Problem 140 - Chemistry and Chemical Reactivity

Question

Which of the following statements about \(57.1 \mathrm{g}\) of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) is (are) not true? (a) \(57.1 \mathrm{g}\) is 0.500 mol of octane. (b) The compound is \(84.1 \%\) C by weight. (c) The empirical formula of the compound is \(\mathbf{C}_{4} \mathbf{H}_{9}\) (d) \(57.1 \mathrm{g}\) of octane contains \(28.0 \mathrm{g}\) of hydrogen atoms.

Step-by-Step Solution

Answer

Statement (d) is not true.

Step 1: Calculate moles from grams

To determine if statement (a) is true, we need to calculate the number of moles of octane. The molar mass of octane (\( \mathrm{C}_{8} \mathrm{H}_{18} \)) is calculated as follows: \(8 \times 12.01 \text{ g/mol} + 18 \times 1.01 \text{ g/mol} = 114.26 \text{ g/mol}. \) Then, the number of moles in 57.1 g of octane is \( \frac{57.1}{114.26} \approx 0.500 \text{ mol}. \) Therefore, statement (a) is true.

Step 2: Evaluate percentage composition

To assess statement (b), calculate the percentage of carbon in octane. The mass of carbon in octane is \( 8 \times 12.01 = 96.08 \) g. The percentage composition is calculated as \( \left( \frac{96.08}{114.26} \right) \times 100 \approx 84.1\%. \) Therefore, statement (b) is true.

Step 3: Empirical formula of the compound

For statement (c), determine the empirical formula of \( \mathrm{C}_{8} \mathrm{H}_{18} \). Simplifying the subscripts by dividing by 2, we have \( \mathbf{C}_{4} \mathbf{H}_{9} \). Therefore, the empirical formula is \( \mathbf{C}_{4} \mathbf{H}_{9} \), so statement (c) is also true.

Step 4: Calculate mass of hydrogen in octane

To verify statement (d), calculate the mass of hydrogen in 57.1 g of octane. The percentage of hydrogen in octane is \( \frac{18 \times 1.01}{114.26} \times 100 \approx 15.9\%. \) Therefore, the mass of hydrogen in 57.1 g of octane is \( 57.1 \times 0.159 \approx 9.1 \) g. Statement (d) is false as the mass is not 28.0 g.

Key Concepts

Molar Mass Calculations

Understanding molar mass is essential when dealing with chemical substances. It's the weight of one mole of a given compound and is expressed in grams per mole (g/mol).
To find the molar mass of a compound like octane (\(\mathrm{C}_{8} \mathrm{H}_{18}\)), you add the atomic masses of each element in the molecule multiplied by their quantities available in the formula.For octane:

• Carbon (C) has an atomic mass of 12.01 g/mol, multiplied by 8 because there are eight carbon atoms. This gives us 96.08 g/mol.
• Hydrogen (H) has an atomic mass of 1.01 g/mol, multiplied by 18 due to eighteen hydrogen atoms. That equates to 18.18 g/mol.

Adding these together, we get the molar mass of octane as 114.26 g/mol.
This is a critical step, as it helps convert grams to moles and vice versa, facilitating calculations in chemical reactions.

Percentage Composition

Percentage composition provides insight into the elemental make-up of a compound. It's the percentage by mass of each element in a compound.
This is calculated by dividing the mass of each element in one mole of the compound by the total molar mass of the compound, then multiplying by 100.For octane,

• The percentage of carbon is calculated as: \(\left(\frac{96.08}{114.26}\right) \times 100 \approx 84.1\%\).
• The percentage of hydrogen is: \(\left(\frac{18.18}{114.26}\right) \times 100 \approx 15.9\%\).

These percentages tell us the fraction of the compound's mass contributed by each element, which is valuable in understanding both the qualitative and quantitative aspects of the chemical structure.

Stoichiometry

Stoichiometry is a concept used to calculate the relationships between reactants and products in chemical equations.
It's based on the principles of the conservation of mass and the definite proportions of elements in a compound.The empirical formula calculation touches on stoichiometry:

• In our example, the molecular formula for octane is \(\mathrm{C}_{8} \mathrm{H}_{18}\). By dividing the subscripts by the greatest common factor, in this case, 2, we achieve the empirical formula \(\mathrm{C}_{4} \mathrm{H}_{9}\).

Empirical formulas represent the simplest whole-number ratio of the elements in a compound.
This aspect of stoichiometry helps chemists infer the right amounts of each element needed to replicate or react specific materials, thereby optimizing reactions and making them more resource-efficient.