Question

A sample of a hydrocarbon is combusted completely in O₂(g) to produce 21.83 g CO₂(g), 4.47 g H₂O(g), and 311 kJ of heat.

(a) What is the mass of the hydrocarbon sample that was combusted?
(b) What is the empirical formula of the hydrocarbon?
(c) Calculate the value of ∆H°f per empirical- formula unit of the hydrocarbon.​

Answer 1

The mass of the hydrocarbon sample is 11.93 g. The empirical formula of the hydrocarbon is CH. The value of ∆H°f per empirical formula unit is -679.3 kJ/mol.

Step-by-step explanation:

(a) To find the mass of the hydrocarbon sample, we need to determine the mass of carbon and hydrogen in the products (CO2 and H2O) and then calculate the mass of the hydrocarbon by subtracting the mass of the oxygen. From the given information, we have:

Mass of CO2 = 21.83 g

Mass of H2O = 4.47 g

From the balanced equation for the combustion of hydrocarbons, we know that:

1 mol of hydrocarbon = 1 mol of CO2 + 1/2 mol of H2O

Using the molar masses of carbon (12.01 g/mol), hydrogen (1.01 g/mol), and oxygen (16.00 g/mol), we can calculate the number of moles of carbon and hydrogen:

Moles of carbon (C): (21.83 g CO2) * (1 mol CO2/44.01 g CO2) * (1 mol C/1 mol CO2) = 0.496 mol C

Moles of hydrogen (H): (4.47 g H2O) * (1 mol H2O/18.02 g H2O) * (2 mol H/1 mol H2O) = 0.496 mol H

Since the molar ratio of C:H in a hydrocarbon is 1:2, we can conclude that the number of moles of carbon and hydrogen are equal.

Therefore, the mass of the hydrocarbon sample is:

Mass of hydrocarbon = (0.496 mol C) * (12.01 g C/mol) + (0.496 mol H) * (1.01 g H/mol) = 11.93 g

(b) To determine the empirical formula of the hydrocarbon, we need to find the simplest whole number ratio of carbon to hydrogen. From the previous step, we know that the number of moles of carbon and hydrogen are equal. Therefore, the empirical formula is CH.

(c) The enthalpy of combustion per empirical formula unit of the hydrocarbon can be calculated by dividing the value of ∆H°f (heat of formation) by the number of empirical formula units. From the given information, we have:

∆H°f for CO2 = -393.5 kJ/mol

∆H°f for H2O = -285.8 kJ/mol

Since the balanced equation for the combustion of hydrocarbons is:

hydrocarbon + O2 → CO2 + H2O

And the coefficients in the equation represent the number of empirical formula units, we can calculate the value of ∆H°f per empirical formula unit:

∆H°f per empirical formula unit = (∆H°f for CO2 + ∆H°f for H2O) / 1 = (-393.5 kJ/mol + -285.8 kJ/mol) / 1 = -679.3 kJ/mol