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# 8.03: Molar Mass

Learning Objectives

• Determine the molar mass for a compound or molecule.
• Convert from moles to grams and grams to moles.

In the previous section we defined molar mass as the mass of one mole of anything, or the mass of 6.022 x 1023 of that thing. In this section we're going to look at how this applied to molecules or compounds.

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

Here are some further examples:

• The mass of a hydrogen atom is 1.0079 amu; the mass of 1 mol of hydrogen atoms is 1.0079 g.
• Elemental hydrogen exists as a diatomic molecule, H2. One molecule has a mass of 1.0079 + 1.0079 = 2.0158 amu, while 1 mol H2 has a mass of 2.0158 g.
• A molecule of H2O has a mass of about 18.01 amu; 1 mol H2O has a mass of 18.01 g.
• A single unit of NaCl has a mass of 58.45 amu; NaCl has a molar mass of 58.45 g.

In each of these moles of substances, there are 6.022 × 1023 units:

• 6.022 × 1023 atoms of H
• 6.022 × 1023 molecules of H2 and H2O,
• 6.022 × 1023 units of NaCl ions.

These relationships give us plenty of opportunities to construct conversion factors for simple calculations.

Example $$\PageIndex{2}$$

What is the molar mass of C6H12O6?

Solution

To determine the molar mass, we simply add the atomic masses of the atoms in the molecular formula but express the total in grams per mole, not atomic mass units. The masses of the atoms can be taken from the periodic table.

 6 C = 6 × 12.011 = 72.066 12 H = 12 × 1.0079 = 12.0948 6 O = 6 × 15.999 = 95.994 TOTAL = 180.155 g/mol

Per convention, the unit grams per mole is written as a fraction.

Exercise $$\PageIndex{2}$$

What is the molar mass of AgNO3?

Answer

169.87 g/mol

Knowing the molar mass of a substance, we can calculate the number of moles in a certain mass of a substance and vice versa, as these examples illustrate. The molar mass is used as the conversion factor.

Example $$\PageIndex{3}$$

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:

$3.56\, \cancel{mol\, HgCl_{2}}\times \frac{271.49\, g\, HgCl_{2}}{\cancel{mol\, HgCl_{2}}}=967\, g\, HgCl_{2}$

Exercise $$\PageIndex{3}$$

What is the mass of 33.7 mol of H2O?

Answer

607 g

Example $$\PageIndex{4}$$

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:

$240.0\, \cancel{g\, H_{2}O}\times \frac{1\, mol\, H_{2}O}{18.015\cancel{g\, H_{2}O}}=13.32\, mol\, H_{2}O$

Exercise $$\PageIndex{4}$$

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

Answer

0.363 mol

Other conversion factors can be combined with the definition of mole-density, for example.

Example $$\PageIndex{5}$$

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.

$100\cancel{ml}\: ethanol\times \frac{0.789\, g}{\cancel{ml}}\times \frac{1\, mol}{46.08\, \cancel{g}}=1.71\, mol\, ethanol$

Exercise $$\PageIndex{5}$$

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

### Converting between mass, moles or atoms of a compound to mass, moles or atoms of its elements

Now that we know how to convert from atoms to molecules, from molecules to moles and from moles to grams we can string these conversion factors together to solve more complicated problems.

Example $$\PageIndex{6}$$: Converting between grams and Atoms

How many atoms of hydrogen are in 4.6 g of CH3OH?

Solution

First we need to determine the mass of one mole of methane (CH3OH).

Using the periodic table to find the mass for each mole of our elements we have:

$1\, mole\, C \,= 1\, \cancel{mole\, C}\,\times \left(\frac{12.011\, g\, C}{1\,\cancel{mole\,C}} \right)\, = 12.011 \, g\, C$

$4\, mole\, H \,= 4\, \cancel{mole\, H}\,\times \left(\frac{1.008\, g\, H}{1\,\cancel{mole\,H}} \right)\, = 4.032 \,g\, H$

$1\, mole\, O \,= 1\, \cancel{mole\, O}\,\times \left(\frac{15.999\, g\, O}{1\,\cancel{mole\,O}} \right)\, = 15.999 \, g\,O$

Adding the masses of our individual elements have:

$12.011\, g \,+ \,4.032\,g\, +\, 15.999\, g\, =\,32.042\, g\, CH_{3}OH$

As we were calculating the molar mass of CH3OH we have

$32.042\, g\, CH_{3}OH\, = \,1\, mole \,CH_{3}OH$

Which we can use as a conversion factor

$\frac{32.042\, g\, CH_{3}OH}{1\, mole\, CH_{3}OH}\, or\, \frac{1\, mole\, CH_{3}OH}{32.042\, g\, CH_{3}OH}$

We also know that for every molecule of CH3OH we have 4 atoms of H.

