# In methane one part hydrogen combines with three parts carbon by mass. If a sample of a compound containing only carbon and hydrogen contains 32.0g of Carbon and 8.0g of hydrogen, could the sample be methane? If the sample is not methane, show that the law of multiple proportions is followed for both methane and the other substance.

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

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