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Question

Nicotine is a compound that contains C,H,N If a sample 2.50 g is burned in oxygen, 6.78 g CO2, 1.94 g H2O and 0.432 g N2 are the products of combustion. What is the percent coumpound of nicotine?

Answer 1

CxHyNz + O2 ===> CO2 + H2O + N2

moles CO2 = 6.78 g x 1 mole/44 g = 0.154 moles CO2 = 0.154 moles C since 1mole CO2 = 1 mole C

moles H20 = 1.94 g x 1 mole/18 g = 0.108 moles H2O = 0.216 moles H since 1 mole H2O = 2 moles H

moles N2 = 0.432 g N2 x 1 mole/28 g = 0.0154 moles N2 = 0.0309 moles N since 1 mole N2 = 2 moles N

Converting all these to grams:

grams C = 0.154 moles C x 12 g/mole = 1.848 g C

grams H = 0.216 moles H x 1 g/mole = 0.216 g H

grams N = 0.0309 moles x 14g/mole = 0.433 g N

% C = 1.848 g C/2.50 g (x100%) = 73.92%

% H = 0.216 g/2.50 g (x100%) = 8.64%

% N = 0.433 g/2.50 g (x100%) = 17.32%