Which contains more moles, 4 g of methane `(CH_(4))` or 4g of oxyen `(O_(2))` ?
(Atomic masses : `C = 12 u, H = 1 , O = 16 u`)

Explanation

We will convert 4 g of methane and 4 g of oxygen into moles and then compare the two.
Gram molecular mass of methane `(CH_(4))`
=At. mass of `C+4xxAt`. mass of H
`=12+4xx1=16` atoms
As we know, Gram molecular mass of a compound = 1 mole of that compound
`:. 16g` of `CH_(4)=1` mole of `CH_(4)`
4 g of `CH_(4)=(1)/(16)xx4` mole of `CH_(4)`
`=(1)/(4)=0.25` mole of `CH_(4)`
Gram molecular mass of `O_(2)=2xx `At. mass of O
`=2xx16=32g`
32 g of oxygen `(O_(2))=1` mole of `O_(2)`
4g of oxygen `(O_(2))=(1)/(32)xx4`
`=(1)/(8)0.125` mole of `O_(2)`
`:.4g` of `CH_(4)` contains more moles (0.25 moles) than 4 g of `O_(2)` (0.125 moles). Since, 4 g of `CH_(4)` has more moles, so it contains more molecules (than 4 g of `O_(2))`