Question

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

• A. $V_2{O}_3$
• B. $VO$
• C. $V_2{O}_5$
• D. $V_2{O}_7$

Answer 1

The correct option is C $V_2{O}_5$

$\text{Metal oxide = 2.74 g};\text{weight of V}=1.53g$
$\%ofV=\frac{1.53}{2.74}\times 100=55.83$
$\text{Thus,}\%ofO=100-55.83=44.17$
$\text{Number of moles of V}=\frac{55.83}{52}=1.1$
$\text{Number of moles of}O=\frac{44.17}{16}=2.76$
$\text{Simplest ratio of V and O}=1:2.5or2:5$
$\text{Hence, the empirical formula}=V_2{O}_5$