Answer to Question #148614 in Inorganic Chemistry for berkcan

Question

Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass. Find the empirical formula and molecular formula of caffeine since its mole mass is 194.2 g / mol. (C:12 O:16 N:14.01 g/mol)

Answer 1

C = 49.48%

H = 5.15%

N = 28.87%

O = 16.40%

Divide by atomic mass

C = 49.48%/12 = 4.12

H = 5.15%/1 = 5.15

N = 28.87%/14 = 2.06

O = 16.40%/16 = 1.03

Divide by the lowest number

C = 4.12/1.03 = 4

H = 5.15/1.03 = 5

N = 2.06/1.03 = 2

O = 1.03/1.03 = 1

Therefore its empirical formula is C4H5N2O1 which can be re-written as $C_4H_5N_2O$

Its molecular formula is calculated by;

(C4H5N2O)n = 194.2

(12×4 + 1×5 + 14.01×2 + 16)n = 194.2

97.02n = 194.2

n = 2

Therefore the molecular formula of caffeine is

(C4H5N2O)2 = $C_8H_{10}N_4O_2$