Question

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL) with excess benzaldehyde and NaOH to produce 79.4 g of (1E,4E)-1,5-diphenylpenta-1,4-dien-3-one (234.29 g/mol). What is the percent yield of this student's experiment

Answer 1

% yield of the student's experiment is

`(0.34)/(0.60)` ˣ 100 = 56.67%

Step-by-step explanation:

given

volume of acetone= 43.8 mL

molar weight of acetone = 58.08 g/mol

density of acetone = 0.791 g/mL

A student mixes 43.8 mL of acetone (58.08 g/mol, 0.791 g/mL)

43.8 mL = 43.8mL × 0.791g/mL

= 34.6458g ≈34.65g

1 mole of acetone = 58.08g

∴34.65g = 34.65g/58.08g

= 0.60mol

molecular weight of the product 1,5-diphenylpenta-1,4-dien-3-one = 234.29 g/mol

mole = mass/ molar weight

mole = 79.4g/ 234.29g/mol

mole(n) = 0.3389mol ≈ 0.34mol

1 mole of acetone will produce 1 mole of the product

∴0.60mol of acetone will produce 0.60mol of the product

but we get 0.34mol of the product

∴ % yield of the student's experiment is

`(0.34)/(0.60)` ˣ 100 = 56.67%