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