# 10.3: Molar Mass

I want to make a solution that contains 1.8 moles of potassium dichromate. I don't have a balance calibrated in molecules, but I do have one calibrated in grams. If I know the relationship between moles and the number of grams in a mole, I can use my balance to measure out the needed amount of material.

## Molar Mass

Molar mass is defined as the mass of one mole of representative particles of a substance. By looking at a periodic table, we can conclude that the molar mass of lithium is $$6.94 \: \text{g}$$, the molar mass of zinc is $$65.38 \: \text{g}$$, and the molar mass of gold is $$196.97 \: \text{g}$$. Each of these quantities contains $$6.02 \times 10^{23}$$ atoms of that particular element. The units for molar mass are grams per mole or $$\text{g/mol}$$.

### Molar Masses of Compounds

The molecular formula of the compound carbon dioxide is $$\ce{CO_2}$$. One molecule of carbon dioxide consists of 1 atom of carbon and 2 atoms of oxygen. We can calculate the mass of one molecule of carbon dioxide by adding together the masses of 1 atom of carbon and 2 atoms of oxygen.

$12.01 \: \text{amu} + 2 \left( 16.00 \: \text{amu} \right) = 44.01 \: \text{amu}$

The molecular mass of a compound is the mass of one molecule of that compound. The molecular mass of carbon dioxide is $$44.01 \: \text{amu}$$.

The molar mass of any compound is the mass in grams of one mole of that compound. One mole of carbon dioxide molecules has a mass of $$44.01 \: \text{g}$$, while one mole of sodium sulfide formula units has a mass of $$78.04 \: \text{g}$$. The molar masses are $$44.01 \: \text{g/mol}$$ and $$78.04 \: \text{g/mol}$$ respectively. In both cases, that is the mass of $$6.02 \times 10^{23}$$ representative particles. The representative particle of $$\ce{CO_2}$$ is the molecule, while for $$\ce{Na_2S}$$ it is the formula unit.

Example $$\PageIndex{1}$$

Calcium nitrate, $$\ce{Ca(NO_3)_2}$$, is used as a component in fertilizer. Determine the molar mass of calcium nitrate.

Solution

Step 1: List the known and unknown quantities and plan the problem.

Known

• Formula $$= \ce{Ca(NO_3)_2}$$
• Molar mass $$\ce{Ca} = 40.08 \: \text{g/mol}$$
• Molar mass $$\ce{N} = 14.01 \: \text{g/mol}$$
• Molar mass $$\ce{O} = 16.00 \: \text{g/mol}$$

Unknown

• Molar mass $$\ce{Ca(NO_3)_2}$$

First we need to analyze the formula. Since the $$\ce{Ca}$$ lacks a subscript, there is one $$\ce{Ca}$$ atom per formula unit. The 2 outside the parentheses means that there are two nitrate ions per formula unit and each nitrate ion consists of one nitrogen atom and three oxygen atoms per formula unit. Thus, $$1 \: \text{mol}$$ of calcium nitrate contains $$1 \: \text{mol}$$ of $$\ce{Ca}$$ atoms, $$2 \: \text{mol}$$ of $$\ce{N}$$ atoms, and $$6 \: \text{mol}$$ of $$\ce{O}$$ atoms.

Step 2: Calculate

Use the molar masses of each atom together with the number of atoms in the formula and add together.

$1 \: \text{mol} \: \ce{Ca} \times \frac{40.08 \: \text{g} \: \ce{Ca}}{1 \: \text{mol} \: \ce{Ca}} = 40.08 \: \text{g} \: \ce{Ca}$

$2 \: \text{mol} \: \ce{N} \times \frac{14.01 \: \text{g} \: \ce{N}}{1 \: \text{mol} \: \ce{N}} = 28.02 \: \text{g} \: \ce{N}$

$6 \: \text{mol} \: \ce{O} \times \frac{16.00 \: \text{g} \: \ce{O}}{1 \: \text{mol} \: \ce{O}} = 96.00 \: \text{g} \: \ce{O}$

Molar mass of $$\ce{Ca(NO_3)_2} = 40.08 \: \text{g} + 28.02 \: \text{g} + 96.00 \: \text{g} = 164.10 \: \text{g/mol}$$

## Summary

• Calculations are described for the determination of molar mass of an atom or a compound.