Question

An herbicide contains only C, H, Cl, and N. The complete combustion of a 200.0 mg sample of the herbicide in excess oxygen produced 209.2 mL of CO2 and 122.0 mL of H2O vapor at STP. A separate analysis determined the 200.0 mg sample contained 55.14 mg Cl. Determine the percent composition of the herbicide.

Answer 1

Answer:

%C = 56,1%

%H = 5,5%

%Cl = 27,6%

%N = 10,8%

Explanation:

The moles of CO₂ are the same than moles of C in the herbicide.

Moles of H₂O are ¹/₂ of moles of H in the herbicide.

Moles of CO₂ are obtained using:

n = PV/RT

Where, in STP: P is 1 atm; V is 0,2092L; R is 0,082atmL/molK; T is 273 K

moles of CO₂ are: 9,345x10⁻³ mol≡ mol of C×12,01g/mol = 0,1122 g of C ≡ 112,2mg of C

In the same way, moles of H₂O are 5,450x10⁻³mol×2 =0,1090 mol of H×1,01g/mol = 0,0110 g of H ≡ 11,0mg of H

As you have 55,14 mg of Cl, the mg of N are:

200,0mg - 112,2 mg of C - 11,0 mg of H - 55,14 mg of Cl = 21,66 mg of N

Thus, precent composition of the herbicide is:

%C = $\frac{112,2 mgC}{200,0mg}$×100 = 56,1%C

%H = $\frac{11,0 mgH}{200,0mg}$×100 = 5,5%H

%Cl = $\frac{55,14 mgCl}{200,0mg}$×100 = 27,6%Cl

%N = $\frac{21,66 mgN}{200,0mg}$×100 = 10,8%N

I hope it helps!