# How do you calculate the percentage composition of (NH_4)_2CO_3?

Apr 4, 2016

29.16%N, 8.39%H, 12.5%C, 49.95%O

#### Explanation:

Calculate the total mass of the compound by adding together the mass of each atom.

$14.007 g m o {l}^{-} 1 \cdot 2 + 1.008 g m o {l}^{-} 1 \cdot 8 + 12.011 g m o {l}^{-} 1 + 15.999 g m o {l}^{-} 1 \cdot 3 = 96.086 g m o {l}^{-} 1$

To calculate the percentage composition of an element,

M_e/M_t * 100%

where ${M}_{e}$ is the mass of the element and ${M}_{t}$ is the total mass of the compound.

N -> (28.014gmol^-1)/(96.086gmol^-1)*100% = 29.16%

H -> (8.064gmol^-1)/(96.086gmol^-1)*100% = 8.39%

C -> (12.011gmol^-1)/(96.086gmol^-1)*100% = 12.5%

O -> (47.997gmol^-1)/(96.086gmol^-1)*100% = 49.95%