu?\r\n
                \r\n \t
              1. \"A<\/li>\r\n \t
              2. \"A<\/li>\r\n \t
              3. \"A<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
              4. Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.<\/li>\r\n \t
              5. Compare 1 mole of H2<\/sub>, 1 mole of O2<\/sub>, and 1 mole of F2<\/sub>.\r\n
                  \r\n \t
                1. Which has the largest number of molecules? Explain why.<\/li>\r\n \t
                2. Which has the greatest mass? Explain why.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                3. Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2<\/sub>H5<\/sub>OH), 0.60 mol of formic acid (HCO2<\/sub>H), or 1.0 mol of water (H2<\/sub>O)? Explain why.<\/li>\r\n \t
                4. Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C2<\/sub>H5<\/sub>OH), 1 mol of formic acid (HCO2<\/sub>H), or 1 mol of water (H2<\/sub>O)? Explain why.<\/li>\r\n \t
                5. How are the molecular mass and the molar mass of a compound similar and how are they different?<\/li>\r\n \t
                6. Calculate the molar mass of each of the following compounds:\r\n
                    \r\n \t
                  1. hydrogen fluoride, HF<\/li>\r\n \t
                  2. ammonia, NH3 \u00a0<\/sub><\/li>\r\n \t
                  3. nitric acid, HNO3 \u00a0<\/sub><\/li>\r\n \t
                  4. silver sulfate, Ag2<\/sub>SO4 \u00a0<\/sub><\/li>\r\n \t
                  5. boric acid, B(OH)3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                  6. Calculate the molar mass of each of the following:\r\n
                      \r\n \t
                    1. S8 \u00a0\u00a0<\/sub><\/li>\r\n \t
                    2. C5<\/sub>H12 \u00a0\u00a0<\/sub><\/li>\r\n \t
                    3. Sc2<\/sub>(SO4<\/sub>)3 \u00a0\u00a0<\/sub><\/li>\r\n \t
                    4. CH3<\/sub>COCH3<\/sub> (acetone)<\/li>\r\n \t
                    5. C6<\/sub>H12<\/sub>O6<\/sub> (glucose)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                    6. Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:\r\n
                        \r\n \t
                      1. limestone, CaCO3 \u00a0\u00a0<\/sub><\/li>\r\n \t
                      2. halite, NaCl<\/li>\r\n \t
                      3. beryl, Be3<\/sub>Al2<\/sub>Si6<\/sub>O18 \u00a0\u00a0<\/sub><\/li>\r\n \t
                      4. malachite, Cu2<\/sub>(OH)2<\/sub>CO3 \u00a0\u00a0<\/sub><\/li>\r\n \t
                      5. turquoise, CuAl6<\/sub>(PO4<\/sub>)4<\/sub>(OH)8<\/sub>(H2<\/sub>O)4<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                      6. Calculate the molar mass of each of the following:\r\n
                          \r\n \t
                        1. the anesthetic halothane, C2<\/sub>HBrClF3 \u00a0<\/sub><\/li>\r\n \t
                        2. the herbicide paraquat, C12<\/sub>H14<\/sub>N2<\/sub>Cl2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                        3. caffeine, C8<\/sub>H10<\/sub>N4<\/sub>O2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                        4. urea, CO(NH2<\/sub>)2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                        5. a typical soap, C17<\/sub>H35<\/sub>CO2<\/sub>Na<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                        6. Determine the number of moles of compound and the number of moles of each type of atom in each of the following:\r\n
                            \r\n \t
                          1. 25.0 g of propylene, C3<\/sub>H6 \u00a0<\/sub><\/li>\r\n \t
                          2. [latex]3.06\\times {10}^{-3}\\text{g}[\/latex] of the amino acid glycine, C2<\/sub>H5<\/sub>NO2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                          3. 25 lb of the herbicide Treflan, C13<\/sub>H16<\/sub>N2<\/sub>O4<\/sub>F (1 lb = 454 g)<\/li>\r\n \t
                          4. 0.125 kg of the insecticide Paris Green, Cu4<\/sub>(AsO3<\/sub>)2<\/sub>(CH3<\/sub>CO2<\/sub>)2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                          5. 325 mg of aspirin, C6<\/sub>H4<\/sub>(CO2<\/sub>H)(CO2<\/sub>CH3<\/sub>)<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                          6. Determine the mass of each of the following:\r\n
                              \r\n \t
                            1. 0.0146 mol KOH<\/li>\r\n \t
                            2. 10.2 mol ethane, C2<\/sub>H6 \u00a0\u00a0<\/sub><\/li>\r\n \t
                            3. [latex]1.6\\times {10}^{-3}\\text{ mol }{\\text{Na}}_{2}{\\text{SO}}_{4}[\/latex]<\/li>\r\n \t
                            4. [latex]6.854\\times {10}^{3}\\text{ mol glucose},{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}[\/latex]<\/li>\r\n \t
                            5. 2.86 mol Co(NH3<\/sub>)6<\/sub>Cl3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                            6. Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:\r\n
                                \r\n \t
                              1. 2.12 g of potassium bromide, KBr<\/li>\r\n \t
                              2. 0.1488 g of phosphoric acid, H3<\/sub>PO4 \u00a0<\/sub><\/li>\r\n \t
                              3. 23 kg of calcium carbonate, CaCO3 \u00a0<\/sub><\/li>\r\n \t
                              4. 78.452 g of aluminum sulfate, Al2<\/sub>(SO4<\/sub>)3 \u00a0<\/sub><\/li>\r\n \t
                              5. 0.1250 mg of caffeine, C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                              6. Determine the mass of each of the following:\r\n
                                  \r\n \t
                                1. 2.345 mol LiCl<\/li>\r\n \t
                                2. 0.0872 mol acetylene, C2<\/sub>H2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                                3. [latex]3.3\\times {10}^{-2}\\text{ mol }{\\text{Na}}_{2}{\\text{CO}}_{3}[\/latex]<\/li>\r\n \t
                                4. [latex]1.23\\times {10}^{3}\\text{ mol fructose, }{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}[\/latex]<\/li>\r\n \t
                                5. 0.5758 mol FeSO4<\/sub>(H2<\/sub>O)7<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                6. The approximate minimum daily dietary requirement of the amino acid leucine, C6<\/sub>H13<\/sub>NO2<\/sub>, is 1.1 g. What is this requirement in moles?<\/li>\r\n \t
                                7. Determine the mass in grams of each of the following:\r\n
                                    \r\n \t
                                  1. 0.600 mol of oxygen atoms<\/li>\r\n \t
                                  2. 0.600 mol of oxygen molecules, O2 \u00a0\u00a0<\/sub><\/li>\r\n \t
                                  3. 0.600 mol of ozone molecules, O3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                  4. A 55-kg woman has [latex]7.5\\times {10}^{-3}\\text{mol}[\/latex] of hemoglobin (molar mass = 64,456 g\/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?<\/li>\r\n \t
                                  5. Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO4<\/sub>, a semiprecious stone.<\/li>\r\n \t
                                  6. Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH4<\/sub>, 0.6 mol of C6<\/sub>H6<\/sub>, or 0.4 mol of C3<\/sub>H8<\/sub>.<\/li>\r\n \t
                                  7. Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO4<\/sub>, 266 g of A12<\/sub>C16<\/sub>, or 225 g of A12<\/sub>S3<\/sub>.<\/li>\r\n \t
                                  8. Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?<\/li>\r\n \t
                                  9. The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone<\/li>\r\n \t
                                  10. One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?<\/li>\r\n \t
                                  11. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12<\/sub>H22<\/sub>O11<\/sub>) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?<\/li>\r\n \t
                                  12. A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na2<\/sub>PO3<\/sub>F) in 100 mL\r\n
                                      \r\n \t
                                    1. What mass of fluorine atoms in mg was present?<\/li>\r\n \t
                                    2. How many fluorine atoms were present?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                    3. Which of the following represents the least number of molecules?\r\n
                                        \r\n \t
                                      1. 20.0 g of H2<\/sub>O (18.02 g\/mol)<\/li>\r\n \t
                                      2. 77.0 g of CH4<\/sub> (16.06 g\/mol)<\/li>\r\n \t
                                      3. 68.0 g of CaH2<\/sub> (42.09 g\/mol)<\/li>\r\n \t
                                      4. 100.0 g of N2<\/sub>O (44.02 g\/mol)<\/li>\r\n \t
                                      5. 84.0 g of HF (20.01 g\/mol)<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"649796\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"649796\"]\r\n\r\n1. Each\u00a0molecule has the following mass (amu)\u00a0of carbon.\r\n
                                          \r\n \t
                                        1. [latex]1\\times 12.01\\text{ amu}=12.01\\text{ amu}[\/latex]<\/li>\r\n \t
                                        2. [latex]1\\times 12.01\\text{ amu}=12.01\\text{ amu}[\/latex]<\/li>\r\n \t
                                        3. [latex]12\\times 12.01\\text{ amu}=144.12\\text{ amu}[\/latex]<\/li>\r\n \t
                                        4. [latex]5\\times 12.01\\text{ amu}=60.05\\text{ amu}[\/latex]<\/li>\r\n<\/ol>\r\n3. The molecular or formula masses are as follows:\r\n
                                            \r\n \t
                                          1. [latex]4\\times 30.974\\text{ amu}=123.896\\text{ amu}[\/latex]<\/li>\r\n \t
                                          2. [latex]2\\times 1.008\\text{ amu}+15.999\\text{ amu}=18.015\\text{ amu}[\/latex]<\/li>\r\n \t
                                          3. [latex]40\\times 0.078\\text{ amu}+2\\times 14.007\\text{ amu}+6\\times 15.999\\text{ amu}=164.086\\text{ amu}[\/latex]<\/li>\r\n \t
                                          4. [latex]2\\times 12.011\\text{ amu}+4\\times 1.008\\text{ amu}+2\\times 15.999\\text{ amu}=60.052\\text{ amu}[\/latex]<\/li>\r\n \t
                                          5. [latex]12\\times 12.011\\text{ amu}+22\\times 1.008\\text{ amu}\\times 11\\times 15.999\\text{ amu}=342.297\\text{ amu}[\/latex]<\/li>\r\n<\/ol>\r\n5. \u00a0The molecular mass of each compound is as follows:\r\n
                                              \r\n \t
                                            1. C4<\/sub>H8\r\n<\/sub>[latex]\\begin{array}{ll}4\\text{C}\\times 12.011\\hfill & =48.044\\text{amu}\\hfill \\\\ 8\\text{H}\\times 1.0079\\hfill & =\\underline{8.06352\\text{amu}}\\hfill \\\\ \\hfill & =56.107\\text{amu}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                            2. C4<\/sub>H6\r\n<\/sub>[latex]\\begin{array}{ll}4\\text{C}\\times 12.011\\hfill & =48.044\\text{amu}\\hfill \\\\ 6\\text{H}\\times 1.0079\\hfill & =\\underline{6.0474\\text{amu}}\\hfill \\\\ \\hfill & =54.091\\text{amu}\\hfill \\end{array}[\/latex]<\/span><\/li>\r\n \t
                                            3. H2<\/sub>Si2<\/sub>Cl4\r\n<\/sub>[latex]\\begin{array}{ll}\\hfill 2\\text{H}\\times 1.0079& =2.01558\\text{amu}\\hfill \\\\ 2\\text{Si}\\times 28.0855\\hfill & =56.1710\\text{amu}\\hfill \\\\ 4\\text{Cl}\\times 35.4527\\hfill & =\\underline{141.8108\\text{amu}}\\hfill \\\\ \\hfill & =199.9976\\text{amu}\\hfill \\end{array}[\/latex]<\/span><\/li>\r\n \t
                                            4. H3<\/sub>PO4\r\n<\/sub>[latex]\\begin{array}{ll}\\hfill 3\\text{H}\\times 1.0079& =3.0237\\text{amu}\\hfill \\\\ 1\\text{P}\\times 30.973762\\hfill & =30.973762\\text{amu}\\hfill \\\\ \\hfill 4\\text{O}\\times 15.9994& =\\underline{63.9976\\text{amu}}\\hfill \\\\ \\hfill & =97.9950\\text{amu}\\hfill \\end{array}[\/latex]<\/span><\/li>\r\n<\/ol>\r\n7.\u00a0Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.\r\n\r\n9.\u00a0Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.\r\n\r\n11.The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of [latex]6.022\\times {10}^{23}[\/latex] molecules.\r\n\r\n13. \u00a0The molecular mass of each compound is as follows:\r\n
                                                \r\n \t
                                              1. S8<\/sub>:\r\n[latex]8\\text{S}=8\\times 32.066=256.528\\text{g\/mol}[\/latex]<\/li>\r\n \t
                                              2. C5<\/sub>H12<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill 5\\text{C}=5\\times 12.011& =& 60.055\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{12H}=12\\times 1.00794& =& \\underline{12.09528\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 72.150\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                              3. Sc2<\/sub>(SO4<\/sub>)3<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill 2\\text{Sc}=2\\times 44.9559109& =& 89.9118218\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{3S}=3\\times 32.066& =& 96.198\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 12\\text{O}=12\\times 15.99943& =& \\underline{191.99316\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 378.103\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                              4. CH3<\/sub>COCH3<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill 3\\text{C}=3\\times 12.011& =& 36.033\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1O}=1\\times 15.9994& =& 15.9994\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 6\\text{H}=6\\times 1.00794& =& \\underline{6.04764\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 58.080\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                              5. C6<\/sub>H12<\/sub>O6<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{6C}=6\\times 12.011& =& 72.066\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{12H}=12\\times 1.00794& =& 12.09528\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 6\\text{O}=6\\times 15.9994& =& \\underline{95.9964\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 180.158\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n15. The molecular mass of each compound is as follows:\r\n
                                                  \r\n \t
                                                1. C2<\/sub>HBrClF3<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{2C}=2\\times 12.011& =& 24.022\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1H}=1\\times 1.00794& =& 1.00794\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1Br}=1\\times 79.904& =& 79.904\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1Cl}=1\\times 35.453& =& 35.453\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 3\\text{F}=3\\times 18.998403& =& \\underline{56.995209\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 197.682\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                2. C12<\/sub>H14<\/sub>N2<\/sub>Cl2<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{12C}=12\\times 12.011& =& 144.132\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{14H}=14\\times 1.00794& =& 14.111\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2N}=2\\times 14.0067& =& 28.0134\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2Cl}=2\\times 35.453& =& \\underline{70.906\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 257.163\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                3. C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{8C}=8\\times 12.011& =& 96.088\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{10H}=10\\times 1.007& =& 10.079\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{4N}=4\\times 14.0067& =& 56.027\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2O}=2\\times 15.9994& =& \\underline{31.999\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 194.193\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                4. CO(NH2<\/sub>)2<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{1C}=1\\times 12.011& =& 12.011\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1O}=1\\times 15.9994& =& 15.9994\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2N}=2\\times 14.0067& =& 28.0134\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{4H}=4\\times 1.00794& =& \\underline{4.03176\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 60.056\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                5. C17<\/sub>H35<\/sub>CO2<\/sub>Na:\r\n[latex]\\begin{array}{lll}\\hfill \\text{18C}=18\\times 12.011& =& 216.198\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{35H}=35\\times 1.00794& =& 35.2779\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2O}=2\\times 15.9994& =& 31.9988\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1Na}=1\\times 22.98977& =& \\underline{22.98977\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 306.464\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n17. \u00a0The mass of each compound is as follows:\r\n
                                                    \r\n \t
                                                  1. KOH:\r\n[latex]\\begin{array}{lll}\\hfill 1\\text{K}=1\\times 39.0983& =& 39.0983\\hfill \\\\ \\hfill 1\\text{O}=1\\times 15.9994& =& 15.9994\\hfill \\\\ \\hfill \\text{1H}=1\\times 1.00794& =& \\underline{1.00794}\\hfill \\\\ \\hfill \\text{molar mass}& =& 56.1056\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]\r\n[latex]\\text{Mass}=0.0146\\text{mol}\\times 56.1056\\text{g\/mol}=0.819\\text{g}[\/latex]<\/li>\r\n \t
                                                  2. C2<\/sub>H6<\/sub>:\r\n<\/sub>[latex]\\begin{array}{lll}\\hfill 2\\text{C}=2\\times 12.011& =& 24.022\\hfill \\\\ \\hfill 6\\text{H}=6\\times 1.00794& =& \\underline{6.04764}\\hfill \\\\ \\hfill \\text{molar mass}& =& 30.070\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]\r\n[latex]\\text{Mass}=10.2\\text{mol}\\times 30.070\\text{g\/mol}=307\\text{g}[\/latex]<\/li>\r\n \t
                                                  3. Na2<\/sub>SO4<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill 2\\text{Na}=2\\times 22.990& =& 45.98\\hfill \\\\ \\hfill 1\\text{S}=1\\times 32.066& =& 32.066\\hfill \\\\ \\hfill 4\\text{O}=4\\times 15.9994& =& \\underline{63.9976}\\hfill \\\\ \\hfill \\text{molar mass}& =& 142.044\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]\r\n[latex]\\text{Mass}=1.6\\times {10}^{-3}\\text{mol}\\times 142.044\\text{g\/mol}=\\text{0.23 g}[\/latex]<\/li>\r\n \t
                                                  4. C6<\/sub>H12<\/sub>O6<\/sub>:\r\n<\/sub>[latex]\\begin{array}{lll}\\hfill 6\\text{C}=6\\times 12.011& =& 72.066\\hfill \\\\ \\hfill 12\\text{H}=12\\times 1.00794& =& 12.0953\\hfill \\\\ \\hfill 6\\text{O}=6\\times 15.9994& =& \\underline{95.9964}\\hfill \\\\ \\hfill \\text{molar mass}& =& 180.158\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]\r\n[latex]\\text{Mass}=6.854\\times {10}^{3}\\text{mol}\\times 180.158\\text{g\/mol}=1.235\\times {10}^{6}\\text{g (}1235\\text{kg)}[\/latex]<\/li>\r\n \t
                                                  5. Co(NH3<\/sub>)6<\/sub>Cl3<\/sub>:\r\n<\/sub>[latex]\\begin{array}{lll}\\hfill \\text{Co}=1\\times 58.99320& =& 58.99320\\hfill \\\\ \\hfill 6\\text{N}=6\\times 14.0067& =& 84.0402\\hfill \\\\ \\hfill 18\\text{H}=18\\times 1.00794& =& 18.1429\\hfill \\\\ \\hfill 3\\text{Cl}=3\\times 35.4527& =& \\underline{106.358}\\hfill \\\\ \\hfill \\text{molar mass}& =& 267.5344\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]\r\n<\/span>[latex]\\text{Mass}=2.856\\text{mol}\\times 267.5344\\text{g\/mol}=765\\text{g}[\/latex]<\/span><\/li>\r\n<\/ol>\r\n19.\r\n
                                                      \r\n \t
                                                    1. 2.345 mol LiCl:\r\n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left(\\text{LiCl}\\right)=1\\times 6.941+1\\times 35.4527& =& 42.394\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=2.345\\cancel{\\text{mol}}\\times 42.394\\text{g}\\cancel{{\\text{mol}}^{-1}}& =& 99.41\\text{g}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                    2. 0.0872 mol acetylene, C2<\/sub>H2<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left({\\text{C}}_{2}{\\text{H}}_{2}\\right)=2\\times 12.011+2\\times 1.00794& =& 26.038\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=0.0872\\cancel{\\text{mol}}\\times 26.038\\text{g}\\cancel{{\\text{mol}}^{-1}}& =& 2.27\\text{g}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                    3. 3.3\u00a0\u00d7\u00a010\u22122<\/sup>\u00a0mol Na2<\/sub>CO3<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left({\\text{Na}}_{2}{\\text{CO}}_{3}\\right)=2\\times 22.989768+1\\times 12.011+3\\times 15.9994& =& 105.989\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=3.3\\times {10}^{-2}\\cancel{\\text{mol}}\\times 105.989\\text{g}{\\text{mol}}^{-1}& =& 3.5\\text{g}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                    4. 1.23\u00a0\u00d7\u00a0103<\/sup>\u00a0mol fructose, C6<\/sub>H12<\/sub>O6<\/sub>:[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left({\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}\\right)=6\\times 12.011+12\\times 1.00794+6\\times 15.9994& =& 180.158\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=1.23\\times {10}^{3}\\cancel{\\text{mol}}\\times 180.158\\text{g}{\\text{mol}}^{-1}=2.22\\times {10}^{5}\\text{g}& =& 222\\text{kg}\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                    5. 0.5758 mol FeSO4<\/sub>(H2<\/sub>O)7<\/sub>:\r\n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left[{\\text{FeSO}}_{4}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{7}\\right]& =& 1\\times 55.847+1\\times 32.066+4\\times 15.999\\hfill \\\\ \\hfill +7\\left(2\\times 1.00794+15.9994\\right)& =& 278.018\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=0.5758\\cancel{\\text{mol}}\\times 278.018\\text{g}\\cancel{{\\text{mol}}^{-1}}& =\\hfill & 160.1\\text{g}\\hfill \\end{array}[\/latex]<\/li>\r\n<\/ol>\r\n21.\u00a0The mass of each compound is as follows:\r\n
                                                        \r\n \t
                                                      1. [latex]0.600\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=9.60\\text{g}[\/latex]<\/li>\r\n \t
                                                      2. [latex]0.600\\cancel{\\text{mol}}\\times 2\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=19.2\\text{g}[\/latex]<\/li>\r\n \t
                                                      3. [latex]0.600\\cancel{\\text{mol}}\\times 3\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=28.8\\text{g}[\/latex]<\/li>\r\n<\/ol>\r\n23.\u00a0Determine the number of moles of each component. From the moles, calculate the number of atoms and the mass of the elements involved. Zirconium: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times 1023\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 91.224\\text{g\/}\\cancel{\\text{mol}}=30.87\\text{g;}[\/latex] Silicon: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times {10}^{23}\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 28.0855\\text{g\/}\\cancel{\\text{mol}}=9.504\\text{g;}[\/latex] Oxygen: [latex]4\\times 0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=8.151\\times {10}^{23}\\text{atoms;}4\\times 0.3384\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=21.66\\text{g}[\/latex]\r\n\r\n25. Determine the molar mass and, from the grams present, the moles of each substance. The compound with the greatest number of moles of Al has the greatest mass of Al.\r\n
                                                          \r\n \t
                                                        • Molar mass AlPO4<\/sub>: 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g\/mol<\/li>\r\n \t
                                                        • Molar mass Al2<\/sub>Cl6<\/sub>: 2(26.981539) + 6(35.4527) = 266.6793 g\/mol<\/li>\r\n \t
                                                        • Molar mass Al2<\/sub>S3<\/sub>: 2(26.981539) + 3(32.066) = 150.161 g\/mol<\/li>\r\n<\/ul>\r\nAlPO4<\/sub>: [latex]\\frac{122\\cancel{\\text{g}}}{121.9529\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.000\\text{mol.}[\/latex]\r\n\r\n[latex]\\text{mol Al}=1\\times 1.000\\text{mol}=1.000\\text{mol}[\/latex]\r\n\r\nAl2<\/sub>Cl6<\/sub>: [latex]\\frac{266\\text{g}}{266.6793\\text{g}{\\text{mol}}^{-1}}=0.997\\text{mol}[\/latex]\r\n\r\n[latex]\\text{mol Al}=2\\times 0.997\\text{mol}=1.994\\text{mol}[\/latex]\r\n\r\nAl2<\/sub>S3<\/sub>: [latex]\\frac{225\\cancel{\\text{g}}}{150.161\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.50\\text{mol}[\/latex]\r\n\r\n[latex]\\text{mol Al}=2\\times 1.50\\text{mol}=3.00\\text{mol}[\/latex]\r\n\r\n27. \u00a0Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro\u2019s number by the number of moles:\r\n

