Question

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

Answer 1

• 602 mg of CO₂ and 94.8 mg of H₂O

Step-by-step 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)* 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)*10molH_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