Example 5

Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is C₁₄H₁₈N₂O₅.

1. Calculate the mass percentage of each element in aspartame.
2. Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.

Given: molecular formula and mass of sample

Asked for: mass percentage of all elements and mass of one element in sample

Strategy:

A Use atomic masses from the periodic table to calculate the molar mass of aspartame.

B Divide the mass of each element by the molar mass of aspartame; then multiply by 100 to obtain percentages.

C To find the mass of an element contained in a given mass of aspartame, multiply the mass of aspartame by the mass percentage of that element, expressed as a decimal.

Solution:

1. A We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:

   $$\begin{array}{rrrr}\hfill \text{14C}& \hfill \text{(14 mol C)(12}\text{.011 g/mol C)}& \hfill =& \hfill \text{ 168}\text{.154 g}\\ \hfill \text{18H}& \hfill \text{(18 mol H)(1}\text{.008 g/mol H)}& \hfill =& \hfill \text{18}\text{.114 g}\\ \hfill \text{2N}& \hfill \text{(2 mol N)(14}\text{.007 g/mol N)}& \hfill =& \hfill \text{ 28}\text{.014 g}\\ \hfill +\text{5O }& \hfill \text{(5 mol O)(15}\text{.999 g/mol O)}& \hfill =& \hfill \text{79}\text{.995 g}\\ \hfill {\text{C}}_{\text{14}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{O}}_{\text{5}}& \hfill \text{molar mass of aspartame}& \hfill =& \hfill \text{294}\text{.277 g/mol}\end{array}$$

   Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).

   B To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:

   $$\begin{array}{l}\text{mass \% C }=\frac{\text{168}\text{.154 g C}}{\text{294}\text{.277 g aspartame}}\times \text{100 }=\text{ 57}\text{.14\% C}\\ \text{mass \% H }=\frac{\text{18}\text{.114 g H}}{\text{294}\text{.277 g aspartame}}\times \text{100 }=\text{ 6}\text{.16\% H}\\ \text{mass \% N }=\frac{\text{28}\text{.014 g N}}{\text{294}\text{.277 g aspartame}}\times \text{100 }=\text{ 9}\text{.52\% N}\\ \text{mass \% O }=\frac{\text{79}\text{.995 g O}}{\text{294}\text{.277 g aspartame}}\times \text{100 }=\text{ 27}\text{.18\% O}\end{array}$$

   As a check, we can add the percentages together:

   57.14% + 6.16% + 9.52% + 27.18% = 100.00%

   If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation.

2. C The mass of carbon in 1.00 g of aspartame is calculated as follows:

   $$\text{mass of C }=\text{ 1}\text{.00 }\cancel{\text{g aspartame}}\times \frac{57.14\text{ g C}}{\text{100 }\cancel{\text{g aspartame}}}=0.571\text{ g C}$$

Exercise

Calculate the mass percentage of each element in aluminum oxide (Al₂O₃). Then calculate the mass of aluminum in a 3.62 g sample of pure aluminum oxide.

Answer: 52.93% aluminum; 47.08% oxygen; 1.92 g Al

Example 6

Calculate the empirical formula of the ionic compound calcium phosphate, a major component of fertilizer and a polishing agent in toothpastes. Elemental analysis indicates that it contains 38.77% calcium, 19.97% phosphorus, and 41.27% oxygen.

Given: percent composition

Asked for: empirical formula

Strategy:

A Assume a 100 g sample and calculate the number of moles of each element in that sample.

B Obtain the relative numbers of atoms of each element in the compound by dividing the number of moles of each element in the 100 g sample by the number of moles of the element present in the smallest amount.

C If the ratios are not integers, multiply all subscripts by the same number to give integral values.

D Because this is an ionic compound, identify the anion and cation and write the formula so that the charges balance.

