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Chemistry

A compound has O = 61.32%, S = 11.15%, H = 4.88% and Zn = 22.65%.The relative molecular mass of the compound is 287 a.m.u. Find the molecular formula of the compound, assuming that all the hydrogen is present as water of crystallisation.

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Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Zn22.656522.6565\dfrac{22.65}{65} = 0.34840.34840.3484\dfrac{0.3484}{0.3484} = 1
S11.153211.1532\dfrac{11.15}{32} = 0.34840.34840.3484\dfrac{0.3484}{0.3484} = 1
O61.321661.3216\dfrac{61.32 }{16} = 3.8323.8320.3484\dfrac{3.832}{0.3484} = 10.99 = 11
H4.8814.881\dfrac{4.88}{1} = 4.884.880.3484\dfrac{4.88 }{0.3484} = 14

Simplest ratio of whole numbers = Zn : S : O : H = 1 : 1 : 11 : 14

Hence, empirical formula is ZnSO11H14

Molecular weight = 287

Empirical formula weight = 65 + 32 + 11(16) + 14(1) = 65 + 32 + 176 + 14 = 287

n=Molecular weightEmpirical formula weight=287287=1\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{287}{287} = 1

Molecular formula = n[E.F.] = 1[ZnSO11H14] = ZnSO11H14

Since all the hydrogen in the compound is in combination with oxygen as water of crystallization .

Therefore, 14 atoms of hydrogen and 7 atoms of oxygen = 7H2O and hence, 4 atoms of oxygen remain.

∴ Molecular formula is ZnSO4.7H2O.

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