Question

A reaction was performed in which 1.400 g of camphor was reduced by an excess of sodium borohydride to make 1.061 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.

Answer 1

Answer:

The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %

Explanation:

Step 1: Data given

Camphor = C10H16O

sodium borohydride = NaBH4

Isoborneol = C10H18O

Mass of camphor = 1.400 grams

Mass of sodium borohydride = 1.061 grams

Molar mass of camphor = 152.23 g/mol

Molar mass of sodium borohydride = 37.83 g/mol

Molar mass of isoborneol = 154.25 g/mol

sodium borohydride is in excess, this means camphor is the limiting reactant

Step 2: The balanced equation

2C10H16O + NaBH4 → 2C10H18O + NaB

Step 3: Calculate moles of camphor

Moles camphor = mass camphor / molar mass camphor

Moles camphor = 1.400 grams / 152.23 g/mol

Moles camphor = 0.00920 moles

Step 4: Calculate moles of isoborneol

For 2 moles camphor we need 1 mol of NaBH4 to produce 2 moles of isoborneol

For 0.00920 moles camphor we'll have 0.00920 moles isoborneol

Step 5: Calculate mass isoborneol

Mass isoborneol = moles isoborneol * molar mass isoborneol

Mass isoborneol = 0.00920 * 154.25 g/mol

Mass isoborneol =  1.4191 grams (= theoretical yield)

% yield = (actual yield / theoretical yield) *100%

% yield = (1.061 / 1.4191) *100 %

% yield = 74.77 %

The theoretical yield is 1.4191 grams of isoborneol. The % yield is 74.77 %