Solution:

A A 100 g sample of calcium phosphate contains 38.77 g of calcium, 19.97 g of phosphorus, and 41.27 g of oxygen. Dividing the mass of each element in the 100 g sample by its molar mass gives the number of moles of each element in the sample:

$$\begin{array}{l}\text{moles Ca }=\text{ 38}\text{.77 }\cancel{\text{g Ca}}\times \frac{\text{1 mol Ca}}{\text{40}\text{.078 }\cancel{\text{g Ca}}}=\text{ 0}\text{.9674 mol Ca}\\ \text{moles P }=\text{ 19}\text{.97 }\cancel{\text{g P}}\times \frac{\text{1 mol P}}{\text{30}\text{.9738 }\cancel{\text{g P}}}=\text{ 0}\text{.6447 mol P}\\ \text{moles O }=\text{ 41}\text{.27 }\cancel{\text{g O}}\times \frac{\text{1 mol O}}{\text{15}\text{.9994 }\cancel{\text{g O}}}=\text{ 2}\text{.5800 mol O}\end{array}$$

B To obtain the relative numbers of atoms of each element in the compound, divide the number of moles of each element in the 100-g sample by the number of moles of the element in the smallest amount, in this case phosphorus:

$$\begin{array}{lll}\text{P: }\frac{\text{0}\text{.6447 mol P}}{\text{0}\text{.6447 mol P}}=\text{ 1}\text{.000}\hfill & \text{Ca: }\frac{\text{0}\text{.9674}}{\text{0}\text{.6447}}=\text{ 1}\text{.501}\hfill & \text{O: }\frac{\text{2}\text{.5800}}{\text{0}\text{.6447}}=\text{ 4}\text{.002}\hfill \end{array}$$

C We could write the empirical formula of calcium phosphate as Ca_1.501P_1.000O_4.002, but the empirical formula should show the ratios of the elements as small whole numbers. To convert the result to integral form, multiply all the subscripts by 2 to get Ca_3.002P_2.000O_8.004. The deviation from integral atomic ratios is small and can be attributed to minor experimental errors; therefore, the empirical formula is Ca₃P₂O₈.

D The calcium ion (Ca²⁺) is a cation, so to maintain electrical neutrality, phosphorus and oxygen must form a polyatomic anion. We know from Chapter 2 "Molecules, Ions, and Chemical Formulas" that phosphorus and oxygen form the phosphate ion (PO₄³⁻; see Table 2.4 "Common Polyatomic Ions and Their Names"). Because there are two phosphorus atoms in the empirical formula, two phosphate ions must be present. So we write the formula of calcium phosphate as Ca₃(PO₄)₂.

Exercise

Calculate the empirical formula of ammonium nitrate, an ionic compound that contains 35.00% nitrogen, 5.04% hydrogen, and 59.96% oxygen by mass; refer to Table 2.4 "Common Polyatomic Ions and Their Names" if necessary. Although ammonium nitrate is widely used as a fertilizer, it can be dangerously explosive. For example, it was a major component of the explosive used in the 1995 Oklahoma City bombing.

Answer: N₂H₄O₃ is NH₄⁺NO₃⁻, written as NH₄NO₃

Example 7

Naphthalene, the active ingredient in one variety of mothballs, is an organic compound that contains carbon and hydrogen only. Complete combustion of a 20.10 mg sample of naphthalene in oxygen yielded 69.00 mg of CO₂ and 11.30 mg of H₂O. Determine the empirical formula of naphthalene.

Given: mass of sample and mass of combustion products

Asked for: empirical formula

Strategy:

A Use the masses and molar masses of the combustion products, CO₂ and H₂O, to calculate the masses of carbon and hydrogen present in the original sample of naphthalene.

B Use those masses and the molar masses of the elements to calculate the empirical formula of naphthalene.

Solution:

A Upon combustion, 1 mol of CO₂ is produced for each mole of carbon atoms in the original sample. Similarly, 1 mol of H₂O is produced for every 2 mol of hydrogen atoms present in the sample. The masses of carbon and hydrogen in the original sample can be calculated from these ratios, the masses of CO₂ and H₂O, and their molar masses. Because the units of molar mass are grams per mole, we must first convert the masses from milligrams to grams:

