Inthereactionshown,100.0gC6H11OHyielded64.0g C6H10 . (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of C6H11OH would produce 100.0 g C6H10 if the percent yield is that determined in part (b)?
C6H11OH ¡ C6H10 + H2O
Inthereactionshown,100.0gC6H11OHyielded64.0g C6H10 . (a) What is the theoretical yield of the reaction? (b) What is the percent yield? (c) What mass of C6H11OH would produce 100.0 g C6H10 if the percent yield is that determined in part (b)?
C6H11OH ¡ C6H10 + H2O
C6H11OH ==> C6H10 + H2O
Theoretical yield for 100.0 g will produce 82 g C6H10 so TY = 82 grams.
%yield = (64/82)*100 = approx 78%.
C) You want 100.0 g C6H10. mols C6H10 in 100 g = 100/about 1.2 mols so you would need 1.2 mols to start if the reaction were 100%. It isn't so you must start with 1.2/0.78 = about 1.6 mols C6H11OH. Then grams C6H11OH = mols x molar mass = ?
To calculate the theoretical yield, percent yield, and the mass of C6H11OH required for a given yield, we need to use the balanced chemical equation and the given information.
The balanced equation for the reaction is:
C6H11OH → C6H10 + H2O
(a) The theoretical yield is the maximum amount of product that can be obtained from the given reactant. To calculate the theoretical yield, we need to determine the molar masses of C6H11OH and C6H10.
The molar mass of C6H11OH = 6 * atomic mass of carbon + 11 * atomic mass of hydrogen + atomic mass of oxygen
= 6 * 12.01 g/mol + 11 * 1.01 g/mol + 1 * 16.00 g/mol
= 86.16 g/mol
The molar mass of C6H10 = 6 * atomic mass of carbon + 10 * atomic mass of hydrogen
= 6 * 12.01 g/mol + 10 * 1.01 g/mol
= 82.14 g/mol
Since the stoichiometric ratio between C6H11OH and C6H10 is 1:1, the theoretical yield can be calculated using the molar mass of C6H10:
Theoretical yield = (mass of C6H10) / (molar mass of C6H10)
= 64.0 g / 82.14 g/mol
≈ 0.778 mol
(b) The percent yield is the actual yield (given) divided by the theoretical yield (calculated) and multiplied by 100:
Percent yield = (actual yield / theoretical yield) * 100
= (64.0 g / 0.778 mol) * 100
≈ 8231.2 %
(c) To determine the mass of C6H11OH required for a 100.0 g yield of C6H10, we need to rearrange the percent yield formula and solve for the mass of C6H11OH:
Mass of C6H11OH = (Yield of C6H10 / Percent yield) * 100
= (100.0 g / 8231.2%) * 100
≈ 1.21 g
Therefore, approximately 1.21 g of C6H11OH would be required to produce a 100.0 g yield of C6H10.
To determine the answers to these questions, we need to use stoichiometry and the given information. Let's break down each part of the question.
(a) What is the theoretical yield of the reaction?
To calculate the theoretical yield, we need to determine the stoichiometric ratio between C6H11OH and C6H10. Looking at the balanced chemical equation:
C6H11OH → C6H10 + H2O
We see that the stoichiometric ratio between C6H11OH and C6H10 is 1:1. This means that for every mole of C6H11OH, we can expect the same number of moles of C6H10.
To calculate the theoretical yield, we can convert the given mass of C6H11OH to moles and then use the stoichiometric ratio to determine the moles of C6H10. Finally, we can convert the moles of C6H10 back to grams.
1. Convert grams of C6H11OH to moles using the molar mass of C6H11OH (C6H11OH = 86.16 g/mol):
moles of C6H11OH = 100.0 g / 86.16 g/mol = 1.16 moles
2. Since the stoichiometric ratio is 1:1, the moles of C6H10 will be equal to 1.16 moles.
3. Convert moles of C6H10 to grams using the molar mass of C6H10 (C6H10 = 82.14 g/mol):
theoretical yield = 1.16 moles × 82.14 g/mol = 95.26 g
Therefore, the theoretical yield of the reaction is 95.26 grams.
(b) What is the percent yield?
The percent yield is calculated by dividing the actual yield (given) by the theoretical yield (calculated) and multiplying by 100.
Given actual yield = 64.0 g
Theoretical yield = 95.26 g (calculated in part a)
percent yield = (actual yield / theoretical yield) × 100
percent yield = (64.0 g / 95.26 g) × 100 = 67.26%
Therefore, the percent yield of the reaction is 67.26%.
(c) What mass of C6H11OH would produce 100.0 g C6H10 if the percent yield is that determined in part (b)?
We now have the percent yield and the mass of C6H10, and we need to find the mass of C6H11OH that would produce that amount of C6H10.
Using the formula for percent yield:
percent yield = (actual yield / theoretical yield) × 100
Rearranging the formula, we can solve for the actual yield:
actual yield = (percent yield / 100) × theoretical yield
On substituting the values:
actual yield = (67.26 / 100) × 95.26 g = 64.00 g
Since the actual yield is 64.0 g, which is the given value, the mass of C6H11OH required to produce 100.0 g of C6H10 is 100.0 g.
To summarize:
(a) The theoretical yield of the reaction is 95.26 grams.
(b) The percent yield is 67.26%.
(c) The mass of C6H11OH required to produce 100.0 g of C6H10 (using the given percent yield) is also 100.0 g.