Question #39025

1 Answer
anor277
Aug 19, 2017

#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.

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