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 $72gmol^-1$ ?

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Answer 1 · 2017-08-13T10:33:25

I gets $"pentane"$ ...................

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 $C_5H_12$ .

A hydrocarbon is 83.33%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^-172*g*mol^-1?