Question

29. The heat of combustion for a given organic molecule was being measured in a bomb calorimeter (heat capacity of 2.24 J/°C). 0.9168 g of the solid were combusted and resulted in the temperature of the surrounding water (2400 g) to increase by 4.6 °C. a. Given that the compound is 85.71% C and 14.29%H, determine its empirical formula b. The density of the gas at 720 torr and 65°C in a 1.2 L vessel is determined to be 0.956 g/L. Calculate its molar mass and moelcular formula. c. Calculate the heat produced in the combustion reaction in kJ/g. d. Calculate the heat of combustion for the compound in kJ/mol.

Answer 1

Answer:-

This question is solved by using the the simple concept of determination of empirical formula and molecular mass.

The answer is given in the image,

mores 12 Answers Ccal = 2.245/°C AT=4.6°C ти, о : 94vog. On C-85-714. H=14.297. Molesofc = 85.71 – 7.1425 mol MolesqH = 14.29 - 14.290omol, Mole ratio of CiH 1:2 Empirical formula = CH2 b) P = 723terv 2 780atm T= 65+213 = '338% d=8.956 g/l PRENRT pa DRT PAMURT = ela AZ ART = 0.956X0.08214 338 P 720/760 M= 0.956X0,0821x338 7760 720 M=28.0g/mol. c) Empiricel mass=-12+2X12 14 na 2014 - 2

Melecular formula = (Che)n =(CH), = C2H+ (S) Heat released Heat gained =met + Ceal at - 2400X4.18 x406+ 2.24x416 = 46147.2+10.304 =46157.59 = 46.1665 Heat produced in 199-46.16 0.9168 =-50.3561/9. ed) Heat of Cambeestion = -50-35 x28 =-1409.7kJ/mol