02 The Mole and Chemical Composition

Connecting particles to measurable amounts

Chemists need a way to connect atoms and molecules, which are too small to count individually in ordinary laboratory work, to measurable quantities. The mole provides that connection: one mole contains exactly $6.02214076 \times 10^{23}$ specified entities. This number is the Avogadro constant, written $N_A$.

Always be clear about what is being counted. One mole of $\mathrm{H_2O}$ means one mole of water molecules. Because each water molecule contains two hydrogen atoms and one oxygen atom, that amount also contains $2$ mol of hydrogen atoms and $1$ mol of oxygen atoms.

The mole is useful because it links particle counts to masses that can be measured in the laboratory.

Calculating molar mass and converting quantities

The molar mass of a substance is the mass per mole. To calculate it from a chemical formula, add the atomic molar masses for every atom, multiplying each by its formula subscript. For water:

The relationship between mass, amount in moles, and molar mass is:

Here, $n$ is the amount in moles, $m$ is the mass in grams, and $M$ is the molar mass in grams per mole. For example, a $2.50$-mol sample of sodium chloride has a mass of:

The same method applies to ionic compounds. In that case, the molar mass is for a mole of formula units; sodium chloride, $\mathrm{NaCl}$, has a molar mass of about $58.44\ \mathrm{g/mol}$.

Takeaway: Use molar mass to convert between a sample's mass and its amount in moles.

Interpreting percent composition

Percent composition states how much of a compound's total mass comes from each element. Calculate an element's percentage by dividing the mass of that element in one mole of the compound by the compound's molar mass, then multiplying by $100\%$:

For water, the molar mass is $18.02\ \mathrm{g/mol}$. Hydrogen contributes $2.016\ \mathrm{g}$ per mole, while oxygen contributes $16.00\ \mathrm{g}$:

The element percentages should total approximately $100\%$. Small differences can arise from rounding. The same composition information can also be used in reverse: measured mass percentages can help determine the compound's formula.

Finding the simplest formula from composition

An empirical formula gives the simplest whole-number ratio of atoms in a compound. To find one from mass data or mass percentages, convert each element's mass to moles, then reduce the mole amounts to their simplest ratio.

A convenient approach for percentages is to assume a $100$-g sample. Each percentage then has the same numerical value as the corresponding element's mass in grams. For a compound that is $40.0\%$ carbon, $6.7\%$ hydrogen, and $53.3\%$ oxygen, assume $100\ \mathrm{g}$. The masses are $40.0\ \mathrm{g}$ C, $6.7\ \mathrm{g}$ H, and $53.3\ \mathrm{g}$ O. Convert these to moles:

Divide each amount by the smallest amount, $3.33\ \mathrm{mol}$, to get an approximate ratio of $1:2:1$. The empirical formula is $\mathrm{CH_2O}$.

Ratios from measurements may be close to, rather than exactly, simple fractions because of experimental uncertainty and rounding. For example, a ratio near $1:1.5$ can be multiplied by $2$ to give $2:3$. Do not round to whole numbers too early; check whether values are near fractions such as $0.5$, $0.33$, or $0.25$ before choosing a common multiplier.

Takeaway: Convert masses to moles before comparing elements; the simplest whole-number mole ratio gives the empirical formula.

Using molar mass to determine a molecular formula

A molecular formula gives the actual number of each kind of atom in one molecule. It is a whole-number multiple of the empirical formula, so composition data alone may not reveal the molecular formula. A molar mass or other information about molecular size is needed to find the multiplier.

First calculate the empirical-formula mass. Then divide the compound's molecular molar mass by that value to find the multiplier $k$:

Multiply every subscript in the empirical formula by $k$. For the empirical formula $\mathrm{CH_2O}$, the empirical-formula mass is about $30.03\ \mathrm{g/mol}$. If the molecular molar mass is $180.16\ \mathrm{g/mol}$, then $k \approx 6$. Multiplying each subscript by $6$ gives the molecular formula $\mathrm{C_6H_{12}O_6}$.

Takeaway: The empirical formula establishes the simplest ratio; the molar mass determines how many times that ratio occurs in a molecule.