Chapter 3 Chemical Reactions

Example 1

Calculate the molecular mass of ethanol, whose condensed structural formula is CH₃CH₂OH. Among its many uses, ethanol is a fuel for internal combustion engines.

Given: molecule

Asked for: molecular mass

Strategy:

A Determine the number of atoms of each element in the molecule.

B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.

C Add together the masses to give the molecular mass.

Solution:

A The molecular formula of ethanol may be written in three different ways: CH₃CH₂OH (which illustrates the presence of an ethyl group, CH₃CH₂−, and an −OH group), C₂H₅OH, and C₂H₆O; all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.

B Taking the atomic masses from the periodic table, we obtain

$$\begin{array}{l}\text{2 }\times \text{ atomic mass of carbon }=\text{ 2 }\cancel{\text{atoms}}\left(\frac{\text{12}\text{.011 amu}}{\cancel{\text{atom}}}\right)=24.022\text{ amu}\\ \text{6 }\times \text{ atomic mass of hydrogen }=\text{ 6 }\cancel{\text{atoms}}\left(\frac{\text{1}\text{.0079 amu}}{\cancel{\text{atom}}}\right)=6.0474\text{ amu}\\ \text{1 }\times \text{ atomic mass of oxygen }=\text{ 1 }\cancel{\text{atom}}\left(\frac{\text{15}\text{.9994 amu}}{\cancel{\text{atom}}}\right)=15.9994\text{ amu}\end{array}$$

C Adding together the masses gives the molecular mass:

24.022 amu + 6.0474 amu + 15.9994 amu = 46.069 amu

Alternatively, we could have used unit conversions to reach the result in one step, as described in Essential Skills 2 (Section 3.7 "Essential Skills 2"):

$$\left[\text{2 }\cancel{\text{atoms C}}\left(\frac{12.011\text{ amu}}{\text{1 }\cancel{\text{atom C}}}\right)\right]+\left[\text{6 }\cancel{\text{atoms H}}\left(\frac{1.0079\text{ amu}}{\text{1 }\cancel{\text{atom H}}}\right)\right]+\left[\text{1 }\cancel{\text{atoms O}}\left(\frac{15.9994\text{ amu}}{\text{1 }\cancel{\text{atom O}}}\right)\right]=46.069\text{ amu}$$

The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:

$$\begin{array}{rrrr}\hfill \text{2C}& \hfill \text{(2 atoms)(12}\text{.011 amu/atom)}& \hfill =& \hfill \text{24}\text{.022 amu}\\ \hfill \text{6H}& \hfill \text{(6 atoms)(1}\text{.0079 amu/atom)}& \hfill =& \hfill \text{6}\text{.0474 amu}\\ \hfill +\text{1O}& \hfill \text{(1 atom)(15}\text{.9994 amu/atom)}& \hfill =& \hfill \text{15}\text{.9994 amu}\\ \hfill {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}& \hfill \text{molecular mass of ethanol}& \hfill =& \hfill \text{46}\text{.069 amu}\end{array}$$

Exercise

Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is CCl₃F. Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:

Answer: 137.368 amu

Example 2

Calculate the formula mass of Ca₃(PO₄)₂, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.

Given: ionic compound

Asked for: formula mass

Strategy:

A Determine the number of atoms of each element in the empirical formula.

B Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.

C Add together the masses to give the formula mass.

Solution:

A The empirical formula—Ca₃(PO₄)₂—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca²⁺ ions and two PO₄³⁻ ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.

B Taking atomic masses from the periodic table, we obtain

$$\begin{array}{l}\text{3 }\times \text{ atomic mass of calcium }=\text{ 3 }\cancel{\text{atoms}}\left(\frac{\text{40}\text{.078 amu}}{\cancel{\text{atom}}}\right)=120.234\text{ amu}\\ \text{2 }\times \text{ atomic mass of phosphorus }=\text{ 2 }\cancel{\text{atoms}}\left(\frac{\text{30}\text{.973761 amu}}{\cancel{\text{atom}}}\right)=61.947522\text{ amu}\\ \text{8 }\times \text{ atomic mass of oxygen }=\text{ 8 }\cancel{\text{atoms}}\left(\frac{\text{15}\text{.9994 amu}}{\cancel{\text{atom}}}\right)=127.9952\text{ amu}\end{array}$$

