5 Ml Of Bromo-2-Methylpropane Were Used, Density Of Brono-2-Methylpropane Is 1.22 G/ML
Calculate the mass of naphthalene required to react stoichiometrically with the moles of 2- bromo-2-methylpropane that you have placed in the flask. Dispense that quantity of naphthalene into the weighing dish. There is some imprecision in delivering the powder onto the weigh boat, but a mass within 0.040 g of the calculated stoichiometric quantity will be sufficient for this experiment. Note that you can use a combination of the 1 g and the 0.1 g buttons to add the required mass. If you overshoot the goal, you can discard the weighing boat and get a new one. 5 ml of bromo-2-methylpropane RECord the sample mass dispensed below mass of naphthalene (g) 5.9924 Move the used scoopula into the Recycle cabinet.
The Correct Answer and Explanation is :
To calculate the mass of naphthalene required to react stoichiometrically with 5 mL of bromo-2-methylpropane, we need to go through the following steps:
Step 1: Calculate the mass of bromo-2-methylpropane
We are given the volume and density of bromo-2-methylpropane:
- Volume = 5 mL
- Density = 1.22 g/mL
The mass of bromo-2-methylpropane can be calculated using the formula:
[
\text{Mass} = \text{Density} \times \text{Volume}
]
[
\text{Mass of bromo-2-methylpropane} = 1.22 \, \text{g/mL} \times 5 \, \text{mL} = 6.10 \, \text{g}
]
Step 2: Calculate the moles of bromo-2-methylpropane
To find the moles of bromo-2-methylpropane, we need the molar mass of bromo-2-methylpropane. Assuming the chemical formula for bromo-2-methylpropane is ( \text{C}_4\text{H}_9\text{Br} ):
- The molar mass of ( \text{C}_4\text{H}_9\text{Br} ) is calculated as:
- C (12.01 g/mol) = 4 × 12.01 = 48.04 g/mol
- H (1.008 g/mol) = 9 × 1.008 = 9.072 g/mol
- Br (79.904 g/mol) = 79.904 g/mol
So, the molar mass of bromo-2-methylpropane is:
[
\text{Molar mass} = 48.04 + 9.072 + 79.904 = 137.016 \, \text{g/mol}
]
Now, calculate the moles of bromo-2-methylpropane:
[
\text{Moles of bromo-2-methylpropane} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6.10 \, \text{g}}{137.016 \, \text{g/mol}} = 0.0445 \, \text{mol}
]
Step 3: Calculate the moles of naphthalene required
From the balanced chemical equation for the reaction between bromo-2-methylpropane and naphthalene, assuming a 1:1 mole ratio (since no specific stoichiometry is provided):
[
\text{Moles of naphthalene} = 0.0445 \, \text{mol}
]
Step 4: Calculate the mass of naphthalene required
Now, we need the molar mass of naphthalene (( \text{C}_10\text{H}_8 )):
- The molar mass of naphthalene is calculated as:
- C (12.01 g/mol) = 10 × 12.01 = 120.10 g/mol
- H (1.008 g/mol) = 8 × 1.008 = 8.064 g/mol
Thus, the molar mass of naphthalene is:
[
\text{Molar mass of naphthalene} = 120.10 + 8.064 = 128.164 \, \text{g/mol}
]
Now, calculate the mass of naphthalene:
[
\text{Mass of naphthalene} = \text{Moles of naphthalene} \times \text{Molar mass of naphthalene} = 0.0445 \, \text{mol} \times 128.164 \, \text{g/mol} = 5.70 \, \text{g}
]
Conclusion:
The mass of naphthalene required to react with 5 mL of bromo-2-methylpropane is approximately 5.70 g. Since there is some imprecision in delivering the powder, a mass within 0.040 g of this value is acceptable. Therefore, the mass of naphthalene dispensed (5.9924 g) is slightly above the stoichiometric requirement, but this is within an acceptable range.