Chemistry Homework Help?

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A) Paradichlorobenzene, a common ingredient in solid air fresheners, contains only C, H, and Cl and has a molar mass of about 147g. Given that the combustion of 1.68 g of this compound produces 3.02 g CO2 and 0.412 g H2O, determine its empirical and molecular formulas.

B) Using the following information and the fact that the average C-H bond enthalpy is 414 kj/mol, estimate the standard enthalpy of formation of methane (CH4).

C(s) —-> C(g) H=716 kj/mol
2H2(g)—–> 4H(g) H= 872.8 kj/mol

✅ Answers

? Best Answer

  • Of course the name kind of gives it away, you realize. (C6H4Cl2)

    CxHyClz + O2 –> CO2 + H2O + HCl ……….. unbalanced
    1.68 g ……………. 3.02g ..0.412g

    3.02 g CO2 is 0.0686 mol, so there is 0.0686 moles of carbon, which is 0.823 grams of carbon

    0.412g H2O is 0.0229 mol, so there is 0.0458 moles of hydrogen, which is 0.0458 g of hydrogen

    Therefore, the mass of chlorine is 1.68 – 0.823 – 0.0458 = 0.811 g chlorine, which is 0.0228 moles of chlorine.

    Empirical formula:
    0.0686 mol C / 0.0228 = 3
    0.0458 mol H / 0.0228 = 2
    0.0254 mol Cl / 0.0228 = 1
    Empirical formula is C3H2Cl
    The empirical molar mass is 73.5 g/mol
    147 / 73.5 = 2 …. therefore, double all the subscripts to get the molecular formula:
    C6H4Cl2

    =================

    C(s) —-> C(g) H=716 kj/mol
    2H2(g)—–> 4H(g) H= 872.8 kj/mol
    C-H bond energy is 414 kj/mol,

    ……….+716 + 872.8 ………. 4(-414)
    ………_____________……_________
    ………|…………………… |…..| ……………..|
    C(s) + 2H2(g) –> C(g) + 4H(g) –> CH4(g)
    ………|____________ ____________|
    …………………….ΔH(formation)

    Hess’s law of constant heat summation:
    ΔHf = (716 + 872.8 0 + 4(-414) = -67.2 kJ/mol

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