Molecularweight of C = 12g/mol and H= 1g/mol

according to the given data the number of C moles are,

= (32/12) = (8/3)mol

Methane has the formula of CH4.

Hence the number of H moles = (8/3) x 4 = (32/3) mol

weight of H moles = (32/3) g =...

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Molecularweight of C = 12g/mol and H= 1g/mol

according to the given data the number of C moles are,

= (32/12) = (8/3)mol

Methane has the formula of CH4.

Hence the number of H moles = (8/3) x 4 = (32/3) mol

weight of H moles = (32/3) g = 10.67 g

Therefore this cannot be methane as on 8 g of H is there

According to the given data C:H molar ratio is 1:3

Ethane (C2H6) has the same ratio of C:H = 1:3

Hence this can be ethane if it is a pure substance.

or this can be a mixture of ethylene (C2H4) and hydrogen (H2)

So we cannot exactly say what is the substanc there but what we can say is that the mole ratio of C:H is 1:3