A compound has $67.3 \% \rm~C,4.62 \%~N,6.93 \%~ H$ and O. Determine the molecular formula of the compound knowing its molar mass is $283\rm~\frac{g}{mol}$.

My solution : The percentage for the oxygen is $$100-( 67.3+4.62+6.93)=21.15$$ percent oxygen. Lets determine the number of moles for every substance.

For the carbon we have $$n_C = \frac{67.3 \rm~g}{12\rm~\frac{g}{mol}} =5.6~\rm mol$$

We do the same for N, H and O. We have 0.33 atoms of N,6.93 atoms of H and 1.32 atoms of O.

So the ratio is $5.6 : 0.33 : 6.93 : 1.32$.

We divide by $0.33$ and we have $17 : 1 : 21 : 4$. The empirical formula is $\ce{C17H21NO4}$.

Its molar mass is 303 g/mol. Now we find the quotient:

$$K= \frac{\text{Molar mass of the real formula}}{\text{Molar mass of the empirical formula}} =\frac{283}{303}\approx 1$$

So the molecular formula is $\ce{C17H21NO4}$. Is this correct?


1 Answer 1


Rather than starting out with 100 g of unknown, I would start out with 1 mol of unknown, i.e. 283 g. That way, you can directly calculate how many moles of each element are in one mole of compound:

$$ n = 67.3 \% \cdot \frac{\pu{283 g}}{\pu{12 g mol-1}} = \pu{15.86 mol}$$

Or to get the stoichiometric coefficient $\nu_C$ directly, take the percentage and multiply it by the ratio of compound molar mass to atomic mass:

$$ \nu_C = 67.3\% \cdot \frac{\pu{283 g mol-1}}{\pu{12 g mol-1}} = \pu{15.86}$$

If you do that for C, N, O and H, you get the formula:


You can try rounding this to the nearest integers:


This compound has a molar mass of 293 g/mol. Playing around with the formula, you can find coefficients that match the molar mass exactly, e.g. $\ce{C16H13NO4}$ or $\ce{C17H17NO3}$. In any case, neither one of these nor the formula suggested by the OP matches both the molar mass and the mass percentages.

Is this correct?

$\ce{C17H21NO4}$ is a good answer, but the data are not consistent, so it is not the complete answer. I would say depending on whether the errors on the molar mass determination or the elemental composition are larger, you would have a different top candidate, and should report the numerically exact formula with non-integer coefficients. That way, the reader can critically evaluate the meaning of the data themselves.

  • 1
    $\begingroup$ I suspect there is something fishy about the molar mass: if it were 303 g/mol, not 283 g/mol, then $\ce{C17H21NO4}$ would be a perfect fit. If you google the percentage, you find that originally it was a part of bigger problem where a student was supposed to determine molar mass on their own, so probably OP or a person who adapted this problem made a mistake in determining $M_\mathrm{r}$. $\endgroup$
    – andselisk
    Commented Nov 26, 2019 at 19:36
  • 1
    $\begingroup$ @andselisk Yeah, or they were on crack after doing the experimental analysis... $\endgroup$
    – Karsten
    Commented Nov 26, 2019 at 19:47

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