# What is the percentage composition of (NH_4)_2CO_3?

Feb 28, 2016

$\text{29.155% N}$
$\text{8.392% H}$
$\text{12.500% C}$
$\text{49.952% O}$

#### Explanation:

In order to determine percent composition of the elements in a compound, the molar mass (MM) of each element is divided by the MM of the compound and multiplied by 100.

"% composition of element"=("MM of element")/("MM of compound")xx100

The molar mass of ammonium carbonate $\left(\text{(NH"_4")"_2"CO"_3}\right)$ is $\text{96.086 g/mol}$. (Determined by multiplying the subscripts of each element times its MM using its atomic mass in g/mol on the periodic table.)

Determine the MM of each element in the compound by multiplying the subscript of each element by its MM (relative atomic mass in g/mol from the periodic table).

$\text{N} :$ (2xx14.007"g/mol")="28.014 g N"
$\text{H} :$ (8xx1.008"g/mol")="8.064 g H"
$\text{C} :$ (1xx12.011"g/mol")="12.011 g C"
$\text{O} :$ (3xx15.999"g/mol")="47.997 g O"

Now determine the percent composition of each element in the compound.

%"composition N"=("28.014 g/mol")/("96.086 g/mol")xx100="29.155% N"

%"composition H"=("8.064 g/mol")/("96.086 g/mol")xx100="8.392% H"

%"composition C"=("12.011 g/mol")/("96.086 g/mol")xx100="12.500% C"

%"composition O"=("47.997 g/mol")/("96.086 g/mol")xx100="49.952% O"

Add the percentages: 29.155%+8.392%+12.500%+49.952%="99.999%"~~100%"