                                                          [latex]\\frac{3104\\cancel{\\text{carats}}\\times \\frac{200\\cancel{\\text{mg}}}{1\\cancel{\\text{carat}}}\\times \\frac{1\\cancel{\\text{g}}}{1000\\cancel{\\text{mg}}}}{12.011\\cancel{\\text{g}}\\cancel{{\\text{mol}}^{-1}}\\left(6.022\\times {10}^{23}\\cancel{{\\text{mol}}^{-1}}\\right)}=3.113\\times {10}^{25}\\text{C atoms}[\/latex]<\/p>\r\n29. \u00a0Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g\/mol; Then [latex]0.0278\\text{mol}\\times 342.300\\text{g\/mol}=9.52\\text{g sugar.}[\/latex] This 9.52 g of sugar represents [latex]\\frac{11.0}{60.0}[\/latex] of one serving or\r\n

                                                          [latex]\\frac{60.0\\text{g serving}}{11.0\\cancel{\\text{g sugar}}}\\times 9.52\\cancel{\\text{g sugar}}=51.9\\text{g cereal.}[\/latex]<\/p>\r\nThis amount is [latex]\\frac{51.9\\text{g cereal}}{60.0\\text{g serving}}=0.865[\/latex] servings, or about 1 serving.\r\n\r\n31.\u00a0Calculate the number of moles of each species, then remember that 1 mole of anything [latex]=6.022\\times {10}^{23}[\/latex] species.\r\n

