A #10.5*g# mass of a hydrocarbon contains #1.5*g# of hydrogen. If the molecular mass of the hydrocarbon in #210*g*mol^-1#, what are the empirical and molecular formulae of the hydrocarbon?

2 Answers
Apr 23, 2017

Answer:

#1.5g# of H would be #1.5div1=1.5mol#
#9g# of C would be #9div12=0.75mol#

Explanation:

So the mol-ratio of #CdivH=0.75div1.5=1div2#
And the empirical formula is #(CH_2)_n#

One of those units has a mass of #12+2xx1=14u#

So there are #210div14=15# of those units, and the molecular formula will be:

#C_15H_30#

Apr 23, 2017

Answer:

We find #"(i) the empirical formula"#, and then #"(ii) the molecular formula"#.

Explanation:

#"Moles of hydrogen"# #=# #(1.5*g)/(1.00794*g*mol^-1)=1.49*mol#

#"Moles of carbon"# #=# #(9.0*g)/(12.011*g*mol^-1)=0.75*mol#.

We divide thru by the smallest molar quantity, and (clearly) we get an empirical formula of #CH_2#.

But we know that the #"molecular formula"# is whole number multiple of the #"empirical formula"#;

i.e. #"molecular formula"=nxx"empirical formula"#

And thus, #210*g*mol=nxx(12.011+1.00794)*g*mol^-1#, and we solve for #n#, to get #n=16.15#. This value should be closer to an integer, however, this gives a molecular formula of #C_16H_32# based on the figures you quoted.