Question

A mixture of methane \(\left(\mathrm{CH}_{4}\right)\) and ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) of mass \(13.43 \mathrm{~g}\) is completely burned in oxygen. If the total mass of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(64.84 \mathrm{~g},\) calculate the fraction of \(\mathrm{CH}_{4}\) in the mixture.

Short answer

The fraction of \(\mathrm{CH}_4\) in the mixture is approximately 0.4.

Step-by-step solution

Understand the Reaction

When methane (\(\text{CH}_4\)) and ethane (\(\text{C}_2\text{H}_6\)) burn, they react with oxygen to form carbon dioxide (\text{CO}_2) and water (\text{H}_2\text{O}). The reactions are:\[\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}\]\[\text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O}\]

Define Variables

Let's denote the mass of methane as \(m_{\text{CH}_4}\) and the mass of ethane as \(m_{\text{C}_2\text{H}_6}\). We know that:\[m_{\text{CH}_4} + m_{\text{C}_2\text{H}_6} = 13.43 \, \text{g}\]

Establish Relationship for Products

The masses of the products are given as \(64.84 \, \text{g}\) in total. For each mole of \(\text{CH}_4\), 1 mole of \(\text{CO}_2\) and 2 moles of \(\text{H}_2\text{O}\) are produced. For each mole of \(\text{C}_2\text{H}_6\), 2 moles of \(\text{CO}_2\) and 3 moles of \(\text{H}_2\text{O}\) are produced. The equations help us relate the mole quantities to the masses of the products.

Establish Equation Based on Products

For \(\text{CH}_4\), the molar masses are: \(\text{CH}_4: 16.04 \, \text{g/mol}, \text{CO}_2: 44.01 \, \text{g/mol}, \text{H}_2\text{O}: 18.02 \, \text{g/mol}\). For \(\text{C}_2\text{H}_6\), the molar masses are: \(\text{C}_2\text{H}_6: 30.07 \, \text{g/mol}, \text{CO}_2: 44.01 \, \text{g/mol}, \text{H}_2\text{O}: 18.02 \, \text{g/mol}\). Using these, write the mass balance: \[\frac{44.01}{16.04} \times m_{\text{CH}_4} + \frac{44.01 \times 2 + 18.02 \times 3}{30.07} \times m_{\text{C}_2\text{H}_6} = 64.84 \]

Solve the Equations

Using equations from steps 2 and 4, solve the system of equations:1. \(m_{\text{CH}_4} + m_{\text{C}_2\text{H}_6} = 13.43\)2. \(\frac{44.01}{16.04} \times m_{\text{CH}_4} + \frac{44.01 \times 2 + 18.02 \times 3}{30.07} \times m_{\text{C}_2\text{H}_6} = 64.84\)Solve these to find \(m_{\text{CH}_4}\) and \(m_{\text{C}_2\text{H}_6}\).

Calculate the Fraction of CH4

Find the fraction of methane in the mixture using:\[\text{Fraction of } \text{CH}_4 = \frac{m_{\text{CH}_4}}{13.43}\]

Conclusion

Compute the value using the solutions from the equations to determine the fraction of methane.

Key concepts

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships in chemical reactions. It helps chemists calculate how much of each reactant is needed and how much product will be formed. To understand stoichiometry in combustion reactions, such as the burning of methane and ethane, you need to comprehend the balanced equation and the ratios of reactants to products.
For example, when methane (CH\(_4\)) burns, it reacts with oxygen to produce carbon dioxide and water. In this case, each molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water. This ratio is crucial for understanding how much oxygen is needed and how much carbon dioxide and water are formed.
In the exercise where methane and ethane are burned, the stoichiometry of each reaction helps us determine the mass of carbon dioxide and water produced. This is essential for calculating the fraction of each gas in a mixture after combustion.

Chemical Equations

Chemical equations are a concise way of expressing chemical reactions. They show the reactants (substances you start with) and the products (substances formed by the reaction). The equation is balanced if the number of each type of atom on both sides of the equation is the same. This is a reflection of the law of conservation of mass, which states that matter cannot be created or destroyed in an isolated system.
In the combustion of methane and ethane, two balanced chemical equations are employed:

• For methane: \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)
• For ethane: \(\text{C}_2\text{H}_6 + \frac{7}{2}\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \)

These equations not only help to understand what products are formed but also how many molecules of each product are created per reactant molecule. This is fundamental to work out the compositions of gas mixtures after the reaction has happened.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It allows us to convert between the mass of a substance and the amount in moles, which is crucial for stoichiometric calculations.
To calculate the products from the combustion of methane and ethane, we need the molar masses of each substance:

• Methane (CH\(_4\)): 16.04 g/mol
• Ethane (C\(_2\)H\(_6\)): 30.07 g/mol
• Carbon dioxide (CO\(_2\)): 44.01 g/mol
• Water (H\(_2\)O): 18.02 g/mol

Using these molar masses, we convert the mass of each reactant to moles, apply the stoichiometric ratios from the chemical equations, and then convert the moles of products back to mass. This helps us understand how the masses relate to each other throughout the reaction.

Gas Mixtures

When dealing with reactions involving gases, it's common to end up with a gas mixture at the end. Understanding the composition of gas mixtures is essential in many real-world applications, from environmental science to industrial chemistry.
In the given problem, methane and ethane are burned together, resulting in a mixture of gaseous products — carbon dioxide and water vapor. By establishing the mass of these products from the stoichiometric calculations, we can infer the mass and thus calculate the fraction of initial reactants in the mixture.
A gas mixture's composition can be understood in terms of partial pressures or mass fractions. Mass fractions in this context help us determine the proportion of each original gas in the mixture before combustion, enhancing our understanding of the source gases' behavior in chemical processes.