Question

An unknown compound contains the elements carbon, hydrogen and oxygen. When a 6.85g sample of the compound was analysed by combustion in pure oxygen, 11.93g of CO2 and 3.05g of H2O were obtained. Calculate the empirical formula of the compound.

Answer 1

Solution:

C_xH_yO_z - an unknown compound.

The Molar mass of CO₂ is 44.01 g/mol.

The Molar mass of H₂O is 18.015 g/mol.

Let's begin by finding the moles of CO₂ and H₂O:

n(CO₂) = (11.93 g CO₂) × (1 mol CO₂ / 44.01 g) = 0.2711 mol CO₂

n(H₂O) = (3.05 g H₂O) × ( 1 mol H₂O / 18.015 g) = 0.1693 mol H₂O

According to the law of conservation of mass:

1) CO₂ → C

Therefore,

n(C) = n(CO₂) = 0.2711 mol

n(C) = 0.2711 mol

m(C) = n(C) × M(C) = (0.2711 mol) × (12.01 g/mol) = 3.256 g

2) H₂O → 2H

Therefore,

n(H) = 2 × n(H₂O) = 2 × (0.1693 mol) = 0.3386 mol

n(H) = 0.3386 mol

m(H) = n(H) × M(H) = (0.3386 mol) × (1.008 g/mol) = 0.3413 g

3) m(C_xH_yO_z) = m(C) + m(H) + m(O)

m(O) = m(CxHyOz) - m(C) - m(H)

m(O) = 6.85 g - 3.256 g - 0.3413 g = 3.2527 g

n(O) = m(O) / M(O) = (3.2527 g) / (15.999 g/mol) = 0.2033 mol

n(O) = 0.2033 mol

n(C) : n(H) : n(O) = 0.2711 mol C : 0.3386 mol H : 0.2033 mol O;

Divide each molar amount by the lesser of the three:

n(C) : n(H) : n(O) = 1.333 : 1.666 : 1 = 4 : 5 : 3

Since the resulting ratio is 4 C : 5 H : 3 O, the empirical formula is C₄H₅O₃.

Answer: The empirical formula is C₄H₅O₃.