Question

A crystalline salt when heated becomes anhydrous and loses $51.2 %$ of its weight. The anhydrous salt on analysis gave the percentage composition as:

$Mg = 20.0 %; S = 26.66 %; O = 53.33 %;$ Calculate the molecular formula of the anhydrous salt and the crystalline salt. Molecular mass of the anhydrous salt is $120$ .

Accepted Answer

Given,

$Mg = 20.0 %; S = 26.66 %; O = 53.33 %;$

Element / group Percentage Atomic mass Atomic ratio The simplest ratio Whole number ratio

$Mg$ $20.0$ $24$ $\frac{20.0}{24} = 0.833$ $\frac{0.833}{0.833} = 1$ $1$

$S$ $26.66$ $32$ $\frac{26.66}{32} = 0.833$ $\frac{0.833}{0.833} = 1$ $1$

$O$ $53.33$ $16$ $\frac{53.33}{16} = 3.333$ $\frac{3.333}{0.833} = 4.00$ $4$

Ratio of $Mg : S : O = 1 : 1 : 4$ .

The Simplest formula $= {Mg}_{1} S_{1} O_{4}$ .

The molecular mass of ${MgSO}_{4} = 24 + 32 + (4 \times 16) = 24 + 32 + 64 = 120 g {mol}^{- 1}$ .

Molecular Mass $= 120 u$ .

i.e., Empirical formula and molecular formula are same which is ${MgSO}_{4}$ (anhydrous salt).

Loss of weight $=$ Water of crystallization Ratio.

Percentage of water in the crystalline ${MgSO}_{4}$ $= 51.2 %$ .

Percentage of ${MgSO}_{4}$ (anhydrous) $= 48.8 %$ .

The ratio of ${MgSO}_{4} = \frac{48.8}{120} = 0.40 \Rightarrow \frac{0.40}{0.40} = 1 .$

Therefore, the ratio of water molecules $= \frac{51.2}{18} = 2.84 \Rightarrow \frac{2.84}{0.40} = 7.1 ≃ 7 .$

Thus, seven water molecules are present in the crystalline solid.

Therefore, the molecular formula is ${MgSO}_{4} . 7 H_{2} O$ .

S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Some Basic Concepts of Chemistry, Exercise 14: PROBLEMS FOR PRACTICE

S C Kheterpal Chemistry Solutions for Exercise - S C Kheterpal, S N Dhawan and, P N Kapil Solutions for Chapter: Some Basic Concepts of Chemistry, Exercise 14: PROBLEMS FOR PRACTICE

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