{"id":694,"date":"2018-03-20T15:23:44","date_gmt":"2018-03-20T15:23:44","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/?post_type=chapter&p=694"},"modified":"2018-08-07T18:53:28","modified_gmt":"2018-08-07T18:53:28","slug":"6-5-mole-mass-and-mass-mass-problems","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-orgbiochemistry\/chapter\/6-5-mole-mass-and-mass-mass-problems\/","title":{"raw":"6.5 Mole-Mass and Mass-Mass Problems","rendered":"6.5 Mole-Mass and Mass-Mass Problems"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  1. Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWe have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:\r\n\r\n<\/div>\r\n<\/div>\r\n
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    \"image\"<\/div>\r\n

    Collectively, these conversions are called mole-mass calculations<\/span><\/strong><\/span>.<\/p>\r\n

    As an example, consider the balanced chemical equation<\/p>\r\nFe2<\/sub>O3<\/sub> + 3SO3<\/sub> \u2192 Fe2<\/sub>(SO4<\/sub>)3<\/sub><\/span><\/span>\r\n

    If we have 3.59 mol of Fe2<\/sub>O3<\/sub>, how many grams of SO3<\/sub> can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3<\/sub> in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3<\/sub> needed. Then using the molar mass of SO3<\/sub> as a conversion factor, we determine the mass that this number of moles of SO3<\/sub> has.<\/p>\r\n

    The first step resembles the exercises we did in Section 6.4 \"Mole-Mole Relationships in Chemical Reactions\". As usual, we start with the quantity we were given:<\/p>\r\n [latex]3.59\\,\\rlap{{-----------}}mol\\,Fe_2O_3\\,\u00d7\\,\\frac{3\\,mol\\,SO_3}{1\\,\\rlap{{--------------}}mol\\,Fe_2O_3}\\,=\\,10.77\\,mol\\,SO_3[\/latex] <\/span>\r\n

    The mol Fe2<\/sub>O3<\/sub> units cancel, leaving mol SO3<\/sub> unit. Now, we take this answer and convert it to grams of SO3<\/sub>, using the molar mass of SO3<\/sub> as the conversion factor:<\/p>\r\n [latex]10.77\\,\\rlap{{---------}}mol\\,SO_3\\,\u00d7\\,\\frac{80.06\\,g\\,SO_3}{1\\,\\rlap{{-----------}}mol\\,SO_3}\\,=\\,862\\,g\\,SO_3[\/latex] <\/span>\r\n

    Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3<\/sub> will react with 3.59 mol of Fe2<\/sub>O3<\/sub>. Many problems of this type can be answered in this manner.<\/p>\r\n

    The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:<\/p>\r\n\r\n

    \"image\"<\/div>\r\n

    We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.<\/p>\r\n\r\n

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    Example 8<\/h3>\r\n

    How many grams of CO2<\/sub> are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation?<\/p>\r\nCaCO3<\/sub> + 2HCl \u2192 CaCl2<\/sub> + CO2<\/sub> + H2<\/sub>O<\/span><\/span>\r\n

    Solution<\/p>\r\n

    Our strategy will be to convert from moles of HCl to moles of CO2<\/sub> and then from moles of CO2<\/sub> to grams of CO2<\/sub>. We will need the molar mass of CO2<\/sub>, which is 44.01 g\/mol. Performing these two conversions in a single-line gives 46.0 g of CO2<\/sub>:<\/p>\r\n\r\n

    \"image\"<\/div>\r\n

    The molar ratio between CO2<\/sub> and HCl comes from the balanced chemical equation.<\/p>\r\n\r\n<\/div>\r\n

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    Skill-Building Exercise<\/h3>\r\n
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    1. \r\n
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      How many grams of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) are produced if 17.3 mol of H2<\/sub>O are reacted according to this balanced chemical equation?<\/p>\r\n6CO2<\/sub> + 6H2<\/sub>O \u2192 C6<\/sub>H12<\/sub>O6<\/sub> + 6O2<\/sub><\/span><\/span>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n \r\n\r\n<\/div>\r\n

      It is a small step from mole-mass calculations to mass-mass calculations<\/span><\/strong><\/span>. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance\u2019s molar mass as the conversion factor. Then\u2014and only then\u2014we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:<\/p>\r\n\r\n

