Topic 1.3 Elemental Composition of Pure Substances: calculate percent composition by mass and determine empirical and molecular formulas from experimental data.

What this topic is asking

The College Board (Topic 1.3) wants you to connect the mass of a pure substance to its chemical formula. You must calculate percent composition by mass from a formula, and go the other way to find an empirical formula (the simplest whole-number ratio of atoms) and then a molecular formula (the actual formula) from experimental data such as combustion analysis.

Percent composition by mass

For example, in water $\text{H}_2\text{O}$ ($M = 18.02$ g/mol), the mass of hydrogen per mole is $2(1.008) = 2.016$ g, so hydrogen is $\dfrac{2.016}{18.02} \times 100\% = 11.2\%$ and oxygen is the remaining $88.8\%$. Percent composition is useful for checking purity and for identifying an unknown.

Empirical formula from data

The empirical formula is the simplest whole-number atom ratio. The standard workflow is:

1. Take the mass (or assume 100 g and use the percentages as grams) of each element.
2. Convert each mass to moles using its molar mass.
3. Divide every mole value by the smallest of them.
4. If the results are not all close to whole numbers, multiply them all by a small integer (2 or 3) until they are.

Molecular formula

Once you have the empirical formula, find the empirical-formula mass and divide the compound's actual molar mass by it. The result (rounded to the nearest whole number) is the factor by which every subscript in the empirical formula is multiplied to give the molecular formula.

This is why two quantities are needed to pin down a molecular formula: the mass data fix the ratio, and the molar mass fixes the overall size. Combustion analysis is the classic experimental route: an organic sample is burned, the carbon is captured as $\text{CO}_2$ and the hydrogen as $\text{H}_2\text{O}$, and the masses of those products are worked back to the moles of C and H in the original sample. Any remaining mass is usually oxygen, found by difference.

Significant figures and rounding ratios

When you divide by the smallest mole value, results like $1.33$, $1.5$ or $1.25$ are not rounding errors; they signal a ratio of $\tfrac{4}{3}$, $\tfrac{3}{2}$ or $\tfrac{5}{4}$. Multiply all values by 3, 2 or 4 respectively to clear the fraction. Only round to a whole number when a value is genuinely within about $0.1$ of an integer.

Try this

Q1. Calculate the percent by mass of nitrogen in ammonium nitrate, $\text{NH}_4\text{NO}_3$. [2 points]

• Cue. $M = 80.05$ g/mol; N contributes $2(14.01) = 28.02$ g, so $\dfrac{28.02}{80.05} \times 100\% = 35.0\%$.

Q2. A compound's empirical formula is $\text{CH}$ and its molar mass is $78$ g/mol. Determine the molecular formula. [1 point]

• Cue. Empirical-formula mass $= 13.02$; $78/13.02 = 6$, so the molecular formula is $\text{C}_6\text{H}_6$.