                                                            \r\n \t
                                                          1. 20.0 g = 1.11 mol H2<\/sub>O<\/li>\r\n \t
                                                          2. 77.0 g CH4<\/sub> = 4.79 mol CH4<\/sub><\/li>\r\n \t
                                                          3. 68.0 g CaH2<\/sub> = 1.62 mol CaH2<\/sub><\/li>\r\n \t
                                                          4. 100.0 g N2<\/sub>O = 2.27 mol N2<\/sub>O<\/li>\r\n \t
                                                          5. 84.0 g HF = 4.20 mol HF<\/li>\r\n<\/ol>\r\nTherefore, 20.0 g H2<\/sub>O represents the least number of molecules since it has the least number of moles.\r\n\r\n[\/hidden-answer]\r\n\r\nPart II<\/strong><\/span>\r\n
                                                              \r\n \t
                                                            1. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\r\n
                                                                \r\n \t
                                                              1. The number of moles and the mass of chlorine, Cl2<\/sub>, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.<\/li>\r\n \t
                                                              2. The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.<\/li>\r\n \t
                                                              3. The number of moles and the mass of sodium nitrate, NaNO3<\/sub>, required to produce 128 g of oxygen. (NaNO2<\/sub> is the other product.)<\/li>\r\n \t
                                                              4. The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.<\/li>\r\n \t
                                                              5. The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. (CO2<\/sub> is the other product.)<\/li>\r\n \t
                                                              6. \"This<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                              7. Determine the number of moles and the mass requested for each reaction in Exercise 1.<\/li>\r\n \t
                                                              8. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\r\n
                                                                  \r\n \t
                                                                1. The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl2<\/sub> and H2<\/sub>.<\/li>\r\n \t
                                                                2. The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/li>\r\n \t
                                                                3. The number of moles and the mass of magnesium carbonate, MgCO3<\/sub>, required to produce 283 g of carbon dioxide. (MgO is the other product.)<\/li>\r\n \t
                                                                4. The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, C2<\/sub>H2<\/sub>, in an excess of oxygen.<\/li>\r\n \t
                                                                5. The number of moles and the mass of barium peroxide, BaO2<\/sub>, needed to produce 2.500 kg of barium oxide, BaO (O2<\/sub> is the other product.)<\/li>\r\n \t
                                                                6. \"This<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                7. Determine the number of moles and the mass requested for each reaction in Exercise 3.<\/li>\r\n \t
                                                                8. H2<\/sub> is produced by the reaction of 118.5 mL of a 0.8775-M solution of H3<\/sub>PO4<\/sub> according to the following equation: [latex]2\\text{Cr}+2{\\text{H}}_{3}{\\text{PO}}_{4}\\rightarrow 3{\\text{H}}_{2}+2{\\text{CrPO}}_{4}\\text{.}[\/latex]\r\n
                                                                    \r\n \t
                                                                  1. Outline the steps necessary to determine the number of moles and mass of H2<\/sub>.<\/li>\r\n \t
                                                                  2. Perform the calculations outlined.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                  3. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: [latex]2\\text{Ga}+6\\text{HCl}\\rightarrow 2{\\text{GaCl}}_{3}+3{\\text{H}}_{2}\\text{.}[\/latex]\r\n
                                                                      \r\n \t
                                                                    1. Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/li>\r\n \t
                                                                    2. Perform the calculations outlined.<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                    3. I2<\/sub> is produced by the reaction of 0.4235 mol of CuCl2<\/sub> according to the following equation: [latex]2{\\text{CuCl}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+4\\text{KCl}+{\\text{I}}_{2}\\text{.}[\/latex]\r\n
                                                                        \r\n \t
                                                                      1. How many molecules of I2<\/sub> are produced?<\/li>\r\n \t
                                                                      2. What mass of I2<\/sub> is produced?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                      3. Silver is often extracted from ores as K[Ag(CN)2<\/sub>] and then recovered by the reaction [latex]2\\text{K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\text{(}aq\\text{)}+\\text{Zn}\\text{(}s\\text{)}\\rightarrow 2Ag\\text{(}s\\text{)}+\\text{Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{(}aq\\text{)}+2\\text{KCN}\\text{(}aq\\text{)}[\/latex]\r\n
                                                                          \r\n \t
                                                                        1. How many molecules of Zn(CN)2<\/sub> are produced by the reaction of 35.27 g of K[Ag(CN)2<\/sub>]?<\/li>\r\n \t
                                                                        2. What mass of Zn(CN)2<\/sub> is produced?<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                        3. What mass of silver oxide, Ag2<\/sub>O, is required to produce 25.0 g of silver sulfadiazine, AgC10<\/sub>H9<\/sub>N4<\/sub>SO2<\/sub>, from the reaction of silver oxide and sulfadiazine? [latex]2{\\text{C}}_{10}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{Ag}}_{2}\\text{O}\\rightarrow 2{\\text{AgC}}_{10}{\\text{H}}_{9}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\r\n \t
                                                                        4. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO2<\/sub>, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO2<\/sub> is required to produce 3.00 kg of SiC.<\/li>\r\n \t
                                                                        5. Automotive air bags inflate when a sample of sodium azide, NaN3<\/sub>, is very rapidly decomposed.[latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 2\\text{Na}\\text{(}s\\text{)}+3{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]\u00a0What mass of sodium azide is required to produce 2.6 ft3<\/sup> (73.6 L) of nitrogen gas with a density of 1.25 g\/L?<\/li>\r\n \t
                                                                        6. Urea, CO(NH2<\/sub>)2<\/sub>, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO2<\/sub> produced by combustion of [latex]1.00\\times {10}^{3}\\text{kg}[\/latex] of carbon followed by the reaction? [latex]{\\text{CO}}_{2}\\text{(}g\\text{)}+2{\\text{NH}}_{3}\\text{(}g\\text{)}\\rightarrow\\text{CO}{\\text{(}{\\text{NH}}_{2}\\text{)}}_{2}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\r\n \t
                                                                        7. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2<\/sub>CO3<\/sub> was quickly spread on the area and CO2<\/sub> was released by the reaction. Was sufficient Na2<\/sub>CO3<\/sub> used to neutralize all of the acid?<\/li>\r\n \t
                                                                        8. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g\/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).<\/li>\r\n \t
                                                                        9. What volume of a 0.750 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid? [latex]\\text{NaCl}\\text{(}s\\text{)}+{\\text{H}}_{2}{\\text{SO}}_{4}\\text{(}l\\text{)}\\rightarrow\\text{HCl}\\text{(}g\\text{)}+{\\text{NaHSO}}_{4}\\text{(}s\\text{)}[\/latex]<\/li>\r\n \t
                                                                        10. What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3<\/sub>)2<\/sub> in 43.88 mL of a 0.3842 M solution of Cu(NO3<\/sub>)2<\/sub>? [latex]2\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+{\\text{I}}_{2}+4{\\text{KNO}}_{3}[\/latex]<\/li>\r\n \t
                                                                        11. A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.[latex]2{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow\\text{Ca}{\\text{(}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{)}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex] What mass of Ca(OH)2<\/sub> is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g\/mL and containing 58.0% acetic acid by mass?<\/li>\r\n \t
                                                                        12. The toxic pigment called white lead, Pb3<\/sub>(OH)2<\/sub>(CO3<\/sub>)2<\/sub>, has been replaced in white paints by rutile, TiO2<\/sub>. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO3<\/sub>) by mass?[latex]2{\\text{FeTiO}}_{3}+4\\text{HCl}+{\\text{Cl}}_{2}\\rightarrow 2{\\text{FeCl}}_{3}+2{\\text{TiO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"132180\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"132180\"]\r\n\r\n2. (a)\u00a00.435 mol Na, 0.271 mol Cl2<\/sub>, 15.4 g Cl2<\/sub>;\r\n\r\n(b) 0.005780 mol HgO, 2.890 \u00d7 10\u22123<\/sup> mol O2<\/sub>, 9.248 \u00d7 10\u22122<\/sup> g O2<\/sub>;\r\n\r\n(c) 8.00 mol NaNO3<\/sub>, 6.8 \u00d7 102<\/sup> g NaNO3<\/sub>;\r\n\r\n(d) 1665 mol CO2<\/sub>, 73.3 kg CO2<\/sub>;\r\n\r\n(e) 18.86 mol CuO, 2.330 kg CuCO3<\/sub>;\r\n\r\n(f) 0.4580 mol C2<\/sub>H4<\/sub>Br2<\/sub>, 86.05 g C2<\/sub>H4<\/sub>Br2<\/sub>\r\n\r\n4. \u00a0(a) [latex]\\text{mol Mg}=5.00\\cancel{\\text{g HCl}}\\times \\frac{1\\cancel{\\text{mol HCl}}}{36.4606\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol Mg}}{2\\cancel{\\text{mol HCl}}}=0.0686\\text{mol,}[\/latex] [latex]\\text{g Mg}=0.0686\\cancel{\\text{mol Mg}}\\times \\frac{\\text{24.305 g}}{1\\cancel{\\text{mol Mg}}}=\\text{1.67 g};[\/latex]\r\n(b) [latex]{\\text{mol O}}_{2}=1.252\\cancel{\\text{g}{\\text{Ag}}_{2}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}{231.7358\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}=2.701\\times {10}^{-3}[\/latex], [latex]{\\text{g O}}_{2}=2.701\\times {10}^{-3}\\cancel{{\\text{mol O}}_{2}}\\times \\frac{\\text{31.9988 g}}{1\\cancel{\\text{mol}{\\text{O}}_{2}}}=0.08644\\text{g;}[\/latex]\r\n\r\n(c) [latex]{\\text{mol MgCO}}_{3}=283\\cancel{\\text{g}{\\text{CO}}_{2}}\\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{44.010\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{MgCO}}_{3}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=6.43\\text{mol,}[\/latex] [latex]{\\text{g MgCO}}_{3}=6.43\\cancel{{\\text{mol MgCO}}_{3}}\\times \\frac{\\text{84.314 g}}{1\\cancel{{\\text{mol MgCO}}_{3}}}=542\\text{g;}[\/latex]\r\n\r\n(d) [latex]{\\text{mol H}}_{2}\\text{O}=2.00\\times {10}^{4}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}{28.054\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{H}}_{2}\\text{O}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}=713\\text{mol,}[\/latex] [latex]{\\text{g H}}_{2}\\text{O}=713\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}\\times \\frac{18.01528\\cancel{\\text{g}}}{1\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}}\\times \\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=\\text{12.8 kg}[\/latex]\r\n\r\n(e) [latex]2.500\\cancel{\\text{kg BaO}}\\times \\frac{1000\\cancel{\\text{g BaO}}}{1\\cancel{\\text{kg BaO}}}\\times \\frac{1\\cancel{\\text{mol BaO}}}{153.326\\cancel{\\text{g BaO}}}\\times \\frac{\\text{2 mol}{\\text{BaO}}_{2}}{2\\cancel{\\text{mol BaO}}}=\\text{16.31 mol}{\\text{BaO}}_{2}[\/latex] [latex]16.31\\cancel{\\text{mol}{\\text{BaO}}_{2}}\\times \\frac{\\text{169.326 g}{\\text{BaO}}_{2}}{1\\cancel{\\text{mol}{\\text{BaO}}_{2}}}=\\text{2762 g}{\\text{BaO}}_{2}[\/latex]\r\n\r\n(f) [latex]9.55\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}{46.068\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}\\times \\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}=\\text{0.207 mol}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex] [latex]0.207\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{\\text{28.053 g}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}}=\\text{5.81 g}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex]\r\n\r\n6. (a) [latex]\\text{volume HCl solution}\\rightarrow\\text{mol HCl}\\rightarrow{\\text{mol GaCl}}_{3};[\/latex]\r\n\r\n(b) [latex]2.6\\cancel{\\text{L HCl}}\\times \\frac{1.44\\cancel{\\text{mol HCl}}}{1\\cancel{\\text{L HCl}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}{6\\cancel{\\text{mol HCl}}}\\times \\frac{\\text{180.079 g}{\\text{GaCl}}_{3}}{1\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}=2.3\\times {10}^{2}\\text{g}{\\text{GaCl}}_{3}[\/latex]\r\n\r\n8.\u00a0The development requires the following: [latex]\\text{mass K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\rightarrow\\text{molecules of Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{g Zn}{\\text{(}\\text{CN}\\text{)}}_{2};[\/latex]\r\n\r\n(a) [latex]35.27\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}\\times \\frac{1\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}{199.002\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{2\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{6.022\\times {10}^{23}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=5.337\\times {10}^{22}\\text{molecules}[\/latex]\r\n\r\n(b) [latex]5.337\\times {10}^{22}\\cancel{\\text{molecules}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{6.022\\times {10}^{23}\\cancel{\\text{molecules}}}\\times \\frac{\\text{117.43 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=\\text{10.41 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}[\/latex]\r\n\r\n10. [latex]{\\text{SiO}}_{2}+3\\text{C}\\rightarrow\\text{SiC}+2\\text{CO.}[\/latex] From the balanced equation, 1 mol of SiO2<\/sub> produces 1 mol of SiC. The unknown is the mass of SiO2<\/sub> required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO2<\/sub> required, determine the molar masses of SiO2<\/sub> and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO2<\/sub> to SiC, find the mass of SiO2<\/sub> required. The conversions required are: [latex]\\text{g SiC}\\rightarrow\\text{mol SiC}\\rightarrow\\text{mol}{\\text{SiO}}_{2}\\rightarrow\\text{g}{\\text{SiO}}_{2}[\/latex]\r\nMolar masses: SiO2<\/sub> = 60.0843 g mol\u20131<\/sup>; SiC = 40.0955 g mol\u20131<\/sup>\r\n[latex]\\text{mass SiO}2=3000\\cancel{\\text{g SiC}}\\times \\frac{1\\cancel{\\text{mol SiC}}}{40.955\\cancel{\\text{g SiC}}}\\times \\frac{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}{1\\cancel{\\text{mol SiC}}}\\times \\frac{\\text{60.843 g}{\\text{SiO}}_{2}}{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}=\\text{4496 g}{\\text{SiO}}_{2}=\\text{4.50 kg}{\\text{SiO}}_{2}[\/latex]\r\n\r\n12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol\u20131<\/sup>\r\n[latex]\\text{1 mol C}\\rightarrow 1{\\text{mol CO}}_{2}\\rightarrow 1\\text{mol urea}[\/latex]\r\n[latex]\\begin{array}{ll}\\hfill \\text{mass urea}& =1.00\\times {10}^{3}\\cancel{\\text{kg}}\\times \\frac{1000\\cancel{\\text{g}}}{\\cancel{\\text{kg}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.0\\cancel{\\text{g C}}}\\times \\frac{1\\cancel{\\text{mol urea}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{60.054 g urea}}{1\\cancel{\\text{mol urea}}}\\\\ & =5.00\\times {10}^{6}\\text{g or}5.00\\times {10}^{3}\\text{kg}\\end{array}[\/latex]\r\n\r\n14. \u00a0The balanced chemical equation is [latex]\\text{C}\\text{(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}[\/latex]\r\n[latex]\\begin{array}{l}\\\\ \\\\ \\\\ 500\\cancel{\\text{miles}}\\times \\frac{1\\cancel{\\text{gallon}}}{37.5\\cancel{\\text{miles}}}\\times \\frac{3.785\\cancel{\\text{L}}}{1\\cancel{\\text{gallon}}}\\times \\frac{1000\\cancel{\\text{mL}}}{1\\cancel{\\text{L}}}\\times \\frac{0.8205\\cancel{\\text{g gas}}}{1\\cancel{\\text{mL}}\\text{gas}}\\times \\frac{84.2\\cancel{\\text{g C}}}{100\\cancel{\\text{g gas}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.01\\cancel{\\text{g C}}}\\\\ \\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{44.01 g}{\\text{CO}}_{2}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=1.28\\times {10}^{5}\\text{g}{\\text{CO}}_{2}\\end{array}[\/latex]\r\n\r\n16. Use molarity to convert. This solution involves the following steps:\r\n\r\nConverting the volume of KI to moles of KI\r\n\r\nConverting the moles of KI to moles of Cu(NO3<\/sub>)2<\/sub>\r\n\r\nConverting the moles of [latex]\\text{K}\\rightarrow\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}[\/latex] to a volume of KI. Cu(NO3<\/sub>)2<\/sub> solution\r\n[latex]\\text{43.88 mL}\\times \\frac{\\text{1 L}}{\\text{1000 mL}}\\times \\frac{0.3842\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}{\\text{1 L}}\\times \\frac{4\\cancel{\\text{mol KI}}}{2\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}\\times \\frac{\\text{1 L KI}}{0.2089\\cancel{\\text{mol KI}}}=\\text{161.4 mL}[\/latex]\r\nAll of these steps can be shown together, as follows:\r\n[latex]\\frac{\\text{43.88 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{1}\\times \\frac{\\text{0.3842 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{\\text{1000 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{4 mol KI}}{\\text{2 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{1000 mL KI}}{\\text{0.2089 mol KI}}=\\text{161.40 mL KI solution}[\/latex]\r\n\r\n18. Find from worked example, check your learning problem\r\n[latex]\\text{mass of ilmenite}=379\\cancel{\\text{g ore}}\\times \\frac{\\text{0.883 g}{\\text{FeTiO}}_{3}}{1\\cancel{\\text{g ore}}}=\\text{334.6 g}{\\text{FeTiO}}_{3}[\/latex]\r\n[latex]\\text{mass of rutile}=334.6\\cancel{\\text{g}{\\text{FeTiO}}_{3}}\\times \\frac{1\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}{151.7\\cancel{\\text{g}{\\text{FeTiO}}_{3}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{TiO}}_{2}}}{2\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}\\times \\frac{\\text{79.88 g}{\\text{TiO}}_{2}}{1\\cancel{\\text{mol}{\\text{TiO}}_{2}}}=\\text{176 g}{\\text{TiO}}_{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nPart III<\/strong><\/span>\r\n
                                                                            \r\n \t
                                                                          1. Calculate the following to four significant figures:\r\n
                                                                              \r\n \t
                                                                            1. the percent composition of ammonia, NH3 \u00a0<\/sub><\/li>\r\n \t
                                                                            2. the percent composition of photographic \u201chypo,\u201d Na2<\/sub>S2<\/sub>O3 \u00a0<\/sub><\/li>\r\n \t
                                                                            3. the percent of calcium ion in Ca3<\/sub>(PO4<\/sub>)2<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                            4. Determine the following to four significant figures:\r\n
                                                                                \r\n \t
                                                                              1. the percent composition of hydrazoic acid, HN3 \u00a0<\/sub><\/li>\r\n \t
                                                                              2. the percent composition of TNT, C6<\/sub>H2<\/sub>(CH3<\/sub>)(NO2<\/sub>)3 \u00a0<\/sub><\/li>\r\n \t
                                                                              3. the percent of SO4<\/sub>2\u2013<\/sup> in Al2<\/sub>(SO4<\/sub>)3<\/sub><\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                              4. Determine the percent ammonia, NH3<\/sub>, in Co(NH3<\/sub>)6<\/sub>Cl3<\/sub>, to three significant figures.<\/li>\r\n \t
                                                                              5. Determine the percent water in CuSO4<\/sub>\u22195H2<\/sub>O to three significant figures.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"954320\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"954320\"]\r\n\r\n1. In each of these exercises asking for the percent composition, divide the molecular weight of the desired element or group of elements (the number of times it\/they occur in the formula times the molecular weight of the desired element or elements) by the molecular weight of the compound.\r\n
                                                                                  \r\n \t
                                                                                1. [latex]\\begin{array}{l}\\\\ \\%\\text{N}=\\frac{14.0067\\text{g}{\\text{mol}}^{-1}\\times 100\\%}{\\left[3\\left(1.007940+14.0067\\right)\\right]\\text{g}{\\text{mol}}^{-1}}=\\frac{14.0067\\text{g}{\\text{mol}}^{-1}}{17.0305\\text{g}{\\text{mol}}^{-1}}=82.24\\%\\\\ \\%\\text{H}=\\frac{3\\times 1.00794\\text{g}{\\text{mol}}^{-1}}{17.0305\\text{g}{\\text{mol}}^{-1}}\\times 100\\%=17.76\\%\\end{array}[\/latex]<\/li>\r\n \t
                                                                                2. [latex]\\begin{array}{lll}\\\\ \\%\\text{Na}\\hfill & =\\hfill & \\frac{2\\times 22.989768}{2\\times 22.989768+2\\times 32.066+3\\times 15.9994}\\times 100\\%=\\frac{45.9795}{158.1097}\\times 100=29.08\\%\\hfill \\\\ \\%\\text{S}\\hfill & =\\hfill & \\frac{64.132}{158.1097}\\times 100\\%=40.56\\%\\hfill \\\\ \\%\\text{O}\\hfill & =\\hfill & \\frac{47.9982}{158.1097}\\times 100\\%=30.36\\%\\hfill \\end{array}[\/latex]<\/li>\r\n \t
                                                                                3. [latex]\\%{\\text{Ca}}^{2+}=\\frac{3\\times 40.078}{3\\times 40.078+2\\times 30.973762+8\\times 15.9994}\\times 100\\%=\\frac{120.234}{310.1816}\\times 100\\%=38.76\\%[\/latex]<\/li>\r\n<\/ol>\r\n3.\u00a0[latex]\\%{\\text{NH}}_{3}=\\frac{6\\left(14.007+3\\times 40.078\\right)}{58.933+6\\left(14.007+3\\times 1.008\\right)+3\\left(35.453\\right)}\\times 100\\%=\\frac{102.186}{267.478}\\times 100\\%=38.2\\%[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\nPart IV<\/strong><\/span>\r\n
                                                                                    \r\n \t
                                                                                  1. What information do we need to determine the molecular formula of a compound from the empirical formula?<\/li>\r\n \t
                                                                                  2. Determine the empirical formulas for compounds with the following percent compositions:\r\n
                                                                                      \r\n \t
                                                                                    1. 15.8% carbon and 84.2% sulfur<\/li>\r\n \t
                                                                                    2. 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                                    3. Determine the empirical formulas for compounds with the following percent compositions:\r\n
                                                                                        \r\n \t
                                                                                      1. 43.6% phosphorus and 56.4% oxygen<\/li>\r\n \t
                                                                                      2. 28.7% K, 1.5% H, 22.8% P, and 47.0% O<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                                      3. Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:\r\n
                                                                                          \r\n \t
                                                                                        1. Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O<\/li>\r\n \t
                                                                                        2. Saran; 24.8% C, 2.0% H, 73.1% Cl<\/li>\r\n \t
                                                                                        3. polyethylene; 86% C, 14% H<\/li>\r\n \t
                                                                                        4. polystyrene; 92.3% C, 7.7% H<\/li>\r\n \t
                                                                                        5. Orlon; 67.9% C, 5.70% H, 26.4% N<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
                                                                                        6. A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g\/mol. What is its molecular formula?<\/li>\r\n \t
                                                                                        7. Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g\/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?<\/li>\r\n \t
                                                                                        8. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g\/mol.<\/li>\r\n \t
                                                                                        9. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g\/mol. Determine the molecular formula of the dye.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"265420\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"265420\"]\r\n\r\n2. The empirical formulas\u00a0can be found as follows:\r\n
                                                                                            \r\n \t
                                                                                          1. The percent of an element in a compound indicates the percent by mass. The mass of an element in a 100.0-g sample of a compound is equal in grams to the percent of that element in the sample; hence, 100.0 g of the sample contains 15.8 g of C and 84.2 g of S. The relative number of moles of C and S atoms in the compound can be obtained by converting grams to moles as shown.\r\nStep 1: [latex]\\begin{array}{l}\\\\ \\text{C:}15.8\\text{g}\\times \\frac{1\\text{mol}}{12.011\\text{g}}=1.315\\text{mol}\\\\ \\text{S:}84.2\\text{g}\\times \\frac{1\\text{mol}}{32.066\\text{g}}=2.626\\text{mol}\\end{array}[\/latex]\r\nStep 2: [latex]\\begin{array}{l}\\\\ \\text{C:}\\frac{1.315\\text{mol}}{1.315\\text{mol}}=1.000\\\\ \\text{S:}\\frac{2.626\\text{mol}}{1.315\\text{mol}}=1.997\\end{array}[\/latex]\r\n
                                                                                              \r\n \t
                                                                                            • The empirical formula is CS2<\/sub>.<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
                                                                                            • Step 1: [latex]\\begin{array}{l}\\\\ \\text{C:}40.0\\text{g}\\times \\frac{1\\text{mol}}{12.011\\text{g}}=3.330\\text{mol}\\\\ \\text{H:}6.7\\text{g}\\times \\frac{1\\text{mol}}{1.00794\\text{g}}=6.647\\text{mol}\\\\ \\text{O:}53.3\\text{g}\\times \\frac{1\\text{mol}}{15.9994\\text{g}}=3.331\\text{mol}\\end{array}[\/latex]\r\nStep 2: [latex]\\begin{array}{l}\\\\ \\text{C:}\\frac{3.330\\text{mol}}{3.330\\text{mol}}=1.0\\\\ \\text{H:}\\frac{6.647\\text{mol}}{3.330\\text{mol}}=2\\\\ \\text{O:}\\frac{3.331\\text{mol}}{3.330\\text{mol}}=1.0\\end{array}[\/latex]\r\n
                                                                                                \r\n \t
                                                                                              • The empirical formula is CH2<\/sub>O.<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n5. To determine the empirical formula, a relationship between percent composition and atom composition must be established. The percent composition of each element in a compound can be found either by dividing its mass by the total mass of compound or by dividing the molar mass of that element as it appears in the formula (atomic mass times the number of times the element appears in the formula) by the formula mass of the compound. From this latter perspective, the percent composition of an element can be converted into a mass by assuming that we start with a 100-g sample. Then, multiplying the percentage times 100 g gives the mass in grams of that component. Division of each mass by its respective atomic mass gives the relative ratio of atoms in the formula. From the numbers so obtained, the whole-number ratio of elements in the compound can be found by dividing each ratio by the number representing the smallest ratio. Generally, this process can be done in two simple steps (a third step is needed if the ratios are not whole numbers).\r\n\r\nStep 1: Divide each element\u2019s percentage (converted to grams) by its atomic mass:\r\n

                                                                                                [latex]\\begin{array}{l}\\text{C:}\\frac{92.3\\text{g}}{12.011\\text{g}{\\text{mol}}^{-1}}=7.68\\text{mol}\\\\ \\text{H:}\\frac{7.7\\text{g}}{1.00794\\text{g}{\\text{mol}}^{-1}}=7.6\\text{mol}\\end{array}[\/latex]<\/p>\r\nThis operation established the relative ration of carbon to hydrogen in the formula.\r\n\r\nStep 2: To establish a whole-number ratio of carbon to hydrogen, divide each factor by the smallest factor. In this case, both factors are essentially equal; thus the ration of atoms is 1 to 1:\r\n

