Starting with 100 mL of acetone calculate the theoretical yield of product and grams ( density of acetone = .791g/mol ; molar mass of acetone =58.09 g of moles molar mass of product is 98.16g/mol

Question

C3H6O- —catalytic NaOH—>C6H10O + H2O

Starting with 100 mL of acetone calculate the theoretical yield of product and grams ( density of acetone = .791g/mol ; molar mass of acetone =58.09 g of moles molar mass of product is 98.16g/mol

If you recover 58.9 g of product in this reaction what is your percent yield?

Answer 1

First, you have to have a correctly balanced equation:

2C₃H₆O ==cat==> C₆H₁₀O + H₂O ... balanced equation

Next we find the mols C₃H₆O that we have:

NOTE: density is g/ml not g/mol. So correct density of acetone is 0.791 g / ml.

100 ml C₃H₆O x 0.791 g / ml x 1 mol / 58.09 g = 1.36 mols

Use stoichiometry of the balanced equation to find grams of product:

1.36 mols C₃H₆O x 1 mol C₆H₁₀O / mol C₃H₆O x 98.16 g / mol = 134 g C₆H₁₀O

% yield = actual yield / theoretical yield (x100%) = 58.9 g / 134 g (x100%) = 44.0%