Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated. Calculate the heats of combustion of these alcohols in kJ/mol.
(a) methanol (CH3OH), -22.6 kJ
(b) ethanol (C2H5OH), -29.7 kJ
(c) n-propanol (C3H7OH), -33.4 kJ
I started calculating but found this page on the web before I finished. The table is sown a little from the top and list methaol, ethanol, and propanol, among others.
To calculate the heat of combustion of each alcohol in kJ/mol, we need to convert the given values from grams to moles and then divide the heat liberated by the number of moles.
(a) Methanol (CH3OH):
- Given: 1.00 g of methanol and -22.6 kJ of heat liberated
- Convert grams to moles:
- Molar mass of methanol (CH3OH) = 12.01 g/mol + 1.01 g/mol + 16.00 g/mol = 32.04 g/mol
- Moles of methanol = 1.00 g / 32.04 g/mol = 0.031 mol
- Calculate the heat of combustion per mole:
- Heat of combustion of methanol in kJ/mol = -22.6 kJ / 0.031 mol = -729 kJ/mol
(b) Ethanol (C2H5OH):
- Given: 1.00 g of ethanol and -29.7 kJ of heat liberated
- Convert grams to moles:
- Molar mass of ethanol (C2H5OH) = 2 * (12.01 g/mol) + 6 * (1.01 g/mol) + 16.00 g/mol = 46.07 g/mol
- Moles of ethanol = 1.00 g / 46.07 g/mol = 0.022 mol
- Calculate the heat of combustion per mole:
- Heat of combustion of ethanol in kJ/mol = -29.7 kJ / 0.022 mol = -1350 kJ/mol
(c) n-Propanol (C3H7OH):
- Given: 1.00 g of n-propanol and -33.4 kJ of heat liberated
- Convert grams to moles:
- Molar mass of n-propanol (C3H7OH) = 3 * (12.01 g/mol) + 8 * (1.01 g/mol) + 16.00 g/mol = 60.10 g/mol
- Moles of n-propanol = 1.00 g / 60.10 g/mol = 0.017 mol
- Calculate the heat of combustion per mole:
- Heat of combustion of n-propanol in kJ/mol = -33.4 kJ / 0.017 mol = -1965 kJ/mol
Therefore, the heats of combustion of the three alcohols are:
(a) Methanol: -729 kJ/mol
(b) Ethanol: -1350 kJ/mol
(c) n-Propanol: -1965 kJ/mol
To calculate the heats of combustion of these alcohols in kJ/mol, we need to use the given values and convert them to kJ/mol.
First, let's determine the molar mass of each alcohol:
- Methanol (CH3OH): The molar mass can be calculated by adding up the atomic masses of carbon, hydrogen, and oxygen. The atomic masses are approximately 12.01 g/mol for carbon, 1.01 g/mol for hydrogen, and 16.00 g/mol for oxygen. Therefore, the molar mass of methanol is 12.01 + 3(1.01) + 16.00 = 32.04 g/mol.
- Ethanol (C2H5OH): Similarly, summing up the atomic masses of carbon, hydrogen, and oxygen gives us a molar mass of 2(12.01) + 6(1.01) + 16.00 = 46.07 g/mol.
- n-Propanol (C3H7OH): Adding the atomic masses of carbon, hydrogen, and oxygen results in a molar mass of 3(12.01) + 8(1.01) + 16.00 = 60.10 g/mol.
Now, to calculate the heat of combustion per mole, we divide the given heat values by the molar mass of each alcohol:
(a) Methanol:
Heat of combustion in kJ/mol = -22.6 kJ / (32.04 g/mol) = -0.704 kJ/g
(b) Ethanol:
Heat of combustion in kJ/mol = -29.7 kJ / (46.07 g/mol) = -0.645 kJ/g
(c) n-Propanol:
Heat of combustion in kJ/mol = -33.4 kJ / (60.10 g/mol) = -0.555 kJ/g
Therefore, the heats of combustion of these alcohols in kJ/mol are approximately:
(a) Methanol: -0.704 kJ/g
(b) Ethanol: -0.645 kJ/g
(c) n-Propanol: -0.555 kJ/g