1.25 g of a metal (M) reacts with oxygen completely to produce 1.68 g of metal oxide. The empirical formula of the metal oxide is [molar mass of M and O are 69.7 g mol-1 and 16.0 g mol-1, respectively]

Question

1.25 g of a metal (M) reacts with oxygen completely to produce 1.68 g of metal oxide. The empirical formula of the metal oxide is [molar mass of M and O are 69.7 g mol-1 and 16.0 g mol-1, respectively] (A) M2O (B) M2O3 (C) MO2 (D) M3O4

Tardigrade Admin · Accepted Answer

M+O2arrow MO2 Percentage of M = (1.25/1.68)× 100 = 74.4 % Percentage of oxygen in oxide = 100- 74.4 %= 25.6 % Element % of element At mass of element Moles of element Simplest molar ratio Simplest whole no. M 74.4 69.7 (74.4/69.7) =1.06 (1.06/1.06) =1 1× 2=2 O 25.6 16 (25.6/16) =1.6 (1.6/1.06) =1.50 1.50 × 2=3 ∴ Empirical formula of metal oxide is M2O3

Element	% of element	At mass of element	Moles of element	Simplest molar ratio	Simplest whole no.
$M$	$74.4$	$69.7$	$\frac{74.4}{69.7} = 1.06$	$\frac{1.06}{1.06} = 1$	$1 \times 2 = 2$
$O$	$25.6$	$16$	$\frac{25.6}{16} = 1.6$	$\frac{1.6}{1.06} = 1.50$	$1.50 \times 2 = 3$

Q. $1.25$ g of a metal $(M)$ reacts with oxygen completely to produce $1.68$ g of metal oxide. The empirical formula of the metal oxide is [molar mass of $M$ and $O$ are $69.7 g m o l^{- 1}$ and $16.0 g m o l - 1$ , respectively]

$M_{2} O$

$M_{2} O_{3}$

$M O_{2}$

$M_{3} O_{4}$

Q. 1.25 g of a metal (M) reacts with oxygen completely to produce 1.68 g of metal oxide. The empirical formula of the metal oxide is [molar mass of M and O are 69.7gmol−1 and 16.0gmol−1, respectively]

M2​O

M2​O3​

MO2​

M3​O4​

Q. $1.25$ g of a metal $(M)$ reacts with oxygen completely to produce $1.68$ g of metal oxide. The empirical formula of the metal oxide is [molar mass of $M$ and $O$ are $69.7 g m o l^{- 1}$ and $16.0 g m o l - 1$ , respectively]

$M_{2} O$

$M_{2} O_{3}$

$M O_{2}$

$M_{3} O_{4}$