<p>A compound contains only C, H, and N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1 mg H20. What is the empirical formula of the compound?</p>
<p>zumdahl 6th ed…page 125…#72</p>
<p>First you find the %composition of C in CO2, which is 12.01/(12.01+32)x100%=27.3% and then multiply that by 33.5mg to get 9.14mg of C.
Then you find the %composition of H in H2O, which is (2)/(18) x100%=11.1% and multiply that by 41.1mg to get 4.57mg of H.
Finally, find out how much N there is by subtracting the amount of C and H from 35.0mg, so 35-(4.57+9.14)=21.29mg
Now find out how many moles of each element there are, so
C: 0.00914g/(12.01g/mol) = 0.000761mol
H: 0.00457g/(1.01g/mol)=0.00452mol
N: 0.02129g/(14.01g/mol)=0.00152mol
Now divide each amount of moles by the smallest number to get
C: 0.000761/0.000761=1
H: 0.00452/0.000761=5.94
N: 0.00152/0.000761=1.99
So the empirical formula is CH6N2</p>
<p>Thank You Sooooooooo Much!!!</p>
<p>Combustion analysis…</p>
<p>Yep, the only “trick” is to find the percentage of mass of carbon in CO2 and use that to find how much carbon you actually have and do the same for the hydrogen in H20 then subtract the amounts from the original compound to find mass of nitrogen…</p>
<p>Convert to moles, divide into proportions then voila.</p>
<p>Looks like the problem has been answered. I’m independently studying AP chem with a friend and right now I just did that in notes. We’re using 7th edition I think, and sometimes 6 edition, and sometimes if we want too a different publisher 10 edition…bu usually 7e…cover sucks <_<</p>
<p>but I guess that means you guys are on around chap 3?</p>