Problem 75 - Chemistry and Chemical Reactivity

Question

A mixture of butene, \(C_{4} H_{8},\) and butane, \(C_{4} H_{10},\) is burned in air to give \(\mathrm{CO}_{2}\) and water. Suppose you burn \(2.86 \mathrm{g}\) of the mixture and obtain \(8.80 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(4.14 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the weight percents of butene and butane in the mixture?

Step-by-Step Solution

Answer

Butene: 39.23%; Butane: 60.77%.

Step 1: Determine Moles of CO2 and H2O

First, we calculate the moles of \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \) produced. The molar mass of \( \mathrm{CO}_2 \) is \( 44.01 \ \mathrm{g/mol} \), and for \( \mathrm{H}_2\mathrm{O} \) it is \( 18.02 \ \mathrm{g/mol} \).\[ \text{Moles of } \mathrm{CO}_2 = \frac{8.80 \ \mathrm{g}}{44.01 \ \mathrm{g/mol}} \approx 0.200 \ \mathrm{mol} \]\[ \text{Moles of } \mathrm{H}_2\mathrm{O} = \frac{4.14 \ \mathrm{g}}{18.02 \ \mathrm{g/mol}} \approx 0.230 \ \mathrm{mol} \]

Step 2: Set Up Equations for Combustion

Write the balanced chemical equations for the combustion of butene \((\mathrm{C}_4\mathrm{H}_8)\) and butane \((\mathrm{C}_4\mathrm{H}_{10})\):Butene: \( \mathrm{C}_4\mathrm{H}_8 + 6\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O} \)Butane: \( \mathrm{C}_4\mathrm{H}_{10} + \frac{13}{2}\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 5\mathrm{H}_2\mathrm{O} \)

Step 3: Express Variables for Mole Fractions

Let \( x \) be the moles of butene and \( y \) be the moles of butane. Using the stoichiometry from Step 2, form two equations from the total CO2 and H2O:For \( \mathrm{CO}_2 \): \( 4x + 4y = 0.200 \) For \( \mathrm{H}_2\mathrm{O} \): \( 4x + 5y = 0.230 \)

Step 4: Solve the Linear Equations

We solve the system of linear equations: 1. \( 4x + 4y = 0.200 \)2. \( 4x + 5y = 0.230 \)Subtract the first equation from the second equation to solve for \( y \):\[ y = 0.230 - 0.200 = 0.030 \ \mathrm{mol} \] Substitute \( y \) back into the first equation: \[ 4x + 4(0.030) = 0.200 \] \[ 4x + 0.120 = 0.200 \] \[ 4x = 0.080 \] \[ x = 0.020 \ \mathrm{mol} \]

Step 5: Calculate Masses and Weight Percentages

Calculate the mass of butene and butane using their respective molar masses. Molar mass of butene is \( 56.11 \ \mathrm{g/mol} \) and for butane is \( 58.12 \ \mathrm{g/mol} \):Mass of butene: \[ 0.020 \ \mathrm{mol} \times 56.11 \ \mathrm{g/mol} = 1.122 \ \mathrm{g} \] Mass of butane: \[ 0.030 \ \mathrm{mol} \times 58.12 \ \mathrm{g/mol} = 1.744 \ \mathrm{g} \] Calculate weight percentages: Weight percent of butene: \[ \left( \frac{1.122}{2.86} \right) \times 100 \approx 39.23\% \] Weight percent of butane: \[ \left( \frac{1.744}{2.86} \right) \times 100 \approx 60.77\% \]

Key Concepts

Molar Mass Calculation

Molar mass is an essential concept in chemistry that helps us relate the mass of a substance to the number of moles it contains. It's crucial for figuring out how much of a substance you have in a chemical reaction.

To find the molar mass, you add the atomic masses of all atoms in a molecule. For example, the molar mass of CO₂ (carbon dioxide) is calculated by adding the atomic masses of carbon (about 12.01 g/mol) and oxygen (about 16.00 g/mol each). Thus, the molar mass of CO₂ is approximately 44.01 g/mol.

• Molar mass of H₂O (water) is approximately 18.02 g/mol.
• This allows us to convert from grams to moles using the relation: \( ext{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \).

In the combustion analysis, calculating moles from the given mass helps us understand how much of each substance is involved in a reaction.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's a balancing act that helps predict the amounts of substances consumed and produced.

In combustion reactions, stoichiometry is key to understanding how molecules interact. Using balanced chemical equations, you can determine the ratios in which substances react.

• Example: In the combustion of butene, 1 mole produces 4 moles of CO₂ and 4 moles of H₂O.
• For butane: 1 mole results in 4 moles of CO₂ and 5 moles of H₂O.

These balanced equations are formulated using the law of conservation of mass, ensuring atoms are neither created nor destroyed in a reaction. Stoichiometry allows us to solve for unknowns, like the composition of a mixture or the yield of a product.

Chemical Equations

Writing chemical equations is like laying out a plan for a chemical reaction. It provides a way to visually represent the reactants turning into products.

A balanced chemical equation ensures that the number of atoms for each element is the same on both sides of the equation. This balance keeps the reaction aligned with the law of conservation of mass.

• For example, the equation for butene burning in oxygen is \( \text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O} \).
• Each equation shows how many molecules or moles of substances are involved.

Understanding and writing these equations is crucial, especially when solving complex problems like combustion analysis. These equations are the foundation that supports stoichiometric calculations.

Weight Percent

Weight percent (or weight percent composition) is a way of expressing the concentration of a substance in a mixture. It's useful for understanding the makeup of a sample.

To find the weight percent, you calculate the mass of the substance of interest and divide it by the total mass of the mixture, then multiply by 100 to get a percentage.

• In our example, the mass of butene and butane in a mixture is used to find their weight percentages.
• This helps in understanding the proportion of each compound in the total mass.

Weight percent is an essential tool in chemistry when analyzing the composition of mixtures, as it allows for the quantification of individual components in a clear, understandable way.