Percentage composition

What the percentage actually measures. A compound's formula fixes exactly how many atoms of each element sit in one formula unit. Multiplying that atom count by the element's atomic mass gives the mass that element contributes to one mole of compound; dividing by the compound's total molar mass gives the fraction of the mass that is "that element's share." Multiplying by $100$ turns the fraction into a percentage.

The general formula. For a compound of molar mass $M$ in which element $E$ appears as $x$ atoms per formula unit:

where: $x$ is the number of atoms of element $E$ per formula unit (read directly from the subscript in the formula), atomic mass of $E$ is in $\text{g mol}^{-1}$, and $M$ is the molar mass of the whole compound in $\text{g mol}^{-1}$. The percentage is dimensionless.

Why the percentages must add to $100\%$. Every atom in the formula unit belongs to some element, so the masses contributed by all the elements add up to the full molar mass $M$. Dividing that sum by $M$ and multiplying by $100$ necessarily gives exactly $100\%$; any small deviation seen in practice is only rounding in the atomic masses used.

Bracketed groups multiply through. When a formula has a subscript outside brackets, such as $(\text{NH}_4)_2\text{SO}_4$, that subscript multiplies every atom inside the bracket — $2$ nitrogen and $8$ hydrogen, not $1$ nitrogen and $4$ hydrogen. Getting this distribution right is the single most common source of error in percentage composition problems.

Determine the percentage composition by mass of every element in glucose, $\text{C}_6\text{H}_{12}\text{O}_6$, and use it to find the mass of carbon in a $9.00\ \text{g}$ sample of pure glucose. (Atomic masses: $\text{C} = 12$, $\text{H} = 1$, $\text{O} = 16$.)

Step 1. Find the molar mass of glucose. $M = (6 \times 12) + (12 \times 1) + (6 \times 16) = 72 + 12 + 96 = 180\ \text{g mol}^{-1}$.

Step 2. Find the mass each element contributes per mole. Carbon: $6 \times 12 = 72\ \text{g}$. Hydrogen: $12 \times 1 = 12\ \text{g}$. Oxygen: $6 \times 16 = 96\ \text{g}$.

Step 3. Convert each mass to a percentage of $M = 180\ \text{g mol}^{-1}$.

Step 4. Check the percentages sum close to $100\%$: $40.0 + 6.67 + 53.3 = 99.97\% \approx 100\%$, the small gap being rounding. This self-check catches most arithmetic slips before they reach a final answer.

Step 5. Use the percentage of carbon to find the mass of carbon in a $9.00\ \text{g}$ sample. Since $40.0\%$ of any pure glucose sample's mass is carbon, $m_{\text{C}} = 0.400 \times 9.00\ \text{g} = 3.60\ \text{g}$.

Competitive stretch

Ammonium sulphate fertiliser, $(\text{NH}_4)_2\text{SO}_4$ ($\text{N} = 14$, $\text{H} = 1$, $\text{S} = 32$, $\text{O} = 16$), has molar mass $M = 2(14 + 4) + 32 + 4(16) = 36 + 32 + 64 = 132\ \text{g mol}^{-1}$. There are $2$ nitrogen atoms per formula unit, so $\%\ \text{N} = (2 \times 14)/132 \times 100 = 21.2\%$. In a $500\ \text{g}$ bag, $m_{\text{N}} = 0.212 \times 500 = 106\ \text{g}$ of nitrogen — exactly this kind of calculation is what lets a farmer compare two fertiliser brands directly from their formulas, without weighing out and testing either one.

• Forgetting to multiply the atomic mass by the number of atoms of that element (e.g. treating nitrogen in $(\text{NH}_4)_2\text{SO}_4$ as $1$ atom instead of $2$).
• Mishandling brackets: $(\text{NH}_4)_2$ means $2$ N and $8$ H, not $1$ N and $4$ H — the outer subscript multiplies everything inside.
• Using the atomic mass of an element in the denominator instead of the compound's molar mass, or the reverse.
• Skipping the sum-to-$100\%$ check, which is the fastest way to catch a dropped atom or a wrong molar mass before submitting an answer.

Ask the tutor: Ask the AI tutor for more percentage composition problems on compounds with bracketed polyatomic groups, or to check your working on a fertiliser or mineral-analysis calculation.