Question

An organic compound contains C, H, O. 0.45g of an organic compound gave 0.44g of carbon dioxide and 0.09g of water. The molecular mass of the compound is 90 a.m.u. Calculate the molecular formula.

Answer 1

$$\begin{gathered} \%ofCandH: \\ \%ofC=\frac{12}{44}\times \frac{wtofC{O}_2}{wtoforganiccompound}\times 100 \\ =\frac{12}{44}\times \frac{0.44}{0.45}\times 100=26.67\% \\ \%ofH=\frac{2}{18}\times \frac{0.09}{0.45}\times 100 \\ =2.22\% \\ \%ofOxygen=100-(26.67+2.22) \\ =71.11\% \end{gathered}$$
Next find the empirical formula and molecular formula

Element	%	At mass	% / at mass	Simple ratio	Whole number ratio
C	26.67	12	26.67 / 12 = 2.22	2.22 / 2.22 =1	1
H	2.22	1	2.22 /1 = 2.22	2.22 / 2.22 =1	1
O	71.11	16	71.11 / 16 = 4.44	4.44 / 2.22 =2	2

Thus the empirical formula = CHO₂
The Empirical formula mass = 12 + 1 +32 = 45
n = Molecular mass / Empirical formula mass = 90 /45 =2
Molecular formula = Empirical formula x n = CHO_​2 x 2 = C₂H₂O₄