in figuring the percent yield of isoborneol reduced from camphor...is the ratio 1:1, i can do the calculations if i could determine this
To determine the stoichiometry and the ratio between isoborneol and camphor in this reaction, you can refer to the balanced chemical equation. However, since you haven't provided the specific reaction conditions or the complete chemical equation, I can offer you a general understanding of the conversion of camphor to isoborneol.
The conversion of camphor to isoborneol usually involves a reduction reaction using a reducing agent such as sodium borohydride (NaBH4). This reaction includes the transfer of hydride ions (H-) to the carbonyl group of camphor, resulting in the reduction to isoborneol.
The balanced chemical equation for a typical reduction reaction of camphor to isoborneol using NaBH4 would be:
Camphor + NaBH4 + H2O ⟶ Isoborneol + NaOH + B(OH)3
In this reaction, the stoichiometric ratio between camphor and isoborneol is 1:1, meaning that for every 1 mole of camphor reacted, 1 mole of isoborneol is produced. Therefore, the ratio is indeed 1:1.
To calculate the percent yield, you will need to know the actual amount of isoborneol obtained from the reaction and compare it to the theoretical yield. The theoretical yield represents the maximum amount of isoborneol that can be obtained based on the stoichiometry of the reaction and assuming 100% efficiency.
Percent Yield = (Actual Yield ÷ Theoretical Yield) x 100
Keep in mind that the actual yield can be lower than the theoretical yield due to side reactions, incomplete conversions, or losses during separation/purification steps.