      \"image\"<\/div>\r\n

      This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.<\/p>\r\n\r\n

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      Example 9<\/h3>\r\n

      Methane can react with elemental chlorine to make carbon tetrachloride (CCl4<\/sub>). The balanced chemical equation is as follows:<\/p>\r\nCH4<\/sub> + 4Cl2<\/sub> \u2192 CCl4<\/sub> + 4HCl<\/span><\/span>\r\n

      How many grams of HCl are produced by the reaction of 100.0 g of CH4<\/sub>?<\/p>\r\n

      Solution<\/p>\r\n

      First, let us work the problem in stepwise fashion. We begin by converting the mass of CH4<\/sub> to moles of CH4<\/sub>, using the molar mass of CH4<\/sub> (16.05 g\/mol) as the conversion factor:<\/p>\r\n [latex]100.0\\,\\rlap{{-------}}g\\,CH_4\\,\u00d7\\,\\frac{1\\,mol\\,CH_4}{16.05\\,\\rlap{{--------}}g\\,CH_4}\\,=\\,6.231\\,mol\\,CH_4[\/latex]<\/span><\/span>\r\n

      Note that we inverted the molar mass so that the gram units cancel, giving us an answer in moles. Next, we use the balanced chemical equation to determine the ratio of moles CH4<\/sub> and moles HCl and convert our first result into moles of HCl:<\/p>\r\n[latex]6.231\\,\\rlap{{---------}}mol\\,CH_4\\,\u00d7\\,\\frac{4\\,mol\\,HCl}{1\\,\\rlap{{-----------}}mol\\,CH_4}\\,=\\,24.92\\,mol\\,HCl[\/latex]<\/span><\/span>\r\n

      Finally, we use the molar mass of HCl (36.46 g\/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl:<\/p>\r\n [latex]6.231\\,\\rlap{{---------}}mol\\,CH_4\\,\u00d7\\,\\frac{4\\,mol\\,HCl}{1\\,\\rlap{{-----------}}mol\\,CH_4}\\,=\\,24.92\\,mol\\,HCl[\/latex]<\/span>\r\n

      In each step, we have limited the answer to the proper number of significant figures. If desired, we can do all three conversions on a single line:<\/p>\r\n [latex]100.0\\,\\rlap{{------}}g\\,CH_4\\,\u00d7\\,\\frac{1\\,mol\\,CH_4}{16.05\\,\\rlap{{---------}}g\\,CH_4}\\,\u00d7\\,\\frac{4\\,mol\\,HCl}{1\\,\\rlap{{-----------}}mol\\,CH_4}\\,,\u00d7\\,\\frac{36.46\\,g\\,HCl}{1\\,\\rlap{{-----------}}mol\\,HCl}\\,=\\,908.7\\,g\\,HCl[\/latex]<\/span><\/span>\r\n

      This final answer is slightly different from our first answer because only the final answer is restricted to the proper number of significant figures. In the first answer, we limited each intermediate quantity to the proper number of significant figures. As you can see, both answers are essentially the same.<\/p>\r\n\r\n<\/div>\r\n

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      Skill-Building Exercise<\/h3>\r\n
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        The oxidation of propanal (CH3<\/sub>CH2<\/sub>CHO) to propionic acid (CH3<\/sub>CH2<\/sub>COOH) has the following chemical equation:<\/p>\r\nCH3<\/sub>CH2<\/sub>CHO + 2K2<\/sub>Cr2<\/sub>O7<\/sub> \u2192 CH3<\/sub>CH2<\/sub>COOH + other products<\/span><\/span>\r\n

        How many grams of propionic acid are produced by the reaction of 135.8 g of K2<\/sub>Cr2<\/sub>O7<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

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        Concept Review Exercises<\/h3>\r\n
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          What is the general sequence of conversions for a mole-mass calculation?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

        2. \r\n
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          What is the general sequence of conversions for a mass-mass calculation?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

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          Answers<\/h3>\r\n[reveal-answer q=\"275842\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"275842\"]\r\n
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          1. mol first substance \u2192 mol second substance \u2192 mass second substance<\/li>\r\n \t
          2. mass first substance \u2192 mol first substance \u2192 mol second substance \u2192 mass second substance[\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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            To Your Health: The Synthesis of Taxol<\/h3>\r\n

            Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree (Taxus brevifolia<\/em>). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of C47<\/sub>H51<\/sub>NO14<\/sub>. Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory.<\/p>\r\n

            After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed.<\/p>\r\n

            Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug.<\/p>\r\n\r\n

            \r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1525\"]\"image\" Figure 6.4 The Structure of the Cancer Drug Taxol. <\/em>Because of the complexity of the molecule, hydrogen atoms are not shown, but they are present on every atom to give the atom the correct number of covalent bonds (four bonds for each carbon atom).[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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            Key Takeaway<\/h3>\r\n<\/div>\r\n
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            • A balanced chemical equation can be used to relate masses or moles of different substances in a reaction.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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              Exercises<\/h3>\r\n
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                Given the following unbalanced chemical equation,<\/p>\r\nH3<\/sub>PO4<\/sub> + NaOH \u2192 H2<\/sub>O + Na3<\/sub>PO4<\/sub><\/span><\/span>\r\n

                what mass of H2<\/sub>O is produced by the reaction of 2.35 mol of H3<\/sub>PO4<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              2. \r\n
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                Given the following unbalanced chemical equation,<\/p>\r\nC2<\/sub>H6<\/sub> + Br2<\/sub> \u2192 C2<\/sub>H4<\/sub>Br2<\/sub> + HBr<\/span><\/span>\r\n

                what mass of HBr is produced if 0.884 mol of C2<\/sub>H6<\/sub> is reacted?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

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                Certain fats are used to make soap, the first step being to react the fat with water to make glycerol (also known as glycerin) and compounds called fatty acids. One example is as follows:<\/p>\r\n C3<\/sub>H5<\/sub> (OOC(CH2<\/sub>)14<\/sub>CH3<\/sub>)3<\/sub>+3H2<\/sub>O\u2192C3<\/sub>H5<\/sub>(OH)3<\/sub>+3CH3<\/sub>(CH2<\/sub>)14<\/sub>COOH<\/span>\r\n

                How many moles of glycerol can be made from the reaction of 1,000.0 g of C3<\/sub>H5<\/sub>(OOC(CH2<\/sub>)14<\/sub>CH3<\/sub>)3<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

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                Photosynthesis in plants leads to the general overall reaction for producing glucose (C6<\/sub>H12<\/sub>O6<\/sub>):<\/p>\r\n6CO2<\/sub> + 6H2<\/sub>O \u2192 C6<\/sub>H12<\/sub>O6<\/sub> + 6O2<\/sub><\/span><\/span>\r\n

                How many moles of glucose can be made from the reaction of 544 g of CO2<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              5. \r\n
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                Precipitation reactions, in which a solid (called a precipitate) is a product, are commonly used to remove certain ions from solution. One such reaction is as follows:<\/p>\r\nBa(NO3<\/sub>)2<\/sub>(aq) + Na2<\/sub>SO4<\/sub>(aq) \u2192 BaSO4<\/sub>(s) + 2NaNO3<\/sub>(aq)<\/span><\/span>\r\n

                How many grams of Na2<\/sub>SO4<\/sub> are needed to precipitate all the barium ions produced by 43.9 g of Ba(NO3<\/sub>)2<\/sub>?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              6. \r\n
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                Nitroglycerin [C3<\/sub>H5<\/sub>(ONO2<\/sub>)3<\/sub>] is made by reacting nitric acid (HNO3<\/sub>) with glycerol [C3<\/sub>H5<\/sub>(OH)3<\/sub>] according to this reaction:<\/p>\r\nC3<\/sub>H5<\/sub>(OH)3<\/sub> + 3HNO3<\/sub> \u2192 C3<\/sub>H5<\/sub>(ONO2<\/sub>)3<\/sub> + 3H2<\/sub>O<\/span><\/span>\r\n