                                                                                                [latex]\\begin{array}{l}\\text{C:}\\frac{7.68}{7.6}=1\\\\ \\text{H:}\\frac{7.6}{7.6}=1\\end{array}[\/latex]<\/p>\r\nThe empirical formula is CH.\r\n\r\nSince the molecular mass of the compound is 78.1 amu, some integer times the sum of the mass of 1C and 1H in atomic mass units (12.011 amu + 1.00794 amu = 13.019 amu) must be equal to 78.1 amu. To find this number, divide 78.1 amu by 13.019 amu:\r\n

                                                                                                [latex]\\frac{78.1\\text{amu}}{13.019\\text{amu}}=5.9989\\rightarrow 6[\/latex]<\/p>\r\nThe molecular formula is (CH)6<\/sub> = C6<\/sub>H6<\/sub>.\r\n\r\n7. The formulas can be found as follows:\r\n

                                                                                                [latex]\\begin{array}{ccc}\\left(28.03\\text{g Mg}\\right)\\left(\\frac{1\\text{mol Mg}}{24.30\\text{g}}\\right)=1.153\\text{mol Mg}& & \\frac{1.153}{0.769}=1.512\\text{mol Mg}\\\\\\left(21.60\\text{g Si}\\right)\\left(\\frac{1\\text{mol Si}}{28.09\\text{g Si}}\\right)=0.769\\text{mol Si}& &\\frac{0.769}{0.769}=1.00\\text{mol Si}\\\\\\left(1.16\\text{g H}\\right)\\left(\\frac{1\\text{mol H}}{1.01\\text{g H}}\\right)=1.149\\text{mol H}& & \\frac{1.149}{0.769}=1.49\\text{mol H}\\\\\\left(49.21\\text{g O}\\right)\\left(\\frac{1\\text{mol O}}{16.00\\text{g O}}\\right)=3.076\\text{mol O}& & \\frac{3.076}{0.769}=4.00\\text{mol O}\\end{array}[\/latex]<\/p>\r\n(2)(Mg1.5<\/sub>Si1<\/sub>H1.5<\/sub>O4<\/sub>) = Mg3<\/sub>Si2<\/sub>H3<\/sub>O8<\/sub> (empirical formula), empirical mass of 260.1 g\/unit\r\n\r\n[latex]\\frac{\\text{MM}}{\\text{EM}}=\\frac{520.8}{260.1}=2.00,[\/latex] so (2)(Mg3<\/sub>Si2<\/sub>H3<\/sub>O8<\/sub>) = Mg6<\/sub>Si4<\/sub>H6<\/sub>O16<\/sub>\r\n\r\n8. Assume 100.0 g; the percentages of the elements are then the same as their mass in grams. Divide each mass by the molar mass to find the number of moles.\r\nStep 1: [latex]\\begin{array}{l}\\\\ \\frac{75.95\\cancel{\\text{g}}}{12.011\\cancel{\\text{g}}{\\text{mol}}^{-1}}=6.323\\text{mol C}\\\\ \\frac{17.72\\cancel{\\text{g}}}{14.0067\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.265\\text{mol N}\\\\ \\frac{6.33\\cancel{\\text{g}}}{1.00794\\cancel{\\text{g}}{\\text{mol}}^{-1}}=6.28\\text{mol H}\\end{array}[\/latex]\r\n\r\nStep 2: Divide each by the smallest number. The answers are 5C, 1N, and 5H. The empirical formula is C5<\/sub>H5<\/sub>N, which has a molar mass of 79.10 g\/mol. To find the actual molecular formula, divide 240, the molar mass of the compound, by 79.10 to obtain 3. So the formula is three times the empirical formula, or C15<\/sub>H15<\/sub>N3<\/sub>.\r\n\r\n[\/hidden-answer]\r\n\r\nPart V<\/strong><\/span>\r\n

                                                                                                  \r\n \t
                                                                                                1. \r\n
                                                                                                  \r\n

                                                                                                  What mass of CO2<\/sub>\u00a0is produced by the combustion of 1.00 mol of CH4<\/sub>?<\/p>\r\nCH4<\/sub>(g) +\u00a02O2<\/sub>(g) \u2192\u00a0CO2<\/sub>(g) +\u00a02H2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                2. \r\n
                                                                                                  \r\n

                                                                                                  What mass of H2<\/sub>O is produced by the combustion of 1.00 mol of CH4<\/sub>?<\/p>\r\nCH4<\/sub>(g) +\u00a02O2<\/sub>(g) \u2192\u00a0CO2<\/sub>(g) +\u00a02H2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                3. \r\n
                                                                                                  \r\n

                                                                                                  What mass of HgO is required to produce 0.692 mol of O2<\/sub>?<\/p>\r\n2HgO(s) \u2192\u00a02Hg(\u2113) +\u00a0O2<\/sub>(g)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                4. \r\n
                                                                                                  \r\n

                                                                                                  What mass of NaHCO3<\/sub>\u00a0is needed to produce 2.659 mol of CO2<\/sub>?<\/p>\r\n2NaHCO3<\/sub>(s) \u2192\u00a0Na2<\/sub>CO3<\/sub>(s) +\u00a0H2<\/sub>O(\u2113) +\u00a0CO2<\/sub>(g)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                5. \r\n
                                                                                                  \r\n

                                                                                                  How many moles of Al can be produced from 10.87 g of Ag?<\/p>\r\nAl(NO3<\/sub>)\u00a03<\/sub>(s) +\u00a03Ag \u2192\u00a0Al +\u00a03AgNO3<\/sub><\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                6. \r\n
                                                                                                  \r\n

                                                                                                  How many moles of HCl can be produced from 0.226 g of SOCl2<\/sub>?<\/p>\r\nSOCl2<\/sub>(\u2113) +\u00a0H2<\/sub>O(\u2113) \u2192\u00a0SO2<\/sub>(g) +\u00a02HCl(g)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                7. \r\n
                                                                                                  \r\n

                                                                                                  How many moles of O2<\/sub>\u00a0are needed to prepare 1.00 g of Ca(NO3<\/sub>)2<\/sub>?<\/p>\r\nCa(s) +\u00a0N2<\/sub>(g) +\u00a03O2<\/sub>(g) \u2192\u00a0Ca(NO3<\/sub>)\u00a02<\/sub>(s)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                8. \r\n
                                                                                                  \r\n

                                                                                                  How many moles of C2<\/sub>H5<\/sub>OH are needed to generate 106.7 g of H2<\/sub>O?<\/p>\r\nC2<\/sub>H5<\/sub>OH(\u2113) +\u00a03O2<\/sub>(g) \u2192\u00a02CO2<\/sub>(g) +\u00a03H2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                9. \r\n
                                                                                                  \r\n

                                                                                                  What mass of O2<\/sub>\u00a0can be generated by the decomposition of 100.0 g of NaClO3<\/sub>?<\/p>\r\n2NaClO3<\/sub>\u00a0\u2192\u00a02NaCl(s) +\u00a03O2<\/sub>(g)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                10. \r\n
                                                                                                  \r\n

                                                                                                  What mass of Li2<\/sub>O is needed to react with 1,060 g of CO2<\/sub>?<\/p>\r\nLi2<\/sub>O(aq) +\u00a0CO2<\/sub>(g) \u2192\u00a0Li2<\/sub>CO3<\/sub>(aq)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                11. \r\n
                                                                                                  \r\n

                                                                                                  What mass of Fe2<\/sub>O3<\/sub>\u00a0must be reacted to generate 324 g of Al2<\/sub>O3<\/sub>?<\/p>\r\nFe2<\/sub>O3<\/sub>(s) +\u00a02Al(s) \u2192\u00a02Fe(s) +\u00a0Al2<\/sub>O3<\/sub>(s)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                12. \r\n
                                                                                                  \r\n

                                                                                                  What mass of Fe is generated when 100.0 g of Al are reacted?<\/p>\r\nFe2<\/sub>O3<\/sub>(s) +\u00a02Al(s) \u2192\u00a02Fe(s) +\u00a0Al2<\/sub>O3<\/sub>(s)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                13. \r\n
                                                                                                  \r\n

                                                                                                  What mass of MnO2<\/sub>\u00a0is produced when 445 g of H2<\/sub>O are reacted?<\/p>\r\nH2<\/sub>O(\u2113) +\u00a02MnO4<\/sub>\u2212<\/sup>(aq) +\u00a0Br\u2212<\/sup>(aq) \u2192\u00a0BrO3<\/sub>\u2212<\/sup>(aq) +\u00a02MnO2<\/sub>(s) +\u00a02OH\u2212<\/sup>(aq)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                14. \r\n
                                                                                                  \r\n

                                                                                                  What mass of PbSO4<\/sub>\u00a0is produced when 29.6 g of H2<\/sub>SO4<\/sub>\u00a0are reacted?<\/p>\r\nPb(s) +\u00a0PbO2<\/sub>(s) +\u00a02H2<\/sub>SO4<\/sub>(aq) \u2192\u00a02PbSO4<\/sub>(s) +\u00a02H2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                15. \r\n
                                                                                                  \r\n

                                                                                                  If 83.9 g of ZnO are formed, what mass of Mn2<\/sub>O3<\/sub>\u00a0is formed with it?<\/p>\r\nZn(s) +\u00a02MnO2<\/sub>(s) \u2192\u00a0ZnO(s) +\u00a0Mn2<\/sub>O3<\/sub>(s)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                16. \r\n
                                                                                                  \r\n

                                                                                                  If 14.7 g of NO2<\/sub>\u00a0are reacted, what mass of H2<\/sub>O is reacted with it?<\/p>\r\n3NO2<\/sub>(g) +\u00a0H2<\/sub>O(\u2113) \u2192\u00a02HNO3<\/sub>(aq) +\u00a0NO(g)<\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                17. \r\n
                                                                                                  \r\n

                                                                                                  If 88.4 g of CH2<\/sub>S are reacted, what mass of HF is produced?<\/p>\r\nCH2<\/sub>S +\u00a06F2<\/sub>\u00a0\u2192\u00a0CF4<\/sub>\u00a0+\u00a02HF +\u00a0SF6<\/sub><\/span><\/span>\r\n\r\n<\/div><\/li>\r\n \t

                                                                                                18. \r\n
                                                                                                  \r\n

                                                                                                  If 100.0 g of Cl2<\/sub>\u00a0are needed, what mass of NaOCl must be reacted?<\/p>\r\nNaOCl +\u00a0HCl \u2192\u00a0NaOH +\u00a0Cl2<\/sub><\/span><\/span>\r\n\r\n[reveal-answer q=\"265420\"]Show Selected Answers[\/reveal-answer]\r\n[hidden-answer a=\"265420\"]\r\n

                                                                                                  \r\n

                                                                                                  1.<\/strong>\u00a0 44.0 g<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  3.<\/strong>\u00a0 3.00 \u00d7 102<\/sup>\u00a0g<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  5.<\/strong>\u00a0 0.0336 mol<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  7.<\/strong>\u00a0 0.0183 mol<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  9.<\/strong>\u00a0 45.1 g<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  11.\u00a0<\/strong> 507 g<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  13.<\/strong>\u00a0 4.30 \u00d7 103<\/sup>\u00a0g<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  15.<\/strong>\u00a0 163 g<\/p>\r\n\r\n<\/div>\r\n

                                                                                                  \r\n

                                                                                                  17.<\/strong>\u00a0 76.7 g<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div><\/li>\r\n<\/ol>","rendered":"

                                                                                                  Part I<\/strong><\/span><\/p>\n

                                                                                                    \n
                                                                                                  1. What is the total mass (amu) of carbon in each of the following molecules?\n
                                                                                                      \n
                                                                                                    1. CH4 \u00a0<\/sub><\/li>\n
                                                                                                    2. CHCl3 \u00a0<\/sub><\/li>\n
                                                                                                    3. C12<\/sub>H10<\/sub>O6 \u00a0<\/sub><\/li>\n
                                                                                                    4. CH3<\/sub>CH2<\/sub>CH2<\/sub>CH2<\/sub>CH3<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                    5. What is the total mass of hydrogen in each of the molecules?\n
                                                                                                        \n
                                                                                                      1. CH4 \u00a0<\/sub><\/li>\n
                                                                                                      2. CHCl3 \u00a0<\/sub><\/li>\n
                                                                                                      3. C12<\/sub>H10<\/sub>O6 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                      4. CH3<\/sub>CH2<\/sub>CH2<\/sub>CH2<\/sub>CH3<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                      5. Calculate the molecular or formula mass of each of the following:\n
                                                                                                          \n
                                                                                                        1. P4 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                        2. H2<\/sub>O<\/li>\n
                                                                                                        3. Ca(NO3<\/sub>)2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                        4. CH3<\/sub>CO2<\/sub>H (acetic acid)<\/li>\n
                                                                                                        5. C12<\/sub>H22<\/sub>O11<\/sub> (sucrose, cane sugar).<\/li>\n<\/ol>\n<\/li>\n
                                                                                                        6. Determine the molecular mass of the following compounds:\n
                                                                                                            \n
                                                                                                          1. \"A<\/li>\n
                                                                                                          2. \"A<\/li>\n
                                                                                                          3. \"A<\/li>\n
                                                                                                          4. \"A<\/li>\n<\/ol>\n<\/li>\n
                                                                                                          5. Determine the molecular mass of the following compounds:\n
                                                                                                              \n
                                                                                                            1. \"A<\/li>\n
                                                                                                            2. \"A<\/li>\n
                                                                                                            3. \"A<\/li>\n
                                                                                                            4. \"A<\/li>\n<\/ol>\n<\/li>\n
                                                                                                            5. Which molecule has a molecular mass of 28.05 amu?\n
                                                                                                                \n
                                                                                                              1. \"A<\/li>\n
                                                                                                              2. \"A<\/li>\n
                                                                                                              3. \"A<\/li>\n<\/ol>\n<\/li>\n
                                                                                                              4. Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.<\/li>\n
                                                                                                              5. Compare 1 mole of H2<\/sub>, 1 mole of O2<\/sub>, and 1 mole of F2<\/sub>.\n
                                                                                                                  \n
                                                                                                                1. Which has the largest number of molecules? Explain why.<\/li>\n
                                                                                                                2. Which has the greatest mass? Explain why.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                3. Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2<\/sub>H5<\/sub>OH), 0.60 mol of formic acid (HCO2<\/sub>H), or 1.0 mol of water (H2<\/sub>O)? Explain why.<\/li>\n
                                                                                                                4. Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C2<\/sub>H5<\/sub>OH), 1 mol of formic acid (HCO2<\/sub>H), or 1 mol of water (H2<\/sub>O)? Explain why.<\/li>\n
                                                                                                                5. How are the molecular mass and the molar mass of a compound similar and how are they different?<\/li>\n
                                                                                                                6. Calculate the molar mass of each of the following compounds:\n
                                                                                                                    \n
                                                                                                                  1. hydrogen fluoride, HF<\/li>\n
                                                                                                                  2. ammonia, NH3 \u00a0<\/sub><\/li>\n
                                                                                                                  3. nitric acid, HNO3 \u00a0<\/sub><\/li>\n
                                                                                                                  4. silver sulfate, Ag2<\/sub>SO4 \u00a0<\/sub><\/li>\n
                                                                                                                  5. boric acid, B(OH)3<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                  6. Calculate the molar mass of each of the following:\n
                                                                                                                      \n
                                                                                                                    1. S8 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                    2. C5<\/sub>H12 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                    3. Sc2<\/sub>(SO4<\/sub>)3 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                    4. CH3<\/sub>COCH3<\/sub> (acetone)<\/li>\n
                                                                                                                    5. C6<\/sub>H12<\/sub>O6<\/sub> (glucose)<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                    6. Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:\n
                                                                                                                        \n
                                                                                                                      1. limestone, CaCO3 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                      2. halite, NaCl<\/li>\n
                                                                                                                      3. beryl, Be3<\/sub>Al2<\/sub>Si6<\/sub>O18 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                      4. malachite, Cu2<\/sub>(OH)2<\/sub>CO3 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                      5. turquoise, CuAl6<\/sub>(PO4<\/sub>)4<\/sub>(OH)8<\/sub>(H2<\/sub>O)4<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                      6. Calculate the molar mass of each of the following:\n
                                                                                                                          \n
                                                                                                                        1. the anesthetic halothane, C2<\/sub>HBrClF3 \u00a0<\/sub><\/li>\n
                                                                                                                        2. the herbicide paraquat, C12<\/sub>H14<\/sub>N2<\/sub>Cl2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                        3. caffeine, C8<\/sub>H10<\/sub>N4<\/sub>O2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                        4. urea, CO(NH2<\/sub>)2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                        5. a typical soap, C17<\/sub>H35<\/sub>CO2<\/sub>Na<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                        6. Determine the number of moles of compound and the number of moles of each type of atom in each of the following:\n
                                                                                                                            \n
                                                                                                                          1. 25.0 g of propylene, C3<\/sub>H6 \u00a0<\/sub><\/li>\n
                                                                                                                          2. [latex]3.06\\times {10}^{-3}\\text{g}[\/latex] of the amino acid glycine, C2<\/sub>H5<\/sub>NO2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                          3. 25 lb of the herbicide Treflan, C13<\/sub>H16<\/sub>N2<\/sub>O4<\/sub>F (1 lb = 454 g)<\/li>\n
                                                                                                                          4. 0.125 kg of the insecticide Paris Green, Cu4<\/sub>(AsO3<\/sub>)2<\/sub>(CH3<\/sub>CO2<\/sub>)2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                          5. 325 mg of aspirin, C6<\/sub>H4<\/sub>(CO2<\/sub>H)(CO2<\/sub>CH3<\/sub>)<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                          6. Determine the mass of each of the following:\n
                                                                                                                              \n
                                                                                                                            1. 0.0146 mol KOH<\/li>\n
                                                                                                                            2. 10.2 mol ethane, C2<\/sub>H6 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                            3. [latex]1.6\\times {10}^{-3}\\text{ mol }{\\text{Na}}_{2}{\\text{SO}}_{4}[\/latex]<\/li>\n
                                                                                                                            4. [latex]6.854\\times {10}^{3}\\text{ mol glucose},{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}[\/latex]<\/li>\n
                                                                                                                            5. 2.86 mol Co(NH3<\/sub>)6<\/sub>Cl3<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                            6. Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:\n
                                                                                                                                \n
                                                                                                                              1. 2.12 g of potassium bromide, KBr<\/li>\n
                                                                                                                              2. 0.1488 g of phosphoric acid, H3<\/sub>PO4 \u00a0<\/sub><\/li>\n
                                                                                                                              3. 23 kg of calcium carbonate, CaCO3 \u00a0<\/sub><\/li>\n
                                                                                                                              4. 78.452 g of aluminum sulfate, Al2<\/sub>(SO4<\/sub>)3 \u00a0<\/sub><\/li>\n
                                                                                                                              5. 0.1250 mg of caffeine, C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                              6. Determine the mass of each of the following:\n
                                                                                                                                  \n
                                                                                                                                1. 2.345 mol LiCl<\/li>\n
                                                                                                                                2. 0.0872 mol acetylene, C2<\/sub>H2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                                3. [latex]3.3\\times {10}^{-2}\\text{ mol }{\\text{Na}}_{2}{\\text{CO}}_{3}[\/latex]<\/li>\n
                                                                                                                                4. [latex]1.23\\times {10}^{3}\\text{ mol fructose, }{\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}[\/latex]<\/li>\n
                                                                                                                                5. 0.5758 mol FeSO4<\/sub>(H2<\/sub>O)7<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                6. The approximate minimum daily dietary requirement of the amino acid leucine, C6<\/sub>H13<\/sub>NO2<\/sub>, is 1.1 g. What is this requirement in moles?<\/li>\n
                                                                                                                                7. Determine the mass in grams of each of the following:\n
                                                                                                                                    \n
                                                                                                                                  1. 0.600 mol of oxygen atoms<\/li>\n
                                                                                                                                  2. 0.600 mol of oxygen molecules, O2 \u00a0\u00a0<\/sub><\/li>\n
                                                                                                                                  3. 0.600 mol of ozone molecules, O3<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                  4. A 55-kg woman has [latex]7.5\\times {10}^{-3}\\text{mol}[\/latex] of hemoglobin (molar mass = 64,456 g\/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?<\/li>\n
                                                                                                                                  5. Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO4<\/sub>, a semiprecious stone.<\/li>\n
                                                                                                                                  6. Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH4<\/sub>, 0.6 mol of C6<\/sub>H6<\/sub>, or 0.4 mol of C3<\/sub>H8<\/sub>.<\/li>\n
                                                                                                                                  7. Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO4<\/sub>, 266 g of A12<\/sub>C16<\/sub>, or 225 g of A12<\/sub>S3<\/sub>.<\/li>\n
                                                                                                                                  8. Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?<\/li>\n
                                                                                                                                  9. The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone<\/li>\n
                                                                                                                                  10. One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?<\/li>\n
                                                                                                                                  11. A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12<\/sub>H22<\/sub>O11<\/sub>) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?<\/li>\n
                                                                                                                                  12. A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na2<\/sub>PO3<\/sub>F) in 100 mL\n
                                                                                                                                      \n
                                                                                                                                    1. What mass of fluorine atoms in mg was present?<\/li>\n
                                                                                                                                    2. How many fluorine atoms were present?<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                    3. Which of the following represents the least number of molecules?\n
                                                                                                                                        \n
                                                                                                                                      1. 20.0 g of H2<\/sub>O (18.02 g\/mol)<\/li>\n
                                                                                                                                      2. 77.0 g of CH4<\/sub> (16.06 g\/mol)<\/li>\n
                                                                                                                                      3. 68.0 g of CaH2<\/sub> (42.09 g\/mol)<\/li>\n
                                                                                                                                      4. 100.0 g of N2<\/sub>O (44.02 g\/mol)<\/li>\n
                                                                                                                                      5. 84.0 g of HF (20.01 g\/mol)<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n
                                                                                                                                        Show Selected Answers<\/span><\/p>\n
                                                                                                                                        \n

