Question

Pure bromobenzene, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\), boils at \(156.0^{\circ} \mathrm{C}\) and has a boiling point constant of \(6.26^{\circ} \mathrm{C} / \mathrm{m} .\) A sample of bromobenzene is contaminated by anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10} .\) The boiling point of the impure sample is \(159.2^{\circ} \mathrm{C} .\) How pure is the sample? (Express your answer as mass percent bromobenzene.)

Short answer

Answer: The mass percent of bromobenzene in the impure sample is approximately 8.02%.

Step-by-step solution

Calculate the boiling point elevation of the solution

The boiling point elevation can be calculated by subtracting the boiling point of pure bromobenzene from the boiling point of impure bromobenzene. \(\Delta T_b = T_{b\ impure} - T_{b \ pure}\) \(\Delta T_b = 159.2^\circ C - 156.0 ^\circ C\) \(\Delta T_b = 3.2 ^\circ C\)

Calculate mole fraction of bromobenzene using the boiling point elevation formula

The boiling point elevation formula is: \(\Delta T_b = K_b \cdot m\) where \(m\) is the molality. Rearrange the formula: \(m = \frac{\Delta T_b}{K_b}\) Put the values: \(m = \frac{3.2^\circ C}{6.26^\circ C / mol}\) \(m = 0.511 Mol/kg\)

Calculate the mass fraction of bromobenzene

The mass fraction of bromobenzene in the impure sample can be calculated, considering 1 kg of solution: \(Mass\ of\ bromobenzene = 0.511 \ mol/kg * 1\ kg * Molar\ mass\ of\ bromobenzene\) The molar mass of bromobenzene is: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\) = 6(12.01) + 5(1.01) + 79.90 =\( 157.01 g/mol\) Therefore, \(Mass\ of\ bromobenzene = 0.511 \ mol/kg * 1\ kg * 157.01\ g/mol\simeq 80.2 g\)

Calculate mass of anthracene

The mass of anthracene can be calculated considering that the total mass of the solution is 1 kg. \(Mass\ of\ anthracene = Total\ mass\ of\ solution - Mass\ of\ bromobenzene =1\ kg - 80.2\ g \simeq 919.8\ g\)

Calculate mass percent of bromobenzene

Finally, we can calculate the mass percent of bromobenzene in the impure sample using the following formula: \(Mass\ percent\ of\ bromobenzene = \frac{Mass\ of\ bromobenzene}{Mass\ of\ bromobenzene + Mass\ of\ anthracene} * 100\) \(Mass\ percent\ of\ bromobenzene = \frac{80.2\ g}{80.2\ g + 919.8\ g} * 100 \simeq 8.02 \%\) The purity of the sample is approximately \(8.02\%\) bromobenzene by mass.