C Adding together the masses gives the formula mass of Ca₃(PO₄)₂:

120.234 amu + 61.947522 amu + 127.9952 amu = 310.177 amu

We could also find the formula mass of Ca₃(PO₄)₂ in one step by using unit conversions or a tabular format:

$$\left[\text{3 }\cancel{\text{atoms Ca}}\left(\frac{\text{40}\text{.078 amu}}{\text{1 }\cancel{\text{atom Ca}}}\right)\right]+\left[\text{2 }\cancel{\text{atoms P}}\left(\frac{\text{30}\text{.973761 amu}}{\text{1 }\cancel{\text{atom P}}}\right)\right]+\left[\text{8 }\cancel{\text{atoms O}}\left(\frac{\text{15}\text{.9994 amu}}{\text{1 }\cancel{\text{atom O}}}\right)\right]=310.177\text{ amu}$$ $$\begin{array}{rrrr}\hfill \text{3Ca}& \hfill \text{(3 atoms)(40}\text{.078 amu/atom)}& \hfill =& \hfill \text{120}\text{.234 amu}\\ \hfill \text{2P}& \hfill \text{(2 atoms)(30}\text{.973761 amu/atom)}& \hfill =& \hfill \text{61}\text{.947522 amu}\\ \hfill +\text{8O}& \hfill \text{(8 atoms)(15}\text{.9994 amu/atom)}& \hfill =& \hfill \text{127}\text{.9952 amu}\\ \hfill {\text{Ca}}_3{\text{P}}_2{\text{O}}_8& \hfill {\text{formula mass of Ca}}_3{{\text{(PO}}_4)}_2& \hfill =& \hfill \text{310}\text{.177 amu}\end{array}$$

Exercise

Calculate the formula mass of Si₃N₄, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.

Answer: 140.29 amu

Example 3

For 35.00 g of ethylene glycol (HOCH₂CH₂OH), which is used in inks for ballpoint pens, calculate the number of

1. moles.
2. molecules.

Given: mass and molecular formula

Asked for: number of moles and number of molecules

Strategy:

A Use the molecular formula of the compound to calculate its molecular mass in grams per mole.

B Convert from mass to moles by dividing the mass given by the compound’s molar mass.

C Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.

Solution:

1. A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example 1:

   $$\begin{array}{rrrr}\hfill \text{2C}& \hfill \text{(2 atoms)(12}\text{.011 amu/atom)}& \hfill =& \hfill \text{24}\text{.022 amu}\\ \hfill \text{6H}& \hfill \text{(6 atoms)(1}\text{.0079 amu/atom)}& \hfill =& \hfill \text{6}\text{.0474 amu}\\ \hfill +\text{2O}& \hfill \text{(2 atoms)(15}\text{.9994 amu/atom)}& \hfill =& \hfill \text{31}\text{.9988 amu}\\ \hfill {\text{C}}_2{\text{H}}_6{\text{O}}_2& \hfill \text{molecular mass of ethylene glycol}& \hfill =& \hfill \text{62}\text{.068 amu}\end{array}$$

   The molar mass of ethylene glycol is 62.068 g/mol.

   B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):

   $$\frac{\text{mass of ethylene glycol (g)}}{\text{molar mass (g/mol)}}=\text{ moles ethylene glycol (mol)}$$

   So

   $$\text{35}\text{.00 }\cancel{\text{g ethylene glycol}}\left(\frac{\text{1 mol ethylene glycol}}{\text{62}\text{.068 }\cancel{\text{g ethylene glycol}}}\right)=0.5639\text{ mol ethylene glycol}$$

   It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.

2. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:

   $$\begin{array}{ll}\text{molecules of ethylene glycol }\hfill & =\text{ 0}\text{.5639 }\cancel{\text{mol}}\left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{ molecules}}{\text{1 }\cancel{\text{mol}}}\right)\hfill \\ \hfill & =\text{ 3}\text{.396}\times {\text{10}}^{\text{23}}\text{ molecules}\hfill \end{array}$$

   Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 10²³ molecules, which is indeed the case.

