A formula is a ratio. That is the whole idea behind this lesson, and it is worth saying plainly because the arithmetic can obscure it: when you determine an empirical formula you are recovering the simplest whole-number ratio of atoms from a measurement that only ever gave you masses.

This is also the first place the course asks you to tell a pure substance from a mixture using numbers rather than words. A pure compound has a fixed composition by mass; a mixture does not, and that difference is what the last section here exploits.

Percent composition: mass, not atoms

The percent composition of a compound is the fraction of its molar mass contributed by each element. In water, hydrogen is two atoms out of three — but only 11.2% of the mass, because a hydrogen atom is so light.

Read the question carefully: percent by mass and percent of the atoms are different numbers, and AP asks for both.

$\%\ \text{by mass of X}=\frac{(\text{atoms of X})(\text{molar mass of X})}{\text{molar mass of compound}}\times 100\%$

Empirical formula: the four-step routine

The routine never changes, and knowing that is half the battle under time pressure. Percent to grams, grams to moles, divide by the smallest, clear any fraction.

[Diagram — see the figure in the print workbook.]

A ratio ending in .5 means multiply by 2; .33 or .67 means multiply by 3; .25 or .75 means multiply by 4. Rounding 3.5 to 4 is not allowed and costs the whole question.

Molecular formula = empirical formula × an integer

The empirical formula gives the ratio; it cannot give the size. CH₂O is the empirical formula of formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) alike.

To get from one to the other you need one extra measurement — the molar mass of the actual molecule. Divide it by the empirical formula mass and you get the multiplier, which is always a whole number.

$n=\frac{\text{molar mass of the molecule}}{\text{empirical formula mass}}\qquad\text{molecular formula}=(\text{empirical formula})_n$

Mixtures: composition is not fixed

A compound has one composition by mass, always. A mixture can have any — which is exactly what lets you work out how much of each component is present from a single measurement.

The method is always the same: write the total mass as the sum of the parts, write the measured quantity as the sum of each part’s contribution, and solve the two equations.

[Diagram — see the figure in the print workbook.]

Worked example 1

Calculate the percent composition by mass of NH₄NO₃.

1.Molar mass: N 2(14.01) = 28.02; H 4(1.008) = 4.032; O 3(16.00) = 48.00. Total = 80.05 g/mol.
2.%N = 28.02/80.05 × 100 = 35.00%.
3.%H = 4.032/80.05 × 100 = 5.04%.
4.%O = 48.00/80.05 × 100 = 59.96%.
5.Check they sum to 100%: 35.00 + 5.04 + 59.96 = 100.00%.

Both nitrogens are counted together even though they sit in chemically different places in the compound — percent composition is about mass, and mass does not care about bonding. The final sum-to-100 check catches a dropped subscript instantly.

Answer: 35.00% N, 5.04% H, 59.96% O

Worked example 2

A compound is 40.0% C, 6.7% H and 53.3% O by mass. Find its empirical formula.

1.Assume 100 g: 40.0 g C, 6.7 g H, 53.3 g O.
2.Moles: C 40.0/12.01 = 3.33; H 6.7/1.008 = 6.65; O 53.3/16.00 = 3.33.
3.Divide by the smallest (3.33): C 1.00, H 2.00, O 1.00.
4.Already whole numbers, so the empirical formula is CH₂O.

This composition belongs to a whole family of compounds — formaldehyde, acetic acid, glucose — and percent composition alone cannot tell them apart. That is the next example.

Answer: CH₂O

Worked example 3

The compound in the previous example has a molar mass of 180.2 g/mol. Find its molecular formula.

1.Empirical formula mass of CH₂O: 12.01 + 2(1.008) + 16.00 = 30.03 g/mol.
2.Multiplier: 180.2 / 30.03 = 6.00.
3.Multiply every subscript by 6: C₆H₁₂O₆.

The multiplier must come out as a whole number. If yours is 5.8 or 6.3, the empirical formula is wrong, not the molar mass — go back and check step 3.

Answer: C₆H₁₂O₆ (glucose)

Worked example 4

A compound is 26.6% K, 35.4% Cr and 38.1% O by mass. Determine its empirical formula.

1.Assume 100 g and convert: K 26.6/39.10 = 0.680 mol; Cr 35.4/52.00 = 0.681 mol; O 38.1/16.00 = 2.381 mol.
2.Divide by the smallest (0.680): K 1.000, Cr 1.001, O 3.501.
3.The 3.50 is the signal: multiply everything by 2.
4.K 2, Cr 2, O 7 → K₂Cr₂O₇.

Rounding 3.50 up to 4 would have given KCrO₄, a compound that does not exist. A half in a mole ratio is almost never noise; it is the reason step 4 is in the routine.

Answer: K₂Cr₂O₇ (potassium dichromate)

Worked example 5

Burning 0.3000 g of a compound containing only C, H and O produces 0.4397 g of CO₂ and 0.1800 g of H₂O. Find the empirical formula.

1.All the carbon ends up in the CO₂: n(C) = 0.4397/44.01 = 0.009991 mol, which is 0.1200 g of C.
2.All the hydrogen ends up in the H₂O, two H per molecule: n(H) = 2 × 0.1800/18.02 = 0.01998 mol, which is 0.02014 g of H.
3.Oxygen is found by difference, never by measurement: 0.3000 − 0.1200 − 0.02014 = 0.1599 g, so n(O) = 0.1599/16.00 = 0.009991 mol.
4.Divide by the smallest (0.009991): C 1.00, H 2.00, O 1.00.

Two things here are worth memorizing. The factor of 2 on hydrogen, because each water molecule carries two H atoms. And oxygen by difference — you cannot get it from the CO₂ or H₂O, because the combustion supplied extra oxygen of its own.

Answer: CH₂O

Worked example 6

A 2.000 g sample of a mixture of NaCl and Na₂SO₄ is found to contain 0.7171 g of sodium. Calculate the mass percent of NaCl in the mixture.

1.Sodium fraction in NaCl: 22.99/58.44 = 0.3934.
2.Sodium fraction in Na₂SO₄: 45.98/142.04 = 0.3237.
3.Let x be the mass of NaCl; the rest, (2.000 − x), is Na₂SO₄.
4.0.3934x + 0.3237(2.000 − x) = 0.7171.
5.0.0697x = 0.7171 − 0.6474 = 0.0697, so x = 1.000 g.
6.Mass percent: 1.000/2.000 × 100 = 50.0%.

The structure here is worth more than the answer: two unknowns, two equations — one for total mass, one for the measured element. Every mixture problem in this course has that shape, whatever the substances.

Answer: 50.0% NaCl by mass

Common trap: rounding a ratio that was trying to tell you something

When step 3 returns 1.00 : 1.00 : 3.50, the 3.50 is not experimental noise. It is the compound telling you the true ratio is 2 : 2 : 7. Rounding it to 4 gives KCrO₄, which does not exist.

The working rule: only round when you are within about 0.1 of a whole number. Anything else gets multiplied up. If a value sits at, say, 2.25, multiply everything by 4 — not just that one term.

Hack: assume exactly 100 grams, every time

Percent composition problems give you no mass at all, which makes students hesitate. Assume the sample is exactly 100 g and every percentage becomes a mass in grams with no arithmetic: 40.0% carbon becomes 40.0 g of carbon.

This is legitimate because you are after a ratio, and a ratio does not care what size sample you imagined. Write ‘assume 100 g’ as your first line and the AP reader knows exactly what you are doing.

Percent Composition, Empirical Formulas and Mixtures