Answer to Question #144773 in General Chemistry for Joshua

Question

A 12.62 g sample of a compound containing only C,H and O produces 27.73 g of CO2 and 9.09 g H2O in a combustion analysis. Its molecular mass is found to be 300 g. Determine its a) empirical formula b) molecular formula

Answer 1

Molar mass of CO₂ = 44.01 g/mol

Molar mass of H₂O = 18.01 g/mol

mass of C in CO₂ = $(m 〖CO〗_2 × M of C)/(M 〖CO〗_2 )= (27.73 g ×12.01 g/mol)/(44.01 g/mol)$

= 7.57 g C

mass of H in H₂O = $( m H_2 O ×2 ×M of H)/(M H_2 O) = (9.09 g ×2×1.01)/(18.01 g/mol)$    = 1.02 g H

mass of O = 12.62g - 7.57g - 1.02g = 4.03 g O

7.57g C×(1mol C)/12.01g C) = 0.63 mol C

1.02g H×(1mol H)/(1.01g H) = 1.0 mol H

4.03g O×(1mol O)/(16.00g O) = 0.2525 mol O

Since atoms combine in the same ratio that moles do, we divide all of the numbers of moles by the smallest number to put everything into lowest terms:

C_0.63 H_1.0 O_0.2525 → C_0.63/0.2525 H_1.0/0.2525 O_{0.2525/0.2525} → C_2.5 H_4.0 O_1.0

If the mole ratio is not all whole numbers, we multiply through by the smallest integer which will turn all of the numbers into integers. These numbers are the subscripts of the elements in the empirical formula.

(C_2.5 H_4.0 O_1.0)×2 = C₅ H₈ O₂ (empirical formula)

If we know the molar mass of the compound, we can obtain the molecular formula by dividing the weight of the empirical formula into the molar mass; this will determine the number of empirical formula units in the molecule.

(C₅ H₈ O₂ )× 3 = C₁₅H₂₄O₆

Answer: empirical formula C₅H₈O₂; molecular formula C₁₅H₂₄O₆