1. Each molecule has the following mass (amu) of carbon.

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

3. The molecular or formula masses are as follows:

1. [latex]4\times 30.974\text{ amu}=123.896\text{ amu}[/latex]
2. [latex]2\times 1.008\text{ amu}+15.999\text{ amu}=18.015\text{ amu}[/latex]
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]
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]
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]

5.  The molecular mass of each compound is as follows:

1. C₄H₈
   [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]
2. C₄H₆
   [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]
3. H₂Si₂Cl₄
   [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]
4. H₃PO₄
   [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]

7. Use 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.

9. Formic 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.

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.

13.  The molecular mass of each compound is as follows:

1. S₈:
   [latex]8\text{S}=8\times 32.066=256.528\text{g/mol}[/latex]
2. C₅H₁₂:
   [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]
3. Sc₂(SO₄)₃:
   [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]
4. CH₃COCH₃:
   [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]
5. C₆H₁₂O₆:
   [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]

15. The molecular mass of each compound is as follows:

1. C₂HBrClF₃:
   [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]
2. C₁₂H₁₄N₂Cl₂:
   [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]
3. C₈H₁₀N₄O₂:
   [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]
4. CO(NH₂)₂:
   [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]
5. C₁₇H₃₅CO₂Na:
   [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]

17.  The mass of each compound is as follows:

1. KOH:
   [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]
   [latex]\text{Mass}=0.0146\text{mol}\times 56.1056\text{g/mol}=0.819\text{g}[/latex]
2. C₂H₆:
   [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]
   [latex]\text{Mass}=10.2\text{mol}\times 30.070\text{g/mol}=307\text{g}[/latex]
3. Na₂SO₄:
   [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]
   [latex]\text{Mass}=1.6\times {10}^{-3}\text{mol}\times 142.044\text{g/mol}=\text{0.23 g}[/latex]
4. C₆H₁₂O₆:
   [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]
   [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]
5. Co(NH₃)₆Cl₃:
   [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]
   [latex]\text{Mass}=2.856\text{mol}\times 267.5344\text{g/mol}=765\text{g}[/latex]

19.

1. 2.345 mol LiCl:
   [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]
2. 0.0872 mol acetylene, C₂H₂:
   [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]
3. 3.3 × 10⁻² mol Na₂CO₃:
   [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]
4. 1.23 × 10³ mol fructose, C₆H₁₂O₆:[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]
5. 0.5758 mol FeSO₄(H₂O)₇:
   [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]

21. The mass of each compound is as follows:

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

23. Determine 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]

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.

• Molar mass AlPO₄: 26.981539 + 30.973762 + 4(15.9994) = 121.9529 g/mol
• Molar mass Al₂Cl₆: 2(26.981539) + 6(35.4527) = 266.6793 g/mol
• Molar mass Al₂S₃: 2(26.981539) + 3(32.066) = 150.161 g/mol

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

[latex]\text{mol Al}=1\times 1.000\text{mol}=1.000\text{mol}[/latex]

Al₂Cl₆: [latex]\frac{266\text{g}}{266.6793\text{g}{\text{mol}}^{-1}}=0.997\text{mol}[/latex]

[latex]\text{mol Al}=2\times 0.997\text{mol}=1.994\text{mol}[/latex]

Al₂S₃: [latex]\frac{225\cancel{\text{g}}}{150.161\cancel{\text{g}}{\text{mol}}^{-1}}=1.50\text{mol}[/latex]

[latex]\text{mol Al}=2\times 1.50\text{mol}=3.00\text{mol}[/latex]

27.  Determine the number of grams present in the diamond and from that the number of moles. Find the number of carbon atoms by multiplying Avogadro’s number by the number of moles:

[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]

29.  Determine 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

[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]

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

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

1. 20.0 g = 1.11 mol H₂O
2. 77.0 g CH₄ = 4.79 mol CH₄
3. 68.0 g CaH₂ = 1.62 mol CaH₂
4. 100.0 g N₂O = 2.27 mol N₂O
5. 84.0 g HF = 4.20 mol HF

Therefore, 20.0 g H₂O represents the least number of molecules since it has the least number of moles.

2. (a) 0.435 mol Na, 0.271 mol Cl₂, 15.4 g Cl₂;

(b) 0.005780 mol HgO, 2.890 × 10⁻³ mol O₂, 9.248 × 10⁻² g O₂;

(c) 8.00 mol NaNO₃, 6.8 × 10² g NaNO₃;

(d) 1665 mol CO₂, 73.3 kg CO₂;

(e) 18.86 mol CuO, 2.330 kg CuCO₃;

(f) 0.4580 mol C₂H₄Br₂, 86.05 g C₂H₄Br₂

4.  (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]
(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]

(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]

(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]

(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]

(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]

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

(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]

8. The 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]

(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]

(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]

10. [latex]{\text{SiO}}_{2}+3\text{C}\rightarrow\text{SiC}+2\text{CO.}[/latex] From the balanced equation, 1 mol of SiO₂ produces 1 mol of SiC. The unknown is the mass of SiO₂ required to produce 3.00 kg (3000 g) of SiC. To calculate the mass of SiO₂ required, determine the molar masses of SiO₂ and SiC. Then calculate the number of moles of SiC required, and through the mole relation of SiO₂ to SiC, find the mass of SiO₂ 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]
Molar masses: SiO₂ = 60.0843 g mol^–1; SiC = 40.0955 g mol^–1
[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]

12. Molar mass urea = 12.011 + 15.9994 + 2(14.0067) + 4(1.0079) = 60.054 g mol^–1
[latex]\text{1 mol C}\rightarrow 1{\text{mol CO}}_{2}\rightarrow 1\text{mol urea}[/latex]
[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]

14.  The balanced chemical equation is [latex]\text{C}\text{(}s\text{)}+{\text{O}}_{2}\text{(}g\text{)}\rightarrow{\text{CO}}_{2}\text{(}g\text{)}[/latex]
[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]

16. Use molarity to convert. This solution involves the following steps:

Converting the volume of KI to moles of KI

Converting the moles of KI to moles of Cu(NO₃)₂

Converting the moles of [latex]\text{K}\rightarrow\text{Cu}{\text{(}{\text{NO}}_{3}\text{)}}_{2}[/latex] to a volume of KI. Cu(NO₃)₂ solution
[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]
All of these steps can be shown together, as follows:
[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]

18. Find from worked example, check your learning problem
[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]
[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]

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.

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]
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]
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]

3. [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]

2. The empirical formulas can be found as follows:

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.
   Step 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]
   Step 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]
   • The empirical formula is CS₂.
2. 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]
   Step 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]
   • The empirical formula is CH₂O.

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).

Step 1: Divide each element’s percentage (converted to grams) by its atomic mass:

[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]

This operation established the relative ration of carbon to hydrogen in the formula.

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:

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

The empirical formula is CH.

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:

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

The molecular formula is (CH)₆ = C₆H₆.

7. The formulas can be found as follows:

[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]

(2)(Mg_1.5Si₁H_1.5O₄) = Mg₃Si₂H₃O₈ (empirical formula), empirical mass of 260.1 g/unit

[latex]\frac{\text{MM}}{\text{EM}}=\frac{520.8}{260.1}=2.00,[/latex] so (2)(Mg₃Si₂H₃O₈) = Mg₆Si₄H₆O₁₆

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.
Step 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]

Step 2: Divide each by the smallest number. The answers are 5C, 1N, and 5H. The empirical formula is C₅H₅N, 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 C₁₅H₁₅N₃.