Percentage Composition, Empirical and Molecular Formula

Percentage composition. Once the molar mass of a compound and the contribution of each element to that

molar mass are known, it is straightforward to calculate what percentage of the compound's total mass is

contributed by each element. This is called the percentage composition, and it is calculated as:

For calcium carbonate, $\text{CaCO}_3$ (molar mass $100.09\ \text{g mol}^{-1}$): calcium contributes

$40.08\ \text{g}$, carbon contributes $12.01\ \text{g}$, and oxygen (three atoms) contributes $48.00\ \text{g}$

per mole. So the percentages are $\%\text{Ca} = \frac{40.08}{100.09} \times 100 \approx 40.0\%$,

$\%\text{C} \approx 12.0\%$, and $\%\text{O} \approx 48.0\%$ — and, as a useful check, these three percentages

should always add up to (very close to) $100\%$.

Empirical formula. The empirical formula of a compound is the simplest whole-number ratio of the atoms

of each element present in the compound. It does not necessarily show the true number of atoms in one molecule

— only the smallest ratio those numbers reduce to. For example, glucose has the molecular formula

$\text{C}_6\text{H}_{12}\text{O}_6$, but the ratio $6:12:6$ reduces to $1:2:1$, so its empirical formula is

$\text{CH}_2\text{O}$.

Finding the empirical formula from percentage composition. This is one of the most important calculations

in this chapter, and it follows a fixed procedure:

1. Assume a $100\ \text{g}$ sample of the compound, so that each given percentage becomes a mass in grams.
2. Convert each element's mass into moles by dividing by its atomic mass: $n = \dfrac{\text{mass}}{\text{atomic mass}}$.
3. Divide every mole value obtained by the smallest of those mole values, to get the simplest ratio.
4. If the ratio obtained is not made of whole numbers (e.g. it comes out close to $x.5$), multiply every term by the smallest whole number that clears the fraction (multiplying by 2 turns a ratio ending in $.5$ into whole numbers).
5. Write the empirical formula using these whole-number ratios as subscripts.

Molecular formula from empirical formula. The molecular formula shows the actual number of atoms of

each element in one real molecule of the compound, and is always a whole-number multiple, $n$, of the empirical

formula:

The empirical formula mass is simply the sum of atomic masses in the empirical formula (calculated exactly as

in Section 1.3). Once $n$ is found — it should always come out to be, or round very cleanly to, a whole number

— every subscript in the empirical formula is multiplied by $n$ to obtain the true molecular formula. This is

how the molecular formula of glucose, $\text{C}_6\text{H}_{12}\text{O}_6$, is confirmed from its empirical

formula $\text{CH}_2\text{O}$ (empirical formula mass $\approx 30.03\ \text{g mol}^{-1}$) together with its …