                                                                                                                                        1. Each\u00a0molecule has the following mass (amu)\u00a0of carbon.<\/p>\n

                                                                                                                                          \n
                                                                                                                                        1. [latex]1\\times 12.01\\text{ amu}=12.01\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                        2. [latex]1\\times 12.01\\text{ amu}=12.01\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                        3. [latex]12\\times 12.01\\text{ amu}=144.12\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                        4. [latex]5\\times 12.01\\text{ amu}=60.05\\text{ amu}[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                          3. The molecular or formula masses are as follows:<\/p>\n

                                                                                                                                            \n
                                                                                                                                          1. [latex]4\\times 30.974\\text{ amu}=123.896\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                          2. [latex]2\\times 1.008\\text{ amu}+15.999\\text{ amu}=18.015\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                          3. [latex]40\\times 0.078\\text{ amu}+2\\times 14.007\\text{ amu}+6\\times 15.999\\text{ amu}=164.086\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                          4. [latex]2\\times 12.011\\text{ amu}+4\\times 1.008\\text{ amu}+2\\times 15.999\\text{ amu}=60.052\\text{ amu}[\/latex]<\/li>\n
                                                                                                                                          5. [latex]12\\times 12.011\\text{ amu}+22\\times 1.008\\text{ amu}\\times 11\\times 15.999\\text{ amu}=342.297\\text{ amu}[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                            5. \u00a0The molecular mass of each compound is as follows:<\/p>\n

                                                                                                                                              \n
                                                                                                                                            1. C4<\/sub>H8
                                                                                                                                              \n<\/sub>[latex]\\begin{array}{ll}4\\text{C}\\times 12.011\\hfill & =48.044\\text{amu}\\hfill \\\\ 8\\text{H}\\times 1.0079\\hfill & =\\underline{8.06352\\text{amu}}\\hfill \\\\ \\hfill & =56.107\\text{amu}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                            2. C4<\/sub>H6
                                                                                                                                              \n<\/sub>[latex]\\begin{array}{ll}4\\text{C}\\times 12.011\\hfill & =48.044\\text{amu}\\hfill \\\\ 6\\text{H}\\times 1.0079\\hfill & =\\underline{6.0474\\text{amu}}\\hfill \\\\ \\hfill & =54.091\\text{amu}\\hfill \\end{array}[\/latex]<\/span><\/li>\n
                                                                                                                                            3. H2<\/sub>Si2<\/sub>Cl4
                                                                                                                                              \n<\/sub>[latex]\\begin{array}{ll}\\hfill 2\\text{H}\\times 1.0079& =2.01558\\text{amu}\\hfill \\\\ 2\\text{Si}\\times 28.0855\\hfill & =56.1710\\text{amu}\\hfill \\\\ 4\\text{Cl}\\times 35.4527\\hfill & =\\underline{141.8108\\text{amu}}\\hfill \\\\ \\hfill & =199.9976\\text{amu}\\hfill \\end{array}[\/latex]<\/span><\/li>\n
                                                                                                                                            4. H3<\/sub>PO4
                                                                                                                                              \n<\/sub>[latex]\\begin{array}{ll}\\hfill 3\\text{H}\\times 1.0079& =3.0237\\text{amu}\\hfill \\\\ 1\\text{P}\\times 30.973762\\hfill & =30.973762\\text{amu}\\hfill \\\\ \\hfill 4\\text{O}\\times 15.9994& =\\underline{63.9976\\text{amu}}\\hfill \\\\ \\hfill & =97.9950\\text{amu}\\hfill \\end{array}[\/latex]<\/span><\/li>\n<\/ol>\n

                                                                                                                                              7.\u00a0Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.<\/p>\n

                                                                                                                                              9.\u00a0Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.<\/p>\n

                                                                                                                                              11.The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of [latex]6.022\\times {10}^{23}[\/latex] molecules.<\/p>\n

                                                                                                                                              13. \u00a0The molecular mass of each compound is as follows:<\/p>\n

                                                                                                                                                \n
                                                                                                                                              1. S8<\/sub>:
                                                                                                                                                \n[latex]8\\text{S}=8\\times 32.066=256.528\\text{g\/mol}[\/latex]<\/li>\n
                                                                                                                                              2. C5<\/sub>H12<\/sub>:
                                                                                                                                                \n[latex]\\begin{array}{lll}\\hfill 5\\text{C}=5\\times 12.011& =& 60.055\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{12H}=12\\times 1.00794& =& \\underline{12.09528\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 72.150\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                              3. Sc2<\/sub>(SO4<\/sub>)3<\/sub>:
                                                                                                                                                \n[latex]\\begin{array}{lll}\\hfill 2\\text{Sc}=2\\times 44.9559109& =& 89.9118218\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{3S}=3\\times 32.066& =& 96.198\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 12\\text{O}=12\\times 15.99943& =& \\underline{191.99316\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 378.103\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                              4. CH3<\/sub>COCH3<\/sub>:
                                                                                                                                                \n[latex]\\begin{array}{lll}\\hfill 3\\text{C}=3\\times 12.011& =& 36.033\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1O}=1\\times 15.9994& =& 15.9994\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 6\\text{H}=6\\times 1.00794& =& \\underline{6.04764\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 58.080\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                              5. C6<\/sub>H12<\/sub>O6<\/sub>:
                                                                                                                                                \n[latex]\\begin{array}{lll}\\hfill \\text{6C}=6\\times 12.011& =& 72.066\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{12H}=12\\times 1.00794& =& 12.09528\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 6\\text{O}=6\\times 15.9994& =& \\underline{95.9964\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 180.158\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                                15. The molecular mass of each compound is as follows:<\/p>\n

                                                                                                                                                  \n
                                                                                                                                                1. C2<\/sub>HBrClF3<\/sub>:
                                                                                                                                                  \n[latex]\\begin{array}{lll}\\hfill \\text{2C}=2\\times 12.011& =& 24.022\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1H}=1\\times 1.00794& =& 1.00794\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1Br}=1\\times 79.904& =& 79.904\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1Cl}=1\\times 35.453& =& 35.453\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill 3\\text{F}=3\\times 18.998403& =& \\underline{56.995209\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 197.682\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                2. C12<\/sub>H14<\/sub>N2<\/sub>Cl2<\/sub>:
                                                                                                                                                  \n[latex]\\begin{array}{lll}\\hfill \\text{12C}=12\\times 12.011& =& 144.132\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{14H}=14\\times 1.00794& =& 14.111\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2N}=2\\times 14.0067& =& 28.0134\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2Cl}=2\\times 35.453& =& \\underline{70.906\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 257.163\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                3. C8<\/sub>H10<\/sub>N4<\/sub>O2<\/sub>:
                                                                                                                                                  \n[latex]\\begin{array}{lll}\\hfill \\text{8C}=8\\times 12.011& =& 96.088\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{10H}=10\\times 1.007& =& 10.079\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{4N}=4\\times 14.0067& =& 56.027\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2O}=2\\times 15.9994& =& \\underline{31.999\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 194.193\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                4. CO(NH2<\/sub>)2<\/sub>:
                                                                                                                                                  \n[latex]\\begin{array}{lll}\\hfill \\text{1C}=1\\times 12.011& =& 12.011\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1O}=1\\times 15.9994& =& 15.9994\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2N}=2\\times 14.0067& =& 28.0134\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{4H}=4\\times 1.00794& =& \\underline{4.03176\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 60.056\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                5. C17<\/sub>H35<\/sub>CO2<\/sub>Na:
                                                                                                                                                  \n[latex]\\begin{array}{lll}\\hfill \\text{18C}=18\\times 12.011& =& 216.198\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{35H}=35\\times 1.00794& =& 35.2779\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{2O}=2\\times 15.9994& =& 31.9988\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{1Na}=1\\times 22.98977& =& \\underline{22.98977\\text{g}{\\text{mol}}^{-1}}\\hfill \\\\ \\hfill & =& 306.464\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                                  17. \u00a0The mass of each compound is as follows:<\/p>\n

                                                                                                                                                    \n
                                                                                                                                                  1. KOH:
                                                                                                                                                    \n[latex]\\begin{array}{lll}\\hfill 1\\text{K}=1\\times 39.0983& =& 39.0983\\hfill \\\\ \\hfill 1\\text{O}=1\\times 15.9994& =& 15.9994\\hfill \\\\ \\hfill \\text{1H}=1\\times 1.00794& =& \\underline{1.00794}\\hfill \\\\ \\hfill \\text{molar mass}& =& 56.1056\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]
                                                                                                                                                    \n[latex]\\text{Mass}=0.0146\\text{mol}\\times 56.1056\\text{g\/mol}=0.819\\text{g}[\/latex]<\/li>\n
                                                                                                                                                  2. C2<\/sub>H6<\/sub>:
                                                                                                                                                    \n<\/sub>[latex]\\begin{array}{lll}\\hfill 2\\text{C}=2\\times 12.011& =& 24.022\\hfill \\\\ \\hfill 6\\text{H}=6\\times 1.00794& =& \\underline{6.04764}\\hfill \\\\ \\hfill \\text{molar mass}& =& 30.070\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]
                                                                                                                                                    \n[latex]\\text{Mass}=10.2\\text{mol}\\times 30.070\\text{g\/mol}=307\\text{g}[\/latex]<\/li>\n
                                                                                                                                                  3. Na2<\/sub>SO4<\/sub>:
                                                                                                                                                    \n[latex]\\begin{array}{lll}\\hfill 2\\text{Na}=2\\times 22.990& =& 45.98\\hfill \\\\ \\hfill 1\\text{S}=1\\times 32.066& =& 32.066\\hfill \\\\ \\hfill 4\\text{O}=4\\times 15.9994& =& \\underline{63.9976}\\hfill \\\\ \\hfill \\text{molar mass}& =& 142.044\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]
                                                                                                                                                    \n[latex]\\text{Mass}=1.6\\times {10}^{-3}\\text{mol}\\times 142.044\\text{g\/mol}=\\text{0.23 g}[\/latex]<\/li>\n
                                                                                                                                                  4. C6<\/sub>H12<\/sub>O6<\/sub>:
                                                                                                                                                    \n<\/sub>[latex]\\begin{array}{lll}\\hfill 6\\text{C}=6\\times 12.011& =& 72.066\\hfill \\\\ \\hfill 12\\text{H}=12\\times 1.00794& =& 12.0953\\hfill \\\\ \\hfill 6\\text{O}=6\\times 15.9994& =& \\underline{95.9964}\\hfill \\\\ \\hfill \\text{molar mass}& =& 180.158\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]
                                                                                                                                                    \n[latex]\\text{Mass}=6.854\\times {10}^{3}\\text{mol}\\times 180.158\\text{g\/mol}=1.235\\times {10}^{6}\\text{g (}1235\\text{kg)}[\/latex]<\/li>\n
                                                                                                                                                  5. Co(NH3<\/sub>)6<\/sub>Cl3<\/sub>:
                                                                                                                                                    \n<\/sub>[latex]\\begin{array}{lll}\\hfill \\text{Co}=1\\times 58.99320& =& 58.99320\\hfill \\\\ \\hfill 6\\text{N}=6\\times 14.0067& =& 84.0402\\hfill \\\\ \\hfill 18\\text{H}=18\\times 1.00794& =& 18.1429\\hfill \\\\ \\hfill 3\\text{Cl}=3\\times 35.4527& =& \\underline{106.358}\\hfill \\\\ \\hfill \\text{molar mass}& =& 267.5344\\text{g}{\\text{mol}}^{-1}\\hfill \\end{array}[\/latex]
                                                                                                                                                    \n<\/span>[latex]\\text{Mass}=2.856\\text{mol}\\times 267.5344\\text{g\/mol}=765\\text{g}[\/latex]<\/span><\/li>\n<\/ol>\n

