Consider the sodium borohydride reduction of camphor to isoborneol. A reaction was performed in which 1.000 g of camphor was reduced by an excess of sodium borohydride to make 0.661 g of isoborneol. Calculate the theoretical yield and percent yield for this reaction.

m = 1.0164 g

% = 65.03%

Explanation:

First of all, we need to write the chemical reaction that is taking place here. We have the camphor being reduced to isoborneol:

C₁₀H₁₆O + NaBH₄ > C₁₀H₁₈O + NaBH₂

We have a 1:1 mole ratio between Camphor and isoborneol, so, the moles of the camphor will be the same moles produced of isoborneol.

To get the theorical yield we need to calculate the theorical moles produced of isoborneol, then, the mass and compare it to the given mass. In that way we will get the %yield.

The Molar mass of camphor and isoborneol are:

MM C₁₀H₁₆O = (16*1) + (12*10) + 16 = 152 g/mol

MM C₁₀H₁₈O = (18*1) + (12*10) + 16 = 154 g/mol

The moles of camphor will be:

molesC₁₀H₁₆O = 1 / 152 = 0.0066 moles

The mass produced then of isoborneol should be:

mC₁₀H₁₈O = 0.0066 * 154

Now, the %yield would be:

% = (0.661 / 1.0164) * 100

Hope this helps

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