Exercise

For 75.0 g of CCl₃F (Freon-11), calculate the number of

1. moles.
2. molecules.

Answer:

1. 0.546 mol
2. 3.29 × 10²³ molecules

Example 4

Calculate the mass of 1.75 mol of each compound.

1. S₂Cl₂ (common name: sulfur monochloride; systematic name: disulfur dichloride)
2. Ca(ClO)₂ (calcium hypochlorite)

Given: number of moles and molecular or empirical formula

Asked for: mass

Strategy:

A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).

B Convert from moles to mass by multiplying the moles of the compound given by its molar mass.

Solution:

We begin by calculating the molecular mass of S₂Cl₂ and the formula mass of Ca(ClO)₂.

1. A The molar mass of S₂Cl₂ is obtained from its molecular mass as follows:

   $$\begin{array}{rrrr}\hfill \text{2S}& \hfill \text{(2 atoms)(32}\text{.065 amu/atom)}& \hfill =& \hfill \text{64}\text{.130 amu}\\ \hfill +\text{2Cl}& \hfill \text{(2 atoms)(35}\text{.453 amu/atom)}& \hfill =& \hfill \text{70}\text{.906 amu}\\ \hfill {\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}& \hfill {\text{molecular mass of S}}_2{\text{Cl}}_2& \hfill =& \hfill \text{1}35.036\text{ amu}\end{array}$$

   The molar mass of S₂Cl₂ is 135.036 g/mol.

   B The mass of 1.75 mol of S₂Cl₂ is calculated as follows:

   $$\begin{array}{l}{\text{moles S}}_{\text{2}}{\text{Cl}}_{\text{2}}\left[\text{molar mass}\left(\frac{\text{g}}{\text{mol}}\right)\right]\to {\text{mass of S}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\\ \text{1}\text{.75 }\cancel{{\text{mol S}}_{\text{2}}{\text{Cl}}_{\text{2}}}\left(\frac{\text{135}{\text{.036 g S}}_{\text{2}}{\text{Cl}}_{\text{2}}}{\text{1 }\cancel{{\text{mol S}}_{\text{2}}{\text{Cl}}_{\text{2}}}}\right)={\text{236 g S}}_{\text{2}}{\text{Cl}}_{\text{2}}\end{array}$$

2. A The formula mass of Ca(ClO)₂ is obtained as follows:

   $$\begin{array}{rrrr}\hfill \text{1Ca}& \hfill \text{(1 atom)(40}\text{.078 amu/atom)}& \hfill =& \hfill \text{40}\text{.078 amu}\\ \hfill \text{2Cl}& \hfill \text{(2 atoms)(35}\text{.453 amu/atom)}& \hfill =& \hfill \text{70}\text{.906 amu}\\ \hfill +\text{2O}& \hfill \text{(2 atoms)(15}\text{.9994 amu/atom)}& \hfill =& \hfill \text{31}\text{.9988 amu}\\ \hfill {\text{Ca(ClO)}}_2& \hfill {\text{formula mass of Ca(ClO)}}_2& \hfill =& \hfill \text{1}42.983\text{ amu}\end{array}$$

   The molar mass of Ca(ClO)₂ 142.983 g/mol.

   B The mass of 1.75 mol of Ca(ClO)₂ is calculated as follows:

   $$\begin{array}{l}{\text{moles Ca(ClO)}}_{\text{2}}\left[\frac{{\text{molar mass Ca(ClO)}}_{\text{2}}}{{\text{1 mol Ca(ClO)}}_{\text{2}}}\right]={\text{mass Ca(ClO)}}_{\text{2}}\\ \text{1}\text{.75 }\cancel{{\text{mol Ca(ClO)}}_{\text{2}}}\left[\frac{\text{142}{\text{.983 g Ca(ClO)}}_{\text{2}}}{\text{1 }\cancel{{\text{mol Ca(ClO)}}_{\text{2}}}}\right]={\text{250 g Ca(ClO)}}_{\text{2}}\end{array}$$

   Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.

