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Calorimetry Worksheet 3 Questions with Solutions, Exercises of Chemistry

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Calorimetry Worksheet
W 337
Everett Community College Tutoring Center
Student Support Services Program
Cp (H2O) = 4.184 J / g0C
H = mCpT
1) A compound is burned in a bomb calorimeter that contains 3.00 L of water. If the
combustion of 0.285 moles of this compound causes the temperature of the water to
rise 36.00 C, what is the molar heat of combustion of the compound?
2) When 62.3g of a compound was burned in a bomb calorimeter that contained 0.500
L of water the temperature rise of the water in the calorimeter was 48.00 C. If the
heat of combustion of the compound is 1,160 kJ/mol, what is the molar mass of the
compound?
3) The molar heat of combustion of a compound is 1,350 kJ/mol. If 0.875 moles of this
compound was burned in a bomb calorimeter containing 1.70 L of water, what would
the increase in temperature be?
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Calorimetry Worksheet W 337

Everett Community College Tutoring Center Student Support Services Program

Cp (H 2 O) = 4.184 J / g^0 C

H = mCpT

  1. A compound is burned in a bomb calorimeter that contains 3.00 L of water. If the combustion of 0.285 moles of this compound causes the temperature of the water to rise 36.0^0 C, what is the molar heat of combustion of the compound?

  2. When 62.3g of a compound was burned in a bomb calorimeter that contained 0. L of water the temperature rise of the water in the calorimeter was 48.0^0 C. If the heat of combustion of the compound is 1,160 kJ/mol, what is the molar mass of the compound?

  3. The molar heat of combustion of a compound is 1,350 kJ/mol. If 0.875 moles of this compound was burned in a bomb calorimeter containing 1.70 L of water, what would the increase in temperature be?

  1. A compound is burned in a bomb calorimeter that contains 3.00 L of water. If the combustion of 0.285 moles of this compound causes the temperature of the water to rise 36.0^0 C, what is the molar heat of combustion of the compound? 3.00 L H 2 O = 3000mL H 2 O = 3.00 x 10^3 g H 2 O

H = (3.00 x 10^3 g H 2 O)(4.184 J/g^0 C)(36.0^0 C) = 452000 J H = 452 kJ We now have the heat generated by the combustion of the compound so to find the molar heat of combustion we need only divide this by the number of moles of compound burned.

452 kJ = 1590 kJ/mol 0.285 mol

  1. When 62.3 g of a compound was burned in a bomb calorimeter that contained 0. L of water the temperature rise of the water in the calorimeter was 48.0^0 C. If the heat of combustion of the compound is 1,160 kJ/mol, what is the molar mass of the compound? 0.500 L H 2 O = 500 mL H 2 O = 5.00 x 10^2 g H 2 O H = (5.00 x 10^2 g H 2 O)(4.184 J/g^0 C)(48.0^0 C) = 100000 J H = 100 kJ

Now that we have the heat generated by the combustion of the compound we divide it by the given molar heat of combustion to find the number of moles of the compound that was burned.

  1. kJ = 0.0862 mol 1160 kJ/mol Since we know that 62.3 g of the compound was combusted and that this is equal to the number of moles burned we can divide the mass by the moles to get the molar mass of the compound. 62.3 g = 723 g/mol 0.0862 mol
  1. The molar heat of combustion of a compound is 1,350 kJ/mol. If 0.875 moles of this compound was burned in a bomb calorimeter containing 1.70 L of water, what would the increase in temperature be?

We are given both the number of moles of compound and the molar heat of combustion so by multiplying them together we can find the heat generated. (0.875 mol)(1350 kJ/mol) = 1180 kJ = 118000 J Plug this value into our equation and then rearrange to find the change in the temperature.

118000 J = (1.70 x 10^3 g H 2 O)(4.184 J/g^0 C)( T) T = 118000 J = 16.6 0 C (1.70 x 10^3 g H 2 O)(4.184 J/g^0 C)