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Q. What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53 g of metal?
Given: Take atomic mass of vanadium as $\text{52} \, \text{g/mol} \text{.}$

NTA AbhyasNTA Abhyas 2020Organic Chemistry – Some Basic Principles and Techniques

$V_{2}O_{3}$

$VO$

17%

$V_{2}O_{5}$

83%

$VO_{2}$

Solution:

% of V $=\frac{1.53}{2.74}\times 100=55.83$

So % of $O=44.17$

Element

%

Atomic ratio

Simplest ratio

V

55.83

$\frac{55.83}{52}=1.1$

$\frac{1.1}{1.1}=1$

O

44.17

$\frac{44.17}{16}=2.76$

$\frac{2.76}{1.1}=2.5$

V : O = 2: 5

Thus, empirical formula $=V_{2}O_{5}$

Element	%	Atomic ratio	Simplest ratio
V	55.83	$\frac{55.83}{52}=1.1$	$\frac{1.1}{1.1}=1$
O	44.17	$\frac{44.17}{16}=2.76$	$\frac{2.76}{1.1}=2.5$