Exercise

Calculate the mass of 0.0122 mol of each compound.

1. Si₃N₄ (silicon nitride), used as bearings and rollers
2. (CH₃)₃N (trimethylamine), a corrosion inhibitor

Answer:

1. 1.71 g
2. 0.721 g

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

Example 9

The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:

C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l)

Construct a table showing how to interpret the information in this equation in terms of

1. a single molecule of glucose.
2. moles of reactants and products.
3. grams of reactants and products represented by 1 mol of glucose.
4. numbers of molecules of reactants and products represented by 1 mol of glucose.

Given: balanced chemical equation

Asked for: molecule, mole, and mass relationships

Strategy:

A Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.

B Use the molar masses of the reactants and products to convert from moles to grams.

C Use Avogadro’s number to convert from moles to the number of molecules.

Solution:

This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.

1. A One molecule of glucose reacts with 6 molecules of O₂ to yield 6 molecules of CO₂ and 6 molecules of H₂O.
2. One mole of glucose reacts with 6 mol of O₂ to yield 6 mol of CO₂ and 6 mol of H₂O.

3. B To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O₂, 31.9988; CO₂, 44.010; and H₂O, 18.015.

   $$\begin{array}{c}\text{mass of reactants }=\text{ mass of products}\\ \text{g glucose} + {\text{g O}}_{\text{2}}={\text{g CO}}_{\text{2}}+{\text{g H}}_{\text{2}}\text{O}\\ \text{1 }\cancel{\text{mol glucose}}\left(\frac{\text{180}\text{.16 g}}{\text{1 }\cancel{\text{mol glucose}}}\right)+\text{6 }\cancel{{\text{mol O}}_{\text{2}}}\left(\frac{\text{31}\text{.9988 g}}{\text{1 }\cancel{{\text{mol O}}_{\text{2}}}}\right)=\text{6 }\cancel{{\text{mol CO}}_{\text{2}}}\left(\frac{\text{44}\text{.010 g}}{\text{1 }\cancel{{\text{mol CO}}_{\text{2}}}}\right)+\text{6 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}\left(\frac{\text{18}\text{.015 g}}{\text{1 }\cancel{{\text{mol H}}_{\text{2}}\text{O}}}\right)\\ \text{372}\text{.15 g }=\text{ 372}\text{.15 g}\end{array}$$

4. C One mole of glucose contains Avogadro’s number (6.022 × 10²³) of glucose molecules. Thus 6.022 × 10²³ glucose molecules react with (6 × 6.022 × 10²³) = 3.613 × 10²⁴ oxygen molecules to yield (6 × 6.022 × 10²³) = 3.613 × 10²⁴ molecules each of CO₂ and H₂O.

   In tabular form:

   	C6H12O6(s)	+	6O2(g)	→	6CO2(g)	+	6H2O(l)
   a.	1 molecule		6 molecules		6 molecules		6 molecules
   b.	1 mol		6 mol		6 mol		6 mol
   c.	180.16 g		191.9928 g		264.06 g		108.09 g
   d.	6.022 × 1023 molecules		3.613 × 1024 molecules		3.613 × 1024 molecules		3.613 × 1024 molecules

Exercise

Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas. The explosion resulted from the following reaction:

2NH₄NO₃(s) → 2N₂(g) + 4H₂O(g) + O₂(g)

Construct a table showing how to interpret the information in the equation in terms of

1. individual molecules and ions.
2. moles of reactants and products.
3. grams of reactants and products given 2 mol of ammonium nitrate.
4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.

Answer:

Example 12

Ethyl acetate (CH₃CO₂C₂H₅) is the solvent in many fingernail polish removers and is used to decaffeinate coffee beans and tea leaves. It is prepared by reacting ethanol (C₂H₅OH) with acetic acid (CH₃CO₂H); the other product is water. A small amount of sulfuric acid is used to accelerate the reaction, but the sulfuric acid is not consumed and does not appear in the balanced chemical equation. Given 10.0 mL each of acetic acid and ethanol, how many grams of ethyl acetate can be prepared from this reaction? The densities of acetic acid and ethanol are 1.0492 g/mL and 0.7893 g/mL, respectively.

