EXERCISES
Exercise 1. What is the mass percent of carbon and hydrogen in C10H22?
First, calculate the molar mass of C10H22 The molar mass is:
Molar mass of C10H22 = 10 x 12.011 g/mol + 22 x 1.00794 g/mol = 142.285 g/mol
\(\displaystyle \%\;C\;=\;\frac{10\times\;{12.011\;g\;C}}{142.285\;g\;C_{10}H_{22}}\times\;{100}\;=\;\mathbf{84.42\%\;C}\)
\(\displaystyle \%\;H\;=\;\frac{22\times\;{1.00794\;g\;H}}{142.285\;g\;C_{10}H_{22}}\times\;{100}\;=\;\mathbf{15.58\%\;H}\)
We could also have calculated the %H by subtracting 84.42% C from 100.00%
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Exercise 2. What is the mass percent of nitrogen in 365 grams (NH4)2CO3
It does not matter how much of the compound we have. The percent composition by mass will be the same for 365 grams or 1 mole of (NH4)2CO3. The molar mass of (NH4)2CO3 is 96.099 g/mol. There are 2 moles of nitrogen atoms in one mole of (NH4)2CO3. The molar mass of nitrogen, N, is 14.0067 g/mol.
\(\displaystyle \%\;N\;=\;\frac{2\times\;{14.0067\;g\;N}}{96.099\;g\;(NH_4)_2CO_3}\times\;{100}\;=\;\mathbf{29.15\%\;N}\)
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Exercise 3. A compound is 55.3% K, 14.6% P and 30.1% O. What is the empirical formula?
Assume 100 g of compound. We have 55.3 g of K, 14.6 g P, and 30.1 g O. Look up the molar masses of K, P, and O and convert these to moles.
\(\displaystyle 55.3\;g\;K\times\frac{1\;mol\;K}{39.0983\;g\;K}\;=\;\mathbf{1.4\underline{1}44\;mol\;K}\)
\(\displaystyle 14.6\;g\;P\times\frac{1\;mol\;P}{30.9738\;g\;P}\;=\;\mathbf{0.47\underline{1}37\;mol\;P}\)
\(\displaystyle 30.1\;g\;O\times\frac{1\;mol\;O}{15.9994\;g\;O}\;=\;\mathbf{1.8\underline{8}13\;mol\;O}\)
The formula can be written as:
K1.4144P0.47137O1.8813
Divide the subscripts by the smallest subscript, 0.47137
\(\displaystyle K_{\frac{1.4144}{0.47137}}P_{\frac{0.47137}{0.47137}}O_{\frac{1.8813}{0.47137}}\;=\;\mathbf{K_3PO_4}\)
The formula is K3PO4.
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Exercise 4. A substance contains by mass 7.1% H, 59.0% C, 26.2% O and 7.7% N. The molecular mass of the unknown compound was determined by mass spectrometry to be 180. amu. Determine both the empirical and molecular formula of the compound.
There a different ways to do this problem. We could do it the same way as in the previous problem to determine the empirical formula. Since we are given the molecular mass, we can also do the problem like so. We are given the percentages of each element and we can multiply the molar mass by each fractional amount (percent/100) to determine the number of grams of each element in 1 mole of the compound. We then convert the grams of each element to moles. The calculation is:
0.071 x 180 g = 12.78 g H → 12.68 mol H
0.590 x 180 g = 106.2 g C → 8.84 mol C
0.262 x 180 g = 47.16 g O → 2.95 mol O
0.077 x 180 g = 13.86 g N → 0.99 mol N
The molecular formula is:
\(\displaystyle C_{\frac{8.84}{0.99}}H_{\frac{12.68}{0.99}}N_{\frac{0.99}{0.99}}O_{\frac{2.95}{0.99}}\;=\;C_{8.9}H_{12.81}NO_3\;=\;\mathbf{C_9H_{13}NO_3}\)
The subscripts cannot be reduced further. Both the molecular formula and empirical formula is C9H13NO3.
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Exercise 5. Combustion analysis of 0.960 g of a hormone that contains hydrogen, carbon, and oxygen yielded 2.650 g of CO2 and 0.818 g of H2. Which of the following substances could be the unknown hormone?
First, we find the number of grams and moles of carbon and hydrogen.
\(\displaystyle 2.650\;g\;CO_2\times\frac{1\;mol\;CO_2}{44.01\;g\;CO_2}\times\frac{1\;mol\;C}{1\;mol\;CO_2}\times\frac{12.011\;g\;C}{1\;mol\;C}\;=\;\mathbf{0.72323\;g\;C}\)
\(\displaystyle 0.72323\;g\;C\times\frac{1\;mol\;C}{12.011\;g\;C}\;=\;\mathbf{0.06021\;mol\;C}\)
\(\displaystyle 0.818\;g\;H_2O\times\frac{1\;mol\;H_2O}{18.02\;g\;H_2O}\times\frac{2\;mol\;H}{1\;mol\;H_2O}\times\frac{1.00794\;g\;H}{1\;mol\;H}\;=\;\mathbf{0.09151\;g\;H}\)
\(\displaystyle 0.09151\;g\;H\times\frac{1\;mol\;H}{1.00794\;g\;H}\;=\;\mathbf{0.09079\;mol\;H}\)
Next we find the grams and moles of hydrogen by subtracting the grams of hydrogen and carbon from the mass of the compound.
grams of H = 0.960 g – (0.72323 g + 0.09151 g) = 0.14526 g O
\(\displaystyle 0.14526\;g\;O\times\frac{1\;mol\;O}{15.9994\;g\;O}\;=\;\mathbf{0.009079\;mol\;O}\)
Write the formula and divide each subscript by the lowest value subscript:
The empirical formula is C7H10O and its molar mass is 110.156 g/mol. We can divide the molar masses of the compounds by the empirical formula mass to find the whole number multiple.
362.4662/110.156 = 3.29
360.45032/110.156 = 3.26
330.4674/110.156 = 2.99 ≅ 3
270.37148/110.156 = 2.45
288.38676/110.156 = 2.61
Multiply the subscripts of the empirical formula by 3. The hormone is most likely 11-deoxycorticosterone, C21H30O3
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Exercise 6. There are several oxides of tungsten, W. If 7.56 g of the compound contains 5.99 g of W, what is the empirical formula of the compound?
First, find the percentage of W and from that we can find the percentage of oxygen in the compound.
\(\displaystyle\frac{5.99}{7.56}\times\;{100}\;=\;\mathbf{79.2\%\;W}\)
100.0% – 79.2% = 20.8% O.
Assume 100.0 g of compound that contains 79.2 g of W and 20.8 g of O. Convert to moles.
79.2 g W = 0.4308 mol W
20.8 g O = 1.300 mol O
\(\displaystyle W_{0.4308}O_{1.300}\;=\;W_{\frac{0.4308}{0.4308}}O_{\frac{1.300}{0.4308}}\;=\;WO_3\)
The empirical formula is WO3, tungsten(VI) oxide. This compound is also known as tungsten trioxide.
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