If `1` g of sulphur dioxide contains `x` molecules, what will be the number of molecules in `1` g of methane ? `(S = 32 u, O = 16 u, C = 12 u, H = 1 u)`

Explanation

The ratio of molecules in sulphur dioxide and methane will be the same as the ratio of their moles. So, first of all we should find out the number of moles of sulphur dioxide in 1 gram of sulphur dioxide in 1 gram of sulphur dioxide, and the number of moles of methane in 1 gram of methane. This can be done as follows :
(i) The molecular formula of sulphur dioxide is `SO_(2)`.
So, `1` mole of `SO_(2) = "Mass of S" + "Mass of 2 'O'"`
`= 32 + 2 xx 16`
` = 64` grams
Now, `64 g` of sulphur dioxide `= 1` mole
So, `1g` of sulphur dioxide ` = 1/64` mole
Thus, we have `1/64` mole of sulphur dioxide and it contains molecules in it. Now, since equal moles of all the substance contain equal number of molecules, therefore, `1/64` mole of methane will also contain x molecules of methane.
(ii) Molecular formula of methan is `CH_(4)`
So, 1 mole of `CH_(4)` = Mass of C + Mass of 4 H
` = 12 + 4 xx 12`
Now, 16g of methane = 1 mole
So, 1 g of mathane = `1/16` mole
We know that :
`1/64` mole of methane contains = x molecules
So, `1/16` mole of contains will contain `= (x xx 64)/(16)` molecules
`= 4x` molecules