$$\begin{array}{c}\text{mass of C }=\text{ 69}\text{.00 }\cancel{{\text{mg CO}}_{\text{2}}}\times \frac{\text{1 }\cancel{\text{g}}}{\text{1000 }\cancel{\text{mg}}}\times \frac{\text{1 }\cancel{{\text{mol CO}}_{\text{2}}}}{\text{44}\text{.010 }\cancel{{\text{g CO}}_{\text{2}}}}\times \frac{\text{1 }\cancel{\text{mol C}}}{\text{1 }\cancel{{\text{mol CO}}_{\text{2}}}}\times \frac{\text{12}\text{.011 g}}{\text{1 }\cancel{\text{mol C}}}\\ =\text{ 1}\text{.883}\times {\text{10}}^{-\text{2}}\text{ g C}\\ \text{mass of H }=\text{ 11}\text{.30 }\cancel{{\text{mg H}}_{\text{2}}\text{O}}\times \frac{\text{1 }\cancel{\text{g}}}{\text{1000 }\cancel{\text{mg}}}\times \frac{\text{1 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}}{\text{18}\text{.015 }\cancel{{\text{g H}}_{\text{2}}\text{O}}}\times \frac{\text{2 }\cancel{\text{mol H}}}{\text{1 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}}\times \frac{\text{1}\text{.0079 g}}{\text{1 }\cancel{\text{mol H}}}\\ =\text{ 1}\text{.264}\times {\text{10}}^{-\text{3}}\text{ g H}\end{array}$$

B To obtain the relative numbers of atoms of both elements present, we need to calculate the number of moles of each and divide by the number of moles of the element present in the smallest amount:

$$\begin{array}{l}\text{moles C }=\text{ 1}\text{.883}\times {\text{10}}^{–\text{2}}\cancel{\text{g C}}\times \frac{\text{1 mol C}}{\text{12}\text{.011 }\cancel{\text{g C}}}=\text{ 1}\text{.568}\times {\text{10}}^{–\text{3}}\text{ mol C}\\ \text{moles H }=\text{ 1}\text{.264}\times {\text{10}}^{–\text{3}}\cancel{\text{g H}}\times \frac{\text{1 mol H}}{\text{1}\text{.0079 }\cancel{\text{g H}}}=\text{ 1}\text{.254}\times {\text{10}}^{–\text{3}}\text{ mol H}\end{array}$$

Dividing each number by the number of moles of the element present in the smaller amount gives

$$\begin{array}{ll}\text{H: }\frac{\text{1}\text{.254}\times {\text{10}}^{-\text{3}}}{\text{1}\text{.254}\times {\text{10}}^{-\text{3}}}=\text{ 1}\text{.000}\hfill & \text{C: }\frac{\text{1}\text{.568}\times {\text{10}}^{-\text{3}}}{\text{1}\text{.254}\times {\text{10}}^{-\text{3}}}=\text{ 1}\text{.250}\hfill \end{array}$$

Thus naphthalene contains a 1.25:1 ratio of moles of carbon to moles of hydrogen: C_1.25H_1.0. Because the ratios of the elements in the empirical formula must be expressed as small whole numbers, multiply both subscripts by 4, which gives C₅H₄ as the empirical formula of naphthalene. In fact, the molecular formula of naphthalene is C₁₀H₈, which is consistent with our results.

Exercise

1. Xylene, an organic compound that is a major component of many gasoline blends, contains carbon and hydrogen only. Complete combustion of a 17.12 mg sample of xylene in oxygen yielded 56.77 mg of CO₂ and 14.53 mg of H₂O. Determine the empirical formula of xylene.
2. The empirical formula of benzene is CH (its molecular formula is C₆H₆). If 10.00 mg of benzene is subjected to combustion analysis, what mass of CO₂ and H₂O will be produced?

Answer:

1. The empirical formula is C₄H₅. (The molecular formula of xylene is actually C₈H₁₀.)
2. 33.81 mg of CO₂; 6.92 mg of H₂O

Example 8

Calculate the molecular formula of caffeine, a compound found in coffee, tea, and cola drinks that has a marked stimulatory effect on mammals. The chemical analysis of caffeine shows that it contains 49.18% carbon, 5.39% hydrogen, 28.65% nitrogen, and 16.68% oxygen by mass, and its experimentally determined molar mass is 196 g/mol.

Given: percent composition and molar mass

Asked for: molecular formula

Strategy:

A Assume 100 g of caffeine. From the percentages given, use the procedure given in Example 6 to calculate the empirical formula of caffeine.

B Calculate the formula mass and then divide the experimentally determined molar mass by the formula mass. This gives the number of formula units present.

C Multiply each subscript in the empirical formula by the number of formula units to give the molecular formula.

