# How many moles are there in 397 grams of Na_2SO_4?

Apr 8, 2017

2.80

#### Explanation:

A mole is a unit-molecular weight (grams/mole in this case) of an Avogadro's number of molecules or atoms. It is calculated for a molecule by adding up the individual atomic weights of its constituent atoms.

For $N {a}_{2} S {O}_{4}$ this is $2 x 23 = 46$ for the sodium, plus $4 x 16 = 64$ for the oxygen, plus $32$ for the sulfur. That adds up to a molecular weight of $142 \left(\frac{g}{\text{mole}}\right)$. Now we divide the given number of grams by this value to derive the number of moles.
(397/142)(g/(g/"mol")) = 2.80 "moles"

Apr 8, 2017

${n}_{N {a}_{2} S {O}_{4}} = 2.79 m o l$

#### Explanation:

Molar mass of $N {a}_{2} S {O}_{4} = 142.04 \frac{g}{m o l}$

Use the equation,

${n}_{N {a}_{2} S {O}_{4}} = \frac{m}{M}$

where,

$n =$ number of moles
$m =$mass (in grams)
$M =$molar mass (in $\frac{g r a m s}{m o l}$)

${n}_{N {a}_{2} S {O}_{4}} = \frac{397 g}{142.04 \left(\frac{g}{m o l}\right)}$

${n}_{N {a}_{2} S {O}_{4}} = 2.79 m o l$