Question

What is the theoretical yield of fluorenone if you oxidize 175 mg of fluorene?

Answer 1

Answer:

• 602 mg of CO₂ and 94.8 mg of H₂O

Explanation:

The yield is measured by the amount of each product produced by the reaction.

The chemical formula of fluorene is C₁₃H₁₀, and its molar mass is 166.223 g/mol.

The oxidation, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

        $2C_{13}H_{10}+31O_2\rightarrow 26CO_2+10H_2O$

To calculate the yield follow these steps:

1. Mole ratio

          $2molC_{13}H_{10}:31molO_2:26molCO_2:10molH_2O$

2. Convert 175mg of fluorene to number of moles

• $175mg/times 1g/1,000mg=0.175g$

• Number of moles = mass in grams / molar mass

• $\text{number of moles}=0.175g/166.223g/mol=0.0010528mol$

3. Set a proportion for each product of the reaction

a) For CO₂

i) number of moles

         $2molC_{10}H_{13}/26molCO_2=0.0010528molC_{10}H{13}/x$

$x=0.0010528molC_{10}H_{13}\times 26molCO_2/2molC_{10}H_{13}=0.013686molCO_2$

ii) mass in grams

The molar mass of CO₂ is 44.01g/mol

• mass = number of moles × molar mass

• mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg

b) For H₂O

i) number of moles

$0.0010528molC_{10}H_{13}\times10molH_2O/2molC_{10}H_{13}=0.00526molH_2O$

ii) mass in grams

The molar mass of H₂O is 18.015g/mol

• mass = number of moles × molar mass

• mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg