A researcher has prepared a sample of 1-bromopropane from 10g of 1-propanol. After purification he had made 12g of product. Which of the following is percentage yield?
Correct answer: B. 58%
- A. 60%
- B. 58%
- C. 90%
- D. 50%
Explanation
The above question states that 1-bromopropane i.e CH3-CH2-CH2-Br has a molar mass of 123g is being prepared from 10g of 1-propanol i.e CH3-CH2-CH2-OH whose molar mass is 60g, we need to figure out the percentage yield as the given mass of product formed after purification is 12g, the ACTUAL YIELD. Our first step is to find out the theoretical yield which is the quantity of product calculated from a balanced chemical equation The balanced chemical equation is given below, the following reaction is an example of SN2 reactionCH3CH2CH2OH + HBr ============> CH3CH2CH2Br Since we know, 60g of propanol reacts to form 123g of bromopropane, then how many grams of bromopropane will be formed if 10g of propanol reacts?Using conversion factor:20.5g is the THEORETICAL YIELD. The next step is to find the percentage yield The formula for percentage yield is given as under %yield= actual yield/ theoretical yield x 100Plugging in the values, we get = 12g/ 20.5 x 100 = 58.5% Hence the correct answer is B, as the value 58.5 is closest to option B.
This question appeared on the UHS MDCAT 2017 paper, which you can sit online with every answer explained.
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About Theoretical and Percentage Yield
Theoretical yield is the maximum product predicted from a balanced equation and the amount of limiting reactant. Work includes identifying the limiting reagent, comparing theoretical and actual yield, and calculating percentage yield using stoichiometric mole relationships.
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