Question

1.25g of a metal (M) reacts with oxygen completely to produce 1.68g of metal oxide. The empirical formula of the metal oxide is:
[Molar mass of M and O are 69.7 g/mol and g/mol respectively.]
(a) ${M_2}O$
(b) ${M_2}{O_3}$
(c) $M{O_2}$
(d) ${M_3}{O_4}$

Answer 1

Hint: Empirical formula of a compound is the simplest positive integer ratio of atoms present in a compound. It is obtained by dividing the molecular weight by n (a factor).

Complete step by step answer:
Mass of metal (M) = 1.25g
Metal oxide produced = 1.68g
Molar mass of metal (M) = 69.7 g/mol
Molar mass of oxygen (O) = 16.0 g/mol
1: The reaction is:
$M + {O_2} \to {M_x}{O_y}$
2: To find out the equivalent factor:
$\dfrac{{1.25}}{E} = \dfrac{{1.68}}{{E + 8}}$
Here, E is the equivalent mass of the metal.
3: The equivalent mass of oxygen is 8. We can calculate this by dividing the atomic mass of oxygen by its valency. We know atomic mass of oxygen is 16u and its valency is 2. So equivalent mass is 8u.
4: Cross multiplying the above equation, we get:
$
1.68E = (E + 8)1.25 \\
\Rightarrow 1.68E - 1.25E = 8 \times 1.25 \\
  \\
$ (By simplifying)
$
   \Rightarrow 0.43E = 10 \\
   \Rightarrow E = 23.25 \\
 $
5: Now the n-factor comes out to be:
$
\dfrac{{69.7}}{{23.25}} = 3 \\
  \therefore n = 3 \\
 $
So, the charge on M will be +3. For oxygen, as we know it will be -2
6: So \[{M^{3 + }}{O^{2 - }}\] for ${M_x}{O_y}$
$ \Rightarrow {M_2}{O_3}$ , where x=2 and y=3
Thus, the Empirical Formula of the metal oxide will be ${M_2}{O_3}$.
The correct option is (b)

Note:
The concept of empirical formula must also be clear. Let’s take an example to understand it better. Suppose for sulphur monoxide the empirical formula will be SO. This will also be the empirical formula for disulfur dioxide even though its molecular formula is ${S_2}{O_2}.$
Thus the empirical formula doesn’t represent the actual number of atoms. It represents the simplest positive integer ratio of atoms. When we multiply it by n-factor we obtain the molecular formula with the actual number of atoms.