Question

Which contain more molecules $4$gram of methane or $4$gram of oxygen?

Answer 1

First find the number of moles of methane and number of moles of oxygen using the formula-
Number of moles=$\dfrac{{{\text{Given mass}}}}{{{\text{Molecular mass}}}}$ . We already know the molecular mass of methane is $16$ and the molecular mass of oxygen is$32$.So put the values and solve. Then use the formula-
Number of molecules=no. of moles × Avogadro number to know the number of molecules of methane and oxygen. Here, Avogadro number = $6.023 \times {10^{23}}$
Then take the ratio of the number of molecules of both compounds to get the answer.

Complete step by step answer:
Given, the mass of methane=$4$gram
And the mass of oxygen is=$4$gram
We know that the formula of number of moles is given as-
Number of moles=$\dfrac{{{\text{Given mass}}}}{{{\text{Molecular mass}}}}$
Then we can count the moles of methane and oxygen from this formula.
We already know the molecular mass of methane is $16$ and the molecular mass of oxygen is$32$.
Then applying the formula we get-
$\Rightarrow$ Number of moles of methane=$\dfrac{4}{{16}}$
On division, we get-
$\Rightarrow$ Number of moles of methane=$\dfrac{1}{4}$
And the number of moles of oxygen on applying the formula will be- will be-
$\Rightarrow$ Number of moles of oxygen=$\dfrac{4}{{32}}$
On division we get-
$\Rightarrow$ Number of moles of oxygen=$\dfrac{1}{8}$
Now we know the formula of number of molecules is given as-
Number of molecules=no. of moles × Avogadro number
Now we know that the Avogadro number = $6.023 \times {10^{23}}$
Then on applying the formula, we get-
Number of molecules of methane=$\dfrac{1}{4} \times 6.023 \times {10^{23}}$ --- (i)
And number of molecules of oxygen=$\dfrac{1}{8} \times 6.023 \times {10^{23}}$ --- (ii)
Now dividing eq. (i) by (ii), we get-
$\Rightarrow \dfrac{{{\text{Number of molecules of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Number of molecules of }}{{\text{O}}_2}}} = \dfrac{{\dfrac{1}{4} \times 6.023 \times {{10}^{23}}}}{{\dfrac{1}{8} \times 6.023 \times {{10}^{23}}}}$
Then on simplifying, we get-
$\Rightarrow \dfrac{{{\text{Number of molecules of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Number of molecules of }}{{\text{O}}_2}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{8}}}$
On simplifying further, we get-
$\Rightarrow \dfrac{{{\text{Number of molecules of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Number of molecules of }}{{\text{O}}_2}}} = \dfrac{8}{4}$
On division we get-
$\Rightarrow \dfrac{{{\text{Number of molecules of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Number of molecules of }}{{\text{O}}_2}}} = \dfrac{2}{1}$
Now cross multiplying we get,
$\Rightarrow$ Number of molecules of methane=$2 \times$ number of molecules of oxygen

Hence the number of molecules of methane is twice the number of molecules of oxygen.

Note:
Here the student can also solve the question like this-
We know that the number of molecules of a compound is directly proportional to the number of molecules of that compound. So we can first take the ratio of number of moles of methane and oxygen.
On taking ratio, we get-
$\Rightarrow$ $\dfrac{{{\text{Number of moles of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Number of moles of }}{{\text{O}}_2}}} = \dfrac{{\dfrac{1}{4}}}{{\dfrac{1}{8}}}$
Now on solving, we get-
$\Rightarrow \dfrac{{{\text{Number of moles of C}}{{\text{H}}_{\text{4}}}}}{{{\text{Number of moles of }}{{\text{O}}_2}}} = \dfrac{2}{1}$
So here we can see that the number of moles of methane is twice the number of moles of oxygen and since number of moles is directly proportional to the number of molecules then we can write-
The number of molecules of methane is twice the number of molecules of oxygen.