KEY2CHEM

Conversion between Moles and Other Units

Atoms and molecules are the building blocks of all matter; however, due to their extremely small size, they are difficult to count individually. Chemists use the concept of the mole (the SI unit for amount) to count atoms and molecules by weighing them. The mole allows for the relationship between numbers of particles, amount (moles), mass, and volume. Different conversion factors allow for the interpretation of these relationships.


Example 1.

How many moles of \(\require{mhchem}\ce{NaCl}\) are contained in \(14.7\text { g}\) of \(\require{mhchem}\ce{NaCl}\)?

A. \(0.252\text{ mol}\)

B. \(3.98\text{ mol}\)

C. \(859\text{ mol}\)

 

Solution

A. \(0.252\text{ mol}\)

The molar mass is used as a conversion factor between grams and moles. 

\(14.7\text{ g } \times \frac{1\text{ mol }\require{mhchem}\ce{NaCl}}{58.44\text{ g} \require{mhchem}\ce{NaCl}} = 0.252\text{ mol} \)


Example 2.

What is the molarity of a \(500.0\text{ mL}\) solution that contains \(15.8\text{ g}\) of \(\require{mhchem}\ce{NaOH}\).

A. \(7.92 \times 10^{-4}\text{ M}\)

B. \(0.792\text{ M}\)

C. \(0.0316 \text{ M}\)

 

Solution

B. \(0.792\text{ M}\)

Converting grams to moles using molar mass, we can then divide by the volume (in liters) to get to molarity, which has units of moles per liter. 

\(15.8\text{ g} \times \frac{1\text{ mol }\require{mhchem}\ce{NaOH}}{39.908 \text{ g } \ce{NaOH}} = 0.396\text{ mol } \ce{NaOH} \)

\(\text{Molarity} = \frac{0.396\text{ mol }\require{mhchem}\ce{NaOH}}{0.500\text{ L}} = 0.792\text{ M}\)


Example 3. 

How many molecules of \(\require{mhchem}\ce{CO2}\) are contained in \(95.2\text{ g}\) of \(\require{mhchem}\ce{CO2}\)?

A. \(1.30\times10^{24}\text{ molecules } \)

B. \(2.52\times10^{27}\text{ molecules } \)

C. \(2.16\text{ molecules}\)

 

Solution

A. \(1.30\times10^{24}\text{ molecules } \)

This is a two-step conversion, where grams are converted first to moles using molar mass, and then moles are converted to molecules using Avogadro’s number. 

\(95.2\text{ g }\require{mhchem}\ce{CO2} \times \frac{1\text{ mol }\ce{CO2}}{44.01 \text{ g } \ce{CO2}} \times \frac{6.022\times 10^{23}\text{ molecules}}{1\text{ mol }\ce{CO2}}= 1.30\times10^{24}\text{ molecules } \ce{CO2} \)