The overall balanced equation for the production of ethanol from glucose is as follow
C6H12O6 = 2C2H5OH + 2CO2
a) what is the theoratical yield of ethanol available from 10gram of glucose?
b) If in a particular experiment, 1gram of glucose produce 0.664gram of ethanol,what is the percentage yield?
C6H12O6 = 2C2H5OH + 2CO2
How many moles in 10 g? That's mols = g/molar mass = 10/180 = 0.055
1 mol C6H12O6 produces 2 mols C2H5OH; therefore, 0.055 mols glucose will produce 0.11 mols ethanol. Then grams = mols x molar mass = 0.11 x 46 = approx 5.1 g C2H5OH.
b. If theoretical yield for 10 g is approximately 5.1 g then yield for 1 g would be approximately 0.51 g. Your experiment produced 0.664? That's over 100% yield. You should get a patent on this process. Can't be done.
a) Well, to figure out the theoretical yield of ethanol, we need to use some math magic. We're given that for every 1 mole of glucose (C6H12O6), we get 2 moles of ethanol (C2H5OH). The molar mass of glucose is 180 grams/mol, and the molar mass of ethanol is 46 grams/mol. So, we can set up a proportion:
(2 moles of ethanol / 1 mole of glucose) = (x grams of ethanol / 10 grams of glucose)
Cross-multiplying, we get:
2 moles of ethanol * 10 grams of glucose = 1 mole of glucose * x grams of ethanol
20 grams of ethanol = x grams of ethanol
So, the theoretical yield of ethanol from 10 grams of glucose is 20 grams.
b) Now, to calculate the percentage yield, we use the formula:
Percentage yield = (Actual yield / Theoretical yield) * 100
In this case, the actual yield is given as 0.664 grams of ethanol, and the theoretical yield we calculated earlier is 20 grams of ethanol. Substituting these values into the formula:
Percentage yield = (0.664 grams / 20 grams) * 100
Percentage yield ≈ 3.32%
Looks like this particular experiment didn't quite reach its full potential. Maybe the ethanol needs a pep talk or some motivational quotes to boost its productivity.
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%.
To determine the theoretical yield of ethanol from glucose, we need to use the balanced equation and the molar masses of glucose and ethanol.
a) The molar mass of glucose (C6H12O6) can be calculated as follows:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
Therefore, molar mass of C6H12O6 = (6 * 12.01) + (12 * 1.01) + (6 * 16.00) = 180.18 g/mol
According to the balanced equation, 1 mole of glucose (C6H12O6) produces 2 moles of ethanol (C2H5OH). Hence, the molar mass of ethanol is:
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
Therefore, molar mass of C2H5OH = (2 * 12.01) + (6 * 1.01) + (1 * 16.00) = 46.07 g/mol
Using these values, we can calculate the theoretical yield of ethanol from 10 grams of glucose.
The number of moles of glucose in 10 grams can be calculated using the formula:
Number of moles = Mass (in grams) / Molar mass
Number of moles of glucose = 10 g / 180.18 g/mol
Now, we can determine the theoretical yield of ethanol based on the stoichiometry of the balanced equation:
Theoretical yield of ethanol = Number of moles of glucose * (2 moles of ethanol / 1 mole of glucose) * Molar mass of ethanol
b) To calculate the percentage yield, we need to know the actual yield from the experiment and compare it to the theoretical yield.
Percentage yield = (Actual yield / Theoretical yield) * 100
Let's calculate the percentage yield using the given values.
Actual yield of ethanol = 0.664 gram
Theoretical yield of ethanol can be calculated using the formula we derived in part a.
Percentage yield = (0.664 g / Theoretical yield) * 100
Using these steps, we can easily calculate the theoretical yield and the percentage yield for the production of ethanol from glucose.