A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Nitrogen (by mass). Calculate the empirical formula.

I'm getting $\ce{C7H14N8}$, but my professor says it is $\ce{CH2N}$.

Which is the correct answer?

When brought to moles, 40 g C is 3.33 moles of C, 6.7 g of H is 6.63 moles of H, and 53.3 g of N is 3.80 moles of N.

Divided by the lowest mole count (3.33), you get one carbon, two hydrogen, and 1.14 nitrogen. I am multiplying all by 7, to get $\ce{C7H14N8}$.

He is saying 1.14 is 1, basically, and that is can all be simplified as it is, no need to multiply, and your final answer is $\ce{CH2N}$, marking my answer half off. It's not worth going to battle for, but I want to know who is really right, because I feel my answer is better.

So the real question is, with N, should you get it closer to a whole number, or leave "as is"? Thanks.

  • 1
    $\begingroup$ Could you include your calculation? It will probably clarify the situation. $\endgroup$
    – jerepierre
    Commented Mar 22, 2017 at 23:48
  • $\begingroup$ The professor's answer can't be right. The carbon to nitrogen mass ratio ought to be 6:7 if the mole ratio is 1:1. $\endgroup$
    – Zhe
    Commented Mar 23, 2017 at 0:03
  • $\begingroup$ Incidentally - "homework" is kind of a misnomer, because around here it doesn't mean literally homework. See here for more info. Anyway, with the extra detail you added, your question is fine, so don't worry about it. $\endgroup$ Commented Mar 23, 2017 at 1:21
  • 2
    $\begingroup$ These would be some terrible analytical results for a compound of nothing but C, H and N. The carbon value in particular would fail any audit if the known compound were 42.9% C and the reported value was 40.0%. Nitrogen gets a fail too, though you don't expect quite the accuracy for N as you do for C. Finicky H gets by on this one, but overall, if the emperical formula is CH2N, these are aweful results. What a terrible problem for an exam. $\endgroup$
    – airhuff
    Commented Mar 23, 2017 at 1:29

1 Answer 1


A sad Ph.D. student had been working on the synthesis of a natural product, with empirical formula $\ce{CH2N}$, for several years. He finally succeeded in synthesising it - or so he thought - because his NMR data seemed to fit the target structure. Excited about his breakthrough (in fact, he was so excited that he didn't even bother taking a mass spectrum!), he sent a sample for elemental analysis, expecting to get some results along the lines of 42.8% carbon, 7.2% hydrogen, and 50.0% nitrogen.

Not long after this, the results came back, and he discovered that his compound inexplicably had 40.0% carbon, 6.7% hydrogen, and 53.3% nitrogen. "That's not what I expected," he thought to himself - "I wonder if I can publish it and try to gloss over the data."

Upon checking the Author Guidelines for the Journal of Organic Chemistry (section 2.2.2), he found that they were actually pretty strict.

Found values for carbon, hydrogen, and nitrogen should be within 0.4% of the [calculated] values for the proposed formula.

Thus was the story of the sad Ph.D. student who had to go back to the drawing board. (Thankfully, he had synthesised enough of the immediate precursor and by varying the conditions, he eventually succeeded in making his target.)

TL;DR Rounding down 1.14 to 1 is a gross error and won't ever get published in a reputable journal. Considering that this is not even a real-life scenario, there's absolutely no reason why standards should be lowered.

  • 2
    $\begingroup$ Exactly. As an analytical guy, that's what drove me nuts about this terrible test/homework problem. (see my comment ripping the analytical data on OP if interested). Big +1 for clever presentation! $\endgroup$
    – airhuff
    Commented Mar 23, 2017 at 1:38

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