1. Acetaminophen (commercially known as Tylenol) can be made by reacting p-amino
ID: 545014 • Letter: 1
Question
1. Acetaminophen (commercially known as Tylenol) can be made by reacting p-aminophenol with acetic anhydride (balanced reaction is presented below) HaC HaC HO NH2 O HO NH CH HO p-Aminophenol Molar mass 109.1 g/mo Molar mass- 102.1 g/mol Acetaminophen Molar mass 151.16 g/mol Acetic Anhydride Acetic Acid Density = 1.08 g/mL A student weighs out 0.216 g of p-aminophenol and adds 0.15 mL of acetic anhydride in order to synthesize acetaminophen. If the student recovers 0.207 g of acetaminophen, what is the percent yield for the reaction? The recovered acetaminophen was titrated with 0.500 M NaOH in order to determine the purity of the sample. It was discovered that 2.46 mL of NaOH was required to reach the equivalence point. Assuming the base only reacted with acetaminophen, how many grams of acetaminophern were actually synthesized from the reaction? Which of the following indicators would be best suited for the titration of acetaminophen with sodium hydroxide if the pH at the equivalence point is 8.5? The possible indicators are methyl orange, bromothymol blue, litmus, thymol blue a. b. c. Common Acid-Base Indicators Approximate pH Range IndicatorColor for Color Change Change Methyl Orae 3.2-4.4Red to Yellow Bromothymol Blue 6.0-7.6 Yellow to Blue Phenolphthain 8.2-10.0 Colorless to Pink 5.5-8.2 Litmus Red to Blue Bromocresol Green 3.8-54 Yellow to Blue Thymol Blue Alizarin Yw 11.0-12.0 Yellow to Violet 8.0-9.6 Yellow to BlueExplanation / Answer
Solution:- (a) From balanced equation, there is 1:1 mol ratio between the reactants and the products.
Product depends on the limiting reactant. So, we would calculate the amount of the product for both the reactants and see which one gives us the less amount of the product. It would be the theoretical yield.
0.216 g p-aminophenol x (1mol/109.1 g) x (1mol acetaminophen/1mol p-aminophenol) x (151.16 g/1mol)
= 0.299 g acetaminophen
Now, we would do the calculations for acetaminophen using acetic anhydride..
0.15 mL acetic anhydride x (1.08 g/mL) x (1mol/102.1 g) x (1mol acetaminophen/1mol acitic anhydride) x (151.16g/1mol)
= 0.240 g acetaminophen
From calculations, acetic anhydride gives less amount of acetaminophen, so the theoretical yield is 0.240 g.
percent yield = (actual/theoretical)*100
percent yield = (0.207/0.240)*100 = 86.25%
(b) from given volume and molarity we can calculate the moles of NaOH. NaOH and acetaminophen react in 1:1 ratio
2.46 mL NaOH x (1L/1000 mL) x (0.500 mol/1L) x (1mol acetaminophen/1mol NaOH) x (151.16 g/1mol)
= 0.186 g acetaminophen
(c) The best indicator for this titration is Thymol bule as it's pH range is 8.0 - 9.6, the equivalenc epoint pH falls in this range.
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