Additional Problems: 1. A confiscated white substance, suspected of being cocain

Question

Additional Problems: 1. A confiscated white substance, suspected of being cocaine, was purified by a forensic e yielded chemist and subjected to elemental analysis. Combustion of a 50.85 mg sampl 150.0 mg of co, and 46.05 mg of H,0. Analysis for nitrogen showed that the compou contained 9.39% N by mass. What is the empirical formula of the white substance this information? The chemical formula for cocaine is Ci,Ha,NO,. Can the forensic conclude that the suspected compound is cocaine? using chemist

Explanation / Answer

mass of the compound is given = 50.85 mg

mass of CO2 produced = 150 mg

44 g of CO2 requires 12 f of Carbon

percentage of Carbon = (12/44*150) /50.85 * 100 = 80.45 %

mass of H2O produced = 46.05 mg

18 g of H2O requires 2 g of Hydrogen

percentage of H = (2/18*46.05) /50.85*100 = 10.062%

Percentage of Oxygen = 100 - (80.45 + 10.062 + 9.39) = 0.098 %

From the above percentage of elements, oxygen is very less percentage So amount of oxygen is negligible but in cocaine formula there are 4 oxygen atoms compare to Nitrogen atom.

Hence the analysis compound maynot bt Cocaine

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