Bromine can be added to limonene to created a brominated compound. The balance equation is shown below. When 613 mg of limonene and 3.5 ml of a 10% (v/v) solution of Br2 in dicholomethane (a solvent) are combined, 959 mg of the product is isokated. Calculate the percent yield for this reaction.
Limonene: 136.23 g/mol + Br2: 160 g/mol = 296.04 g/mol
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To calculate the percent yield for this reaction, we need to compare the actual yield (the amount of product obtained) with the theoretical yield (the maximum amount of product that can be obtained based on stoichiometry).
First, let's calculate the number of moles of limonene and Br2 used:
- Limonene: 613 mg = 0.613 g
- Moles of limonene = mass / molar mass = 0.613 g / 136.23 g/mol = 0.0045 mol
- Br2 solution: 3.5 ml = 3.5 cm3
- Moles of Br2 = volume (in liters) × concentration (in moles/liter)
= 3.5 cm^3 × (10/1000) mol/cm^3 = 0.035 mol
The balanced equation shows that the molar ratio between limonene and product is 1:1. Therefore, the moles of product formed should be the same as the moles of limonene used.
Now, let's calculate the theoretical yield:
- Theoretical yield = Moles of product × molar mass of product
- The molar mass of the product is given as 296.04 g/mol
- Theoretical yield = 0.0045 mol × 296.04 g/mol = 1.332 g
Lastly, we can calculate the percent yield:
- Percent yield = (Actual yield / Theoretical yield) × 100%
- The actual yield is given as 959 mg = 0.959 g
- Percent yield = (0.959 g / 1.332 g) × 100% = 72.0%
Therefore, the percent yield for this reaction is 72.0%.