Question

Question #39025

Answer 1

`CH_4`

Explanation:

As always, it is useful to assume `100*g` of unknown compound, and then we interrogate its atomic composition:

`"Moles of C"` `=` `(75*g)/(12.011*g*mol^-1)=6.24*mol*C`

`"Moles of H"` `=` `(25*g)/(100794*g*mol^-1)=24.8*mol*H`

And we divide thru by the SMALLEST molar quantity to get......

`C_((6.24*mol)/(6.24*mol))H_((24.8*mol)/(6.24*mol))=CH_4`.

Clearly, the question was devised purely for pedagogy, as the unknown compound is methane on the basis of these data - it would be hard to collect combustion data on a gas.