A compound has O = 61. 32 % S = 11.15 % H = 4.88 % and Zn = 22.65 % The relative molecular mass fo the compound is 287 a.m.u. Find hte molecular formula of the compound assuming that all the hydrogen is present as water of crystallisation.

Open in App

Solution

Ans:

Zn S H O

22.65 11.15
4.88
61.32

Step-1(Divide each element percentage with its atomic mass) $\frac{22.65}{65} = 0.34$ $\frac{11.65}{32} = 0.36$ $\frac{4.88}{1} = 4.88$ $\frac{61.32}{16} = 3.83$

Step-2 (divide all the obtained values with lowest value & make it nearest round figure value) $\frac{0.34}{0.34} = 1$ $\frac{0.36}{0.34} = 1$ $\frac{4.88}{0.34} = 14.35 = 14$ $\frac{3.83}{0.34} = 11.26 = 11$

$Z n_{1}$ $S_{1}$ $7 H_{2}$ (Since all hydrogens are in the form water of crystallization) $7 O$ (Since all hydrogens are in the form water of crystallization. So it requires 7 Oxygen atoms)
&
$O_{4}$

So, the molecular formula of the compound is $Z n S O_{4} .7 H_{2} O$
Its molecular weight is $65 + 32 + 64 + (7 \times 18) = 161 + 126 = 287$

Suggest Corrections

Similar questions

Z n =

S =

O =

H =

Z n = 65, S = 32, O = 16, H = 1

N a = 14.31 %, S = 9.97 %, H = 6.22 %, O = 69.5 %

322

N a = 18.60 %, S = 25.80 %, H = 4.02 %

O = 51.58 %

Join BYJU'S Learning Program

	Zn	S	H	O
	22.65	11.15	4.88	61.32
Step-1(Divide each element percentage with its atomic mass)	$\frac{22.65}{65} = 0.34$	$\frac{11.65}{32} = 0.36$	$\frac{4.88}{1} = 4.88$	$\frac{61.32}{16} = 3.83$
Step-2 (divide all the obtained values with lowest value & make it nearest round figure value)	$\frac{0.34}{0.34} = 1$	$\frac{0.36}{0.34} = 1$	$\frac{4.88}{0.34} = 14.35 = 14$	$\frac{3.83}{0.34} = 11.26 = 11$
	$Z n_{1}$	$S_{1}$	$7 H_{2}$ (Since all hydrogens are in the form water of crystallization)	$7 O$ (Since all hydrogens are in the form water of crystallization. So it requires 7 Oxygen atoms) & $O_{4}$