AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/1/1-3)
Unit 1 · Topic 1.3
1.3 Elemental Composition of Pure Substances
A pure compound always has the same elements in the same ratio by mass, no matter where the sample came from. Using that fixed ratio, you can calculate percent composition and work out the empirical formula, the simplest whole-number ratio of atoms.
Key terms
- pure substance
- law of definite proportions
- percent composition
- empirical formula
- formula unit
Pure substances
A pure substance has only one kind of particle. Some are made of separate molecules, such as H₂O or CO₂. Others, like NaCl, are huge 3-D arrays of ions with no separate molecules; for these, the formula describes a formula unit, the simplest ratio of ions (one Na⁺ for every Cl⁻).
The law of definite proportions says any pure sample of a compound has the same mass ratio of its elements. Water from a glacier and water made in a lab are both 11.19% hydrogen and 88.81% oxygen by mass. If a sample's mass ratio is different, it's either a different compound or not pure.
Percent composition
Percent by mass of an element = (mass of that element in the compound ÷ total mass) × 100.
Using a formula, take the mass of the element in one mole of the compound and divide by the molar mass. For glucose, C₆H₁₂O₆ (180.16 g/mol), carbon is 6(12.01) ÷ 180.16 × 100 = 40.00% C.
From mass data to an empirical formula
The empirical formula is the lowest whole-number ratio of atoms. Subscripts count atoms, not grams, so you always have to convert masses to moles before comparing.
- If you're given percentages, assume a 100 g sample so each percent becomes grams.
- Convert each mass to moles using the atomic mass.
- Divide every mole value by the smallest one.
- If a ratio ends in about .5, .33 or .25, multiply all ratios by 2, 3 or 4 to get whole numbers. Don't round 1.5 to 2.
- Write the formula with those whole numbers as subscripts.
Using lab data
In the lab, you rarely start with percentages. A common experiment heats a hydrate, a salt with water molecules built into its crystals, such as CuSO₄·5H₂O. The mass lost on heating is the water. Convert the mass of water and the mass of dry salt left behind to moles, and their ratio gives the number of water molecules per formula unit.
The same idea works for any data: find the mass of each element, convert to moles, and compare. If some water stays behind because the sample wasn't heated long enough, the measured water mass is too small, and the calculated ratio of water to salt comes out too low.
Empirical versus molecular formula
Different compounds can share an empirical formula. Formaldehyde (CH₂O), acetic acid (C₂H₄O₂) and glucose (C₆H₁₂O₆) all reduce to CH₂O, so they have identical percent compositions. Percent composition alone can't tell them apart.
If you also know the molar mass, you can find the molecular formula: divide the molar mass by the empirical formula's mass, then multiply every subscript by that whole number. For ionic compounds, the empirical formula is the formula you write.
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Empirical formula from percent composition
A compound is 40.0% carbon, 6.71% hydrogen and 53.3% oxygen by mass. Find its empirical formula.
Show the solutionHide the solution
- Step 1: Assume 100 g: 40.0 g C, 6.71 g H, 53.3 g O.
- Step 2: Moles: C = 40.0 ÷ 12.01 = 3.331; H = 6.71 ÷ 1.008 = 6.657; O = 53.3 ÷ 16.00 = 3.331.
- Step 3: Divide by the smallest (3.331): C = 1.00, H = 2.00, O = 1.00.
Answer: CH₂O
- Example 2Calculator allowed
A ratio that ends in .5 (classic trap)
An oxide of iron is 69.94% Fe and 30.06% O by mass. Find its empirical formula.
Show the solutionHide the solution
- Step 1: Moles in 100 g: Fe = 69.94 ÷ 55.85 = 1.252; O = 30.06 ÷ 16.00 = 1.879.
- Step 2: Divide by the smaller: Fe = 1.000, O = 1.879 ÷ 1.252 = 1.50.
- Step 3: 1.50 is not close to a whole number, so don't round it to 2. Multiply both by 2: Fe = 2, O = 3.
- Step 4: Check with charges: Fe³⁺ and O²⁻ give Fe₂O₃, which matches.
Answer: Fe₂O₃
- Example 3Calculator allowed
Empirical formula from lab masses
A 2.50 g sample of a compound of nitrogen and oxygen contains 0.761 g of nitrogen. What is its empirical formula?
Show the solutionHide the solution
- Step 1: Mass of oxygen = 2.50 − 0.761 = 1.739 g.
- Step 2: Moles: N = 0.761 ÷ 14.01 = 0.05432 mol; O = 1.739 ÷ 16.00 = 0.1087 mol.
- Step 3: Ratio: O ÷ N = 0.1087 ÷ 0.05432 = 2.00, so there are 2 O atoms per N atom.
Answer: NO₂
Common mistakes
- Comparing grams instead of moles. Subscripts are atom ratios, so convert every mass to moles first.
- Rounding ratios like 1.50 or 1.33 to the nearest whole number. Multiply through by 2 or 3 instead.
- Assuming the empirical formula is the molecular formula. You need the molar mass to know the true molecular formula.
- Forgetting to find the mass of the last element by subtraction when only some masses are given.
On the exam
- Free-response questions often give lab data (mass of a sample before and after heating, or mass of one element) and ask for an empirical formula. Show moles of each element and the ratio; that's where the points are.
- Multiple-choice questions may ask which formula matches a given percent composition. You can estimate: the element with the bigger share of the mass often, but not always, has the bigger subscript, so check with moles.
Connected topics
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Check yourself
4 questions on 1.3 Elemental Composition of Pure Substances. Pick an answer to see if you got it, and why.
A 4.00 g sample of a pure iron oxide contains 2.80 g of iron. What is the empirical formula of the compound?
A 50.0 g sample of pure CaCO₃ from a limestone quarry contains 20.0 g of calcium. How many grams of calcium are in a 125 g sample of pure CaCO₃ made in a laboratory?
Ammonium nitrate, NH₄NO₃ (molar mass 80.05 g/mol), is used as a fertilizer. What is the percent by mass of nitrogen in NH₄NO₃?
A sugar has the empirical formula CH₂O and a molar mass of about 180 g/mol. What is its molecular formula?
0 of 4 answered