Question

If 2.21g of a compound contains 1.28g ofsulphur and remaining is phosphorus then theempirical formula is :(2) PAS, (3) PAS (4) P2S(1) Ps​

Answer 1

Explanation:

In 2.21 g of compound 1.28 g is sulphur and remaining is phosphorus i.e, (2.21 - 1.28) = 0.93 g.

No. of moles of each element:

\footnotesize\implies n_{(sulphur)} = \dfrac{given \: mass}{atomic \: mass}⟹n

(sulphur)

=

atomicmass

givenmass

\footnotesize\implies n_{(sulphur)} = \dfrac{1.28}{32}⟹n

(sulphur)

=

32

1.28

\footnotesize\implies \rm \purple{ n_{(sulphur)} = 0.04}⟹n

(sulphur)

=0.04

\footnotesize\implies n_{(phosphorus)} = \dfrac{given \: mass}{atomic\: mass}⟹n

(phosphorus)

=

atomicmass

givenmass

\footnotesize\implies n_{(phosphorus)} = \dfrac{0.93}{31}⟹n

(phosphorus)

=

31

0.93

\footnotesize\implies \rm \orange{ n_{(phosphorus)} = 0.03}⟹n

(phosphorus)

=0.03

Divide no. of moles of each element by