To find the theoretical yield of ethanol from 10 grams of glucose, we can use stoichiometry.
a) Start by finding the molar mass of glucose (C6H12O6):
C: 6 × 12.01 g/mol = 72.06 g/mol
H: 12 × 1.01 g/mol = 12.12 g/mol
O: 6 × 16.00 g/mol = 96.00 g/mol
Adding them up, the molar mass of glucose is:
72.06 g/mol + 12.12 g/mol + 96.00 g/mol = 180.18 g/mol
Now, we can calculate the number of moles of glucose in 10 grams:
10 g / 180.18 g/mol = 0.0555 mol
According to the balanced equation, 1 mole of glucose produces 2 moles of ethanol. Therefore, the moles of ethanol produced from 0.0555 mol of glucose will be:
0.0555 mol × 2 mol ethanol / 1 mol glucose = 0.111 mol ethanol
Next, we need to convert the moles of ethanol to grams. The molar mass of ethanol (C2H5OH) is:
C: 2 × 12.01 g/mol = 24.02 g/mol
H: 5 × 1.01 g/mol = 5.05 g/mol
O: 1 × 16.00 g/mol = 16.00 g/mol
Adding them up, the molar mass of ethanol is:
24.02 g/mol + 5.05 g/mol + 16.00 g/mol = 45.07 g/mol
The theoretical yield of ethanol from 10 grams of glucose will be:
0.111 mol × 45.07 g/mol = 4.99 grams
Therefore, the theoretical yield of ethanol from 10 grams of glucose is approximately 4.99 grams.
b) To calculate the percentage yield, we compare the actual yield (given as 0.664 grams) to the theoretical yield we calculated in part a (4.99 grams).
The percentage yield can be calculated using the formula:
% yield = (actual yield / theoretical yield) × 100
Substituting the values we have:
% yield = (0.664 g / 4.99 g) × 100 = 13.3%
Therefore, the percentage yield of ethanol in this particular experiment is approximately 13.3%.