Question

A 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 are the mass percentages of butene and butane in the mixture?

Short answer

Butene: 39.2%, Butane: 60.8%

Step-by-step solution

Write the combustion reactions

Butene (C_{4}H_{8}) combusts according to the reaction: \[\text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O}\]Butane (C_{4}H_{10}) combusts according to the reaction: \[\text{C}_4\text{H}_{10} + 6.5\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}\]

Find moles of produced gases

Calculate moles of CO₂: \[\text{Moles of CO}_2 = \frac{8.80 \text{ g}}{44.01 \text{ g/mol}} \approx 0.20 \text{ mol}\]Calculate moles of H₂O: \[\text{Moles of H}_2\text{O} = \frac{4.14 \text{ g}}{18.02 \text{ g/mol}} \approx 0.23 \text{ mol}\]

Set up equations from stoichiometry

Let X be moles of butene combusted and Y be moles of butane combusted. From butene, 4X moles of CO₂ and from butane, 4Y moles of CO₂ form.The equation for CO₂ is: \[4X + 4Y = 0.20\]

Solve the system of equations

From butene, 4X moles of H₂O form, and from butane, 5Y moles of H₂O form. Equation for H₂O is: \[4X + 5Y = 0.23\]Solve:1. \(4X + 4Y = 0.20\)2. \(4X + 5Y = 0.23\)Subtract (1) from (2):\[\begin{align*} 0.23 &- 0.20 = 5Y - 4Y \ 0.03 &= Y \end{align*}\]

Find moles of butene

Substitute Y = 0.03 into equation \(4X + 4Y = 0.20\):\[\begin{align*} 4X + 4(0.03) &= 0.20 \ 4X + 0.12 &= 0.20 \ 4X &= 0.08 \ X &= 0.02 \end{align*}\]

Calculate mass percentages

Calculate the mass of butene: \[\text{Mass of C}_4\text{H}_8 = 0.02 \times 56.11 = 1.12 \text{ g}\]Calculate the mass of butane: \[\text{Mass of C}_4\text{H}_{10} = 0.03 \times 58.12 = 1.74 \text{ g}\]Calculate mass percentages:\[\text{Mass \, \% C}_4\text{H}_8 = \frac{1.12}{2.86} \times 100 \approx 39.2\%%\]\[\text{Mass \, \% C}_4\text{H}_{10} = \frac{1.74}{2.86} \times 100 \approx 60.8\%%\]

Key concepts

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In simpler terms, it helps you understand how much of each substance is involved in a reaction.
In a combustion analysis like this one, stoichiometry is used to determine the exact amounts of materials reacted and produced.
First, you write balanced chemical reactions. For butene, the balanced equation is \[\text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O}\]For butane, it is \[\text{C}_4\text{H}_{10} + 6.5\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}\]The coefficients in the balanced equations tell you the mole ratios of the substances.
The stoichiometry calculations start by determining the amount of products (in moles) produced from a known mass of reactants.
This exercise uses the stoichiometric relationships from the balanced equations to relate the moles of carbon dioxide and water back to the original amounts of butene and butane.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into others. In combustion reactions, a substance reacts with oxygen, releasing energy in the form of heat and producing oxides. Here, both butene and butane undergo combustion in the presence of oxygen.

• The reaction of butene with oxygen produces carbon dioxide and water, with the overall equation: \[\text{C}_4\text{H}_8 + 6\text{O}_2 \rightarrow 4\text{CO}_2 + 4\text{H}_2\text{O}\]
• For butane, the equation is: \[\text{C}_4\text{H}_{10} + 6.5\text{O}_2 \rightarrow 4\text{CO}_2 + 5\text{H}_2\text{O}\]

These reactions are exothermic, meaning they release heat.
Understanding the specifics of these reactions, like the amount of oxygen required and the products formed, is key to solving problems about combustion analysis and calculating composition, as seen in this example.

Mole Concept

The mole is a fundamental concept in chemistry, serving as a bridge between the atomic and macroscopic worlds. One mole of any substance contains Avogadro's number of entities, usually atoms or molecules, which is \(6.022 \times 10^{23}\).
Using the mole allows chemists to count these entities by weighing, making it practical for lab work.
In combustion analysis problems, the mole concept helps convert masses of reactants and products into moles, allowing you to harness stoichiometric ratios.
The conversion formulas are:

• For \(\text{CO}_2\), use: \[\text{Moles of CO}_2 = \frac{\text{mass of CO}_2}{44.01 \, \text{g/mol}}\]
• For \(\text{H}_2\text{O}\), use: \[\text{Moles of H}_2\text{O} = \frac{\text{mass of H}_2\text{O}}{18.02 \, \text{g/mol}}\]

By applying the mole concept, you can determine the exact amounts of each component in a reaction, helping to solve for unknowns such as mass percentages in mixtures.