Answer :
Answer:
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
mC₁₀H₁₈O = 1.0164 g
Now, the %yield would be:
% = (0.661 / 1.0164) * 100
% = 65.03%
Hope this helps