                If 87.4 g of HNO3<\/sub> are reacted with excess glycerol, what mass of nitroglycerin can be made?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              7. \r\n
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                Antacids are bases that neutralize acids in the digestive tract. Magnesium hydroxide [Mg(OH)2<\/sub>] is one such antacid. It reacts with hydrochloric acid in the stomach according to the following reaction:<\/p>\r\nMg(OH)2<\/sub> + 2HCl \u2192 MgCl2<\/sub> + 2H2<\/sub>O<\/span><\/span>\r\n

                How many grams of HCl can a 200 mg dose of Mg(OH)2<\/sub> neutralize?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              8. \r\n
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                Acid rain is caused by the reaction of nonmetal oxides with water in the atmosphere. One such reaction involves nitrogen dioxide (NO2<\/sub>) and produces nitric acid (HNO3<\/sub>):<\/p>\r\n3NO2<\/sub> + H2<\/sub>O \u2192 2HNO3<\/sub> + NO<\/span><\/span>\r\n

                If 1.82 \u00d7 1013<\/sup> g of NO2<\/sub> enter the atmosphere every year due to human activities, potentially how many grams of HNO3<\/sub> can be produced annually?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              9. \r\n
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                A simplified version of the processing of iron ore into iron metal is as follows:<\/p>\r\n2Fe2<\/sub>O3<\/sub> + 3C \u2192 4Fe + 3CO2<\/sub><\/span><\/span>\r\n

                How many grams of C are needed to produce 1.00 \u00d7 109<\/sup> g of Fe?<\/p>\r\n\r\n<\/div><\/li>\r\n \t

              10. \r\n
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                The SS Hindenburg<\/em> contained about 5.33 \u00d7 105<\/sup> g of H2<\/sub> gas when it burned at Lakehurst, New Jersey, in 1937. The chemical reaction is as follows:<\/p>\r\n2H2<\/sub> + O2<\/sub> \u2192 2H2<\/sub>O<\/span><\/span>\r\n

                How many grams of H2<\/sub>O were produced?<\/p>\r\n\r\n<\/div><\/li>\r\n<\/ol>\r\n<\/div>\r\n

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                Answers<\/h3>\r\n[reveal-answer q=\"135917\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"135917\"]\r\n\r\n1. 127 g\r\n\r\n3.1.236 mol\r\n\r\n5.23.9 g\r\n\r\n7.0.251 g\r\n\r\n9.1.61 \u00d7 108<\/sup> g \u00a0[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"
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                Learning Objectives<\/h3>\n
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                1. Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                  We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:<\/p>\n<\/div>\n<\/div>\n

                  \n
                  \n
                  \"image\"<\/div>\n

                  Collectively, these conversions are called mole-mass calculations<\/span><\/strong><\/span>.<\/p>\n

                  As an example, consider the balanced chemical equation<\/p>\n

                  Fe2<\/sub>O3<\/sub> + 3SO3<\/sub> \u2192 Fe2<\/sub>(SO4<\/sub>)3<\/sub><\/span><\/span><\/p>\n

                  If we have 3.59 mol of Fe2<\/sub>O3<\/sub>, how many grams of SO3<\/sub> can react with it? Using the mole-mass calculation sequence, we can determine the required mass of SO3<\/sub> in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO3<\/sub> needed. Then using the molar mass of SO3<\/sub> as a conversion factor, we determine the mass that this number of moles of SO3<\/sub> has.<\/p>\n

                  The first step resembles the exercises we did in Section 6.4 “Mole-Mole Relationships in Chemical Reactions”. As usual, we start with the quantity we were given:<\/p>\n

                  [latex]3.59\\,\\rlap{{-----------}}mol\\,Fe_2O_3\\,\u00d7\\,\\frac{3\\,mol\\,SO_3}{1\\,\\rlap{{--------------}}mol\\,Fe_2O_3}\\,=\\,10.77\\,mol\\,SO_3[\/latex] <\/span><\/p>\n

                  The mol Fe2<\/sub>O3<\/sub> units cancel, leaving mol SO3<\/sub> unit. Now, we take this answer and convert it to grams of SO3<\/sub>, using the molar mass of SO3<\/sub> as the conversion factor:<\/p>\n

                  [latex]10.77\\,\\rlap{{---------}}mol\\,SO_3\\,\u00d7\\,\\frac{80.06\\,g\\,SO_3}{1\\,\\rlap{{-----------}}mol\\,SO_3}\\,=\\,862\\,g\\,SO_3[\/latex] <\/span><\/p>\n

