1
$\begingroup$

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?

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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:

$$\ce{C_{15.86}H_{19.4}N_{0.93}O_{3.80}}$$

You can try rounding this to the nearest integers:

$$\ce{C16H19NO4}$$

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.

$\endgroup$
2
  • 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
    Nov 26, 2019 at 19:36
  • 1
    $\begingroup$ @andselisk Yeah, or they were on crack after doing the experimental analysis... $\endgroup$
    – Karsten
    Nov 26, 2019 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.