Problem 166 - Chemistry An Atoms First Approach

Question

A \(9.780-\mathrm{g}\) gaseous mixture contains ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Complete combustion to form carbon dioxide and water requires 1.120 mole of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Step-by-Step Solution

Answer

The mass percent of ethane in the original gaseous mixture is approximately 21.14%.

Step 1: Write the balanced combustion equations for ethane and propane

The balanced combustion equations for ethane and propane are \(C_2H_6 + \dfrac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O\) \(C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O\) These equations tell us that 1 mole of ethane requires \(\dfrac{7}{2}\) moles of oxygen for combustion, while 1 mole of propane requires 5 moles of oxygen.

Step 2: Set up a system of equations to find the moles of ethane and propane

Let x represent the moles of ethane and y represent the moles of propane. Then, we have the following system of equations: \(x\left(\dfrac{7}{2}\right) + 5y = 1.120\) moles of oxygen (from the balanced equations) \(30x + 44y = 9.780\) g (total mass of the ethane and propane mixture)

Step 3: Solve the system of equations

To solve the system of equations, we can multiply the first equation by -30 and then add the two equations to eliminate x: \(-15x - 30y = -3.360\) \(-30x - 14y = -9.780\) Now, add the equations: \(-44y = -13.140\) Divide by -44 to solve for y: \(y = 0.29886\) moles of propane Now, substitute y back into the first equation to solve for x: \(x\left(\dfrac{7}{2}\right) + 5(0.29886) = 1.120\) \(x = 0.11762\) moles of ethane

Step 4: Convert moles to mass and find the mass percent of ethane

Now that we have the moles of ethane and propane, we can convert them to mass and find the mass percent of ethane. Mass of ethane = moles of ethane × molar mass of ethane \(Mass_{ethane} = 0.11762 \times 30 = 3.5286\;\mathrm{g}\) Mass of propane = moles of propane × molar mass of propane \(Mass_{propane} = 0.29886 \times 44 = 13.14984\;\mathrm{g}\) Now, we can find the mass percent of ethane: Mass percent of ethane = \(\dfrac{Mass_{ethane}}{Mass_{ethane} + Mass_{propane}} \times 100\) Mass percent of ethane = \(\dfrac{3.5286}{3.5286 + 13.14984} \times 100 \approx 21.14\%\) Therefore, the mass percent of ethane in the original mixture is approximately 21.14%.

Key Concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen and releases energy in the form of heat and light. This process is fundamental in stoichiometry, as it helps determine the reactants and products involved in burning substances like hydrocarbons. For hydrocarbons such as ethane - (\(C_2H_6\)) and propane - (\(C_3H_8\)), the products of complete combustion are carbon dioxide (\(CO_2\)) and water (\(H_2O\)).

In the given problem, ethane and propane react with oxygen. The balanced equations are:

• For ethane: \(C_2H_6 + \dfrac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O\)
• For propane: \(C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O\)

These equations indicate that combustion processes are consistent in their stoichiometry, utilizing specific ratios of reactants and products.

Mass Percent

Mass percent is a way of expressing the concentration of a component in a mixture by comparing the mass of the component to the total mass of the mixture. It is a common technique in chemistry for expressing the composition of mixtures.

To find the mass percent of ethane in a mixture, you first need the mass of ethane and the total mass of the mixture. In the exercise provided, mass calculations for each gas were performed after determining the moles:

• Mass of ethane: calculated using \(0.11762\) moles, the molar mass (\(30\) g/mol), resulting in \(3.5286\) g.
• Mass of propane is \(13.14984\) g, calculated similarly from \(0.29886\) moles and molar mass \(44\) g/mol.

Finally, the mass percent of ethane is determined by dividing the mass of ethane by the total mass of the mixture. Multiplication by 100 converts it to a percentage:
\[ Mass \; percent \; of \; ethane = \dfrac{3.5286}{3.5286 + 13.14984} \times 100 \approx 21.14\% \]

This showed ethane made up approximately \(21.14\%\) of the original mixture.

Balanced Chemical Equations

Balanced chemical equations are crucial in tracking the quantities of reactants and products in a chemical reaction. They ensure that the law of conservation of mass is adhered to, meaning the number of atoms of each element is the same on both sides of the equation.

In the problem of burning ethane and propane, each component's combustion reaction must be balanced. This involves:

• Assigning correct stoichiometric coefficients to each reactant and product.
• Ensuring that both moles and types of atoms are the same on both sides.

For ethane, the balanced equation is: \[C_2H_6 + \frac{7}{2} O_2 \rightarrow 2 CO_2 + 3 H_2O\]And for propane: \[C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O\]

This guarantees that the correct amount of oxygen is used and carbon dioxide with water is formed at precise amounts. Balanced reactions are the foundation for stoichiometric calculations that follow, ensuring accurate quantification of substances involved.