A hydrocarbon is #83.33%# by mass with respect to carbon. What is (i) its empirical formula, and (ii) its molecular formula if the hydrocarbon has a mass of #72*g*mol^-1#?

1 Answer
Aug 13, 2017

Answer:

I gets #"pentane"#...................

Explanation:

As always with these problems it is useful to ASSUME a mass of #100*g# of sample, and then interrogate its ATOMIC composition.

#"Moles of carbon"=(83.33*g)/(12.011*g*mol^-1)=6.94*mol*C#

The balance #100*g-83.33*g# MUST represent the percentage of hydrogen, because we were told that we analyzed an HYDROCARBON.

#"Moles of hydrogen"=(100.0*g-83.33*g)/(1.00794*g*mol^-1)=16.54*mol*H#

We divide thru by the lowest molar quantity (#C#) to get a trial formula of:

#CH_(2.38)#; but we require the simplest WHOLE number ratio, and let us try #5xxCH_(2.38)~=C_5H_12#.

Now the MOLECULAR FORMULA is always a WHOLE number multiple of the EMPIRICAL FORMULA, and since we know the molecular mass:

#nxx(5xx12.011+12xx1.00794)*g*mol^-1=72.11*g*mol^-1#.

Clearly #n=1#, and we can quote the molecular formula as isomeric #C_5H_12#.