Given: reactants, products, and volumes and densities of reactants

Asked for: mass of product

Strategy:

A Balance the chemical equation for the reaction.

B Use the given densities to convert from volume to mass. Then use each molar mass to convert from mass to moles.

C Using mole ratios, determine which substance is the limiting reactant. After identifying the limiting reactant, use mole ratios based on the number of moles of limiting reactant to determine the number of moles of product.

D Convert from moles of product to mass of product.

Solution:

A We always begin by writing the balanced chemical equation for the reaction:

C₂H₅OH(l) + CH₃CO₂H(aq) → CH₃CO₂C₂H₅(aq) + H₂O(l)

B We need to calculate the number of moles of ethanol and acetic acid that are present in 10.0 mL of each. Recall from Chapter 1 "Introduction to Chemistry" that the density of a substance is the mass divided by the volume:

$$\text{density }=\frac{\text{mass}}{\text{volume}}$$

Rearranging this expression gives mass = (density)(volume). We can replace mass by the product of the density and the volume to calculate the number of moles of each substance in 10.0 mL (remember, 1 mL = 1 cm³):

$$\begin{array}{ll}{\text{moles C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH }\hfill & =\frac{{\text{mass C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{{\text{molar mass C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\hfill \\ \hfill & =\frac{{\text{volume C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\times {\text{density C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{{\text{molar mass C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\hfill \\ \hfill & =\text{ 10}\text{.0 }\cancel{{\text{mL C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\times \frac{\text{0}\text{.7893 }\cancel{{\text{g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}{\text{1 }\cancel{{\text{mL C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}\times \frac{1{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}{\text{46}\text{.07 }\cancel{{\text{g C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}\hfill \\ \hfill & =\text{ 0}{\text{.171 mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\hfill \\ {\text{moles CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H }\hfill & =\frac{{\text{mass CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}{{\text{molar mass CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}\hfill \\ \hfill & =\frac{{\text{volume CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\times {\text{density CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}{{\text{molar mass CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}\hfill \\ \hfill & =\text{ 10}\text{.0 }\cancel{{\text{mL CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}\times \frac{\text{1}\text{.0492 }\cancel{{\text{g CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}}{\text{1 }\cancel{{\text{mL CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}}\times \frac{1{\text{ mol CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}{\text{60}\text{.05 }\cancel{{\text{g CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}}}\hfill \\ \hfill & =\text{ 0}{\text{.175 mol CH}}_{\text{3}}{\text{CO}}_{\text{2}}\text{H}\hfill \end{array}$$

C The number of moles of acetic acid exceeds the number of moles of ethanol. Because the reactants both have coefficients of 1 in the balanced chemical equation, the mole ratio is 1:1. We have 0.171 mol of ethanol and 0.175 mol of acetic acid, so ethanol is the limiting reactant and acetic acid is in excess. The coefficient in the balanced chemical equation for the product (ethyl acetate) is also 1, so the mole ratio of ethanol and ethyl acetate is also 1:1. This means that given 0.171 mol of ethanol, the amount of ethyl acetate produced must also be 0.171 mol:

$$\begin{array}{ll}\text{moles ethyl acetate }\hfill & =\text{ mol ethanol}\times \frac{\text{1 mol ethyl acetate}}{\text{1 mol ethanol}}\hfill \\ \hfill & =\text{ 0}\text{.171 }\cancel{{\text{mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}\times \frac{{\text{1 mol CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}{\text{1 }\cancel{{\text{mol C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}}}\hfill \\ \hfill & =\text{ 0}{\text{.171 mol CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\hfill \end{array}$$

D The final step is to determine the mass of ethyl acetate that can be formed, which we do by multiplying the number of moles by the molar mass:

$$\begin{array}{ll}\text{mass of ethyl acetate }\hfill & =\text{ mol ethyl acetate }\times \text{ molar mass ethyl acetate}\hfill \\ \hfill & =\text{ 0}\text{.171 }\cancel{{\text{mol CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}\times \frac{\text{88}{\text{.11 g CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}{\text{1 }\cancel{{\text{mol CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}}\hfill \\ \hfill & =\text{ 15}{\text{.1 g CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\hfill \end{array}$$