                                                                                                                                                    19.<\/p>\n

                                                                                                                                                      \n
                                                                                                                                                    1. 2.345 mol LiCl:
                                                                                                                                                      \n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left(\\text{LiCl}\\right)=1\\times 6.941+1\\times 35.4527& =& 42.394\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=2.345\\cancel{\\text{mol}}\\times 42.394\\text{g}\\cancel{{\\text{mol}}^{-1}}& =& 99.41\\text{g}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                    2. 0.0872 mol acetylene, C2<\/sub>H2<\/sub>:
                                                                                                                                                      \n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left({\\text{C}}_{2}{\\text{H}}_{2}\\right)=2\\times 12.011+2\\times 1.00794& =& 26.038\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=0.0872\\cancel{\\text{mol}}\\times 26.038\\text{g}\\cancel{{\\text{mol}}^{-1}}& =& 2.27\\text{g}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                    3. 3.3\u00a0\u00d7\u00a010\u22122<\/sup>\u00a0mol Na2<\/sub>CO3<\/sub>:
                                                                                                                                                      \n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left({\\text{Na}}_{2}{\\text{CO}}_{3}\\right)=2\\times 22.989768+1\\times 12.011+3\\times 15.9994& =& 105.989\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=3.3\\times {10}^{-2}\\cancel{\\text{mol}}\\times 105.989\\text{g}{\\text{mol}}^{-1}& =& 3.5\\text{g}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                    4. 1.23\u00a0\u00d7\u00a0103<\/sup>\u00a0mol fructose, C6<\/sub>H12<\/sub>O6<\/sub>:[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left({\\text{C}}_{6}{\\text{H}}_{12}{\\text{O}}_{6}\\right)=6\\times 12.011+12\\times 1.00794+6\\times 15.9994& =& 180.158\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=1.23\\times {10}^{3}\\cancel{\\text{mol}}\\times 180.158\\text{g}{\\text{mol}}^{-1}=2.22\\times {10}^{5}\\text{g}& =& 222\\text{kg}\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                    5. 0.5758 mol FeSO4<\/sub>(H2<\/sub>O)7<\/sub>:
                                                                                                                                                      \n[latex]\\begin{array}{lll}\\hfill \\text{molar mass}\\left[{\\text{FeSO}}_{4}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{7}\\right]& =& 1\\times 55.847+1\\times 32.066+4\\times 15.999\\hfill \\\\ \\hfill +7\\left(2\\times 1.00794+15.9994\\right)& =& 278.018\\text{g}{\\text{mol}}^{-1}\\hfill \\\\ \\hfill \\text{mass}=0.5758\\cancel{\\text{mol}}\\times 278.018\\text{g}\\cancel{{\\text{mol}}^{-1}}& =\\hfill & 160.1\\text{g}\\hfill \\end{array}[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                                      21.\u00a0The mass of each compound is as follows:<\/p>\n

                                                                                                                                                        \n
                                                                                                                                                      1. [latex]0.600\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=9.60\\text{g}[\/latex]<\/li>\n
                                                                                                                                                      2. [latex]0.600\\cancel{\\text{mol}}\\times 2\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=19.2\\text{g}[\/latex]<\/li>\n
                                                                                                                                                      3. [latex]0.600\\cancel{\\text{mol}}\\times 3\\times 15.994\\text{g\/}\\cancel{\\text{mol}}=28.8\\text{g}[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                                        23.\u00a0Determine the number of moles of each component. From the moles, calculate the number of atoms and the mass of the elements involved. Zirconium: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times 1023\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 91.224\\text{g\/}\\cancel{\\text{mol}}=30.87\\text{g;}[\/latex] Silicon: [latex]0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=2.038\\times {10}^{23}\\text{atoms;}0.3384\\cancel{\\text{mol}}\\times 28.0855\\text{g\/}\\cancel{\\text{mol}}=9.504\\text{g;}[\/latex] Oxygen: [latex]4\\times 0.3384\\cancel{\\text{mol}}\\times 6.022\\times {10}^{23}{\\cancel{\\text{mol}}}^{\\cancel{-1}}=8.151\\times {10}^{23}\\text{atoms;}4\\times 0.3384\\cancel{\\text{mol}}\\times 15.9994\\text{g\/}\\cancel{\\text{mol}}=21.66\\text{g}[\/latex]<\/p>\n

                                                                                                                                                        25. Determine the molar mass and, from the grams present, the moles of each substance. The compound with the greatest number of moles of Al has the greatest mass of Al.<\/p>\n

                                                                                                                                                          \n
                                                                                                                                                        • Molar mass AlPO4<\/sub>: 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g\/mol<\/li>\n
                                                                                                                                                        • Molar mass Al2<\/sub>Cl6<\/sub>: 2(26.981539) + 6(35.4527) = 266.6793 g\/mol<\/li>\n
                                                                                                                                                        • Molar mass Al2<\/sub>S3<\/sub>: 2(26.981539) + 3(32.066) = 150.161 g\/mol<\/li>\n<\/ul>\n

                                                                                                                                                          AlPO4<\/sub>: [latex]\\frac{122\\cancel{\\text{g}}}{121.9529\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.000\\text{mol.}[\/latex]<\/p>\n

                                                                                                                                                          [latex]\\text{mol Al}=1\\times 1.000\\text{mol}=1.000\\text{mol}[\/latex]<\/p>\n

                                                                                                                                                          Al2<\/sub>Cl6<\/sub>: [latex]\\frac{266\\text{g}}{266.6793\\text{g}{\\text{mol}}^{-1}}=0.997\\text{mol}[\/latex]<\/p>\n

                                                                                                                                                          [latex]\\text{mol Al}=2\\times 0.997\\text{mol}=1.994\\text{mol}[\/latex]<\/p>\n

                                                                                                                                                          Al2<\/sub>S3<\/sub>: [latex]\\frac{225\\cancel{\\text{g}}}{150.161\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.50\\text{mol}[\/latex]<\/p>\n

                                                                                                                                                          [latex]\\text{mol Al}=2\\times 1.50\\text{mol}=3.00\\text{mol}[\/latex]<\/p>\n

                                                                                                                                                          27. \u00a0Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro\u2019s number by the number of moles:<\/p>\n

                                                                                                                                                          [latex]\\frac{3104\\cancel{\\text{carats}}\\times \\frac{200\\cancel{\\text{mg}}}{1\\cancel{\\text{carat}}}\\times \\frac{1\\cancel{\\text{g}}}{1000\\cancel{\\text{mg}}}}{12.011\\cancel{\\text{g}}\\cancel{{\\text{mol}}^{-1}}\\left(6.022\\times {10}^{23}\\cancel{{\\text{mol}}^{-1}}\\right)}=3.113\\times {10}^{25}\\text{C atoms}[\/latex]<\/p>\n

                                                                                                                                                          29. \u00a0Determine the molar mass of sugar. 12(12.011) + 22(1.00794) + 11(15.9994) = 342.300 g\/mol; Then [latex]0.0278\\text{mol}\\times 342.300\\text{g\/mol}=9.52\\text{g sugar.}[\/latex] This 9.52 g of sugar represents [latex]\\frac{11.0}{60.0}[\/latex] of one serving or<\/p>\n

                                                                                                                                                          [latex]\\frac{60.0\\text{g serving}}{11.0\\cancel{\\text{g sugar}}}\\times 9.52\\cancel{\\text{g sugar}}=51.9\\text{g cereal.}[\/latex]<\/p>\n

                                                                                                                                                          This amount is [latex]\\frac{51.9\\text{g cereal}}{60.0\\text{g serving}}=0.865[\/latex] servings, or about 1 serving.<\/p>\n

                                                                                                                                                          31.\u00a0Calculate the number of moles of each species, then remember that 1 mole of anything [latex]=6.022\\times {10}^{23}[\/latex] species.<\/p>\n

                                                                                                                                                            \n
                                                                                                                                                          1. 20.0 g = 1.11 mol H2<\/sub>O<\/li>\n
                                                                                                                                                          2. 77.0 g CH4<\/sub> = 4.79 mol CH4<\/sub><\/li>\n
                                                                                                                                                          3. 68.0 g CaH2<\/sub> = 1.62 mol CaH2<\/sub><\/li>\n
                                                                                                                                                          4. 100.0 g N2<\/sub>O = 2.27 mol N2<\/sub>O<\/li>\n
                                                                                                                                                          5. 84.0 g HF = 4.20 mol HF<\/li>\n<\/ol>\n

                                                                                                                                                            Therefore, 20.0 g H2<\/sub>O represents the least number of molecules since it has the least number of moles.<\/p>\n<\/div>\n<\/div>\n

                                                                                                                                                            Part II<\/strong><\/span><\/p>\n

                                                                                                                                                              \n
                                                                                                                                                            1. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n
                                                                                                                                                                \n
                                                                                                                                                              1. The number of moles and the mass of chlorine, Cl2<\/sub>, required to react with 10.0 g of sodium metal, Na, to produce sodium chloride, NaCl.<\/li>\n
                                                                                                                                                              2. The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide.<\/li>\n
                                                                                                                                                              3. The number of moles and the mass of sodium nitrate, NaNO3<\/sub>, required to produce 128 g of oxygen. (NaNO2<\/sub> is the other product.)<\/li>\n
                                                                                                                                                              4. The number of moles and the mass of carbon dioxide formed by the combustion of 20.0 kg of carbon in an excess of oxygen.<\/li>\n
                                                                                                                                                              5. The number of moles and the mass of copper(II) carbonate needed to produce 1.500 kg of copper(II) oxide. (CO2<\/sub> is the other product.)<\/li>\n
                                                                                                                                                              6. \"This<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                              7. Determine the number of moles and the mass requested for each reaction in Exercise 1.<\/li>\n
                                                                                                                                                              8. Write the balanced equation, then outline the steps necessary to determine the information requested in each of the following:\n
                                                                                                                                                                  \n
                                                                                                                                                                1. The number of moles and the mass of Mg required to react with 5.00 g of HCl and produce MgCl2<\/sub> and H2<\/sub>.<\/li>\n
                                                                                                                                                                2. The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of silver(I) oxide.<\/li>\n
                                                                                                                                                                3. The number of moles and the mass of magnesium carbonate, MgCO3<\/sub>, required to produce 283 g of carbon dioxide. (MgO is the other product.)<\/li>\n
                                                                                                                                                                4. The number of moles and the mass of water formed by the combustion of 20.0 kg of acetylene, C2<\/sub>H2<\/sub>, in an excess of oxygen.<\/li>\n
                                                                                                                                                                5. The number of moles and the mass of barium peroxide, BaO2<\/sub>, needed to produce 2.500 kg of barium oxide, BaO (O2<\/sub> is the other product.)<\/li>\n
                                                                                                                                                                6. \"This<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                7. Determine the number of moles and the mass requested for each reaction in Exercise 3.<\/li>\n
                                                                                                                                                                8. H2<\/sub> is produced by the reaction of 118.5 mL of a 0.8775-M solution of H3<\/sub>PO4<\/sub> according to the following equation: [latex]2\\text{Cr}+2{\\text{H}}_{3}{\\text{PO}}_{4}\\rightarrow 3{\\text{H}}_{2}+2{\\text{CrPO}}_{4}\\text{.}[\/latex]\n
                                                                                                                                                                    \n
                                                                                                                                                                  1. Outline the steps necessary to determine the number of moles and mass of H2<\/sub>.<\/li>\n
                                                                                                                                                                  2. Perform the calculations outlined.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                  3. Gallium chloride is formed by the reaction of 2.6 L of a 1.44 M solution of HCl according to the following equation: [latex]2\\text{Ga}+6\\text{HCl}\\rightarrow 2{\\text{GaCl}}_{3}+3{\\text{H}}_{2}\\text{.}[\/latex]\n
                                                                                                                                                                      \n
                                                                                                                                                                    1. Outline the steps necessary to determine the number of moles and mass of gallium chloride.<\/li>\n
                                                                                                                                                                    2. Perform the calculations outlined.<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                    3. I2<\/sub> is produced by the reaction of 0.4235 mol of CuCl2<\/sub> according to the following equation: [latex]2{\\text{CuCl}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+4\\text{KCl}+{\\text{I}}_{2}\\text{.}[\/latex]\n
                                                                                                                                                                        \n
                                                                                                                                                                      1. How many molecules of I2<\/sub> are produced?<\/li>\n
                                                                                                                                                                      2. What mass of I2<\/sub> is produced?<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                      3. Silver is often extracted from ores as K[Ag(CN)2<\/sub>] and then recovered by the reaction [latex]2\\text{K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\text{(}aq\\text{)}+\\text{Zn}\\text{(}s\\text{)}\\rightarrow 2Ag\\text{(}s\\text{)}+\\text{Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{(}aq\\text{)}+2\\text{KCN}\\text{(}aq\\text{)}[\/latex]\n
                                                                                                                                                                          \n
                                                                                                                                                                        1. How many molecules of Zn(CN)2<\/sub> are produced by the reaction of 35.27 g of K[Ag(CN)2<\/sub>]?<\/li>\n
                                                                                                                                                                        2. What mass of Zn(CN)2<\/sub> is produced?<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                        3. What mass of silver oxide, Ag2<\/sub>O, is required to produce 25.0 g of silver sulfadiazine, AgC10<\/sub>H9<\/sub>N4<\/sub>SO2<\/sub>, from the reaction of silver oxide and sulfadiazine? [latex]2{\\text{C}}_{10}{\\text{H}}_{10}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{Ag}}_{2}\\text{O}\\rightarrow 2{\\text{AgC}}_{10}{\\text{H}}_{9}{\\text{N}}_{4}{\\text{SO}}_{2}+{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\n
                                                                                                                                                                        4. Carborundum is silicon carbide, SiC, a very hard material used as an abrasive on sandpaper and in other applications. It is prepared by the reaction of pure sand, SiO2<\/sub>, with carbon at high temperature. Carbon monoxide, CO, is the other product of this reaction. Write the balanced equation for the reaction, and calculate how much SiO2<\/sub> is required to produce 3.00 kg of SiC.<\/li>\n
                                                                                                                                                                        5. Automotive air bags inflate when a sample of sodium azide, NaN3<\/sub>, is very rapidly decomposed.[latex]2{\\text{NaN}}_{3}\\text{(}s\\text{)}\\rightarrow 2\\text{Na}\\text{(}s\\text{)}+3{\\text{N}}_{2}\\text{(}g\\text{)}[\/latex]\u00a0What mass of sodium azide is required to produce 2.6 ft3<\/sup> (73.6 L) of nitrogen gas with a density of 1.25 g\/L?<\/li>\n
                                                                                                                                                                        6. Urea, CO(NH2<\/sub>)2<\/sub>, is manufactured on a large scale for use in producing urea-formaldehyde plastics and as a fertilizer. What is the maximum mass of urea that can be manufactured from the CO2<\/sub> produced by combustion of [latex]1.00\\times {10}^{3}\\text{kg}[\/latex] of carbon followed by the reaction? [latex]{\\text{CO}}_{2}\\text{(}g\\text{)}+2{\\text{NH}}_{3}\\text{(}g\\text{)}\\rightarrow\\text{CO}{\\text{(}{\\text{NH}}_{2}\\text{)}}_{2}\\text{(}s\\text{)}+{\\text{H}}_{2}\\text{O}\\text{(}l\\text{)}[\/latex]<\/li>\n
                                                                                                                                                                        7. In an accident, a solution containing 2.5 kg of nitric acid was spilled. Two kilograms of Na2<\/sub>CO3<\/sub> was quickly spread on the area and CO2<\/sub> was released by the reaction. Was sufficient Na2<\/sub>CO3<\/sub> used to neutralize all of the acid?<\/li>\n
                                                                                                                                                                        8. A compact car gets 37.5 miles per gallon on the highway. If gasoline contains 84.2% carbon by mass and has a density of 0.8205 g\/mL, determine the mass of carbon dioxide produced during a 500-mile trip (3.785 liters per gallon).<\/li>\n
                                                                                                                                                                        9. What volume of a 0.750 M solution of hydrochloric acid, a solution of HCl, can be prepared from the HCl produced by the reaction of 25.0 g of NaCl with an excess of sulfuric acid? [latex]\\text{NaCl}\\text{(}s\\text{)}+{\\text{H}}_{2}{\\text{SO}}_{4}\\text{(}l\\text{)}\\rightarrow\\text{HCl}\\text{(}g\\text{)}+{\\text{NaHSO}}_{4}\\text{(}s\\text{)}[\/latex]<\/li>\n
                                                                                                                                                                        10. What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3<\/sub>)2<\/sub> in 43.88 mL of a 0.3842 M solution of Cu(NO3<\/sub>)2<\/sub>? [latex]2\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}+4\\text{KI}\\rightarrow 2\\text{CuI}+{\\text{I}}_{2}+4{\\text{KNO}}_{3}[\/latex]<\/li>\n
                                                                                                                                                                        11. A mordant is a substance that combines with a dye to produce a stable fixed color in a dyed fabric. Calcium acetate is used as a mordant. It is prepared by the reaction of acetic acid with calcium hydroxide.[latex]2{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}+\\text{Ca}{\\text{(}\\text{OH}\\text{)}}_{2}\\rightarrow\\text{Ca}{\\text{(}{\\text{CH}}_{3}{\\text{CO}}_{2}\\text{)}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex] What mass of Ca(OH)2<\/sub> is required to react with the acetic acid in 25.0 mL of a solution having a density of 1.065 g\/mL and containing 58.0% acetic acid by mass?<\/li>\n
                                                                                                                                                                        12. The toxic pigment called white lead, Pb3<\/sub>(OH)2<\/sub>(CO3<\/sub>)2<\/sub>, has been replaced in white paints by rutile, TiO2<\/sub>. How much rutile (g) can be prepared from 379 g of an ore that contains 88.3% ilmenite (FeTiO3<\/sub>) by mass?[latex]2{\\text{FeTiO}}_{3}+4\\text{HCl}+{\\text{Cl}}_{2}\\rightarrow 2{\\text{FeCl}}_{3}+2{\\text{TiO}}_{2}+2{\\text{H}}_{2}\\text{O}[\/latex]<\/li>\n<\/ol>\n
                                                                                                                                                                          Show Selected Answers<\/span><\/p>\n
                                                                                                                                                                          \n

                                                                                                                                                                          2. (a)\u00a00.435 mol Na, 0.271 mol Cl2<\/sub>, 15.4 g Cl2<\/sub>;<\/p>\n

                                                                                                                                                                          (b) 0.005780 mol HgO, 2.890 \u00d7 10\u22123<\/sup> mol O2<\/sub>, 9.248 \u00d7 10\u22122<\/sup> g O2<\/sub>;<\/p>\n

