Problem 102 - 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 \(\mathrm{C}_{4} \mathrm{H}_{9}\) (d) 57.1 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 of Octane

To determine if 57.1 g of octane is equivalent to 0.500 moles, we start by calculating the molar mass of octane. The molecular formula for octane is \(C_8H_{18}\). Calculate its molar mass:\[8 \times 12.01 + 18 \times 1.01 = 114.22 \, \text{g/mol}.\]Now, convert grams to moles:\[\frac{57.1}{114.22} \approx 0.500 \, \text{mol}.\]Thus, statement (a) is true.

Step 2: Verify Percentage Composition of Carbon

Next, calculate the percentage of carbon in octane. The mass of carbon in the molecular formula is:\[8 \times 12.01 = 96.08 \, \text{g}.\]The percentage by mass of carbon is calculated as:\[\left(\frac{96.08}{114.22}\right) \times 100\% \approx 84.1\%.\]Thus, statement (b) is true.

Step 3: Determine the Empirical Formula

The empirical formula is the simplest whole-number ratio of elements in the molecule. Divide the number of atoms in the molecular formula \(C_8H_{18}\) by their greatest common factor, which is 2:\[C_4H_9.\]Thus, statement (c) is true.

Step 4: Calculate Mass of Hydrogen in 57.1 g of Octane

Determine the mass of hydrogen in octane using the molecular formula weight of hydrogen:\[18 \times 1.01 = 18.18 \, \text{g}.\]Calculate the mass of hydrogen in 57.1 g of octane:\[\left(\frac{18.18}{114.22}\right) \times 57.1 \approx 9.08 \, \text{g}.\]Since 9.08 g ≠ 28.0 g, statement (d) is false.

Key Concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms present in a compound. It is an essential concept in chemistry, helping us understand the basic composition of molecules.
To determine the empirical formula, consider a compound's molecular formula. For instance, octane's molecular formula is \( C_8H_{18} \). To derive the empirical formula, find the greatest common divisor for the number of moles of each element.
In the case of octane:

• Carbon has 8 atoms per molecule.
• Hydrogen has 18 atoms per molecule.

Divide these numbers by their greatest common factor, which is 2, giving us the empirical formula \( C_4H_9 \).
This simplified representation helps in comparing different compounds and understanding their formulas without unnecessary repetition of elemental symbols.

Percentage Composition

The percentage composition shows what fraction of a compound's mass is made up by each constituent element.
Here's how you calculate it:
First, find the total molar mass of the compound. For octane \( C_8H_{18} \), it is calculated as:

• Carbon contributes \( 8 \times 12.01 = 96.08 \) g/mol.
• Hydrogen contributes \( 18 \times 1.01 = 18.18 \) g/mol.
• Total molar mass = 114.22 g/mol.

Now, to find the percentage composition of carbon:

• Calculate as \( \left(\frac{96.08}{114.22}\right) \times 100\% \approx 84.1\% \).

This value tells us that approximately 84.1% of octane's mass is due to carbon. Calculating percentage composition is an effective way to determine how much of each element is present in a compound, aiding in both chemical analysis and synthesis.

Moles Calculation

Moles are a fundamental unit in chemistry that allows us to quantify the amount of substance present. Understanding how to calculate moles is crucial for stoichiometry and chemical reactions.
To convert grams to moles, use the formula:
\[ \text{Moles} = \frac{\text{mass in grams}}{\text{molar mass}} \]
For octane \( C_8H_{18} \):
First, determine the molar mass:

• Calculated as 114.22 g/mol.

Now, convert 57.1 g of octane to moles by:

• \( \frac{57.1}{114.22} \approx 0.500 \text{ mol} \)

This conversion shows how much of octane you have in moles when given a specific mass. This concept links mass, moles, and volume, helping chemists understand the scale of chemical reactions and the quantities of products and reactants involved.