Q.
A compound, on analysis, gave the following percentage composition:
$ Na = 14.31\%, S = 9.97\%, H = 6.22\%, O = 69.5\%$
What would be the molecular formula of the compound assuming that all the hydrogen in the compound is reset in combination with oxygen as water of crystallisation. Molecular weight of the compound is $322$
Some Basic Concepts of Chemistry
Solution:
Element
%
Atomic ratio
Least ratio
Empirical formula
Na
14.31
$\frac{14.31}{23} = 0.622$
$\frac{0.622}{0.311} = 2$
S
9.97
$\frac{9.97}{32}=0.311$
$\frac{0.311}{0.311}=1$
$Na_{2}SH_{20}O_{14}$
H
6.22
$\frac{6.22}{1}=6.22$
$\frac{6.22}{0.311} = 20$
O
69.5
$\frac{69.5}{16}=4.34$
$\frac{4.34}{0.311}=14$
Empirical Formula mass $= 46 + 32 + 20 + 224 = 322$
$n= \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{322}{322} = 1$
Hence, molecular formula $= Na_{2}SH_{20}O_{14}$
Since, all the hydrogen is in the form of water, thus there are $10H_{2}O$ molecules.
Hence, the molecular formula is $ Na_{2}SO_{4} \cdot 10H_{2}O$
| Element | % | Atomic ratio | Least ratio | Empirical formula |
|---|---|---|---|---|
| Na | 14.31 | $\frac{14.31}{23} = 0.622$ | $\frac{0.622}{0.311} = 2$ | |
| S | 9.97 | $\frac{9.97}{32}=0.311$ | $\frac{0.311}{0.311}=1$ | $Na_{2}SH_{20}O_{14}$ |
| H | 6.22 | $\frac{6.22}{1}=6.22$ | $\frac{6.22}{0.311} = 20$ | |
| O | 69.5 | $\frac{69.5}{16}=4.34$ | $\frac{4.34}{0.311}=14$ |