                                                                                                                                                                          (c) 8.00 mol NaNO3<\/sub>, 6.8 \u00d7 102<\/sup> g NaNO3<\/sub>;<\/p>\n

                                                                                                                                                                          (d) 1665 mol CO2<\/sub>, 73.3 kg CO2<\/sub>;<\/p>\n

                                                                                                                                                                          (e) 18.86 mol CuO, 2.330 kg CuCO3<\/sub>;<\/p>\n

                                                                                                                                                                          (f) 0.4580 mol C2<\/sub>H4<\/sub>Br2<\/sub>, 86.05 g C2<\/sub>H4<\/sub>Br2<\/sub><\/p>\n

                                                                                                                                                                          4. \u00a0(a) [latex]\\text{mol Mg}=5.00\\cancel{\\text{g HCl}}\\times \\frac{1\\cancel{\\text{mol HCl}}}{36.4606\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol Mg}}{2\\cancel{\\text{mol HCl}}}=0.0686\\text{mol,}[\/latex] [latex]\\text{g Mg}=0.0686\\cancel{\\text{mol Mg}}\\times \\frac{\\text{24.305 g}}{1\\cancel{\\text{mol Mg}}}=\\text{1.67 g};[\/latex]
                                                                                                                                                                          \n(b) [latex]{\\text{mol O}}_{2}=1.252\\cancel{\\text{g}{\\text{Ag}}_{2}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}{231.7358\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{O}}_{2}}{2\\cancel{\\text{mol}{\\text{Ag}}_{2}\\text{O}}}=2.701\\times {10}^{-3}[\/latex], [latex]{\\text{g O}}_{2}=2.701\\times {10}^{-3}\\cancel{{\\text{mol O}}_{2}}\\times \\frac{\\text{31.9988 g}}{1\\cancel{\\text{mol}{\\text{O}}_{2}}}=0.08644\\text{g;}[\/latex]<\/p>\n

                                                                                                                                                                          (c) [latex]{\\text{mol MgCO}}_{3}=283\\cancel{\\text{g}{\\text{CO}}_{2}}\\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{44.010\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{MgCO}}_{3}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=6.43\\text{mol,}[\/latex] [latex]{\\text{g MgCO}}_{3}=6.43\\cancel{{\\text{mol MgCO}}_{3}}\\times \\frac{\\text{84.314 g}}{1\\cancel{{\\text{mol MgCO}}_{3}}}=542\\text{g;}[\/latex]<\/p>\n

                                                                                                                                                                          (d) [latex]{\\text{mol H}}_{2}\\text{O}=2.00\\times {10}^{4}\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}{28.054\\cancel{\\text{g}}}\\times \\frac{\\text{1 mol}{\\text{H}}_{2}\\text{O}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{2}}}=713\\text{mol,}[\/latex] [latex]{\\text{g H}}_{2}\\text{O}=713\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}\\times \\frac{18.01528\\cancel{\\text{g}}}{1\\cancel{\\text{mol}{\\text{H}}_{2}\\text{O}}}\\times \\frac{\\text{1 kg}}{1000\\cancel{\\text{g}}}=\\text{12.8 kg}[\/latex]<\/p>\n

                                                                                                                                                                          (e) [latex]2.500\\cancel{\\text{kg BaO}}\\times \\frac{1000\\cancel{\\text{g BaO}}}{1\\cancel{\\text{kg BaO}}}\\times \\frac{1\\cancel{\\text{mol BaO}}}{153.326\\cancel{\\text{g BaO}}}\\times \\frac{\\text{2 mol}{\\text{BaO}}_{2}}{2\\cancel{\\text{mol BaO}}}=\\text{16.31 mol}{\\text{BaO}}_{2}[\/latex] [latex]16.31\\cancel{\\text{mol}{\\text{BaO}}_{2}}\\times \\frac{\\text{169.326 g}{\\text{BaO}}_{2}}{1\\cancel{\\text{mol}{\\text{BaO}}_{2}}}=\\text{2762 g}{\\text{BaO}}_{2}[\/latex]<\/p>\n

                                                                                                                                                                          (f) [latex]9.55\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}\\times \\frac{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}{46.068\\cancel{\\text{g}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}\\times \\frac{\\text{1 mol}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{6}\\text{O}}}=\\text{0.207 mol}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex] [latex]0.207\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}\\times \\frac{\\text{28.053 g}{\\text{C}}_{2}{\\text{H}}_{4}}{1\\cancel{\\text{mol}{\\text{C}}_{2}{\\text{H}}_{4}}}=\\text{5.81 g}{\\text{C}}_{2}{\\text{H}}_{4}[\/latex]<\/p>\n

                                                                                                                                                                          6. (a) [latex]\\text{volume HCl solution}\\rightarrow\\text{mol HCl}\\rightarrow{\\text{mol GaCl}}_{3};[\/latex]<\/p>\n

                                                                                                                                                                          (b) [latex]2.6\\cancel{\\text{L HCl}}\\times \\frac{1.44\\cancel{\\text{mol HCl}}}{1\\cancel{\\text{L HCl}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}{6\\cancel{\\text{mol HCl}}}\\times \\frac{\\text{180.079 g}{\\text{GaCl}}_{3}}{1\\cancel{\\text{mol}{\\text{GaCl}}_{3}}}=2.3\\times {10}^{2}\\text{g}{\\text{GaCl}}_{3}[\/latex]<\/p>\n

                                                                                                                                                                          8.\u00a0The development requires the following: [latex]\\text{mass K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]\\rightarrow\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\rightarrow\\text{molecules of Zn}{\\text{(}\\text{CN}\\text{)}}_{2}\\text{g Zn}{\\text{(}\\text{CN}\\text{)}}_{2};[\/latex]<\/p>\n

                                                                                                                                                                          (a) [latex]35.27\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}\\times \\frac{1\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}{199.002\\cancel{\\text{g K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{2\\cancel{\\text{mol K}\\left[\\text{Ag}{\\text{(}\\text{CN}\\text{)}}_{2}\\right]}}\\times \\frac{6.022\\times {10}^{23}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=5.337\\times {10}^{22}\\text{molecules}[\/latex]<\/p>\n

                                                                                                                                                                          (b) [latex]5.337\\times {10}^{22}\\cancel{\\text{molecules}}\\times \\frac{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}{6.022\\times {10}^{23}\\cancel{\\text{molecules}}}\\times \\frac{\\text{117.43 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}{1\\cancel{\\text{mol Zn}{\\text{(}\\text{CN}\\text{)}}_{2}}}=\\text{10.41 g Zn}{\\text{(}\\text{CN}\\text{)}}_{2}[\/latex]<\/p>\n

                                                                                                                                                                          10. [latex]{\\text{SiO}}_{2}+3\\text{C}\\rightarrow\\text{SiC}+2\\text{CO.}[\/latex] From the balanced equation, 1 mol of SiO2<\/sub> produces 1 mol of SiC. The unknown is the mass of SiO2<\/sub> required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO2<\/sub> required, determine the molar masses of SiO2<\/sub> and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO2<\/sub> to SiC, find the mass of SiO2<\/sub> required. The conversions required are: [latex]\\text{g SiC}\\rightarrow\\text{mol SiC}\\rightarrow\\text{mol}{\\text{SiO}}_{2}\\rightarrow\\text{g}{\\text{SiO}}_{2}[\/latex]
                                                                                                                                                                          \nMolar masses: SiO2<\/sub> = 60.0843 g mol\u20131<\/sup>; SiC = 40.0955 g mol\u20131<\/sup>
                                                                                                                                                                          \n[latex]\\text{mass SiO}2=3000\\cancel{\\text{g SiC}}\\times \\frac{1\\cancel{\\text{mol SiC}}}{40.955\\cancel{\\text{g SiC}}}\\times \\frac{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}{1\\cancel{\\text{mol SiC}}}\\times \\frac{\\text{60.843 g}{\\text{SiO}}_{2}}{1\\cancel{\\text{mol}{\\text{SiO}}_{2}}}=\\text{4496 g}{\\text{SiO}}_{2}=\\text{4.50 kg}{\\text{SiO}}_{2}[\/latex]<\/p>\n

                                                                                                                                                                          12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol\u20131<\/sup>
                                                                                                                                                                          \n[latex]\\text{1 mol C}\\rightarrow 1{\\text{mol CO}}_{2}\\rightarrow 1\\text{mol urea}[\/latex]
                                                                                                                                                                          \n[latex]\\begin{array}{ll}\\hfill \\text{mass urea}& =1.00\\times {10}^{3}\\cancel{\\text{kg}}\\times \\frac{1000\\cancel{\\text{g}}}{\\cancel{\\text{kg}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.0\\cancel{\\text{g C}}}\\times \\frac{1\\cancel{\\text{mol urea}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{60.054 g urea}}{1\\cancel{\\text{mol urea}}}\\\\ & =5.00\\times {10}^{6}\\text{g or}5.00\\times {10}^{3}\\text{kg}\\end{array}[\/latex]<\/p>\n

                                                                                                                                                                          14. \u00a0The balanced chemical equation is [latex]\\text{C}\\text{(}s\\text{)}+{\\text{O}}_{2}\\text{(}g\\text{)}\\rightarrow{\\text{CO}}_{2}\\text{(}g\\text{)}[\/latex]
                                                                                                                                                                          \n[latex]\\begin{array}{l}\\\\ \\\\ \\\\ 500\\cancel{\\text{miles}}\\times \\frac{1\\cancel{\\text{gallon}}}{37.5\\cancel{\\text{miles}}}\\times \\frac{3.785\\cancel{\\text{L}}}{1\\cancel{\\text{gallon}}}\\times \\frac{1000\\cancel{\\text{mL}}}{1\\cancel{\\text{L}}}\\times \\frac{0.8205\\cancel{\\text{g gas}}}{1\\cancel{\\text{mL}}\\text{gas}}\\times \\frac{84.2\\cancel{\\text{g C}}}{100\\cancel{\\text{g gas}}}\\times \\frac{1\\cancel{\\text{mol C}}}{12.01\\cancel{\\text{g C}}}\\\\ \\times \\frac{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}{1\\cancel{\\text{mol C}}}\\times \\frac{\\text{44.01 g}{\\text{CO}}_{2}}{1\\cancel{\\text{mol}{\\text{CO}}_{2}}}=1.28\\times {10}^{5}\\text{g}{\\text{CO}}_{2}\\end{array}[\/latex]<\/p>\n

                                                                                                                                                                          16. Use molarity to convert. This solution involves the following steps:<\/p>\n

                                                                                                                                                                          Converting the volume of KI to moles of KI<\/p>\n

                                                                                                                                                                          Converting the moles of KI to moles of Cu(NO3<\/sub>)2<\/sub><\/p>\n

                                                                                                                                                                          Converting the moles of [latex]\\text{K}\\rightarrow\\text{Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}[\/latex] to a volume of KI. Cu(NO3<\/sub>)2<\/sub> solution
                                                                                                                                                                          \n[latex]\\text{43.88 mL}\\times \\frac{\\text{1 L}}{\\text{1000 mL}}\\times \\frac{0.3842\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}{\\text{1 L}}\\times \\frac{4\\cancel{\\text{mol KI}}}{2\\cancel{\\text{mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}}\\times \\frac{\\text{1 L KI}}{0.2089\\cancel{\\text{mol KI}}}=\\text{161.4 mL}[\/latex]
                                                                                                                                                                          \nAll of these steps can be shown together, as follows:
                                                                                                                                                                          \n[latex]\\frac{\\text{43.88 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{1}\\times \\frac{\\text{0.3842 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}{\\text{1000 mL Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{4 mol KI}}{\\text{2 mol Cu}{\\text{(}{\\text{NO}}_{3}\\text{)}}_{2}}\\times \\frac{\\text{1000 mL KI}}{\\text{0.2089 mol KI}}=\\text{161.40 mL KI solution}[\/latex]<\/p>\n

                                                                                                                                                                          18. Find from worked example, check your learning problem
                                                                                                                                                                          \n[latex]\\text{mass of ilmenite}=379\\cancel{\\text{g ore}}\\times \\frac{\\text{0.883 g}{\\text{FeTiO}}_{3}}{1\\cancel{\\text{g ore}}}=\\text{334.6 g}{\\text{FeTiO}}_{3}[\/latex]
                                                                                                                                                                          \n[latex]\\text{mass of rutile}=334.6\\cancel{\\text{g}{\\text{FeTiO}}_{3}}\\times \\frac{1\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}{151.7\\cancel{\\text{g}{\\text{FeTiO}}_{3}}}\\times \\frac{2\\cancel{\\text{mol}{\\text{TiO}}_{2}}}{2\\cancel{\\text{mol}{\\text{FeTiO}}_{3}}}\\times \\frac{\\text{79.88 g}{\\text{TiO}}_{2}}{1\\cancel{\\text{mol}{\\text{TiO}}_{2}}}=\\text{176 g}{\\text{TiO}}_{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                                                                                                                                                                          Part III<\/strong><\/span><\/p>\n

                                                                                                                                                                            \n
                                                                                                                                                                          1. Calculate the following to four significant figures:\n
                                                                                                                                                                              \n
                                                                                                                                                                            1. the percent composition of ammonia, NH3 \u00a0<\/sub><\/li>\n
                                                                                                                                                                            2. the percent composition of photographic \u201chypo,\u201d Na2<\/sub>S2<\/sub>O3 \u00a0<\/sub><\/li>\n
                                                                                                                                                                            3. the percent of calcium ion in Ca3<\/sub>(PO4<\/sub>)2<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                            4. Determine the following to four significant figures:\n
                                                                                                                                                                                \n
                                                                                                                                                                              1. the percent composition of hydrazoic acid, HN3 \u00a0<\/sub><\/li>\n
                                                                                                                                                                              2. the percent composition of TNT, C6<\/sub>H2<\/sub>(CH3<\/sub>)(NO2<\/sub>)3 \u00a0<\/sub><\/li>\n
                                                                                                                                                                              3. the percent of SO4<\/sub>2\u2013<\/sup> in Al2<\/sub>(SO4<\/sub>)3<\/sub><\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                              4. Determine the percent ammonia, NH3<\/sub>, in Co(NH3<\/sub>)6<\/sub>Cl3<\/sub>, to three significant figures.<\/li>\n
                                                                                                                                                                              5. Determine the percent water in CuSO4<\/sub>\u22195H2<\/sub>O to three significant figures.<\/li>\n<\/ol>\n
                                                                                                                                                                                Show Selected Answers<\/span><\/p>\n
                                                                                                                                                                                \n

                                                                                                                                                                                1. In each of these exercises asking for the percent composition, divide the molecular weight of the desired element or group of elements (the number of times it\/they occur in the formula times the molecular weight of the desired element or elements) by the molecular weight of the compound.<\/p>\n

                                                                                                                                                                                  \n
                                                                                                                                                                                1. [latex]\\begin{array}{l}\\\\ \\%\\text{N}=\\frac{14.0067\\text{g}{\\text{mol}}^{-1}\\times 100\\%}{\\left[3\\left(1.007940+14.0067\\right)\\right]\\text{g}{\\text{mol}}^{-1}}=\\frac{14.0067\\text{g}{\\text{mol}}^{-1}}{17.0305\\text{g}{\\text{mol}}^{-1}}=82.24\\%\\\\ \\%\\text{H}=\\frac{3\\times 1.00794\\text{g}{\\text{mol}}^{-1}}{17.0305\\text{g}{\\text{mol}}^{-1}}\\times 100\\%=17.76\\%\\end{array}[\/latex]<\/li>\n
                                                                                                                                                                                2. [latex]\\begin{array}{lll}\\\\ \\%\\text{Na}\\hfill & =\\hfill & \\frac{2\\times 22.989768}{2\\times 22.989768+2\\times 32.066+3\\times 15.9994}\\times 100\\%=\\frac{45.9795}{158.1097}\\times 100=29.08\\%\\hfill \\\\ \\%\\text{S}\\hfill & =\\hfill & \\frac{64.132}{158.1097}\\times 100\\%=40.56\\%\\hfill \\\\ \\%\\text{O}\\hfill & =\\hfill & \\frac{47.9982}{158.1097}\\times 100\\%=30.36\\%\\hfill \\end{array}[\/latex]<\/li>\n
                                                                                                                                                                                3. [latex]\\%{\\text{Ca}}^{2+}=\\frac{3\\times 40.078}{3\\times 40.078+2\\times 30.973762+8\\times 15.9994}\\times 100\\%=\\frac{120.234}{310.1816}\\times 100\\%=38.76\\%[\/latex]<\/li>\n<\/ol>\n

                                                                                                                                                                                  3.\u00a0[latex]\\%{\\text{NH}}_{3}=\\frac{6\\left(14.007+3\\times 40.078\\right)}{58.933+6\\left(14.007+3\\times 1.008\\right)+3\\left(35.453\\right)}\\times 100\\%=\\frac{102.186}{267.478}\\times 100\\%=38.2\\%[\/latex]<\/p>\n<\/div>\n<\/div>\n

