Question

Determine the empirical and molecular formula given the percent compositions or masses.

26% carbon, 3.3% Hydrogen, 70.3%Oxygen Molar mass=91.0g/mol

Answer 1

M(C) = 12.01 amu

M(H) = 1.0078 amu

M(O) = 15.99 amu

Ratio(C) $= \frac{\%}{M} = \frac{26}{12.01} = 2.16$

Ratio (H) $= \frac{3.3}{1.0078} = 3.27$

Ratio(O) $= \frac{70.3}{15.99} = 4.39$

Simple ratio (C) $= \frac{2.16}{2.16} = 1$

SR(H) $= \frac{3.27}{2.16} = 1.5$

SR(O) $= \frac{4.39}{2.16} = 2.0$

Proportion: 1: 1.5 : 2.0 = 2 : 3 : 4

Empirical formula: C₂H₃O₄

M(C₂H₃O₄) =

$= 2 \times 12.01 + 3 \times 1.0078 + 4 \times 15.99 \\ = 24.02 + 3.0234 + 63.96 \\ = 91.00 \;g/mol$

Molecular formula: C₂H₃O₄