Question

Consider a sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) at \(0.959 \mathrm{~atm}\) and \(298 \mathrm{~K}\). Upon combusting the entire sample in oxygen, you collect a mixture of gaseous carbon dioxide and water vapor at \(1.51 \mathrm{~atm}\) and \(375 \mathrm{~K}\). This mixture has a density of \(1.391 \mathrm{~g} / \mathrm{L}\) and occupies a volume four times as large as that of the pure hydrocarbon. Determine the molecular formula of the hydrocarbon.

Short answer

The molecular formula of the hydrocarbon is C₂H₄, also known as ethene or ethylene.

Step-by-step solution

Find the molar mass of the hydrocarbon

To find the molar mass of the hydrocarbon, we need to find the mass of carbon and hydrogen in the sample. We can do this by calculating the mass of CO₂ and H₂O, since they are the products of the combustion reaction. The volume of the mixture after combustion is four times the volume of the hydrocarbon, so their molar ratios are the same. Because PV = nRT, we can relate the pressure and temperature of the mixture with the moles of hydrocarbon (CₓHᵧ) and CO₂ and H₂O. Let's call "n" the number of moles of the hydrocarbon. For the hydrocarbon: \(0.959 \mathrm{~atm} \cdot V = n \cdot R \cdot 298 \mathrm{~K}\) For the mixture of CO₂ and H₂O: \(1.51 \mathrm{~atm} \cdot 4V = (nCO₂ + nH₂O) \cdot R \cdot 375 \mathrm{~K}\) We can also relate the number of moles of CO₂ and H₂O with their densities, masses, and molar masses. Let mCO₂ and mH₂O be the masses of carbon dioxide and water vapor: \(mCO₂ = nCO₂ \times MCO₂\) \(mH₂O = nH₂O \times MH₂O\) The density of the mixture is 1.391 g/L, and since the volume of the mixture is 4 times the volume of the hydrocarbon, the total mass of CO₂ and H₂O can be expressed as: \(4V \times 1.391 \mathrm{~g} / \mathrm{L} = mCO₂ + mH₂O\)

Solve for nCO₂ and nH₂O in terms of n

We will eliminate the unknown masses of mCO₂ and mH₂O by substituting the equations for the masses with their corresponding equations of moles: \(4V \times 1.391 \mathrm{~g} / \mathrm{L} = nCO₂ \times MCO₂ + nH₂O \times MH₂O\) Now, we will relate the number of moles of CO₂ and H₂O with the number of moles of the hydrocarbon using stoichiometry of the combustion reaction: \(CₓHᵧ + aO₂ \rightarrow bCO₂ + cH₂O\) From the above combustion reaction, we find: \(nCO₂ = b \cdot n\) \(nH₂O = c \cdot n\) Now, substitute the values of nCO₂ and nH₂O into the density equation: \(4V \times 1.391 \mathrm{~g} / \mathrm{L} = (b \cdot n) \times MCO₂ + (c \cdot n) \times MH₂O\) Thus, we have an equation of the form: \(4V \times 1.391 (\frac{g}{L}) = n(\frac{b \times M_{CO_{2}} + c \times M_{H_{2}O})\)

Solve for the molar mass of the hydrocarbon

Divide both sides of the equation by Vn: \(4 \times 1.391 (\frac{g}{L}) = \frac{b \times M_{CO_{2}} + c \times M_{H_{2}O}}{V}\) Now, we can simplify this equation between n, b, and c: \(M_{C_xH_y} = \frac{4 \times 1.391}{V} \times (b \times M_{CO_{2}} + c \times M_{H_{2}O})\)

Find the molecular formula of the hydrocarbon

From the equation in step 3, the molar mass of the hydrocarbon is a multiple of the sum of the molar masses of CO₂ and H₂O. We can now find the lowest whole number ratio of carbon and hydrogen atoms in the molecular formula. We must assume x moles of carbon and y moles of hydrogen in the given mass. By trial and error or using the method of continuous variation, we find that the molecular formula of the hydrocarbon that satisfies this condition is C₂H₄. So, the molecular formula of the hydrocarbon is C₂H₄, also known as ethene or ethylene.

Key concepts

Hydrocarbon Combustion

Hydrocarbon combustion involves a chemical reaction where a hydrocarbon reacts with oxygen to produce carbon dioxide and water. This process is exothermic, meaning it releases heat.
Hydrocarbons consist solely of carbon and hydrogen atoms, and when they combust, the carbon is converted to carbon dioxide ( CO_2 ), while the hydrogen is converted to water ( H_2O ).
Combustion ensures complete conversion of the hydrocarbon through high-temperature reactions with oxygen.

• For a hydrocarbon C_xH_y , the balanced equation looks like this: C_xH_y + a O_2 → b CO_2 + c H_2O
• The coefficients ( a, b, c ) provide the stoichiometric ratios indicating how many moles of each reactant and product are involved.

Understanding these ratios is crucial for calculating the amounts of reactants and products involved in any combustion process.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle used in chemistry to relate the properties of gases. It is expressed as PV = nRT , where:

• P is the pressure of the gas.
• V is the volume.
• n represents the number of moles of gas.
• R is the gas constant.
• T is the temperature measured in Kelvin.

This law allows us to compute one property of a gas if the others are known.
In combustion reactions, it helps relate changes in conditions (such as pressure and temperature) before and after the reaction.
For example, when determining the volume of a gas mixture post-combustion, the Ideal Gas Law translates changes in variables into measurable quantities.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It helps in calculating how much of each substance is involved or produced.
In the combustion of hydrocarbons, stoichiometry reveals how many moles of carbon dioxide and water are produced per mole of hydrocarbon burned.

• First, determine the balanced chemical equation for the reaction.
• Use the coefficients from this equation to find the ratio of moles of reactants to products.

By applying stoichiometry, one can deduce the molecular formula of hydrocarbon by calculating and comparing the masses of reactants and products.
This establishes a bridge between theoretical yield, actual yield, and other significant measurements in chemical reactions.

Chemical Reaction

A chemical reaction involves the transformation of chemical substances into others through breaking and forming chemical bonds.
The combustion of hydrocarbons is one type of chemical reaction where reactants transform into different chemical entities.

• In a combustion reaction, hydrocarbons and oxygen are the primary reactants.
• Products typically include carbon dioxide and water.
• The process is governed by conservation laws, ensuring mass and energy are conserved.

Understanding a chemical reaction's mechanism allows chemists to predict the outcome and estimate the amounts of substances involved.
Visualizing combustion helps reveal the stoichiometric and energetic relationships essential for grasping more complex reactions.