Question

What is the empirical formula of vandium oxide, if $2.73$g of the metal oxide contains $1.53$ g of metal?

[Given atomic mass of $V=51$]

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

Answer 1

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

metal oxide $=2.73gm$

metal $=1.53gm$

$\therefore$ oxygen $=1.20gm$

% of metal $=\frac{1.53}{2.73}\times 100\%=55.84\%$

% of oxygen $=\frac{1.20}{2.73}\times 100\%=44.16\%$

metal :- $1.53$ males $=\frac{1.53}{51}=0.03$ simplest ration $\frac{0.03}{0.03}=1$

oxygen $:-1.20$ moles $=\frac{1.20}{16}=0.075$ $\frac{0.073}{0.03}=2.5$

$\therefore$ empirical formula $=V_2{O}_5$