Question

Calculate the mass percent of different elements present in sodium sulphate ( Na2SO4). ​

Answer 1

Solution :

The molecular formula of sodium sulphate is $\sf{(Na_{2}SO_{4})}$.

So :

1) Sodium (Na): 2 × 23 = 46 g

2) Sulphur (S): 32 g

3) Oxygen (O): 4 × 16 = 64 g

Molar mass of sodium sulphate :

$\implies$ 46 + 32 + 64

$\implies$ 78 + 64

$\implies$ 142 g

Now,

We know that:

$\boxed{\sf{Mass\:\% = \frac{Mass\:of\:element}{Molar\:mass\:of\:compound} \times 100}}$

Let us calculate the mass % of different elements :

1) Mass percent of Sodium:

$\implies$ $\sf{\frac{46}{142} \times 100}$

$\implies$ $\sf{32.4\%}$

2) Mass percent of Sulphur:

$\implies$ $\sf{\frac{32}{142} \times 100}$

$\implies$ $\sf{22.6\%}$

3) Mass percent of Oxygen:

$\implies$ $\sf{\frac{64}{142} \times 100}$

$\implies$ $\sf{45.05\%}$

_________________

Answered by: Niki Swar, Goa❤️

Answer 2

Answer:

Molar mass of Na2SO4 = [(2 × 23.0) + (32.00) + 4 (16.00)] = 142 g

Mass percent of an element = (Mass of that element in compound/Molar mass of that compound) × 100

∴ Mass percent of sodium (Na): (46/142) × 100 = 32.39%

Mass percent of sulphur(S): (32/142) × 100 = 22.54%

Mass percent of oxygen:(O): (64/142) × 100 = 45.07%