Methanol (CH4O) is converted to bromomethane (CH3Br) as follows:
CH4O + HBr --> CH3Br + H2O
If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percent yield?
To calculate the percent yield, we need to compare the actual yield (the amount of bromomethane obtained) with the theoretical yield (the maximum amount of bromomethane that can be produced based on stoichiometry).
Step 1: Find the molar masses
- Methanol (CH4O): C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol
- Bromomethane (CH3Br): C = 12.01 g/mol, H = 1.01 g/mol, Br = 79.90 g/mol
Step 2: Convert grams to moles
- Methanol (CH4O): 5.00 g / (12.01 g/mol + 4 * 1.01 g/mol + 16.00 g/mol) = 0.304 mol (rounded to three decimal places)
- Bromomethane (CH3Br): 12.23 g / (12.01 g/mol + 3 * 1.01 g/mol + 79.90 g/mol) = 0.0855 mol (rounded to four decimal places)
Step 3: Determine the stoichiometry
- From the balanced chemical equation: 1 mol CH4O reacts to produce 1 mol CH3Br
- Therefore, the theoretical yield of CH3Br is 0.304 mol
Step 4: Calculate the percent yield
- Percent yield = (actual yield / theoretical yield) * 100
- Percent yield = (0.0855 mol / 0.304 mol) * 100 ≈ 28.09%
So, the percent yield of bromomethane in this reaction is approximately 28.09%.
To find the percent yield of the reaction, we need to compare the actual yield with the theoretical yield.
First, let's calculate the theoretical yield of bromomethane (CH3Br) from 5.00 g of methanol (CH4O).
Step 1: Calculate the molar mass of methanol (CH4O).
Molar mass of CH4O = (1 * 12.01 g/mol) + (4 * 1.01 g/mol) + (1 * 16.00 g/mol)
= 32.04 g/mol
Step 2: Convert the mass of methanol to moles.
Moles of CH4O = (5.00 g CH4O) / (32.04 g/mol)
= 0.156 moles CH4O
Step 3: According to the balanced chemical equation, the stoichiometric ratio between methanol and bromomethane is 1:1. Therefore, the moles of bromomethane produced will be the same as the moles of methanol used.
Moles of CH3Br = 0.156 moles CH3Br
Step 4: Calculate the mass of bromomethane using its molar mass.
Molar mass of CH3Br = (1 * 12.01 g/mol) + (3 * 1.01 g/mol) + (1 * 79.90 g/mol)
= 94.94 g/mol
Mass of CH3Br = (0.156 moles CH3Br) * (94.94 g/mol)
= 14.8 g CH3Br
The theoretical yield of bromomethane (CH3Br) is 14.8 g.
To find the percent yield, we can use the following formula:
Percent yield = (Actual yield / Theoretical yield) * 100
Given that the actual yield is 12.23 g of bromomethane, we can substitute the values into the formula:
Percent yield = (12.23 g CH3Br / 14.8 g CH3Br) * 100
Calculating this gives us:
Percent yield = 82.7%
Therefore, the percent yield of bromomethane is approximately 82.7%.
Here is a worked example of how to solve simple stoichiometry problems, including percent yield.
http://www.jiskha.com/science/chemistry/stoichiometry.html