# AP Chem Find Empirical Formula

1. Sep 22, 2011

### Phyzwizz

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 whats 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.

2. Sep 23, 2011

### chemisttree

If there's a problem, I can't find it.

3. Sep 23, 2011

### Phyzwizz

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.