Question

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is $1:4$. The ratio of the number of their molecule is___________.
A) $1:4$.
B) $7:32$.
C) $1:8$.
D) $3:16$.

Answer 1

We know that,
The number of moles can be calculated using the formula,
$Moles = \dfrac{{Mass\left( g \right)}}{{Molecular\,Mass\left( {g/mol} \right)}}$

Complete step by step answer:
As we know the mole ratio is equal to the number of molecules. Thus, by finding the number of moles ratio we can calculate the number of molecules ratio.
The number of moles of oxygen ${n_{{O_2}}} = \dfrac{{{m_{{O_2}}}}}{{{M_{{O_2}}}}}$
The mass of oxygen is ${m_{{O_2}}}$and ${M_{{O_2}}}$ is the molecular mass of oxygen.
The number of moles of nitrogen ${n_{{N_2}}} = \dfrac{{{m_{{N_2}}}}}{{{M_{{N_2}}}}}$
The mass of nitrogen is ${m_{{N_2}}}$ and ${M_{{N_2}}}$ is the molecular mass of nitrogen.
The ratio of number of moles is calculated as,
$\dfrac{{{n_{{O_2}}}}}{{{n_{{N_2}}}}} = \dfrac{{\dfrac{{{m_{{O_2}}}}}{{{M_{{O_2}}}}}}}{{\dfrac{{{m_{{N_2}}}}}{{{M_{{N_2}}}}}}} \to 1$
We know,
The molecular weight of oxygen is $16$ and the molecular weight of nitrogen is $14$. Since oxygen and nitrogen are diatomic molecules their molar masses are $32\ ,\& 28$ respectively.
Given: The ratio of masses of oxygen and nitrogen $\left( {\dfrac{{{m_{{O_2}}}}}{{{m_{{N_2}}}}}} \right)$ is $\dfrac{1}{4}$.
Substituting these values in equation $1$ we get,
$\dfrac{{{n_{{O_2}}}}}{{{n_{{N_2}}}}} = \dfrac{1}{4} \times \dfrac{{28}}{{32}} = \dfrac{7}{{32}}$
The number of molecules ratio is $7:32$.

So, the correct answer is Option B.

Note:
The number of molecules ratio can also be calculated as follows.
Given: The ratio of masses of oxygen and nitrogen $\left( {\dfrac{{{m_{{O_2}}}}}{{{m_{{N_2}}}}}} \right)$ is $\dfrac{1}{4}$.
Let, take the Mass of oxygen as $W$.
Take the mass of nitrogen as $4W$.
Now, the number of molecules in oxygen is given as,
Molecules of oxygen ${O_2} = \dfrac{w}{{32 \times {N_A}}}$
The number of molecules in nitrogen is given as,
Molecules of oxygen ${N_2} = \dfrac{{4w}}{{28 \times {N_A}}}$
Where ${N_A}$ is Avogadro’s number.
We know,
The molecular weight of oxygen is $16$ and the molecular weight of nitrogen is $14$. Since oxygen and nitrogen are diatomic molecules their molar masses are $32\,\& 28$ respectively.
$\dfrac{{{N_{{O_2}}}}}{{{N_{{N_2}}}}} = \dfrac{w}{{32}} \times \dfrac{{28}}{{4w}} = \dfrac{7}{{32}}$
The number of molecules ratio is $7:32$.