Solution:

A We begin by dividing the mass of each element in 100.0 g of caffeine (49.18 g of carbon, 5.39 g of hydrogen, 28.65 g of nitrogen, 16.68 g of oxygen) by its molar mass. This gives the number of moles of each element in 100 g of caffeine.

$$\begin{array}{c}\text{moles C }=\text{ 49}\text{.18 }\cancel{\text{g C}}\times \frac{\text{1 mol C}}{\text{12}\text{.011 }\cancel{\text{g C}}}=\text{ 4}\text{.095 mol C}\\ \text{moles H }=\text{ 5}\text{.39 }\cancel{\text{g H}}\times \frac{\text{1 mol H}}{\text{1}\text{.0079 }\cancel{\text{g H}}}=\text{ 5}\text{.35 mol H}\\ \text{moles N }=\text{ 28}\text{.65 }\cancel{\text{g N}}\times \frac{\text{1 mol N}}{\text{14}\text{.0067 }\cancel{\text{g N}}}=\text{ 2}\text{.045 mol N}\\ \text{moles O }=\text{ 16}\text{.68 }\cancel{\text{g O}}\times \frac{\text{1 mol O}}{\text{15}\text{.9994 }\cancel{\text{g O}}}=\text{ 1}\text{.043 mol O}\end{array}$$

To obtain the relative numbers of atoms of each element present, divide the number of moles of each element by the number of moles of the element present in the least amount:

$$\begin{array}{llll}\text{O: }\frac{\text{1}\text{.043}}{\text{1}\text{.043}}=\text{ 1}\text{.000}\hfill & \text{C: }\frac{\text{4}\text{.095}}{\text{1}\text{.043}}=\text{ 3}\text{.926}\hfill & \text{H: }\frac{\text{5}\text{.35}}{\text{1}\text{.043}}=\text{ 5}\text{.13}\hfill & \text{N: }\frac{\text{2}\text{.045}}{\text{1}\text{.043}}=\text{ 1}\text{.960}\hfill \end{array}$$

These results are fairly typical of actual experimental data. None of the atomic ratios is exactly integral but all are within 5% of integral values. Just as in Example 6, it is reasonable to assume that such small deviations from integral values are due to minor experimental errors, so round to the nearest integer. The empirical formula of caffeine is thus C₄H₅N₂O.

B The molecular formula of caffeine could be C₄H₅N₂O, but it could also be any integral multiple of this. To determine the actual molecular formula, we must divide the experimentally determined molar mass by the formula mass. The formula mass is calculated as follows:

$$\begin{array}{rrrr}\hfill \text{4C}& \hfill \text{(4 atoms C)(12}\text{.011 g/atom C)}& \hfill =& \hfill \text{48}\text{.044 g}\\ \hfill \text{5H}& \hfill \text{(5 atoms H)(1}\text{.0079 g/atom H)}& \hfill =& \hfill \text{5}\text{.0395 g}\\ \hfill \text{2N}& \hfill \text{(2 atoms N)(14}\text{.0067 g/atom N)}& \hfill =& \hfill \text{28}\text{.0134 g}\\ \hfill +\text{1O}& \hfill \text{(1 atom O)(15}\text{.9994 g/atom O)}& \hfill =& \hfill \text{15}\text{.9994 g}\\ \hfill {\text{C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{N}}_{\text{2}}\text{O}& \hfill \text{formula mass of caffeine}& \hfill =& \hfill 97.096\text{ g}\end{array}$$

Dividing the measured molar mass of caffeine (196 g/mol) by the calculated formula mass gives

$$\frac{\text{196 g/mol}}{\text{97}{\text{.096 g/C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{N}}_{\text{2}}\text{O}}=2.02\approx 2{\text{ C}}_{\text{4}}{\text{H}}_{\text{5}}{\text{N}}_{\text{2}}\text{O empirical formula units}$$

C There are two C₄H₅N₂O formula units in caffeine, so the molecular formula must be (C₄H₅N₂O)₂ = C₈H₁₀N₄O₂. The structure of caffeine is as follows:

Exercise

Calculate the molecular formula of Freon-114, which has 13.85% carbon, 41.89% chlorine, and 44.06% fluorine. The experimentally measured molar mass of this compound is 171 g/mol. Like Freon-11, Freon-114 is a commonly used refrigerant that has been implicated in the destruction of the ozone layer.

Answer: C₂Cl₂F₄