Question

2.2g of a compound of phosphorus and sulphur has 1.24 g of P in it. Its empirical formula is:
A. ${{P}_{2}}{{S}_{3}}$
B. ${{P}_{3}}{{S}_{2}}$
C. ${{P}_{3}}{{S}_{4}}$
D. ${{P}_{4}}{{S}_{3}}$

Answer 1

Hint: As we know that empirical formula gives us the idea about the lowest whole number ratio in which the atoms are present in the compound. It is found that the empirical formula does not give any idea about the arrangement of the atoms in the compound.

Complete Step by step solution:
- As we know that the empirical formula is the simplest ratio in which atoms combine. The number of atoms in the compound is basically explained by considering the empirical formula. We must always remember that the relative ratio is always the whole number.
- As we know that the Atomic weight of phosphorus is 31, and that of sulphur is 32.
2.2g of a compound of phosphorus and sulphur has 1.24 g of P in it.
- That is given weight of phosphorous= 1.24 and that of sulphur is equal to 2.2-1.24=0.96
As we know that the ratio of weights of P:S is 1.24:[2.2-1.24]=1.24:0.96
- Therefore, we can write the empirical formula as:
(1.24/31) : (0.96/32) = 4:3
The empirical formula is ${{P}_{4}}{{S}_{3}}$.

- Hence, we can conclude that the correct option is (d), that is the empirical formula is ${{P}_{4}}{{S}_{3}}$.

Note: - We should not get confused in between the empirical formula and molecular formula. As, molecular formula represents the arrangement of the atoms in the compounds, whereas empirical formula does not.