Problem 49 - General Chemistry: Principles and Modern Applications

Question

A \(1.562 \mathrm{g}\) sample of the alcohol \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}\) is burned in an excess of oxygen. What masses of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) should be obtained?

Step-by-Step Solution

Answer

The masses of CO2 and H2O that should be obtained are approximately 3.43 g and 1.87 g respectively.

Step 1: Balancing the reaction

The combustion reaction of the alcohol CH3CHOHCH3 in oxygen would give carbon dioxide (CO2) and water (H2O). Therefore, the unbalanced reaction would be: CH3CHOHCH3 + O2 → CO2 + H2O. The balanced chemical equation is: 2CH3CHOHCH3 + 9O2 → 6CO2 + 8H2O.

Step 2: Convert mass of alcohol to moles

From the periodic table, the molar mass of CH3CHOHCH3 (using average atomic masses) is approximately 60.095 g/mol. Therefore, the number of moles of CH3CHOHCH3 burned is \(1.562 \mathrm{g} / 60.095 \mathrm{g/mol} = 0.026 \mathrm{mol}\).

Step 3: Using stoichiometry to find moles of products

From the balanced equation, it can be seen that for every 2 moles of CH3CHOHCH3 combusted, 6 moles of CO2 and 8 moles of H2O are produced. So, the moles of CO2 produced is \(0.026 \mathrm{mol} CH3CHOHCH3 × (6 \mathrm{mol} CO2 / 2 \mathrm{mol} CH3CHOHCH3) = 0.078 \mathrm{mol} CO2\). Likewise, the moles of H2O produced is \(0.026 \mathrm{mol} CH3CHOHCH3 × (8 \mathrm{mol} H2O / 2 \mathrm{mol} CH3CHOHCH3) = 0.104 \mathrm{mol} H2O\).

Step 4: Convert moles of products to mass

Using the molar masses of CO2 and H2O (approximately 44.01 g/mol and 18.015 g/mol respectively), the mass of CO2 and H2O produced are \(0.078 \mathrm{mol} CO2 × 44.01 \mathrm{g/mol} = 3.43 \mathrm{g}\) and \(0.104 \mathrm{mol} H2O × 18.015 \mathrm{g/mol} = 1.87 \mathrm{g}\) respectively.

Key Concepts

Combustion Reaction

In chemistry, a combustion reaction is a specific type of chemical reaction where a substance reacts with oxygen to produce heat and light.
It usually involves organic compounds like hydrocarbons or alcohols, with products typically being carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)).
For instance, when the alcohol \(\mathrm{CH}_3\mathrm{CHOHCH}_2\mathrm{CH}_3\) is burned, it reacts with excess oxygen.
This results in the formation of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\).

• The standard form of any combustion reaction is: Fuel + \(\mathrm{O}_2\) → \(\mathrm{CO}_2\) + \(\mathrm{H}_2\mathrm{O}\).
• The combustion must have sufficient oxygen, referred to as excess oxygen, to complete the reaction thoroughly without producing any monoxide gases.

Understanding combustion reactions is essential, especially when dealing with energy production and understanding environmental impacts.

Chemical Equation Balancing

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is observed.
That means there must be an equal number of each type of atom on both sides of the equation.
With our alcohol combustion: initially, we have:\[\mathrm{CH}_3\mathrm{CHOHCH}_2\mathrm{CH}_3 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}.\]To balance this, you adjust coefficients in front of each compound:

• Given: \(2\mathrm{CH}_3\mathrm{CHOHCH}_2\mathrm{CH}_3 + 9\mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}\).
• Every element (C, H, O) has a matching count across reactants and products.

Balancing involves trial and error, so patience and systematic checking ensure successful balancing.

Molar Mass

Molar mass is the mass of one mole of a substance, expressed in grams per mole (\(g/mol\)).
It's derived from the atomic masses of the elements as stated in the periodic table.
For \(\mathrm{CH}_3\mathrm{CHOHCH}_2\mathrm{CH}_3\) (an alcohol):

• Carbon (C): Atomic mass ≈ 12.01 \(g/mol \times 4 = 48.04 g/mol\).
• Hydrogen (H): Atomic mass ≈ 1.008 \(g/mol \times 10 = 10.08 g/mol\).
• Oxygen (O): Atomic mass ≈ 16.00 \(g/mol \times 1 = 16.00 g/mol\).

Adding these gives a molar mass of precisely around 60.095 \(g/mol\).Knowing molar mass is critical for converting between mass and moles, aiding in stoichiometric calculations.

Mole Conversion

Mole conversion is a key chemistry concept that allows conversion between mass and moles using the formula: \[\text{number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\].
For our alcohol: \(1.562\mathrm{g}\) is divided by its molar mass, 60.095 \(g/mol\), giving about \(0.026\text{ mol}\).
This conversion allows further stoichiometric calculations, such as determining the amounts of \(\mathrm{CO}_2\) and \(\mathrm{H}_2\mathrm{O}\) produced.

• Use mole conversion to link reactant and product amounts in a reaction.
• This reflects how molecules interact in given conditions.

Mastery of mole conversion is quintessential for both practical and theoretical chemistry applications.