Question

A compound on analysis gave the following percentage composition Na =14.31% , S =9.97% , H = 6.22% and O = 69.50%. The molecular mass of the compound is 322. Determine the molecular formula of the compound assuming that all the hydrogen in the compound is present as water.

Answer 1

Here is your answer:

Divide the percentage by molar mass

14.31/23 = 0.62

9.97/32 = 0.31

6.22/1 = 6.2

69.50/16 = 4.34

All are approximate values

taking ratio, we get:

2 : 1 : 20 : 14

empirical mass = 2×23 + 32+ 20×1 + 14 × 16 = 322g

Given:

Molecular mass = 322 g

∴ molecular formula = empirical formula = NaSH₂₀O₁₄

Given that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation

If water of crystallization are nH₂O

then, 2n = 20

or n=10

Crystallized water is 10H₂O

remaining atoms are Na₂SO₄

therefore, the molecule is Na₂SO₄.10H₂ O

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