Now we can go back to the starting value given in the question:

$4.6\,\cancel{g\, CH_{3}OH}\,\times\,\left(\frac{1\, \cancel{mole\, CH_{3}OH}}{32.042\,\cancel{g\,CH_{3}OH}}\right)\times\left(\frac{6.022\, x\, 10^{23}\,\cancel{\,molecules\, CH_{3}OH}}{1 \,\cancel{mole\, CH_{3}OH}}\right)\times\left(\frac{4\, atoms\, H}{1 \cancel{molecule \,CH_{3}OH}}\right) = 3.5\,x\,10^{23}\, atoms\, H$

Exercise $$\PageIndex{6}$$

How many atoms of H are there in 2.06 grams of $$\ce{H_{2}O}$$?

Answer

1.38 x 1023 atoms H

Example $$\PageIndex{7}$$: Converting from Grams to grams

How many grams of oxygen are in 3.45 g of H3PO4

Solution

First we need to determine the mass of one mole of phosphoric acid $$\ce{H_{3}PO_{4}}$$

We have:

$3\, mole\, H \,= 3\, \cancel{mole\, H}\,\times \left(\frac{1.008\, g\, H}{1\,\cancel{mole\,H}} \right)\, = 3.024\,g\, H$

$1\, mole\, P \,= 1\, \cancel{mole\, P}\,\times \left(\frac{30.974\, g\, P}{1\,\cancel{mole\,P}} \right)\, = 30.974 \,g\, P$

$4\, mole\, O \,= 4\, \cancel{mole\, O}\,\times \left(\frac{15.999\, g\, O}{1\,\cancel{mole\,O}} \right)\, = 63.996 \,g\, O$

Adding the masses of our individual elements have:

$3.024\, g \,+ \,30.974\,g\, +\, 63.996\, g\, =\,97.994\, g\, H_{3}PO_{4}$

As we were calculating the molar mass of CH3OH we have

$97.994\, g\, H_{3}PO_{4}\, = \,1\, mole \,H_{3}PO_{4}$

Which we can use as a conversion factor

$\frac{97.994\, g\, H_{3}PO_{4}}{1\, mole\, H_{3}PO_{4}}\, or\, \frac{1\, mole\, H_{3}PO_{4}}{97.994\, g\, H_{3}PO_{4}}$

To calculate the molar mass of $$\ce{H_{3}PO_{4}})\ we used the idea that we have 3 moles of H, 1 mole of P and 4 moles of O for every one mole of \(\ce{H_{3}PO_{4}}). We recall that we can make our own conversion factors as long as the top and bottom are equal to each other. So much the same way I can give you four quarter or 1 dollar, if I hand you a mole of \(\ce{H_{3}PO_{4}}) I've handed you 3 mole of H, 1 mole of P and 4 moles of O. As this is true there is an additional set of conversion factors we can use: $\left(\frac{3\, moles\, H}{1\, mole\, H_{3}PO_{4}}\right)\, and \, \left(\frac{1\, mole\, P}{1\, mole\, H_{3}PO_{4}}\right)\, and\, \left(\frac{4\,moles\, O}{1\, mole\, H_{3}PO_{4}}\right)$ Now we can go back to the starting value given in the question: $3.45\,\cancel{g\,H_{3}PO_{4}}\,\times\,\left(\frac{1\,\cancel{mole\,H_{3}PO_{4}}}{97.994\,\cancel{g\,H_{3}PO_{4}}}\right)\times\left(\frac{4\,\cancel{mole\,O}}{1\,\cancel{\,mole\, H_{3}PO_{4}}}\right)\times\left(\frac{15.999\, g\, O}{1 \cancel{mole \,O}}\right) = 2.25\, g\, O$ Exercise \(\PageIndex{6}$$

How many grams of H are there in 3.45 grams of $$\ce{H_{3}PO_{4}}$$?

Answer

0.106 grams H

## Summary

• The molecular mass (in amu) is the mass of one molecule of a compound
• The molar mass (in grams) is the mass of one mole of a compound or element.
• The molar mass is a useful conversion factor, which can be used to convert from grams to moles or from moles to grams.

## Contributors

• CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.

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