                  Our final answer is expressed to three significant figures. Thus, in a two-step process, we find that 862 g of SO3<\/sub> will react with 3.59 mol of Fe2<\/sub>O3<\/sub>. Many problems of this type can be answered in this manner.<\/p>\n

                  The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:<\/p>\n

                  \"image\"<\/div>\n

                  We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.<\/p>\n

                  \n

                  Example 8<\/h3>\n

                  How many grams of CO2<\/sub> are produced if 2.09 mol of HCl are reacted according to this balanced chemical equation?<\/p>\n

                  CaCO3<\/sub> + 2HCl \u2192 CaCl2<\/sub> + CO2<\/sub> + H2<\/sub>O<\/span><\/span><\/p>\n

                  Solution<\/p>\n

                  Our strategy will be to convert from moles of HCl to moles of CO2<\/sub> and then from moles of CO2<\/sub> to grams of CO2<\/sub>. We will need the molar mass of CO2<\/sub>, which is 44.01 g\/mol. Performing these two conversions in a single-line gives 46.0 g of CO2<\/sub>:<\/p>\n

                  \"image\"<\/div>\n

                  The molar ratio between CO2<\/sub> and HCl comes from the balanced chemical equation.<\/p>\n<\/div>\n

                  \n
                  \n

                  Skill-Building Exercise<\/h3>\n
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                  1. \n
                    \n

                    How many grams of glucose (C6<\/sub>H12<\/sub>O6<\/sub>) are produced if 17.3 mol of H2<\/sub>O are reacted according to this balanced chemical equation?<\/p>\n

                    6CO2<\/sub> + 6H2<\/sub>O \u2192 C6<\/sub>H12<\/sub>O6<\/sub> + 6O2<\/sub><\/span><\/span><\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

                     <\/p>\n<\/div>\n

                    It is a small step from mole-mass calculations to mass-mass calculations<\/span><\/strong><\/span>. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance\u2019s molar mass as the conversion factor. Then\u2014and only then\u2014we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:<\/p>\n

                    \"image\"<\/div>\n

                    This three-part process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.<\/p>\n

                    \n

                    Example 9<\/h3>\n

                    Methane can react with elemental chlorine to make carbon tetrachloride (CCl4<\/sub>). The balanced chemical equation is as follows:<\/p>\n

                    CH4<\/sub> + 4Cl2<\/sub> \u2192 CCl4<\/sub> + 4HCl<\/span><\/span><\/p>\n

                    How many grams of HCl are produced by the reaction of 100.0 g of CH4<\/sub>?<\/p>\n

                    Solution<\/p>\n

                    First, let us work the problem in stepwise fashion. We begin by converting the mass of CH4<\/sub> to moles of CH4<\/sub>, using the molar mass of CH4<\/sub> (16.05 g\/mol) as the conversion factor:<\/p>\n

                    [latex]100.0\\,\\rlap{{-------}}g\\,CH_4\\,\u00d7\\,\\frac{1\\,mol\\,CH_4}{16.05\\,\\rlap{{--------}}g\\,CH_4}\\,=\\,6.231\\,mol\\,CH_4[\/latex]<\/span><\/span><\/p>\n

                    Note that we inverted the molar mass so that the gram units cancel, giving us an answer in moles. Next, we use the balanced chemical equation to determine the ratio of moles CH4<\/sub> and moles HCl and convert our first result into moles of HCl:<\/p>\n

                    [latex]6.231\\,\\rlap{{---------}}mol\\,CH_4\\,\u00d7\\,\\frac{4\\,mol\\,HCl}{1\\,\\rlap{{-----------}}mol\\,CH_4}\\,=\\,24.92\\,mol\\,HCl[\/latex]<\/span><\/span><\/p>\n

                    Finally, we use the molar mass of HCl (36.46 g\/mol) as a conversion factor to calculate the mass of 24.92 mol of HCl:<\/p>\n

                    [latex]6.231\\,\\rlap{{---------}}mol\\,CH_4\\,\u00d7\\,\\frac{4\\,mol\\,HCl}{1\\,\\rlap{{-----------}}mol\\,CH_4}\\,=\\,24.92\\,mol\\,HCl[\/latex]<\/span><\/p>\n

