Example 9.2-3

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 9.2-4

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 9.2-5

What is the mass of 3.56 mol of HgCl₂? The molar mass of HgCl₂ is 271.49 g/mol.

Solution

Use the molar mass as a conversion factor between moles and grams. Because we want to cancel the mole unit and introduce the gram unit, we can use the molar mass as given:

Test Yourself

What is the mass of 33.7 mol of H₂O?

Answer

607 g

Example 9.2-6

How many moles of H₂O are present in 240.0 g of water (about the mass of a cup of water)?

Solution

Use the molar mass of H₂O as a conversion factor from mass to moles. The molar mass of water is (1.0079 + 1.0079 + 15.999) = 18.015 g/mol. However, because we want to cancel the gram unit and introduce moles, we need to take the reciprocal of this quantity, or 1 mol/18.015 g:

Test Yourself

How many moles are present in 35.6 g of H₂SO₄ (molar mass = 98.08 g/mol)?

Answer

0.363 mol

Example 9.2-7

The density of ethanol is 0.789 g/mL. How many moles are in 100.0 mL of ethanol? The molar mass of ethanol is 46.08 g/mol.

Solution

Here, we use density to convert from volume to mass and then use the molar mass to determine the number of moles.

Test Yourself

If the density of benzene, C₆H₆, is 0.879 g/mL, how many moles are present in 17.9 mL of benzene?

Answer

0.201 mol