What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53 g of metal?

Question

What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53 g of metal? (A)  V2O3  (B)  VO  (C)  V2O5  (D)  V2O7

Tardigrade Admin · Accepted Answer

Percentage of V=(1.53/2.74)× 100=55.83 ∴ of O=44.17 Element Percentage 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?

$V_{2} O_{3}$

$V O$

$V_{2} O_{5}$

$V_{2} O_{7}$

Percentage of $V = \frac{1.53}{2.74} \times 100 = 55.83$ $∴$ of $O = 44.17$
Element
Percentage
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 $= V_{2} O_{5}$

Q. What is the empirical formula of vanadium oxide if 2.74 g of metal oxide contains 1.53 g of metal?

V2​O3​

VO

V2​O5​

V2​O7​

Percentage of V=2.741.53​×100=55.83 ∴ of O=44.17 Element Percentage Atomic ratio Simplest ratio V 55.83 5255.83​=1.1 1.11.1​=1 O 44.17 1644.17​=2.76 1.12.76​=2.5 V:O=2:5 Thus, empirical formula =V2​O5​

$V_{2} O_{3}$

$V O$

$V_{2} O_{5}$

$V_{2} O_{7}$

Percentage of $V = \frac{1.53}{2.74} \times 100 = 55.83$ $∴$ of $O = 44.17$
Element
Percentage
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 $= V_{2} O_{5}$