Problem 101 - Chemistry and Chemical Reactivity

Question

A mixture of butene, \(\mathrm{C}_{4} \mathrm{H}_{8},\) and butane, \(\mathrm{C}_{4} \mathrm{H}_{10}\) is burned in air to give \(\mathrm{CO}_{2}\) and water. Suppose you burn 2.86 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 are the mass percentages of butene and butane in the mixture?

Step-by-Step Solution

Answer

39.2% butene, 60.8% butane.

Step 1: Determine Moles of Products

First, calculate the moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \) produced. For \( \mathrm{CO}_{2} \):\[ \text{Molar mass of } \mathrm{CO}_{2} = 12 + 2(16) = 44 \, \text{g/mol} \]\[ \text{Moles of } \mathrm{CO}_{2} = \frac{8.80 \, \text{g}}{44 \, \text{g/mol}} = 0.200 \, \text{mol} \]For \( \mathrm{H}_{2} \mathrm{O} \):\[ \text{Molar mass of } \mathrm{H}_{2} \mathrm{O} = 2(1) + 16 = 18 \, \text{g/mol} \]\[ \text{Moles of } \mathrm{H}_{2} \mathrm{O} = \frac{4.14 \, \text{g}}{18 \, \text{g/mol}} = 0.230 \, \text{mol} \]

Step 2: Write Balanced Combustion Reactions

The balanced reaction equations for the combustion of butene (\(\mathrm{C}_{4} \mathrm{H}_{8}\)) and butane (\(\mathrm{C}_{4} \mathrm{H}_{10}\)) are:1. For butene:\[ \mathrm{C}_{4} \mathrm{H}_{8} + 6\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 4\mathrm{H}_{2} \mathrm{O} \]2. For butane:\[ \mathrm{C}_{4} \mathrm{H}_{10} + 6.5\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 5\mathrm{H}_{2} \mathrm{O} \]

Step 3: Formulate System of Equations

Let \( x \) be the moles of butene and \( y \) be the moles of butane burned. According to the balanced equations, create equations based on moles of \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \):1. \( 4x + 4y = 0.200 \) (for \( \mathrm{CO}_{2} \))2. \( 4x + 5y = 0.230 \) (for \( \mathrm{H}_{2} \mathrm{O} \))

Step 4: Solve the System of Equations

Solve the equations from Step 3 to find \( x \) and \( y \):- Subtract the first equation from the second: \ \( (4x + 5y) - (4x + 4y) = 0.230 - 0.200 \)- Simplify to: \( y = 0.030 \, \text{mol} \)- Substitute \( y = 0.030 \) into the first equation: \ \( 4x + 4(0.030) = 0.200 \) \ \( 4x + 0.120 = 0.200 \)- Solve for \( x \): \ \( 4x = 0.080 \) \ \( x = 0.020 \, \text{mol} \)

Step 5: Calculate Mass of Each Reactant

Find the mass of butene and butane:- Molar mass of butene (\( \mathrm{C}_{4} \mathrm{H}_{8} \)) is 56 g/mol. \[ \text{Mass of butene} = 0.020 \, \text{mol} \times 56 \, \text{g/mol} = 1.12 \, \text{g} \]- Molar mass of butane (\( \mathrm{C}_{4} \mathrm{H}_{10} \)) is 58 g/mol. \[ \text{Mass of butane} = 0.030 \, \text{mol} \times 58 \, \text{g/mol} = 1.74 \, \text{g} \]

Step 6: Determine Mass Percentages

Calculate the mass percentages of butene and butane in the mixture:- Total mass of the mixture is 2.86 g.- Mass percent of butene: \[ \frac{1.12 \, \text{g}}{2.86 \, \text{g}} \times 100 = 39.2\% \]- Mass percent of butane: \[ \frac{1.74 \, \text{g}}{2.86 \, \text{g}} \times 100 = 60.8\% \]

Key Concepts

Combustion Reaction

A combustion reaction involves a substance reacting with oxygen to produce heat, light, and products like carbon dioxide (⁠\(\mathrm{CO}_{2}⁠\)) and water (⁠\(\mathrm{H}_{2} \mathrm{O}⁠\)). It’s a common chemical process used in engines and for energy production. In this exercise, both butene (⁠\(\mathrm{C}_{4} \mathrm{H}_{8}⁠\)) and butane(⁠\(\mathrm{C}_{4} \mathrm{H}_{10}⁠\)) undergo combustion.

• For the combustion of butene: \ \( \mathrm{C}_{4} \mathrm{H}_{8} + 6\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 4\mathrm{H}_{2} \mathrm{O} \).
• For the combustion of butane: \( \mathrm{C}_{4} \mathrm{H}_{10} + 6.5\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 5\mathrm{H}_{2} \mathrm{O} \).

These reactions show that both hydrocarbons produce water and carbon dioxide, but their stoichiometry differs. Understanding the difference in stoichiometry is key to analyzing reactions like these, as it determines the amounts of each product formed.
Being able to balance the chemical equation accurately ensures the conservation of mass, indicating that the same number of each type of atom is present before and after the reaction.

Mass Percentage

Mass percentage helps you understand how much of each component is present in a mixture. In this context, it shows us the compositions of butene and butane in the mixture which combusted.
To calculate the mass percentage, use the formula:

• Mass percentage of a component = \( \left( \frac{\text{mass of component}}{\text{total mass of mixture}} \right) \times 100 \).

In the exercise, we use:

• Mass of total mixture: 2.86 g.
• Mass of butene: 1.12 g \( \rightarrow \) Mass percentage: \( \frac{1.12}{2.86} \times 100 = 39.2\% \).
• Mass of butane: 1.74 g \( \rightarrow \) Mass percentage: \( \frac{1.74}{2.86} \times 100 = 60.8\% \).

Mass percentages provide a simple way to communicate the composition of mixtures,making them incredibly useful in various fields, including chemistry, geology, and medicine to name a few.
This calculation shows us the relative contribution of each hydrocarbon in the original mixture. It's insightful for experimental analysis and understanding the properties of the mixture overall.

Mole Calculation

Mole calculation is central in chemistry as it allows for the conversion between the mass of a substance and the number of particles or units. It's crucial for quantitatively analyzing reactions based on balanced chemical equations.
For instance, finding the moles of products involves the formula:

• Moles = \( \frac{\text{mass (in grams)}}{\text{molar mass (in g/mol)}} \).

In the context of the original problem:

• For \(\mathrm{CO}_{2}\): Molar mass is 44 g/mol.Moles = \( \frac{8.80 \text{ g}}{44 \text{ g/mol}} = 0.200 \text{ mol} \).
• For \(\mathrm{H}_{2} \mathrm{O}\):Molar mass is 18 g/mol.Moles = \( \frac{4.14 \text{ g}}{18 \text{ g/mol}} = 0.230 \text{ mol} \).

Understanding mole calculations enables us to translate the mass and amounts of substances in chemical reactions,allowing for precise and meaningful experimentation.
This method supports the stoichiometry of reactions, ensuring the quantitative relationships of substances are maintained and understood, paving the way for deeper chemical comprehension.