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

Explanation

In this problem, we should convert 4g of methane into moles and 4 g of oxygen should also be converted into moles by using the respective gram molecules masses. The substances having more moles will contains more molecules in it.
(i) Here, Mass of methane `(CH_(4)) = 4` grams
Grams molecular mass of `CH_(4) = "Mass of" C + "Mass of" 4H`
`= 12 + 4 + 1`
`= 16` grams
Now, Number of moles of `CH_(4) = ("Mass of" CH_(4) "in grams")/("Grams molecular mass of" CH_(4))`
`= 4/16`
`= 1/4`
`= 0.250` mole
(ii) Here, Mass of oxygen `(O_(2)) = 4` grams
Gram molecular of `O_(2) =` Mass of `2 'O'` atoms
`= 2 xx 16`
`= 32` grams
Now, Number of moles of `O_(2) = ("Mass of" O_(2) "in grams")/("Gram molecular mass of" O_(2))`
`= 4/32`
`= 1/8`
`= 0.125` mole
We find that 4 g of `CH_(4)` contains more moles of molecules (`0.250` moles), whereas `4 g` of `O_(2)` contains (than `4 g` of oxygen).