# What is the molar mass of (NH4)2CO3?

Nov 3, 2015

96.09 g/mol

#### Explanation:

You just need to first get the atomic weights of the elements involved. You can easily get these from your periodic table.

If you are going to do this properly, please use the weight with at least two decimal places for accuracy (e.g. 15.99 g/mol).

Also, please take note that I will be using the unit g/mol for all the weights. Thus,

Step 1

$\text{N = 14.01 g/mol}$
$\text{H = 1.008 g/mol}$
$\text{O = 16.00 g/mol}$
$\text{C = 12.01 g/mol}$

Since your compound is ${\left(N {H}_{4}\right)}_{2} C {O}_{3}$, you need to multiply the atomic weights by their subscripts. Therefore,

Step 2

$\text{N = 14.01 g/mol × 2 =" bb"28.02 g/mol}$
$\text{H = 1.008 g/mol × (4×2) =" bb"8.064 g/mol}$
$\text{O = 16.00 g/mol × 3 =" bb"48.00 g/mol}$
$\text{C = 12.01 g/mol × 1 =" bb"12.00 g/mol}$

To get the mass of the substance, we need to add all the weights from Step 2.

Step 3

$\text{molar mass of" ("NH"_4)_2"CO"_3 = "(28.02 + 8.064 + 48.00 + 12.01) g/mol" = bb"96.09 g/mol}$