Question

Calculate the percentage of elements present in compound K2SO4:
k=39, S=32, O=16​

Answer 1

Explanation:

think so 39*2+32*16*4

it implise 2, 136

Answer 2

The percentage of K = 44.8 %, The percentage of S = 18.391 % and

The percentage of O = 36.781 %

Explanation:

To find, the percentage of elements present in compound $K_2SO_4$.

The molecular weight of $K_2SO_4$

= 2 × 39 + 32 + 16 × 4

= 78 + 32 + 64

= 174

We know that,

Mass percent (%) = $\dfrac{Mass of component}{Total mass} \times 100$

The percentage of K = $\dfrac{78}{174} \times 100$

= 44.8 %

The percentage of S = $\dfrac{32}{174} \times 100$

= 18.391 %

The percentage of O = $\dfrac{64}{174} \times 100$

= 36.781 %

∴ The percentage of K = 44.8 %, The percentage of S = 18.391 % and

The percentage of O = 36.781 %