Chemistry 1.50 - 1.51 - Moles, mass and particles

What one mole counts

Atoms and ions are so small that even a tiny laboratory sample contains an enormous number of them. Chemists therefore count particles in large, fixed groups. It is the same idea as using the word dozen for 12 objects, except the chemist's group is vastly larger.

The phrase specified particles matters. Depending on the substance and the question, the count might refer to:

• atoms, such as Mg atoms;
• molecules, such as $\ce{O2}$ or $\ce{H2O}$ molecules;
• formula units, such as $\ce{NaCl}$ formula units in an ionic substance;
• ions, such as $\ce{Na^+}$ ions or $\ce{Cl^-}$ ions.

For example, one mole of Mg contains $6.02 \times 10^{23}$ Mg atoms, while one mole of $\ce{H2O}$ contains $6.02 \times 10^{23}$ $\ce{H2O}$ molecules. These samples contain the same number of specified particles, but they do not have the same mass because an $\ce{H2O}$ molecule and an Mg atom have different relative masses.

The mass of one mole has the same numerical value as the substance's relative particle mass, but is measured in grams. For a single atom, use relative atomic mass, $A_r$. For a molecule or formula unit, add all the $A_r$ values in the formula to obtain relative formula mass, $M_r$.

Do not describe the mole as "a fixed mass in grams". A mole is an amount containing a fixed number of specified entities; its mass depends on what those entities are. Also keep formula subscripts separate from amounts: the $2$ in $\ce{H2O}$ says that each molecule contains two H atoms, not that the sample contains two moles.

Moving between mass and moles

A balance measures mass, not particle number. The mass of one mole provides the bridge between the macroscopic measurement in grams and the sub-microscopic amount of particles.

Relative atomic mass and relative formula mass have no units. The mass of a sample is measured in $\,\mathrm{g}$, and amount of substance is measured in $\,\mathrm{mol}$.

Worked example 1: mass to moles

Calculate the amount of sulfuric acid, $\ce{H2SO4}$, in $9.80 \,\mathrm{g}$. Use $A_r: \ce{H} = 1.0, \ce{S} = 32.0, \ce{O} = 16.0$.

1. Find the mass of one mole.

$M_r(\ce{H2SO4}) = (2 \times 1.0) + 32.0 + (4 \times 16.0) = 98.0$

Therefore, one mole of $\ce{H2SO4}$ has a mass of $98.0 \,\mathrm{g}$.

2. Substitute into the relationship.

$n = 9.80 \div 98.0 = 0.100 \,\mathrm{mol}$

3. Sense-check.

The sample is one tenth of $98.0 \,\mathrm{g}$, so an answer of one tenth of a mole is sensible. Multiplying by 98 instead would give far more than one mole even though the sample mass is much less than the mass of one mole.

Worked example 2: moles to mass

Calculate the mass of $0.350 \,\mathrm{mol}$ of calcium carbonate, $\ce{CaCO3}$. Use $A_r: \ce{Ca} = 40.0, \ce{C} = 12.0, \ce{O} = 16.0$.

$M_r(\ce{CaCO3}) = 40.0 + 12.0 + (3 \times 16.0) = 100.0$

$m = 0.350 \times 100.0 = 35.0 \,\mathrm{g}$

This is sensible: $0.350 \,\mathrm{mol}$ is 35% of a mole, so its mass should be 35% of $100.0 \,\mathrm{g}$.