The density of pentanol, C5H12O, is 0.8110 g/mL. Calculate the volume occupied by 5.805 moles of pentanol.
The Correct Answer and Explanation is:
To calculate the volume occupied by 5.805 moles of pentanol (C₅H₁₂O), we follow these steps:
✅ Step-by-Step Solution
1. Find the molar mass of pentanol (C₅H₁₂O):
- Carbon (C): 12.01 g/mol × 5 = 60.05 g/mol
- Hydrogen (H): 1.008 g/mol × 12 = 12.096 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol
Total molar mass = 60.05 + 12.096 + 16.00 = 88.15 g/mol
2. Convert moles to grams:
Mass=Moles×Molar Mass=5.805 mol×88.15 g/mol=511.73 g\text{Mass} = \text{Moles} \times \text{Molar Mass} = 5.805 \text{ mol} \times 88.15 \text{ g/mol} = 511.73 \text{ g}
3. Use the density to find volume:
Density=MassVolume⇒Volume=MassDensity=511.73 g0.8110 g/mL=630.89 mL\text{Density} = \frac{\text{Mass}}{\text{Volume}} \Rightarrow \text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{511.73 \text{ g}}{0.8110 \text{ g/mL}} = 630.89 \text{ mL}
✅ Final Answer:
630.89 mL\boxed{630.89 \text{ mL}}
📘 Explanation
To determine the volume that 5.805 moles of pentanol occupy, we use basic principles from chemistry involving moles, molar mass, and density.
First, we must convert the number of moles into mass. The molar mass of pentanol (C₅H₁₂O) is calculated by summing the atomic masses of all atoms in the molecule: 5 carbon atoms, 12 hydrogen atoms, and 1 oxygen atom. This results in a molar mass of 88.15 g/mol. By multiplying the molar mass by the number of moles (5.805), we find the total mass of the substance, which is approximately 511.73 grams.
Next, we use the definition of density, which is mass divided by volume. Rearranging the formula to solve for volume, we divide the mass by the density. Given that the density of pentanol is 0.8110 g/mL, dividing 511.73 g by 0.8110 g/mL gives us approximately 630.89 mL.
This approach is a practical application of the relationship between mass, volume, and density. It’s commonly used in chemistry to relate the physical properties of substances with their measurable quantities. The precision of values and proper unit conversions are critical in arriving at the correct answer. In real-world laboratory settings, such calculations are essential for preparing chemical solutions, predicting outcomes, and ensuring safety.