                                                                                                                                                                                  Part IV<\/strong><\/span><\/p>\n

                                                                                                                                                                                    \n
                                                                                                                                                                                  1. What information do we need to determine the molecular formula of a compound from the empirical formula?<\/li>\n
                                                                                                                                                                                  2. Determine the empirical formulas for compounds with the following percent compositions:\n
                                                                                                                                                                                      \n
                                                                                                                                                                                    1. 15.8% carbon and 84.2% sulfur<\/li>\n
                                                                                                                                                                                    2. 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                                    3. Determine the empirical formulas for compounds with the following percent compositions:\n
                                                                                                                                                                                        \n
                                                                                                                                                                                      1. 43.6% phosphorus and 56.4% oxygen<\/li>\n
                                                                                                                                                                                      2. 28.7% K, 1.5% H, 22.8% P, and 47.0% O<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                                      3. Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:\n
                                                                                                                                                                                          \n
                                                                                                                                                                                        1. Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O<\/li>\n
                                                                                                                                                                                        2. Saran; 24.8% C, 2.0% H, 73.1% Cl<\/li>\n
                                                                                                                                                                                        3. polyethylene; 86% C, 14% H<\/li>\n
                                                                                                                                                                                        4. polystyrene; 92.3% C, 7.7% H<\/li>\n
                                                                                                                                                                                        5. Orlon; 67.9% C, 5.70% H, 26.4% N<\/li>\n<\/ol>\n<\/li>\n
                                                                                                                                                                                        6. A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g\/mol. What is its molecular formula?<\/li>\n
                                                                                                                                                                                        7. Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g\/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?<\/li>\n
                                                                                                                                                                                        8. Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g\/mol.<\/li>\n
                                                                                                                                                                                        9. A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g\/mol. Determine the molecular formula of the dye.<\/li>\n<\/ol>\n
                                                                                                                                                                                          Show Selected Answers<\/span><\/p>\n
                                                                                                                                                                                          \n

                                                                                                                                                                                          2. The empirical formulas\u00a0can be found as follows:<\/p>\n

                                                                                                                                                                                            \n
                                                                                                                                                                                          1. The percent of an element in a compound indicates the percent by mass. The mass of an element in a 100.0-g sample of a compound is equal in grams to the percent of that element in the sample; hence, 100.0 g of the sample contains 15.8 g of C and 84.2 g of S. The relative number of moles of C and S atoms in the compound can be obtained by converting grams to moles as shown.
                                                                                                                                                                                            \nStep 1: [latex]\\begin{array}{l}\\\\ \\text{C:}15.8\\text{g}\\times \\frac{1\\text{mol}}{12.011\\text{g}}=1.315\\text{mol}\\\\ \\text{S:}84.2\\text{g}\\times \\frac{1\\text{mol}}{32.066\\text{g}}=2.626\\text{mol}\\end{array}[\/latex]
                                                                                                                                                                                            \nStep 2: [latex]\\begin{array}{l}\\\\ \\text{C:}\\frac{1.315\\text{mol}}{1.315\\text{mol}}=1.000\\\\ \\text{S:}\\frac{2.626\\text{mol}}{1.315\\text{mol}}=1.997\\end{array}[\/latex]<\/p>\n
                                                                                                                                                                                              \n
                                                                                                                                                                                            • The empirical formula is CS2<\/sub>.<\/li>\n<\/ul>\n<\/li>\n
                                                                                                                                                                                            • Step 1: [latex]\\begin{array}{l}\\\\ \\text{C:}40.0\\text{g}\\times \\frac{1\\text{mol}}{12.011\\text{g}}=3.330\\text{mol}\\\\ \\text{H:}6.7\\text{g}\\times \\frac{1\\text{mol}}{1.00794\\text{g}}=6.647\\text{mol}\\\\ \\text{O:}53.3\\text{g}\\times \\frac{1\\text{mol}}{15.9994\\text{g}}=3.331\\text{mol}\\end{array}[\/latex]
                                                                                                                                                                                              \nStep 2: [latex]\\begin{array}{l}\\\\ \\text{C:}\\frac{3.330\\text{mol}}{3.330\\text{mol}}=1.0\\\\ \\text{H:}\\frac{6.647\\text{mol}}{3.330\\text{mol}}=2\\\\ \\text{O:}\\frac{3.331\\text{mol}}{3.330\\text{mol}}=1.0\\end{array}[\/latex]<\/p>\n
                                                                                                                                                                                                \n
                                                                                                                                                                                              • The empirical formula is CH2<\/sub>O.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n

                                                                                                                                                                                                5. To determine the empirical formula, a relationship between percent composition and atom composition must be established. The percent composition of each element in a compound can be found either by dividing its mass by the total mass of compound or by dividing the molar mass of that element as it appears in the formula (atomic mass times the number of times the element appears in the formula) by the formula mass of the compound. From this latter perspective, the percent composition of an element can be converted into a mass by assuming that we start with a 100-g sample. Then, multiplying the percentage times 100 g gives the mass in grams of that component. Division of each mass by its respective atomic mass gives the relative ratio of atoms in the formula. From the numbers so obtained, the whole-number ratio of elements in the compound can be found by dividing each ratio by the number representing the smallest ratio. Generally, this process can be done in two simple steps (a third step is needed if the ratios are not whole numbers).<\/p>\n

                                                                                                                                                                                                Step 1: Divide each element\u2019s percentage (converted to grams) by its atomic mass:<\/p>\n

                                                                                                                                                                                                [latex]\\begin{array}{l}\\text{C:}\\frac{92.3\\text{g}}{12.011\\text{g}{\\text{mol}}^{-1}}=7.68\\text{mol}\\\\ \\text{H:}\\frac{7.7\\text{g}}{1.00794\\text{g}{\\text{mol}}^{-1}}=7.6\\text{mol}\\end{array}[\/latex]<\/p>\n

                                                                                                                                                                                                This operation established the relative ration of carbon to hydrogen in the formula.<\/p>\n

                                                                                                                                                                                                Step 2: To establish a whole-number ratio of carbon to hydrogen, divide each factor by the smallest factor. In this case, both factors are essentially equal; thus the ration of atoms is 1 to 1:<\/p>\n

                                                                                                                                                                                                [latex]\\begin{array}{l}\\text{C:}\\frac{7.68}{7.6}=1\\\\ \\text{H:}\\frac{7.6}{7.6}=1\\end{array}[\/latex]<\/p>\n

                                                                                                                                                                                                The empirical formula is CH.<\/p>\n

                                                                                                                                                                                                Since the molecular mass of the compound is 78.1 amu, some integer times the sum of the mass of 1C and 1H in atomic mass units (12.011 amu + 1.00794 amu = 13.019 amu) must be equal to 78.1 amu. To find this number, divide 78.1 amu by 13.019 amu:<\/p>\n

                                                                                                                                                                                                [latex]\\frac{78.1\\text{amu}}{13.019\\text{amu}}=5.9989\\rightarrow 6[\/latex]<\/p>\n

                                                                                                                                                                                                The molecular formula is (CH)6<\/sub> = C6<\/sub>H6<\/sub>.<\/p>\n

                                                                                                                                                                                                7. The formulas can be found as follows:<\/p>\n

                                                                                                                                                                                                [latex]\\begin{array}{ccc}\\left(28.03\\text{g Mg}\\right)\\left(\\frac{1\\text{mol Mg}}{24.30\\text{g}}\\right)=1.153\\text{mol Mg}& & \\frac{1.153}{0.769}=1.512\\text{mol Mg}\\\\\\left(21.60\\text{g Si}\\right)\\left(\\frac{1\\text{mol Si}}{28.09\\text{g Si}}\\right)=0.769\\text{mol Si}& &\\frac{0.769}{0.769}=1.00\\text{mol Si}\\\\\\left(1.16\\text{g H}\\right)\\left(\\frac{1\\text{mol H}}{1.01\\text{g H}}\\right)=1.149\\text{mol H}& & \\frac{1.149}{0.769}=1.49\\text{mol H}\\\\\\left(49.21\\text{g O}\\right)\\left(\\frac{1\\text{mol O}}{16.00\\text{g O}}\\right)=3.076\\text{mol O}& & \\frac{3.076}{0.769}=4.00\\text{mol O}\\end{array}[\/latex]<\/p>\n

                                                                                                                                                                                                (2)(Mg1.5<\/sub>Si1<\/sub>H1.5<\/sub>O4<\/sub>) = Mg3<\/sub>Si2<\/sub>H3<\/sub>O8<\/sub> (empirical formula), empirical mass of 260.1 g\/unit<\/p>\n

                                                                                                                                                                                                [latex]\\frac{\\text{MM}}{\\text{EM}}=\\frac{520.8}{260.1}=2.00,[\/latex] so (2)(Mg3<\/sub>Si2<\/sub>H3<\/sub>O8<\/sub>) = Mg6<\/sub>Si4<\/sub>H6<\/sub>O16<\/sub><\/p>\n

                                                                                                                                                                                                8. Assume 100.0 g; the percentages of the elements are then the same as their mass in grams. Divide each mass by the molar mass to find the number of moles.
                                                                                                                                                                                                \nStep 1: [latex]\\begin{array}{l}\\\\ \\frac{75.95\\cancel{\\text{g}}}{12.011\\cancel{\\text{g}}{\\text{mol}}^{-1}}=6.323\\text{mol C}\\\\ \\frac{17.72\\cancel{\\text{g}}}{14.0067\\cancel{\\text{g}}{\\text{mol}}^{-1}}=1.265\\text{mol N}\\\\ \\frac{6.33\\cancel{\\text{g}}}{1.00794\\cancel{\\text{g}}{\\text{mol}}^{-1}}=6.28\\text{mol H}\\end{array}[\/latex]<\/p>\n

                                                                                                                                                                                                Step 2: Divide each by the smallest number. The answers are 5C, 1N, and 5H. The empirical formula is C5<\/sub>H5<\/sub>N, which has a molar mass of 79.10 g\/mol. To find the actual molecular formula, divide 240, the molar mass of the compound, by 79.10 to obtain 3. So the formula is three times the empirical formula, or C15<\/sub>H15<\/sub>N3<\/sub>.<\/p>\n<\/div>\n<\/div>\n

                                                                                                                                                                                                Part V<\/strong><\/span><\/p>\n

                                                                                                                                                                                                  \n
                                                                                                                                                                                                1. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of CO2<\/sub>\u00a0is produced by the combustion of 1.00 mol of CH4<\/sub>?<\/p>\n

                                                                                                                                                                                                  CH4<\/sub>(g) +\u00a02O2<\/sub>(g) \u2192\u00a0CO2<\/sub>(g) +\u00a02H2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                2. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of H2<\/sub>O is produced by the combustion of 1.00 mol of CH4<\/sub>?<\/p>\n

                                                                                                                                                                                                  CH4<\/sub>(g) +\u00a02O2<\/sub>(g) \u2192\u00a0CO2<\/sub>(g) +\u00a02H2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                3. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of HgO is required to produce 0.692 mol of O2<\/sub>?<\/p>\n

                                                                                                                                                                                                  2HgO(s) \u2192\u00a02Hg(\u2113) +\u00a0O2<\/sub>(g)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                4. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of NaHCO3<\/sub>\u00a0is needed to produce 2.659 mol of CO2<\/sub>?<\/p>\n

                                                                                                                                                                                                  2NaHCO3<\/sub>(s) \u2192\u00a0Na2<\/sub>CO3<\/sub>(s) +\u00a0H2<\/sub>O(\u2113) +\u00a0CO2<\/sub>(g)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                5. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  How many moles of Al can be produced from 10.87 g of Ag?<\/p>\n

                                                                                                                                                                                                  Al(NO3<\/sub>)\u00a03<\/sub>(s) +\u00a03Ag \u2192\u00a0Al +\u00a03AgNO3<\/sub><\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                6. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  How many moles of HCl can be produced from 0.226 g of SOCl2<\/sub>?<\/p>\n

                                                                                                                                                                                                  SOCl2<\/sub>(\u2113) +\u00a0H2<\/sub>O(\u2113) \u2192\u00a0SO2<\/sub>(g) +\u00a02HCl(g)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                7. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  How many moles of O2<\/sub>\u00a0are needed to prepare 1.00 g of Ca(NO3<\/sub>)2<\/sub>?<\/p>\n

                                                                                                                                                                                                  Ca(s) +\u00a0N2<\/sub>(g) +\u00a03O2<\/sub>(g) \u2192\u00a0Ca(NO3<\/sub>)\u00a02<\/sub>(s)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                8. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  How many moles of C2<\/sub>H5<\/sub>OH are needed to generate 106.7 g of H2<\/sub>O?<\/p>\n

                                                                                                                                                                                                  C2<\/sub>H5<\/sub>OH(\u2113) +\u00a03O2<\/sub>(g) \u2192\u00a02CO2<\/sub>(g) +\u00a03H2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                9. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of O2<\/sub>\u00a0can be generated by the decomposition of 100.0 g of NaClO3<\/sub>?<\/p>\n

                                                                                                                                                                                                  2NaClO3<\/sub>\u00a0\u2192\u00a02NaCl(s) +\u00a03O2<\/sub>(g)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                10. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of Li2<\/sub>O is needed to react with 1,060 g of CO2<\/sub>?<\/p>\n

                                                                                                                                                                                                  Li2<\/sub>O(aq) +\u00a0CO2<\/sub>(g) \u2192\u00a0Li2<\/sub>CO3<\/sub>(aq)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                11. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of Fe2<\/sub>O3<\/sub>\u00a0must be reacted to generate 324 g of Al2<\/sub>O3<\/sub>?<\/p>\n

                                                                                                                                                                                                  Fe2<\/sub>O3<\/sub>(s) +\u00a02Al(s) \u2192\u00a02Fe(s) +\u00a0Al2<\/sub>O3<\/sub>(s)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                12. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of Fe is generated when 100.0 g of Al are reacted?<\/p>\n

                                                                                                                                                                                                  Fe2<\/sub>O3<\/sub>(s) +\u00a02Al(s) \u2192\u00a02Fe(s) +\u00a0Al2<\/sub>O3<\/sub>(s)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                13. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of MnO2<\/sub>\u00a0is produced when 445 g of H2<\/sub>O are reacted?<\/p>\n

                                                                                                                                                                                                  H2<\/sub>O(\u2113) +\u00a02MnO4<\/sub>\u2212<\/sup>(aq) +\u00a0Br\u2212<\/sup>(aq) \u2192\u00a0BrO3<\/sub>\u2212<\/sup>(aq) +\u00a02MnO2<\/sub>(s) +\u00a02OH\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                14. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  What mass of PbSO4<\/sub>\u00a0is produced when 29.6 g of H2<\/sub>SO4<\/sub>\u00a0are reacted?<\/p>\n

                                                                                                                                                                                                  Pb(s) +\u00a0PbO2<\/sub>(s) +\u00a02H2<\/sub>SO4<\/sub>(aq) \u2192\u00a02PbSO4<\/sub>(s) +\u00a02H2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                15. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  If 83.9 g of ZnO are formed, what mass of Mn2<\/sub>O3<\/sub>\u00a0is formed with it?<\/p>\n

                                                                                                                                                                                                  Zn(s) +\u00a02MnO2<\/sub>(s) \u2192\u00a0ZnO(s) +\u00a0Mn2<\/sub>O3<\/sub>(s)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                16. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  If 14.7 g of NO2<\/sub>\u00a0are reacted, what mass of H2<\/sub>O is reacted with it?<\/p>\n

                                                                                                                                                                                                  3NO2<\/sub>(g) +\u00a0H2<\/sub>O(\u2113) \u2192\u00a02HNO3<\/sub>(aq) +\u00a0NO(g)<\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                17. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  If 88.4 g of CH2<\/sub>S are reacted, what mass of HF is produced?<\/p>\n

                                                                                                                                                                                                  CH2<\/sub>S +\u00a06F2<\/sub>\u00a0\u2192\u00a0CF4<\/sub>\u00a0+\u00a02HF +\u00a0SF6<\/sub><\/span><\/span><\/p>\n<\/div>\n<\/li>\n

                                                                                                                                                                                                18. \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  If 100.0 g of Cl2<\/sub>\u00a0are needed, what mass of NaOCl must be reacted?<\/p>\n

                                                                                                                                                                                                  NaOCl +\u00a0HCl \u2192\u00a0NaOH +\u00a0Cl2<\/sub><\/span><\/span><\/p>\n

                                                                                                                                                                                                  Show Selected Answers<\/span><\/p>\n
                                                                                                                                                                                                  \n
                                                                                                                                                                                                  \n

                                                                                                                                                                                                  1.<\/strong>\u00a0 44.0 g<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  3.<\/strong>\u00a0 3.00 \u00d7 102<\/sup>\u00a0g<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  5.<\/strong>\u00a0 0.0336 mol<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  7.<\/strong>\u00a0 0.0183 mol<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  9.<\/strong>\u00a0 45.1 g<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  11.\u00a0<\/strong> 507 g<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  13.<\/strong>\u00a0 4.30 \u00d7 103<\/sup>\u00a0g<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  15.<\/strong>\u00a0 163 g<\/p>\n<\/div>\n

                                                                                                                                                                                                  \n

                                                                                                                                                                                                  17.<\/strong>\u00a0 76.7 g<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n<\/ol>\n","protected":false},"author":359083,"menu_order":6,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1181","chapter","type-chapter","status-publish","hentry"],"part":799,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1181","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/users\/359083"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1181\/revisions"}],"predecessor-version":[{"id":1406,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1181\/revisions\/1406"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/parts\/799"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapters\/1181\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/media?parent=1181"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/pressbooks\/v2\/chapter-type?post=1181"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/contributor?post=1181"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-dutchess-introbio2\/wp-json\/wp\/v2\/license?post=1181"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}