19. Aspirin is produced commercially from salicylic acid, C,O,He. A large shipme
ID: 706303 • Letter: 1
Question
19. Aspirin is produced commercially from salicylic acid, C,O,He. A large shipment of salicylic acid is contaminated with boric oxide, which like salicylic acid is a white powder. The heat of combustion of salicylic acid at constant volume is known to be -3.00 x 10 kJ/mol. Boric oxide, because it is fully oxidized, does not burn. When a 3.556 g sample of contaminated salicylic acid is burned in a bomb calorimeter, the temperature increases 2.556°C. From previous measurements, the heat capacity of the calorimeter is known to be 13.62 kJ/K. What is the amount of boric oxide in the sample, in terms of mass percent?Explanation / Answer
19) The combustion of salicylic acid releases heat which is taken up by the bomb calorimeter and increases the temperature of the calorimeter. The heat capacity of the calorimeter is 13.62 kJ/K and the rise in temperature of the calorimeter = 2.556°C = 2.556 K (the Celsius and the Kelvin scales show the same rise in temperature). Therefore,
Heat absorbed by the calorimeter = (heat capacity of calorimeter)*(rise in temperature of the calorimeter) = (13.62 kJ/K)*(2.556 K) = 34.81272 kJ.
The heat released by the combustion of salicylic acid is equal and opposite to the heat absorbed by the calorimeter, i.e, the heat released by the combustion of salicylic acid = -34.81272 kJ.
The heat of combustion of salicylic acid is -3.00*103 kJ/mol; therefore, the moles of salicylic acid combusted = (heat released by combustion)/(heat of combustion) = (-34.81272 kJ)/(-3.00*103 kJ/mol) = 0.0116 mole.
The molar mass of salicylic acid is (7*12.011 + 3*15.999 + 6*1.008) g/mol = 138.122 g/mol.
The mass of salicylic acid combusted = (moles of salicylic acid combusted)*(molar mass of salicylic acid) = (0.0116 mole)*(138.122 g/mol) = 1.6022 g ? 1.602 g.
The weight of the sample of salicylic acid and boric acid = 3.556 g; therefore, the weight of boric oxide in the sample = (3.556 – 1.602) g = 1.954 g.
Mass percent boric oxide in the sample = (mass of boric oxide)/(mass of sample)*100 = (1.954 g)/(3.556 g)*100 = 54.9494% ? 54.95% (ans).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.