Question

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

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

Answer 1

Answer: (C)

$V_2{O}_5$

2.74 g metal oxide contains 1.53g metal and 1.21 g oxygen

% metal =$\frac{1.53}{2.74}\times 100$ = 55.8%

% oxygen =44.2%

no. of atoms of V= $\frac{55.8}{51}$ =1.09

no. of atoms of O= $\frac{44.2}{16}$ =2.76

simplest ratio : for V =$\frac{1.09}{1.09}$=1

for O= $\frac{2.76}{1.09}$= 2.5

whole no. ratio : V= 1 X2=2 , O=2.5X 2=5

Empirical formula is $V_2{O}_5$