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6 May, 17:01

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.

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  1. 6 May, 19:00
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    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%
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