[xc630]

Hi I would appreciate any help with this problem. I don't know where to start besides changing everything to grams.

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

Thanks.

[kcn_10081991]

Hi I would appreciate any help with this problem. I don't know where to start besides changing everything to grams.

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

Thanks.

It's very simple.The answer is CH6N2

[Albert]

That's not the right answer.

Moreover, I think xc630 to be more interested in how to solve this kind of questions rather than having a straight answer. Am I wrong?

First you have to write a useful chemical equation, such as C_xH_yN_z + n O₂ -> x CO₂ + y/2 H₂O + z/2 N₂
Thanks to this chemical equation, you can now work out the moles of oxygen and hydrogen. Once you have done it, having the weight of your sample and being able to calculate the weight of oxygen and hydrogen, you calculate the weight and moles of nitrogen.

Finally, with the moles of the three elements, you divide them by the smallest one (http://www.towson.edu/~ladon/empiric.html).

[xiankai]

did u mistake O and N? because in your products listed there doesnt seem to be N, unless u missed N₂ like Albert added in.

and most of all because such questions are usually posed with C, H, and O sometimes, but seldom N, speaking from experience

[Albert]

I had the same impression, xiankai. However, I tried to explain how to solve this kind of problem.

[Borek]

did u mistake O and N? because in your products listed there doesnt seem to be N, unless u missed N₂ like Albert added in.

Nothing unusual - I will read it as "nitrogen was the rest".

[Albert]

Yes, it's not that weird. Moreover, the answer is a quite famous molecule.

[kayamusty]

the answer is CH6N2.

[xc630]

Hi Albert i followed the procedure found on the webpage you hyperlinked but I still did not get CH6N2 so could anyone point out what i did wrong.

33.5 mg CO2 x 1 g/1000mg x 1 mol CO2/44.01 g CO2 x 1 mol C/ 1mol CO2 x 12.01 g C/ 1 mol C = 0.009g C

0.009 g C/ 0.035 g compund = 25.7 % C

41.1 mg H2O x 1 g/1000mg x 1 mol H2O/18g H2O x 1mol H/ 1 mol H2O x 1g H / 1 mol H = 0.00228 g H

0.00228 g H/ 0.035 g compund = 6.5 % H

100-25.7-6.5 = 67.8% N

25.7 g C X 1mol C/12.01 g C = 2.14 mol C
6.5 g H x 1mol H/ 1 g H = 6.5 mol H
67.8 g N x 1 mol N/ 14.01g N = 4.84 mol N

2.14 mol/2.14 = 1 for C
6.5 mol H/2.14 mol = 3 for H
4.84 mol N/ 2.14= 2.26

I can't figure out what I did wrong. 

[Donaldson Tan]

41.1 mg H2O x 1 g/1000mg x 1 mol H2O/18g H2O x 1mol H/ 1 mol H2O x 1g H / 1 mol H = 0.00228 g H

There are 2 moles of H atom per mole of water molecule

mass of hydrogen in C_xH_yN_z = 1.15mg

The mass of hydrogen in C_xH_yN_z is the same mass in water, ie. mass of hydrogen in C_xH_yN_z = 2/18 * 41.1mg = 4.567mg

Amount of C in C_xH_yN_z = Amount of C in CO₂ formed = 12/44 * 33.5mg = 9.136mg = 0.761mmol

Amount of H in C_xH_yN_z = Amount of H in H₂O formed = 2/18 * 41.1mg = 4.567mg = 4.567mmol

Amount of N in C_xH_yN_z = 35 - 9.136 - 4.567 = 21.297mg = 1.521mmol

molar ratio of C to H to N = 0.761 : 4.56 : 1.521 = 1 : 6 : 2

Emperical Formula is CH₆N₂

[Albert]

Yes, you are right. My apologies: the ratio moles of hydrogen/moles of water is obiously 2:1. This means you have 4.563*10^-3 moles of hydrogen.

[xc630]

I don't understand. Don't you have to change the 12/44 and 2/18 ratio to grams?

[Albert]

Now that Geo showed me my stupid mistake (many thanks  ), here is how you solve the problem:

C_xH_yN_z + n O₂ -> x CO₂ + y/2 H₂O + z/2 N₂
35.0 mg                  33.5 mg     41.1 mg


moles of carbon = moles of CO₂ = 33.5 mg/MW of CO₂ = 7.612*10^-4
moles of hydrogen = moles of water X2 = 41.1 mg/MW of H₂O = 4.563*10^-3

mass of carbon in C_xH_yN_z = 9.136 mg
mass of hydrogen in C_xH_yN_z = 4.567 mg

hence, to quote geodome

Amount of N in CxHyNz = 35 - 9.136 - 4.567 = 21.297mg = 1.521mmol

Now, you divide the moles of each element by the smallest one:

x = 7.612*10^-4/7.612*10^-4 = 1

y = 4.563*10^-3/7.612*10^-4 = 5.99... => 6

z = 1.521*10^-3/7.612*10^-4 = 1.99... => 2

Once again, I apologize for my silly mistake and many thanks indeed, geodome.

[Donaldson Tan]

I don't understand. Don't you have to change the 12/44 and 2/18 ratio to grams?

12/44 and 2/18 are the composition by mass of carbon in carbon dioxide (molar mass 44g/mol) and hydrogen in water (molar mass 18g/mol) respectively.