Calculate the Percent Composition of an Herbicide from Analysis Data?

[6021023]

Homework Statement

An herbicide is found to contain only C, H, N, and Cl. The complete combustion of a 100.0-mg sample of the herbicide in excess oxygen produces 83.16 mL of CO2 and 73.30 mL of H2O vapor at STP. A separate analysis shows that the sample also contains 16.44 mg of Cl.

Determine the percent composition of the substance.

Homework Equations

pV = nRT

The Attempt at a Solution

moles of Cl = .01644/35.453

 

[Borek]

You don't need moles of chlorine. Percent composition means mass/mass.

 

[6021023]

So how do I get the mass of the other elements?

 

[Borek]

From moles. In the case of chlorine mass was given straight away.

 

[6021023]

How do I figure out the number of moles for the other elements?

 

[Borek]

Using equation that you already posted.

Thats enough spoonfeeding, you either start to think on your own, or I am not going to help any further.

 

[6021023]

moles of CO2 = pV/RT = 0.08316/(0.0821 * 273) = 0.003710297 moles
Moles of C = (12/44) * 0.003710297 = 0.00101199 moles
mg of C = (0.00101199)*(12000) = 12.14mg

moles of H2O = pV/RT = 0.07330/(0.0821 * 273) = 0.00327038 moles
moles of H = (2/18) * 0.00327038 = 0.000363376 moles
mg of H = (0.000363376)*(1000) = 0.363375516 mg

mg of N = 100mg - 0.363375516 mg - 12.14mg - 16.44mg = 71.05 mg

So the % composition is
C 12.14%
H 0.3634%
N 71.05%
Cl 16.44%

Is that right?

 

[symbolipoint]

The method looks good, ( I did not check for arithmetic computation mistakes ).

 

[Borek]

moles of CO2 = pV/RT = 0.08316/(0.0821 * 273) = 0.003710297 moles
Moles of C = (12/44) * 0.003710297 = 0.00101199 moles

No. 1 mole of CO₂ contains 1 mole of C.