Empirical Formula Detective

Why it matters

A mystery compound burns in oxygen, leaving behind only carbon dioxide and water. Can you deduce its empirical formula from those products?

Predict first

If 3.0 g of a compound containing only C and H produces 8.8 g of CO₂ and 5.4 g of H₂O, what is the mass of Carbon in the original sample?

Carbon is 12/44 of CO₂'s mass: 8.8 × 12/44 = 2.4 g. Hydrogen is 2/18 of H₂O: 5.4 × 2/18 = 0.6 g. Together 2.4 + 0.6 = 3.0 g — the whole sample, so the compound held only C and H, in mole ratio 2.4/12 : 0.6/1 = 1 : 3.

• 2.4 g — correct
• 8.8 g
• 3.0 g
• 1.2 g

What you do

Now run the same detective work on a mineral: 52.9% Al, 47.1% O. Moles — Al 52.9/27 ≈ 1.96, O 47.1/16 ≈ 2.94, ratio 2 : 3. Snap the ions together to build that formula.

Check yourself

What is the empirical formula of a compound with 40% C, 6.67% H, and 53.33% O by mass?

Moles: C=40/12=3.33, H=6.67/1=6.67, O=53.33/16=3.33. Ratio 1:2:1.

If the empirical formula is CH2O and the molar mass is 180 g/mol, what is the molecular formula?

Empirical mass = 30. 180/30 = 6. Multiply subscripts by 6.

• CH2O — correct
• C2H4O2
• CHO
• C6H12O6
• C6H12O6 — correct
• C3H6O3
• CH2O
• C12H24O12