Question 21A 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.

Question

Accepted Answer

Element% compositionAt. wt.Relative no. of atomsSimplest ratioZn22.656522.6565\dfrac{22.65}{65}6522.65​ = 0.34840.34840.3484\dfrac{0.3484}{0.3484}0.34840.3484​ = 1S11.153211.1532\dfrac{11.15}{32}3211.15​ = 0.34840.34840.3484\dfrac{0.3484}{0.3484}0.34840.3484​ = 1O61.321661.3216\dfrac{61.32 }{16}1661.32​ = 3.8323.8320.3484\dfrac{3.832}{0.3484}0.34843.832​ = 10.99 = 11H4.8814.881\dfrac{4.88}{1}14.88​ = 4.884.880.3484\dfrac{4.88 }{0.3484}0.34844.88​ = 14Simplest ratio of whole numbers = Zn : S : O : H = 1 : 1 : 11 : 14Hence, empirical formula is ZnSO11H14Molecular weight = 287Empirical formula weight = 65 + 32 + 11(16) + 14(1) = 65 + 32 + 176 + 14 = 287n=Molecular weightEmpirical formula weight=287287=1\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \[0.5em] = \dfrac{287}{287} = 1n=Empirical formula weightMolecular weight​=287287​=1Molecular formula = n[E.F.] = 1[ZnSO11H14] = ZnSO11H14Since 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.

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

Mole Concept and Stoichiometry — Question 23

Question 21