Question

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

Answer 1

Answer:

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:

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

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

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

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

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

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

Divide no. of moles of each element by 0.03:

$\footnotesize\implies n_{(sulphur)} = \dfrac{0.04}{0.03} = 1.333 \approx1$

$\footnotesize\implies n_{(phosphorus)} = \dfrac{0.03}{0.03} = 1$

Therefore, Empirical Formula of compound is:

$\footnotesize\implies \underline{ \boxed{\bf \red{ Empirical \: Formula = P_1S_1}}}$