posted by Monique on .
An organic compound was synthesized and found to contain only C, H, N, O, and Cl. It was observed that when .150g sample of the compound was burned, it produced .138g of CO2 and .0566g of H2O. All the Nitrogen in a different .200g sample of the compound was converted to NH3, which was found to have a mass of .0238g. Finally, the chlorine in a .125g sample of the compound was converted to AgCl. The AgCl, when dried, was found to weigh .251g.
a. Calculate the percent by mass of each element in the compound.
b. Determine the empirical formula for the compound.
I did B first so I could could do A. As much as I would like to post my work, it is A LOT. However, I do have answers so that you can check if it's right (although I doubt this), or if I'm in the right direction.
Thus, for the empirical formula I got C3HN8O8CL10. For the molar mass to calculate percent composition, I got 631 grams.
As for percent composition, here's the list of the following elements and my answer:
The problem with this is that I get about 104% when I add the percentages all together, which means the empirical formula is probably a little off.
I know it's a little hard to help me with this without much work, but any suggestions/advice for this? Any help would be great. Thank you very much!
Let me try to get you started in the right direction. If you had shown your work I could have found the error. No work means I don't know where you went wrong. For example,
%Cl = (mass Cl/mass sample) *100
mass Cl = 0.251 g AgCl x (1 mole Cl/1 mole AgCl)*100 = 0.251 x (35.453/143.321)= 0.06209 g Cl
%Cl = 0.06209/0.125 = 49.67%
Similarly, the N must be
%N = [0.0238 x (14/17)]/0.2 = 9.8%
etc. You need to check out the molar masses and atomic masses and I used those in my memory and they may be off a little. As for the percents adding up to more than 100, that can't be for % oxygen is determined by adding C, H, N, and Cl and subtracting from 100 so they MUST add to 100.
I'll let you work with this for awhile. Post your work if you get stuck.