# How do you calculate the molar mass of each of the following compounds?

## $C u \left(O H\right) 2$ $M g O$ $M g {\left(N {O}_{3}\right)}_{2}$ $N a C l$ ${\left(N {H}_{4}\right)}_{2} S {O}_{4}$ ${H}_{2} S {O}_{4}$ $C {a}_{3} {\left(P {O}_{4}\right)}_{2}$ $N {H}_{4} N {O}_{3}$

Sep 12, 2016

I will answer one, and from the answer you can use the same method tom work out the rest: $C {a}_{3} {\left(P {O}_{4}\right)}_{2}$ = 310.177 g/mol

#### Explanation:

You need to look up the atomic weight of each element (these can be found by asking google for the atomic weights):

$C {a}_{3}$ means there are three Ca; $P {O}_{4}$ means there is one P and 4 oxygen, the ${\left(\right)}_{2}$ means there two $P {O}_{4}$. So, substituting in the values:

$C {a}_{3} {\left(P {O}_{4}\right)}_{2}$
(3 x 40.078) + ((30.973762 + (4 x 15.9994)) x 2)
120.234 + ((30.973762 + 63.9976) x 2)
120.234 + (94.971362 x 2)
120.234 + 189.942724
310.177

So the answer is 310.177 g/mol

You should now be able to use the above approach to work out the answers for the other compounds.

Sep 12, 2016

Simply add the molar masses of the individual elements.

#### Explanation:

For calcium phosphate, $C {a}_{3} {\left(P {O}_{4}\right)}_{2}$ we do the sum: $\left\{3 \times 40.08 \left(C a\right) , + 2 \times 31.00 \left(P\right) , + 8 \times 16.00 \left(O\right)\right\} \cdot g \cdot m o {l}^{-} 1$ $=$  310.18*g·mol^-1.

If we had lithium phosphate, $L {i}_{3} P {O}_{4}$, what would be the formula mass? I invite you to do the rest of the salts on your list. You will need a Periodic Table to get the individual molar masses of the elements.