Question

A liquid sample that weighs 2.585 g is a mixture of butane (c4h10) and hexane (c6h14). When the sample is combusted in an excess of oxygen, 3.929 g of water are formed.

Answer 1

Missing question: How much butane and hexane (answer in g) were in the original sample?

Answer is: 1.138 grams of butane and 2.61 grams hexane.

Balanced chemical reactions:

1) C₄H₁₀ + 13/2O₂ → 4CO₂ + 5H₂O.

2) C₆H₁₄ + 19/2O₂ → 6CO₂ + 7H₂O.

m(H₂O) = 3.929 g.

n(H₂O) = 3.929 g ÷ 18 g/mol.

n(H₂O) = 0.218 mol.

Amount of water in both reactions is 0.218 mol, five parts from butane, seven parts from hexane.

n(H₂O) = (0.218 mol ÷ 12) · 5 = 0.091 mol; water from butane.

n(C₄H₁₀) : n(H₂O) = 1 : 4.

n(C₄H₁₀) = 0.023 mol.

m(C₄H₁₀) = 0.023 mol · 58 g/mol = 1.318 g.