Given the following information, calculate the mass in grams of 2-methyl-1-butene produced from 6.28 g of 2-methyl-2-butanol: 8 OH H2SO4 H2O 90% Yield C5H12O C5H10 C5H10
The Correct Answer and Explanation is:
Correct Answer:
Step 1: Calculate Molar Masses
- Molar mass of 2-methyl-2-butanol (C₅H₁₂O)
= (5 × 12.01) + (12 × 1.008) + (16.00)
= 60.05 + 12.096 + 16.00
= 88.15 g/mol - Molar mass of 2-methyl-1-butene (C₅H₁₀)
= (5 × 12.01) + (10 × 1.008)
= 60.05 + 10.08
= 70.13 g/mol
Step 2: Calculate Moles of 2-methyl-2-butanol
Given mass = 6.28 g
Molar mass = 88.15 g/mol
Moles = Mass / Molar mass = 6.28 g ÷ 88.15 g/mol = 0.07126 moles
Step 3: Calculate Theoretical Mass of 2-methyl-1-butene
The reaction is:
C₅H₁₂O → C₅H₁₀ + H₂O
The mole ratio is 1:1, so 0.07126 moles of alcohol produce 0.07126 moles of alkene.
Mass of alkene = Moles × Molar mass
= 0.07126 moles × 70.13 g/mol
= 4.997 g (Theoretical mass)
Step 4: Apply the 90 Percent Yield
Actual mass produced = Theoretical mass × Percent yield
= 4.997 g × 0.90
= 4.497 g
Final Answer:
4.50 grams of 2-methyl-1-butene produced
Explanation:
This question requires calculating the mass of product formed from a known mass of reactant, considering the percent yield. The reaction involves dehydration of 2-methyl-2-butanol using sulfuric acid to produce 2-methyl-1-butene and water.
First, the molar masses of both the reactant and the product are calculated using atomic weights of carbon, hydrogen, and oxygen. The molar mass of 2-methyl-2-butanol is 88.15 g/mol, and the molar mass of 2-methyl-1-butene is 70.13 g/mol.
Next, using the given mass of the alcohol (6.28 grams), the number of moles is found by dividing the mass by the molar mass. Since the mole ratio between the alcohol and the alkene is 1:1, the same number of moles of alkene is expected.
To find the theoretical mass of 2-methyl-1-butene, multiply the number of moles by its molar mass. This gives 4.997 grams, which represents the maximum amount that can be formed if the reaction were 100 percent efficient.
However, the reaction proceeds with a 90 percent yield, meaning only 90 percent of the theoretical mass is actually obtained. Therefore, multiplying the theoretical mass by 0.90 gives the actual mass of product formed, which is 4.50 grams.
This method is common in stoichiometry to predict real-life chemical production outcomes where reactions are rarely perfectly efficient. Percent yield reflects practical limitations like side reactions, incomplete conversions, or experimental losses.
