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 empirical formula.
A. Determine the percentage of the composition of the substance.
B. Calculate its empirical formula.
C. What other informtion would you need to know about this compound to calculate its true molecular formula?

a) 44.59 % C

6.61 % H

16.44% Cl

32.36% N

b)The empirical formula is then: C8H14N5Cl

c)To know the molecular formula we have to know the molar mass of the compound.

Explanation:

Step 1: Data given

Mass of the compound = 100 mg = 0.100 grams

Volume of CO2 = 83.16 mL

Volume of H2O = 73.30 mL

Mass of Cl = 16.44 mg = 0.01644 grams

Atomic mass of Cl = 35.45 g/mol

Atomic mass of C = 12.01 g/mol

atomic mass of O = 16.0 g/mol

atomic mass of N = 14.0 g/mol

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: Calculate moles CO2

1 mol = 22.4L

moles CO2 = 0.8316 L / 22.4 L 0.003713 moles CO2

In 1 mol CO2 we have 1 mol C

Moles C = 0.003713 moles

Step 3: Calculate mass C

Mass C = moles C * molar mass C

Mass C = 0.003713 moles * 12.01 g/mol

Mass C = 0.04459 grams

= 44.59 % C

Step 4: Calculate moles H2O

1 mol = 22.4L

moles H2O = 0.07330L / 22.4 L 0.003272 moles H2O

In 1 mol H2O we have 2 mol H

Moles H = 2* 0.003272 = 0.006544 moles H

Step 5: Calculate mass H

Mass H = 0.006544 moles * 1.01 g/mol = 0.006609 grams H

=6.61 % H

Step 6: Calculate moles Cl

Moles Cl = mass Cl / molar mass Cl

Moles Cl = 0.01644 grams / 35.45 g/mol

Moles Cl = 0.0004638 moles Cl

= 16.44% Cl

Step 7: Calculate mass N

Mass N = 0.1000 grams  - 0.04459 grams - 0.006609 grams - 0.01644 grams = 0.03236 grams N

Step 8: Calculate moles N

Moles N = mass N / molar mass N

Moles N = 0.03236 grams / 14.0 g/mol

Moles N = 0.002311 moles N

= 32.36% N

Step 9: Calculate the mol ratio

We divide each moles by smallest moles

N: 0.002311 moles / 0.0004638 moles = 5

Cl: 0.0004638 / 0.0004638 = 1

C: 0.003713 / 0.0004638 = 8

H: 0.006609 / 0.0004638 = 14

The empirical formula is then: C8H14N5Cl

To know the molecular formula we have to know the molar mass of the compound.