Question

The manufacture of ethyl alcohol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) yields diethyl ether, \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) as a by-product. The complete combustion of a \(1.005 \mathrm{g}\) sample of the product of this process yields \(1.963 \mathrm{g} \mathrm{CO}_{2} .\) What must be the mass percents of \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) and of \(\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}\) in this sample?

Short answer

The mass percents of \(CH_3CH_2OH\) and \((C_2H_5)_2O\) in the given sample should be equivalent to the carbon mass percentage in the sample, calculated by using the mass of carbon obtained from the combustion process.

Step-by-step solution

Calculate moles of carbon from obtained carbon dioxide

The mass of carbon dioxide obtained is \(1.963 g\). The molar mass of carbon dioxide is \(\(44.01 g/mol \)\). By dividing the mass of carbon dioxide by its molar mass, the number of moles of carbon can be calculated using the equation: \(\number of moles = mass/molar mass\)

Calculate mass of carbon in the original sample

The obtained moles in Step 1 are the moles of carbon in the sample. Since every mole of carbon has a mass of \(12 g\), the mass of carbon in the original sample can be determined by multiplying the moles of carbon obtained in Step 1 by \(12 g/mol\)

Calculate the mass percent of ethyl alcohol and diethyl ether

Essentially, both ethyl alcohol (\(CH_3CH_2OH\)) and diethyl ether (\((C_2H_5)_2O\)) have the same proportion of carbon. Hence, the carbon mass percentage in the original sample should be equal to the carbon mass percentage in each compound. The mass percentage can be calculated by dividing the mass of carbon in the original sample by the total mass of the sample, then multiplied by 100. Mass percentages of both compounds in the sample can thus be obtained.

Key concepts

Ethyl Alcohol Combustion

When ethyl alcohol, also known as ethanol, undergoes combustion, it reacts with oxygen to produce carbon dioxide and water. This is a common reaction and is exothermic, meaning that it releases heat. The general equation for the combustion of ethanol is:

• \[C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\]

This equation shows that each molecule of ethyl alcohol combusted produces two molecules of carbon dioxide. To analyze the chemical composition of a mixture that includes ethanol, capturing the amount of produced carbon dioxide is crucial. This helps in calculating the mass percentage of ethanol within the mixture by understanding how CH bonds in ethanol are converted into CO bonds in carbon dioxide during combustion.

Diethyl Ether Production

Diethyl ether is often produced as a by-product when creating ethyl alcohol. The general reaction involves the dehydration of ethanol, where water is removed, forming an ether linkage. This reaction can be expressed as:

• \[2C_2H_5OH \xrightarrow[H_2SO_4]{heat} (C_2H_5)_2O + H_2O\]

Here, two ethanol molecules combine to produce one diethyl ether and a molecule of water. This equilibrium reaction can lead to mixtures containing both diethyl ether and unreacted ethanol. Analyzing and determining the proportions of these substances in a given sample is often necessary to judge the efficiency of the production process and to resolve the composition of the mixture.

Mass Percent Calculation

Mass percent is a way to express the concentration of a component in a mixture. It shows how much of the total mass is contributed by the element or compound in question. The formula for calculating mass percent is:

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

In the context of our problem, after determining the mass of carbon from the produced carbon dioxide, we use this to calculate the mass percent of ethyl alcohol and diethyl ether in the sample. Since both compounds have similar carbon signatures, understanding their contributions through accurate calculations helps determine the composition effectively. This is a crucial step in chemical analysis, allowing us to understand the distribution of different ingredients in an experimental or industrial setup.