What is the mass of 0.042 moles C8H18? a. 0.0178 b. 0.370 c. 0.500 d. 4.80 e. 921


What is the mass of 0.042 moles C8H18? a. 0.0178 b. 0.370 c. 0.500 d. 4.80 e. 921

The Correct Answer and Explanation is:

To find the mass of 0.042 moles of C₈H₁₈ (octane), we use the formula: Mass=Moles×Molar Mass\text{Mass} = \text{Moles} \times \text{Molar Mass}

Step 1: Determine the molar mass of C₈H₁₈

Octane has 8 carbon atoms and 18 hydrogen atoms.

  • Carbon (C): 12.01 g/mol
  • Hydrogen (H): 1.008 g/mol

Molar Mass of C₈H₁₈=(8×12.01)+(18×1.008)=96.08+18.144=114.224 g/mol\text{Molar Mass of C₈H₁₈} = (8 \times 12.01) + (18 \times 1.008) = 96.08 + 18.144 = 114.224 \, \text{g/mol}

Step 2: Multiply by number of moles

Mass=0.042 mol×114.224 g/mol≈4.80 g\text{Mass} = 0.042 \, \text{mol} \times 114.224 \, \text{g/mol} \approx 4.80 \, \text{g}

Correct Answer: d. 4.80


Explanation

To solve this problem, we apply a fundamental concept in chemistry: the relationship between moles and mass. Moles serve as a bridge between the number of particles in a substance and its measurable mass. The formula for converting moles to mass is: Mass (g)=Moles×Molar Mass (g/mol)\text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)}

Here, we’re given 0.042 moles of C₈H₁₈, which is octane, a common hydrocarbon found in gasoline. Our task is to convert this quantity to grams. To do this, we must first calculate the molar mass of octane.

Octane consists of 8 carbon atoms and 18 hydrogen atoms. Using the periodic table, we find that the atomic mass of carbon is approximately 12.01 g/mol and that of hydrogen is about 1.008 g/mol. The molar mass of octane is then the sum of the masses of its constituent atoms: (8×12.01)+(18×1.008)=114.224 g/mol(8 \times 12.01) + (18 \times 1.008) = 114.224 \, \text{g/mol}

With the molar mass known, we multiply by the number of moles: 0.042 mol×114.224 g/mol=4.80 g0.042 \, \text{mol} \times 114.224 \, \text{g/mol} = 4.80 \, \text{g}

Thus, 0.042 moles of octane has a mass of 4.80 grams, making option d the correct answer. This type of calculation is essential in stoichiometry, allowing chemists to relate quantities of substances in reactions.

Scroll to Top