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
MeneerNask
Apr 23, 2017

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

anor277
Apr 23, 2017

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.

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