Problem 113 - General Chemistry: Principles and Modern Applications

Question

An organic liquid is either methyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) ethyl alcohol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right),\) or a mixture of the two. A 0.220-g sample of the liquid is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) and yields \(0.352 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}) .\) Is the liquid a pure alcohol or a mixture of the two?

Step-by-Step Solution

Answer

The liquid is a mixture of methyl alcohol and ethyl alcohol.

Step 1: Write down the balanced chemical equations for the combustion of methyl alcohol and ethyl alcohol

Balanced chemical equation for the combustion of methyl alcohol: \(2 CH_3OH + 3 O_2 \rightarrow 2 CO_2 + 4 H_2O\) Balanced chemical equation for the combustion of ethyl alcohol: \(C_2H_5OH + 3 O_2 \rightarrow 2 CO_2 + 3 H_2O\)

Step 2: Calculate the molar mass of CO2, methyl alcohol and ethyl alcohol

The molar mass of CO2 = 12.01 g/mol (for C) + 2*16.00 g/mol (for 2 O) = 44.01 g/mol The molar mass of methyl alcohol (CH3OH) = 12.01 g/mol (for C) + 3*1.01 g/mol (for 3 H) + 16.00 g/mol (for O)+ 1.01 g/mol (for H) = 32.04 g/mol The molar mass of ethyl alcohol (C2H5OH) =2*12.01 g/mol (for 2 C) + 5*1.01 g/mol (for 5 H) + 16.00 g/mol (for O) + 1.01 g/mol (for H) = 46.07 g/mol

Step 3: Determine the moles of CO2 and the alcohol on combustion

The moles of CO2 formed = mass of CO2/molar mass = 0.352 g/44.01 g/mol = 0.008 mol The moles of methyl alcohol based on the balanced equation for its combustion = moles of CO2 / 2 = 0.008 mol / 2 = 0.004 mol The moles of ethyl alcohol based on the balanced equation for its combustion = moles of CO2 / 2 = 0.008 mol / 2 = 0.004 mol

Step 4: Compare the calculated moles of the alcohol with the moles of the alcohol in the sample

To get the moles of the alcohol in the sample, we divide the sample mass by the molar mass: If the sample was methyl alcohol, the moles would be sample mass/molar mass= 0.220 g/32.04 g/mol = 0.007mol. If the sample was ethyl alcohol, the moles would be sample mass/molar mass = 0.220 g/46.07 g/mol = 0.005 mol. Comparing these values to the moles of alcohol on combustion, it can be seen that neither match. Thus, the liquid is a mixture of the two alcohols.

Key Concepts

Combustion Reactions

In organic chemistry, combustion reactions are vital, showcasing how substances, like alcohols, react with oxygen to produce carbon dioxide and water. When methyl alcohol (CH₃OH) and ethyl alcohol (C₂H₅OH) undergo combustion, they follow distinct reactions, releasing energy and forming new compounds. For example:

• Methyl alcohol combustion: 2 CH₃OH + 3 O₂ → 2 CO₂ + 4 H₂O.
• Ethyl alcohol combustion: C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O.

When either alcohol burns, it converts into the same byproducts, CO₂ and H₂O, but follows different stoichiometric pathways. Recognizing balanced chemical equations is crucial for predicting amounts of products formed. It's integral to calculate the reactants' and products' moles correctly, which helps determine the initial sample substance's nature when unknown, as seen in combustion reactions of mixtures, like the sample problem above.

Molar Mass Calculations

Molar mass calculations serve as a foundation in chemical analysis, providing essential data for measuring reactants and products. To determine an unknown compound or validate a mixture, molar masses of the substances involved are necessary.
For instance, to calculate the molar mass:

• The molar mass of CO₂: 12.01 g/mol (C) + 2 × 16.00 g/mol (O) = 44.01 g/mol.
• Methyl alcohol (CH₃OH): 12.01 g/mol (C) + 3 × 1.01 g/mol (H) + 16.00 g/mol (O) = 32.04 g/mol.
• Ethyl alcohol (C₂H₅OH): 2 × 12.01 g/mol (C) + 5 × 1.01 g/mol (H) + 16.00 g/mol (O) = 46.07 g/mol.

Using these calculations, we can analyze the moles of each reactant or product, compare them with actual measured values, and deduce the substance in question.
This method allows chemistry students to form connections between theoretical chemical equations and real-world laboratory data, aiding in identification tasks, like determining alcohol mixtures.

Alcohols in Chemistry

Alcohols play an essential role in chemistry due to their unique properties and roles in various reactions, including combustion. They are organic compounds featuring a hydroxyl group (–OH) attached to a carbon atom.
In the realm of chemistry, methyl and ethyl alcohols are prime examples:

• Methyl alcohol (CH₃OH), also known as methanol, is a simple alcohol used as a solvent, antifreeze, and as an industrial precursor.
• Ethyl alcohol (C₂H₅OH), or ethanol, serves as a recreational substance in beverages and has applications in thermometers and as a vehicle fuel additive.

Understanding alcohols’ structure aids in predicting their behavior in chemical processes, like combustion. Their chemical combative actions make them crucial in analytical techniques.
The ability to analyze reactions, consider chemical compositions and their implications helps students grasp the practical applications and theoretical underpinnings of chemistry, essential for tackling complex mixtures or tasks like the given exercise.