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Percent composition

  • Thread starter 6021023
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  • #1
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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
 

Answers and Replies

  • #2
Borek
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You don't need moles of chlorine. Percent composition means mass/mass.
 
  • #3
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So how do I get the mass of the other elements?
 
  • #4
Borek
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From moles. In the case of chlorine mass was given straight away.
 
  • #5
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How do I figure out the number of moles for the other elements?
 
  • #6
Borek
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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.
 
  • #7
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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?
 
  • #8
symbolipoint
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The method looks good, ( I did not check for arithmetic computation mistakes ).
 
  • #9
Borek
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2,716
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.
 

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