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.

The empirical formula of the compound is C8H14ClN5

C = 44.5%

H = 6.6 %

Cl = 16.4%

N = 32.5%

Explanation:

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. Calculate its empirical formula. Determine the percentage of the composition of the substance.

Step 1: Data given

Mass of the sample = 100.0 mg

Volume of CO2 = 83.16 mL

Volule of H2O = 73.30 mL

A separate analysis shows that the sample also contains 16.44 mg of Cl.

Step 2: The balanced equation

CwHxNyClz + O2 ⇒ CO2 + H2O

Step 3: Calculate moles of CO2

Moles CO2 = (p*V) / (R*T)

Moles CO2 = (1*0.08316) / (0.08206*273)

Moles CO2 = 0.003712 moles

Step 4: Calculate moles of H2O

Moles H2O = (1*0.07330) / (0.08206*273)

Moles H2O = 0.003272

Step 5: Calculate moles of Cl

Moles Cl = mass Cl / molar mass CL

Moles Cl = 0.01644 grams / 35.45 g/mol

Moles Cl = 4.64 * 10^-4 moles

Step 6: Calculate moles C

CO2 contains 1 mol C

For 0.003712 mol of CO2 we have 0.003712 mol of C

Step 7: Calculate moles of H

H2O contains 2 mole of H

For 0.00372 moles of H2O we have 2*0.003272 = 0.006544 mol H

Step 8: Calculate mass of C and H

0.003712 mol C * 12.01g/mol = 0.04458g C

0.006545 mol H * 1.01g/mol = 0.006610g H

Step 9: Calculate mass of N

0.1 g - (0.04458g C + 0.006610g H+0.01644g Cl) = 0.03237g N

moles N = 0.03237 / 14.00 g/mol = 0.002312

Step 10: Calculate mol ratio

C: 0.003712 / 0.000464 = 8

H: 0.006545 / 0.000464 = 14

Cl: 0.000464 / 0.000464 = 1

N: 0.002312 / 0.000464 = 5

The empirical formula of the compound is C8H14ClN5

The molar mass is 215.59 g/mol

Step 11: Determine the percentage of the composition of the substance.

C = ((8*12) / 215.59) * 100% = 44.5%

H = ((14*1.01) / 215.59) * 100% = 6.6 %

Cl = (35.45 / 215.59) * 100% = 16.4%

N = ((5*14) / 215.59) * 100% = 32.5%