What's Wrong with My Empirical Formula Calculation?

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SUMMARY

The empirical formula calculation for a compound containing only carbon (C), hydrogen (H), and nitrogen (N) was incorrectly derived due to a unit conversion error. The user calculated the masses of C and H from the combustion products, CO2 and H2O, respectively, but failed to convert the total mass from milligrams to grams. This oversight led to an incorrect final empirical formula of C2H12N5 instead of the correct formula. The correct approach requires ensuring all mass values are in grams before proceeding with mole calculations.

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  • Understanding of combustion analysis for empirical formula determination
  • Knowledge of molar mass calculations for C, H, and N
  • Familiarity with unit conversions, specifically mg to g
  • Ability to calculate percentages from mass data
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  • Review combustion analysis techniques for empirical formula calculations
  • Learn about molar mass and its significance in stoichiometry
  • Practice unit conversion methods, focusing on mass units
  • Explore common pitfalls in empirical formula calculations and how to avoid them
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Chemistry students, educators, and anyone involved in analytical chemistry or empirical formula calculations will benefit from this discussion.

Phyzwizz
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So this problem was a part of my homework and I did the whole thing but apparently I got it wrong. Could someone point out what's wrong with my work?

Question:A compound contains only C, H, and N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound.

Attempt: Found percent C in CO2: 2(16.00g)+12.01g=44.01g. 12.01g/44.01gx100%=27.29%C.
Found percent H in H2O:2(1.008g)+16.00g=18.016g. 2.016g/18.016x100%=11.9%H
Found mg of C: 27.29%x33.5mgCO2x1mgC/1mgCO2=9.14mgC
Found mg of H: 11.19%x41.1mgH2Ox1mgH/1mgH2O=4.60mgH

At this point I assumed that because the combusted compound contains only C, H, and N that I could just take the remaining 35-(9.14+4.60)=21.26 and that would be the mg of Nitrogen (If this is where the mistake is made how can I correct it?)

took a 100mg sample
H: 4.60mg/35.0mgx100%=13.1%H 13.1gHx1molH/1.008gH=12.996molH
C:9.14mg/35.0mgx100%=26.1%C 26.1gCx1molC/12.01gC=2.17molC
N:21.26mg/35.0mgx100%=75.0% 75.0gNx1molN/14.01gN=5.355molN

H:12.996/2.17=5.99=6 C: 2.17/2.17=1 N:5.355/2.17=2.47=2.5​
At this point because N turns out to be 2.5 I assume the empirical formula must be doubled and so I end with my final answer of C2H12N5 which is wrong, according to my teacher's work.

I hope all the work shown doesn't overwhelm people into not answering this.
 
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If there's a problem, I can't find it.
 
I was looking over it some more, and apparently I forgot to convert from mg to g and that's why the empirical formula is off.
 

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