Problem 145 - General Chemistry: Principles and Modern Applications

Question

A \(1.000 \mathrm{g}\) sample of a mixture of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) is analyzed by burning it completely in \(\mathrm{O}_{2}\), yielding \(2.776 \mathrm{g} \mathrm{CO}_{2} .\) What is the percentage by mass of \(\mathrm{CH}_{4}\) in the mixture? (a) \(93 \% ;\) (b) \(82 \% ;\) (c) \(67 \% ;\) (d) \(36 \%\) (e) less than \(36 \%\)

Step-by-Step Solution

Answer

The percentage mass of methane (\(\mathrm{CH}_{4}\)) obtained after solving the equations is 82%. Thus, the answer is (b) 82%.

Step 1 Calculation of total mass of carbon in CO2

The subscript beside the C in \(\mathrm{CO}_{2}\) indicates that one molecule of \(\mathrm{CO}_{2}\) contains one atom of carbon. Therefore, the mass of carbon in 2.776g of \(\mathrm{CO}_{2}\) is equal to (mass of \(\mathrm{CO}_{2}\) * ratio of atomic mass of carbon to molecular mass of \(\mathrm{CO}_{2}\)). Which can be found as \(2.776g * \frac{12.011 g/mol}{44.010 g/mol}\ = 0.757 g.\)

Step 2 Calculation of the mass of carbon in methane and ethane

The molecule \(\mathrm{CH}_{4}\) contains 1 atom of carbon and \(\mathrm{C}_{2} \mathrm{H}_{6}\) contains 2 atoms of carbon. We can therefore write 2 equations based on this, where x represents the mass of \(\mathrm{CH}_{4}\) in the mixture and y represents the mass \(\mathrm{C}_{2} \mathrm{H}_{6}\) in the mixture. Equation 1: \(x + y = 1.000g\) (from the total mass of the mixture). Equation 2: \(x * ( \frac{12.011 g/mol}{16.043 g/mol}) + y * (2 * \frac{12.011 g/mol}{30.069 g/mol}) = 0.757g\) (from the total mass of carbon). Solve the simultaneous equations for x, e.g., by substitution or elimination method.

Step 3 Determination of the mass percentage

We can find the percentage mass of methane (\(\mathrm{CH}_{4}\)) by the formula \(( \frac{mass of methane}{mass of total mixture})*100\), where the mass of methane equals to the number found in step 2, and the mass of total mixture equals to 1.000g.

Key Concepts

Chemical Stoichiometry

Chemical stoichiometry is integral to understanding how elements and compounds react with one another in set proportions. It is rooted in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Thus, the quantity of each element must be the same in the reactants as in the products.

To discuss stoichiometry, we first need to know about moles, which are the standard unit for counting the number of molecules or atoms in a sample. Stoichiometry calculations usually involve balanced chemical equations, where coefficients represent the relative amounts of reactants and products. For example, if you're given a chemical equation like \( a \textnormal{A} + b \textnormal{B} \rightarrow c \textnormal{C} \), the coefficients \(a\), \(b\), and \(c\) tell you the proportional amounts of compound A, B, and product C that react and form.

Using stoichiometry, we can calculate the amount of reactants needed to produce a certain amount of product or the amount of product formed from certain reactants. It is fundamental in ensuring the right proportions are used for reactions to proceed efficiently.

Combustion Analysis

Combustion analysis is a laboratory method used to determine the elemental composition of a substance by completely burning it. When organic compounds, which are made primarily of carbon and hydrogen, combust in the presence of oxygen, they form carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2O\)). By measuring the amount of \(\mathrm{CO}_2\) and \(\mathrm{H}_2O\) produced, we can deduce the amount of carbon and hydrogen in the original compound.

In the given exercise, a mixture of \(\mathrm{CH}_4\) (methane) and \(\mathrm{C}_2\mathrm{H}_6\) (ethane) is burned, and the mass of \(\mathrm{CO}_2\) is measured to determine the amount of carbon. This data, along with the law of conservation of mass, allows us to perform the stoichiometric calculations necessary to solve for the percentage mass of methane in the mixture. The problem, therefore, represents a practical application of combustion analysis to find the percentage composition of an unknown mixture using stoichiometry.

Percentage Composition

Percentage composition is a measure of the relative amount of each element within a compound or a mixture. It is calculated as the mass of a particular element divided by the total mass of the compound or mixture, multiplied by 100. This concept is crucial in fields such as chemistry and material science, where it's important to know the precise makeup of substances.

Referring to our exercise, the percentage by mass of \(\mathrm{CH}_4\) in the mixture can be found using the formula: \[\textnormal{Percentage mass of } \mathrm{CH}_4 = \left( \frac{\textnormal{mass of } \mathrm{CH}_4}{\textnormal{mass of total mixture}} \right) \times 100\]
From step 2 of the solution, we determine the mass of \(\mathrm{CH}_4\) (represented as 'x') within the mixture. Knowing the total mass of the mixture (1.000 g), we can then calculate the percentage composition. By mastering this concept, students can better understand how to determine the purity of compounds or the concentrations of different substances in mixtures.