0.717 g of a compound containing carbon, hydrogen, and oxygen is burned and found to produce 1.02 g CO2 and 0.624 g H2O. The molar mass of the compound is 124 g/mol.
(a) Write a chemical equation for the combustion of this unknown hydrocarbon.
(b) Determine the # of g and moles of carbon, hydrogen, and oxygen in the unknown hydrocarbon.
(c) Calculate the empirical formula.
(d) Calculate the molecular formula.
(e) Write the balanced equation for the combustion reaction.0.717 g of a compound containing carbon, hydrogen, and oxygen is burned and found to produce 1.02 g CO2 and 0.624 g H2O. The molar mass of the compound is 124 g/mol.
(a) Write a chemical equation for the combustion of this unknown hydrocarbon.
(b) Determine the # of g and moles of carbon, hydrogen, and oxygen in the unknown hydrocarbon.
(c) Calculate the empirical formula.
(d) Calculate the molecular formula.
(e) Write the balanced equation for the combustion reaction.
Franklin or Anonymous--We prefer that you use the same screen name. It helps us keep everyone straight in our minds. See your post and my response below.
Please do not change names for your posts. DrBob already responded to you.
(a) To write the chemical equation for the combustion of the unknown hydrocarbon, we need to first determine the products formed during the combustion reaction. From the given information, we know that the combustion produces CO2 and H2O. This suggests that the hydrocarbon is reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The balanced chemical equation for the combustion of the hydrocarbon can be written as:
C(H2O) + O2 -> CO2 + H2O
(b) In order to determine the mass and number of moles of carbon, hydrogen, and oxygen in the unknown hydrocarbon, we need to use the given information. From the combustion reaction, we know that 0.717 g of the hydrocarbon produces 1.02 g of CO2 and 0.624 g of H2O. We can use the molar masses of CO2 and H2O to help us calculate the amount of carbon and hydrogen present in the hydrocarbon.
First, determine the number of moles of CO2 produced:
moles of CO2 = mass of CO2 / molar mass of CO2
moles of CO2 = 1.02 g / 44.01 g/mol
moles of CO2 = 0.0232 mol
Next, determine the number of moles of H2O produced:
moles of H2O = mass of H2O / molar mass of H2O
moles of H2O = 0.624 g / 18.02 g/mol
moles of H2O = 0.0346 mol
Since the combustion of 1 mole of hydrocarbon produces 1 mole of CO2 and 0.5 moles of H2O, we can compare the moles of CO2 and H2O to determine the number of moles of carbon and hydrogen in the hydrocarbon.
moles of carbon = moles of CO2
moles of carbon = 0.0232 mol
moles of hydrogen = moles of H2O / 0.5
moles of hydrogen = 0.0346 mol / 0.5
moles of hydrogen = 0.0692 mol
To calculate the mass of carbon and hydrogen, we can use the given molar mass of the hydrocarbon (124 g/mol).
mass of carbon = moles of carbon * molar mass of carbon
mass of carbon = 0.0232 mol * 12.01 g/mol
mass of carbon = 0.2787 g
mass of hydrogen = moles of hydrogen * molar mass of hydrogen
mass of hydrogen = 0.0692 mol * 1.01 g/mol
mass of hydrogen = 0.0699 g
The mass of oxygen can be calculated by subtracting the masses of carbon and hydrogen from the total mass of the hydrocarbon:
mass of oxygen = total mass of hydrocarbon - mass of carbon - mass of hydrogen
mass of oxygen = 0.717 g - 0.2787 g - 0.0699 g
mass of oxygen = 0.3684 g
(c) To calculate the empirical formula of the hydrocarbon, we need to determine the molar ratios of carbon, hydrogen, and oxygen in the compound. Divide the number of moles of each element by the smallest number of moles to get the simplest whole number ratio.
Moles of carbon = 0.0232 mol
Moles of hydrogen = 0.0692 mol
Moles of oxygen = 0.3684 mol
Dividing by the smallest number of moles (0.0232 mol), we get:
Moles of carbon = 1.00
Moles of hydrogen = 2.99
Moles of oxygen = 15.86
Rounding these ratios to the nearest whole number, we find the empirical formula of the hydrocarbon is C1H3O8.
(d) To calculate the molecular formula of the hydrocarbon, we need to compare its empirical formula mass to the given molar mass of 124 g/mol.
Empirical formula mass = (1 * molar mass of carbon) + (3 * molar mass of hydrogen) + (8 * molar mass of oxygen)
Empirical formula mass = (1 * 12.01 g/mol) + (3 * 1.01 g/mol) + (8 * 16.00 g/mol)
Empirical formula mass = 124.15 g/mol
Since the empirical formula mass matches the given molar mass, the empirical formula is also the molecular formula. So, the molecular formula of the hydrocarbon is C1H3O8.
(e) The balanced equation for the combustion reaction is:
C(H2O) + 13O2 -> 8CO2 + 9H2O