A student prepared 3.5g of camphor by dissolving 5 g of isoborneol in 15 mL of g
ID: 496671 • Letter: A
Question
A student prepared 3.5g of camphor by dissolving 5 g of isoborneol in 15 mL of glacier acetic acid and then added 50 mL of Clorox (which contains 5% NaOCI, d 1.11g mL). What is the percent yield of the reaction? b. What is the purpose of adding NaHSO_3 at the end of the oxidation of isoborneol to camphor? Write a balanced equation for the reaction that it undergoes. a. Write the mechanism to show how the electrophile (nitronium ion-NO_2^+) is generated in the electrophilic aromatic substitution of methyl benzoate. Below is the reaction equation:
Explanation / Answer
1 a) The balanced chemical equation is
Isoborneol + NaOCl -------> Camphor
As per the balanced equation,
1 mole isoborneol = 1 mole NaOCl = 1 mole camphor
We need to determine the limiting reactant; the limiting reactant will form the product.
Molar mass of isoborneol = 154 g/mol.
Molar mass of NaOCl = 74.44 g/mol.
Molar mass of camphor = 152 g/mol.
Find out the moles of isoborneol = (5 g)/(154 g/mol) = 0.03247 mole.
Find out the moles of NaOCl. We have 50 mL of Clorax containing 5% NaOCl). Density of Clorax solution = 1.11 g/mL.
Therefore, mass of 50 mL of Clorax solution = (50 mL)*(1.11 g/mL) = 55.5 g.
Therefore, mass of NaOCl in 55.5 g Clorax = (5 g NaOCl/100 g Clorax)*(55.5 g Clorax) = 2.775 g NaOCl.
Moles of NaOCl = (2.775 g)/(74.44 g/mol) = 0.03728 mole.
Since there is a 1:1 molar ratio between isoborneol and NaOCl and we have fewer moles of isoborneol (0.03247 mole vs 0.03728 mole of NaOCl), therefore, isoborneol is the limiting reactant.
Moles of camphor expected to be formed = (0.03247 mole isoborneol)*(1 mole camphor/1 mole isoborneol) = 0.03427 mole.
Mass of camphor expected to be formed = (0.03427 mole)*(152 g/1 mole) = 5.209 g 5.21 g.
However, mass of camphor obtained = 3.5 g.
Therefore, percent yield = (3.5 g)/(5.21 g)*100% = 67.178% 67.2% (ans).
b) NaHSO3 is added to remove the excess NaOCl from the solution. NaOCl is a good oxidizing agent and can further oxidize the ketone formed; therefore, it must be removed. The reaction taking place can be shown as:
NaHSO3 + NaOCl ------> NaCl + NaHSO4
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