<u>Answer:</u> The percent yield of the 1-bromobutane is 48.65 %
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For NaBr:</u>
Given mass of NaBr = 11.1 g
Molar mass of NaBr = 103 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of 1-butanol and NaBr is:
By Stoichiometry of the reaction
1 mole of NaBr produces 1 mole of 1-bromobutane
So, 0.108 moles of NaBr will produce = moles of 1-bromobutane
- Now, calculating the mass of 1-bromobutane from equation 1, we get:
Molar mass of 1-bromobutane = 137 g/mol
Moles of 1-bromobutane = 0.108 moles
Putting values in equation 1, we get:
- To calculate the percentage yield of 1-bromobutane, we use the equation:
Experimental yield of 1-bromobutane = 7.2 g
Theoretical yield of 1-bromobutane = 14.80 g
Putting values in above equation, we get:
Hence, the percent yield of the 1-bromobutane is 48.65 %