Problem 1

A compound Q (molar mass 122.0 g mol

^{-1}

) consists of carbon, hydrogen and oxygen. PART A. The standard enthalpies of formation of

\ce{CO2(g)}

and

\ce{H2O(l)}

25.00\ ^\circ\mathrm{C}

are

-393.51

and

-285.83

kJ mol

^{-1}

;

R = 8.314

J K

^{-1}

mol

^{-1}

(atomic masses: H = 1.0, C = 12.0, O = 16.0). A 0.6000 g sample of solid Q is combusted in excess oxygen in a bomb calorimeter initially containing 710.0 g of water at

25.000\ ^\circ\mathrm{C}

. After reaction the temperature is

27.250\ ^\circ\mathrm{C}

and 1.5144 g of

\ce{CO2(g)}

and 0.2656 g of

\ce{H2O(l)}

are produced. (1.1) Determine the molecular formula of Q and write a balanced equation, with states, for its combustion. (1.2) Given the specific heat of water 4.184 J g

^{-1}

^{-1}

and the internal energy change of the reaction

\Delta U^\circ = -3079

kJ mol

^{-1}

, calculate the heat capacity of the calorimeter (excluding the water). (1.3) Calculate the standard enthalpy of formation

\Delta H_f^\circ

of Q. PART B. The distribution of Q between benzene and water at

6\ ^\circ\mathrm{C}

gave the equilibrium concentrations

(c_B, c_W)

in mol dm

^{-3}

: (0.0118, 0.00281), (0.0478, 0.00566), (0.0981, 0.00812), (0.156, 0.0102). Assume only one species of Q exists in benzene and that Q is a monomer in water. (1.4) Show by calculation whether Q is a monomer or a dimer in benzene. (1.5) For an ideal dilute solution,

\Delta T_f = R(T_f^0)^2 X_s/\Delta H_f

; the molar mass of benzene is 78.0 g mol

^{-1}

, pure benzene freezes at

5.40\ ^\circ\mathrm{C}

at 1 atm and

\Delta H_f = 9.89

kJ mol

^{-1}

. Calculate the freezing point of a solution of 0.244 g of Q in 5.85 g of benzene.

Step 1 of 5: Molecular formula from combustion

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n(\ce{C}):n(\ce{H}):n(\ce{O}) = \frac{1.5144 \times 12.0/44.0}{12.0}:\frac{0.2656 \times 2.0/18.0}{1.0}:\frac{0.1575}{16.0} = 7:6:2

Analysis

The CO $_2$ and H $_2$ O masses fix the C and H amounts; oxygen comes from the mass difference (0.6000 − 0.4128 − 0.0295 = 0.1575 g). The mole ratio 0.0344:0.0295:0.0098 ≈ 7:6:2 gives $\ce{C7H6O2}$ , whose formula mass 122 matches the molar mass. Combustion: $\ce{C7H6O2(s) + 15/2 O2(g) -> 7 CO2(g) + 3 H2O(l)}$ .

Common pitfall. Do not forget the oxygen bound in Q itself — it must be found by difference from the sample mass.