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 text52 textg/mol text.

Question

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 text52 textg/mol text. (A) V2O3 (B) VO (C) V2O5 (D) VO2

Tardigrade Admin · Accepted Answer

% of V =(1.53/2.74)× 100=55.83 So % of O=44.17 Element % Atomic ratio Simplest ratio V 55.83 (55.83/52)=1.1 (1.1/1.1)=1 O 44.17 (44.17/16)=2.76 (2.76/1.1)=2.5 V: O = 2: 5 Thus, empirical formula =V2O5

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 $52 g/mol .$

$V_{2} O_{3}$

$V O$

$V_{2} O_{5}$

$V O_{2}$

% 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	$\frac{44.17}{16} = 2.76$	$\frac{2.76}{1.1} = 2.5$

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 52g/mol.

V2​O3​

VO

V2​O5​

VO2​