Calculate the percentage elemental composition of the confiscated substance.

Question

A white substance had been confiscated on suspicion that it was cocaine. A forensic chemist purified the substance and subjected it to elemental analysis. Combustion of a 50.86 mg sample yielded 150.0 mg of CO2 and 46.05 mg of H2O. Analysis for nitrogen showed that the compound contained 9.39% by mass. The molecular

formula of cocaine is C17H21NO4.

Calculate the percentage elemental composition of the confiscated substance. Is it

cocaine?

Answer 1

150 mg CO2 x 1 g / 1000 mg x 1 mol CO2 / 44 g x 1 mol C / mol CO2 x 12 g / mol = 0.0409 g C

46.05 mg H2O x 1 g / 1000 mg x 1 mol H2O / 18 g x 2 mol H / mol H2O x 1 g H/ mol = 0.00512 g H

9.39% x 50.86 mg = 4.776 mg N x 1 g / 1000 mg = 0.00478 g N

grams O = 0.05086 g - 0.0408 g - 0.00512 g - 0.0048 g = 0.00014 g O

%composition of the unknown:

%C = 0.0409 g / 0.05086 g (x100%) = 80.42%

%H = 0.00512 g / 0.05086 g (x100%) = 10.1%

%N = 0.00478 g / 0.05086 g (x100%) = 9.39%

%O = 0.00014 g / 0.05086 g (x100%) = 0.3%

molar mass coke (C₁₇H₂₁NO₄) = (17x12) + (21x1) + 14 + (4x16) = 204 + 21 + 14 + 64 = 303 g / mol

% composition of coke:

%C = 204 / 303 (x100%) = 67.3%

%H = 21 / 303 (x100%) = 6.9%

%N = 14 / 303 (x100%) = 4.6%

%O = 64 / 303 (x100%) = 21.1%

Since the % composition of the unknown is different from that for coke, the unknown substance is NOT coke