Question

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

(Atomic mass: $V=$ 52, $O=$ 16)

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

Answer 1

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

Amount of $O=2.74-1.53=1.21$

Weight	Moles	Simplest ratio
Metal $=1.53$	$\frac{1.53}{52}=0.03$	$\frac{0.03}{0.03}=1$
$O=1.21$	$\frac{1.21}{16}=0.075$	$\frac{0.075}{0.03}=2.5$

Hence, the oxide is $V_2{O}_5$ .