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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Element	% composition	At. wt.	Relative no. of atoms	Simplest ratio
Zn	22.65	65	$\dfrac{22.65}{65}$ = 0.3484	$\dfrac{0.3484}{0.3484}$ = 1
S	11.15	32	$\dfrac{11.15}{32}$ = 0.3484	$\dfrac{0.3484}{0.3484}$ = 1
O	61.32	16	$\dfrac{61.32 }{16}$ = 3.832	$\dfrac{3.832}{0.3484}$ = 10.99 = 11
H	4.88	1	$\dfrac{4.88}{1}$ = 4.88	$\dfrac{4.88 }{0.3484}$ = 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

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

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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