Thus 15.1 g of ethyl acetate can be prepared in this reaction. If necessary, you could use the density of ethyl acetate (0.9003 g/cm³) to determine the volume of ethyl acetate that could be produced:

$$\begin{array}{ll}\text{volume of ethyl acetate }\hfill & =\text{ 15}\text{.1 }\cancel{{\text{g CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}\times \frac{{\text{1 mL CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}{\text{0}\text{.9003 }\cancel{{\text{g CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}}}\hfill \\ \hfill & =\text{ 16}{\text{.8 mL CH}}_{\text{3}}{\text{CO}}_{\text{2}}{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\hfill \end{array}$$

Exercise

Under appropriate conditions, the reaction of elemental phosphorus and elemental sulfur produces the compound P₄S₁₀. How much P₄S₁₀ can be prepared starting with 10.0 g of P₄ and 30.0 g of S₈?

Answer: 35.9 g

Example 13

Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H₂SO₄ (indicated above the arrow) by the reaction

$$\underset{p\text{-aminobenzoic acid}}{{\text{C}}_{\text{7}}{\text{H}}_{\text{7}}{\text{NO}}_{\text{2}}}+\underset{\text{2-diethylaminoethanol}}{{\text{C}}_{\text{6}}{\text{H}}_{\text{15}}\text{NO}}\stackrel{{\text{H}}_{\text{2}}{\text{SO}}_4}{\to }\underset{\text{procaine}}{{\text{C}}_{\text{13}}{\text{H}}_{\text{20}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}}+{\text{H}}_{\text{2}}\text{O}$$

If we carried out this reaction using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and we isolated 15.7 g of procaine, what was the percent yield?

Given: masses of reactants and product

Asked for: percent yield

Strategy:

A Write the balanced chemical equation.

B Convert from mass of reactants and product to moles using molar masses and then use mole ratios to determine which is the limiting reactant. Based on the number of moles of the limiting reactant, use mole ratios to determine the theoretical yield.

C Calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100.

Solution:

A From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water.

B To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C₇H₇NO₂), 137.14 g/mol; 2-diethylaminoethanol (C₆H₁₅NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants:

$$\begin{array}{l}\text{moles }p\text{-aminobenzoic acid }=\text{ 10}\text{.0 }\cancel{\text{g}}\times \frac{\text{1 mol}}{\text{137}\text{.14 }\cancel{\text{g}}}=\text{ 0}\text{.0729 mol }p\text{-aminobenzoic acid}\\ \text{moles 2-diethylaminoethanol }=\text{ 10}\text{.0 }\cancel{\text{g}}\times \frac{\text{1 mol}}{\text{117}\text{.19 }\cancel{\text{g}}}=\text{ 0}\text{.0853 mol 2-diethylaminoethanol}\end{array}$$

The reaction requires a 1:1 mole ratio of the two reactants, so p-aminobenzoic acid is the limiting reactant. Based on the coefficients in the balanced chemical equation, 1 mol of p-aminobenzoic acid yields 1 mol of procaine. We can therefore obtain only a maximum of 0.0729 mol of procaine. To calculate the corresponding mass of procaine, we use its structural formula (C₁₃H₂₀N₂O₂) to calculate its molar mass, which is 236.31 g/mol.

$$\text{theoretical yield of procaine }=\text{ 0}\text{.0729 }\cancel{\text{mol}}\times \frac{\text{236}\text{.31 g}}{\text{1 }\cancel{\text{mol}}}=\text{ 17}\text{.2 g}$$

C The actual yield was only 15.7 g of procaine, so the percent yield was

$$\text{percent yield }=\frac{\text{15}\text{.7 g}}{\text{17}\text{.2 g}}\times 100=91.3\%$$

(If the product were pure and dry, this yield would indicate that we have very good lab technique!)

Exercise

Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows:

PbO(s) + C(s) → Pb(l) + CO(g)

If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield?

Answer: 89.2%