Question

A 9.780 -g gaseous mixture contains ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .\) Complete combustion to form carbon dioxide and water requires 1.120 \(\mathrm{mole}\) of oxygen gas. Calculate the mass percent of ethane in the original mixture.

Short answer

The mass percent of ethane in the original mixture is found to be 67.63%.

Step-by-step solution

1. Write down the balanced chemical equations for the combustion of ethane and propane

For the complete combustion of ethane and propane, the balanced chemical equations are: Ethane combustion: $$\mathrm{C}_{2} \mathrm{H}_{6} + \frac{7}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2} + 3 \mathrm{H}_{2} \mathrm{O}$$ Propane combustion: $$\mathrm{C}_{3} \mathrm{H}_{8} + 5\mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2} + 4 \mathrm{H}_{2} \mathrm{O}$$

2. Determine moles of ethane and propane in the mixture

Let x moles of ethane and y moles of propane be present in the mixture. According to the given data, the moles of oxygen required for complete combustion of both ethane and propane is 1.120 moles. Based on the balanced chemical equations, we can set up the following equation: $$\frac{7}{2}x + 5y = 1.120$$

3. Find the mass of ethane and propane in the mixture

The given mixture has a mass of 9.780 g. We know the molar masses of ethane and propane, which are 30.07 g/mol and 44.10 g/mol, respectively. If x moles of ethane and y moles of propane are present in the mixture, their total mass can be represented by: $$30.07x + 44.10y = 9.780$$

4. Solve the system of equations to find the moles of ethane and propane

We now have a system of two equations with two variables: $$\frac{7}{2}x + 5y = 1.120$$ $$30.07x + 44.10y = 9.780$$ Solve this system of equations to find the values of x and y. You may use the method of substitution or elimination. Using substitution, we can solve for x in terms of y in the second equation: $$x = \frac{9.780 - 44.10y}{30.07}$$ Now substitute this expression for x into the first equation: $$\frac{7}{2}\left(\frac{9.780 - 44.10y}{30.07}\right) + 5y = 1.120$$ Solve for y to get: $$y = 0.120$$ To find x, substitute the value of y back into the expression for x: $$x = \frac{9.780 - 44.10\cdot 0.120}{30.07}$$ $$x = 0.220$$ Now we know that the mixture contains 0.220 moles of ethane and 0.120 moles of propane.

5. Calculate the mass percent of ethane in the mixture

To find the mass percent of ethane, first find the mass of ethane in the mixture by multiplying the moles of ethane by its molar mass: Mass of ethane = 0.220 moles * 30.07 g/mol = 6.6154 g Next, divide the mass of ethane by the total mass of the mixture and multiply by 100 to obtain the mass percent: Mass percent of ethane = (6.6154 g / 9.780 g) * 100 = 67.63% Hence, the mass percent of ethane in the original mixture is 67.63%.

Key concepts

Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance reacts with oxygen gas to release energy, usually in the form of heat and light. These reactions are vital in everyday life, as they are responsible for processes like burning fuel in car engines or wood in fireplaces.
In our exercise, the combustion involves hydrocarbons, specifically ethane and propane, which react with oxygen to form carbon dioxide and water. This is a typical result of complete combustion reactions, where the hydrocarbons are fully oxidized.
These reactions can be represented by balanced chemical equations, which show the exact proportions of reactants and products involved. For example, ethane combusts as: - Ethane: \( \mathrm{C}_2\mathrm{H}_6 + \frac{7}{2}\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2\mathrm{O} \)- Propane: \( \mathrm{C}_3\mathrm{H}_8 + 5\mathrm{O}_2 \rightarrow 3\mathrm{CO}_2 + 4\mathrm{H}_2\mathrm{O} \) Understanding these equations is essential because they provide the foundation for further calculations in stoichiometry, enabling us to determine how much reactant is needed and what products are formed.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. Essentially, it is the math behind chemistry, allowing us to calculate how much of each substance participates in a reaction.
In the given exercise, stoichiometry helps us determine the moles of ethane and propane that make up the mixture. We start with the balanced equations for the combustion of these gases and use them to understand their consumption of oxygen.
By setting up equations based on the number of moles of oxygen required for combustion, we can solve for the unknowns - the moles of ethane ( - 0.220 ) and propane ( - 0.120 ). This step is crucial because it bridges the information given to the mass of substances involved, further enabling us to calculate the mixture's properties. Mastering stoichiometry facilitates our understanding of reactions in both qualitative and quantitative ways.

Mass Percent Calculation

The mass percent of a component in a mixture is a measure of its concentration, expressed as the ratio of its mass to the total mass of the mixture, multiplied by 100. It's a useful way to express how much of a particular substance is in a compound.
In our case, we need the mass percent of ethane in the mixture. After finding the number of moles of ethane, we calculate its mass using its molar mass (30.07 g/mol). This gives us the mass of ethane in the mixture: - Mass of ethane = 0.220 moles * 30.07 g/mol = 6.6154 g.
By dividing this mass by the total mass of the mixture (9.780 g) and multiplying by 100, we find the mass percent: - Mass percent = (6.6154 g / 9.780 g) * 100 = 67.63% This result tells us that ethane makes up approximately 67.63% of the original mixture by mass. Mass percent is a straightforward way to express how significant a particular substance is in a mixture, simplifying comparisons and calculations.

Chemical Equations

Chemical equations provide a representation of chemical reactions, showing reactants changing into products with their respective quantities, in moles. Each element is balanced on both sides of the equation, maintaining the law of conservation of mass.
For ethane and propane combustion in our exercise, the chemical equations are a guide to understanding what transpires at the molecular level. They show that: - Ethane and propane - react with a certain amount of oxygen to produce carbon dioxide and water.
These balanced equations are essential beyond merely representing a reaction: they allow us to predict the outcomes of reactions in terms of the amount of reactants consumed and products formed, as is true for ethane and propane combustion. By understanding and using chemical equations, we gain a fundamental insight into the stoichiometric and quantitative aspects of chemical reactions. This mastery is critical for accurately conducting and predicting the results of chemical processes, whether in a laboratory or industrial setting.