Question

A compound of phosphorous and sulphur contains $1.24$ g of phosphorous in $2.2$ g of the compound. The formula of the compound is:

• A. $P_4{S}_3$
• B. $P_3{S}_4$
• C. $P_3{S}_2$
• D. $P_2{S}_3$

Answer 1

The correct option is A $P_4{S}_3$

Given,

mass of compound = 2.2g

mass of phosphorous = 1.24

so, mass of sulphur = 2.2-1.24 = 0.96g

moles of phosphorous = $\frac{\text{given weight}}{\text{molecular weight}}$

molecular weight of phosphorous =31

so, moles of phosphorous = $\frac{1.24}{31}=0.04$

molecular weight of sulphur = 32

so, moles of sulphur = $\frac{0.96}{32}=0.03$

so, ratio of phosphorous and sulphur atoms become = $0.04:0.03=4:3$

So, the formula of compound becomes $P_4{S}_3$