                    In each step, we have limited the answer to the proper number of significant figures. If desired, we can do all three conversions on a single line:<\/p>\n

                    [latex]100.0\\,\\rlap{{------}}g\\,CH_4\\,\u00d7\\,\\frac{1\\,mol\\,CH_4}{16.05\\,\\rlap{{---------}}g\\,CH_4}\\,\u00d7\\,\\frac{4\\,mol\\,HCl}{1\\,\\rlap{{-----------}}mol\\,CH_4}\\,,\u00d7\\,\\frac{36.46\\,g\\,HCl}{1\\,\\rlap{{-----------}}mol\\,HCl}\\,=\\,908.7\\,g\\,HCl[\/latex]<\/span><\/span><\/p>\n

                    This final answer is slightly different from our first answer because only the final answer is restricted to the proper number of significant figures. In the first answer, we limited each intermediate quantity to the proper number of significant figures. As you can see, both answers are essentially the same.<\/p>\n<\/div>\n

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                    Skill-Building Exercise<\/h3>\n
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                      The oxidation of propanal (CH3<\/sub>CH2<\/sub>CHO) to propionic acid (CH3<\/sub>CH2<\/sub>COOH) has the following chemical equation:<\/p>\n

                      CH3<\/sub>CH2<\/sub>CHO + 2K2<\/sub>Cr2<\/sub>O7<\/sub> \u2192 CH3<\/sub>CH2<\/sub>COOH + other products<\/span><\/span><\/p>\n

                      How many grams of propionic acid are produced by the reaction of 135.8 g of K2<\/sub>Cr2<\/sub>O7<\/sub>?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

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                      Concept Review Exercises<\/h3>\n
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                      1. \n
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                        What is the general sequence of conversions for a mole-mass calculation?<\/p>\n<\/div>\n<\/li>\n

                      2. \n
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                        What is the general sequence of conversions for a mass-mass calculation?<\/p>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n

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                        Answers<\/h3>\n
                        Show Answer<\/span><\/p>\n
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                        1. mol first substance \u2192 mol second substance \u2192 mass second substance<\/li>\n
                        2. mass first substance \u2192 mol first substance \u2192 mol second substance \u2192 mass second substance<\/div>\n<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n
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                          To Your Health: The Synthesis of Taxol<\/h3>\n

                          Taxol is a powerful anticancer drug that was originally extracted from the Pacific yew tree (Taxus brevifolia<\/em>). As you can see from the accompanying figure, taxol is a very complicated molecule, with a molecular formula of C47<\/sub>H51<\/sub>NO14<\/sub>. Isolating taxol from its natural source presents certain challenges, mainly that the Pacific yew is a slow-growing tree, and the equivalent of six trees must be harvested to provide enough taxol to treat a single patient. Although related species of yew trees also produce taxol in small amounts, there is significant interest in synthesizing this complex molecule in the laboratory.<\/p>\n

                          After a 20-year effort, two research groups announced the complete laboratory synthesis of taxol in 1994. However, each synthesis required over 30 separate chemical reactions, with an overall efficiency of less than 0.05%. To put this in perspective, to obtain a single 300 mg dose of taxol, you would have to begin with 600 g of starting material. To treat the 26,000 women who are diagnosed with ovarian cancer each year with one dose, almost 16,000 kg (over 17 tons) of starting material must be converted to taxol. Taxol is also used to treat breast cancer, with which 200,000 women in the United States are diagnosed every year. This only increases the amount of starting material needed.<\/p>\n

                          Clearly, there is intense interest in increasing the overall efficiency of the taxol synthesis. An improved synthesis not only will be easier but also will produce less waste materials, which will allow more people to take advantage of this potentially life-saving drug.<\/p>\n

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                          \"image\"<\/p>\n

                          Figure 6.4 The Structure of the Cancer Drug Taxol. <\/em>Because of the complexity of the molecule, hydrogen atoms are not shown, but they are present on every atom to give the atom the correct number of covalent bonds (four bonds for each carbon atom).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                          Key Takeaway<\/h3>\n<\/div>\n
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