Answer to Question #133644 in General Chemistry for CK

Question

0.50 g of an organic compound containing carbon, hydrogen and oxygen gives 0.6875 g of carbon dioxide and 0.5625 g of water on combustion. Find the empirical formula of the compound.
(Relative atomic mass: H:1; C: 12; O: 16)

Answer 1

C_xH_yO_z + O₂ → CO₂ + H₂O

m(C_xH_yO_z) = 0.5 g

m(CO₂) = 0.6875 g

m(H₂O) = 0.5626 g

M(CO₂) = 44 g/mol

M(H₂O) = 18 g/mol

n = m/M

n(CO₂) = 0.6875/44 = 0.0156 mol

Since one mole of CO₂ is made up of one mole of C and two moles of O, if we have 0.0156 moles of CO₂ in our sample, then we know we have 0.0156 moles of C in the sample.

m(C) = 0.0156*12 = 0.1872 g

n(H₂O) = 0.5626/18 = 0.0312 mol

Since one mole of H₂O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.0312 moles of H₂O, then we have 2*(0.0312) = 0.0624 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.0624 grams of hydrogen in our original sample.

0.1872 g of C + 0.0624 g of H = 0.2496 g

m(O) = 0.5 – 0.2496 = 0.2504 g

n(O) = 0.2504/16 = 0.0156 mol

Proportion: C:H:O = 0.0156:0.0624:0.0156=1:4:1

The empirical formula for an organic compound would be CH₄O.

Answer: CH₄O.