Challenging chemistry problem help!?

Question

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g CO2 and 0.0310 g H2O. In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3. What is the empirical formula of the compound? Hint: Combustion involves reacting with excess O2. Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O. Also assume that all the nitrogen ends up in the NH3 in the second experiment.

I know the answer, C7H5N3O6, but i don't know how to find it. help, please?

Answer 1

First you have to find the composition by weight.

there are x number of moles of carbon in your initial compound, all of which are converted into CO2. Find the moles of CO2

Molar Mass CO2=12+16*2=44g/mol
.213g/(44g/mol) =.00484mol C
Multiply by molar mass of C
12g/mol * .00484 mol C = .05808 g C

weight percent C = .05808/.157=37%

Molar Mass of H2O is 2*1+16=18g/mol
.031g/(18g/mol)=.001722mol H2O, .00344mol H
at about 1 gram per mole
~.00344 g H

weight percent H = .00344/.157=2.2%

Molar Mass of NH3 is 14+3*1=17g/mol
.023g/(17g/mol)=.00135 mol N
.00135 mol N * 14g/mol=.0189g N

Weight Percent N: .0189g/.103g=18.3%

The remaining weight percent is Oxygen: O=42.5%

Now to convert to molar ratios: divide each percent by its respective molar mass.

C: 37/12= 3.08/1.307= 2.36*3=7.08
H: 2.2/1= 2.2/1.307= 1.68*3=5.04
N: 18.3/14= 1.307/1.307=1*3 =3
O: 42.5/16= 2.66/1.307= 2.03*3=6.09

Divide all by the least number: 1.307
See that by multiplying by 3, you get all near whole numbers
Those are your molecular ratios!


Please give me best answer that was a lot of work :)

Answer 2

Divide 0.213 by the molecular mass of CO2. The result divide by 0.157g you will find the number of moles of C contained in 1 gram of compound. Call it a

multiply 0.0310 by two, then divide the result by the molecular mass of H2O and then divide it by 0.157, you will find the number of atomic hydrogen moles contained in 1 gram of compound, Call it b

Divide 0.0230 by the molecular mass of NH3, and then divide by 0.103 you will find the number of N moles contained in one gram of compound, call it c

multiply each of the results above by its atomic mass to obtain the mass of each element C, H and N. Substract these to 1 gram and you will find the mass of oxygen contained in 1 gr of substance. Divide it by 16 which is the atomic mass of oxygen and you will find the moles of O contained in 1 g of substance, call it d.

Then write
CaHbNcOd

Divide all a, b, c and d by a and call them w, x, y and z respectively (w must be equal to 1). If the results are not all integers, multiply these results by 2, if not all integers then multiply w, x, y by 3, then by 4 and so on until you get a set of integer numbers, or at least very close. You would round up a number very close to an integer. ( If your answer is correct, then you will have to multiply w, x, y and z by 7. Easy to do but difficult to explain.... hope it helps.