chemicals used : S0dium sulate, m-louic acid, oxay dimethylformamide, NaHCO3 and

Question

chemicals used : S0dium sulate, m-louic acid, oxay dimethylformamide, NaHCO3 and HCI. REFER TO MSDS BEFORE STARTING LAB! cat. DMP OH +CI ether ?.w./concertation | Molar ratio mole Amount 2M 1.1 0.48 mL ?.w Molar ratio mole Amount 2.4 Place 0.4g of m-toluic acid, 20 mL of ether and two drops of dimethylformamide in a rb flask with magnetic stirrer. Add mL of Oxalyl chloride dropwise to the mixture above. Cool the solution in an ice bath 1. for 5 minutes and add mL of dimethylamine dropwise. 3. Allow the reaction to react for 5 minutes.

Explanation / Answer

Toluic acid + Oxalylchloride ---------> Toluolyl Chloride

Toluic Acid:

Mass of toluic acid Used = 0.40 g

Molar mass of toluic acid = 136.2 g/mol

Moles of Toluic acid used = mass/molar mass = 0.40 g/136.2 g/mol = 2.9 x 10-3 moles

Oxalylchloride:

Molarity of Oxalylchloride = 2M = 2 moles/Litre

Mole ratio = 1.1 of moles of Toluic acid = 1.1 x 2.9 x 10-3 = 3.2 x 10-3 moles

Volume of Oxalylchloride = Moles /Molarity = 3.2 x 10-3 moles/ 2 moles/Litre = 1.6 x 10-3 L = 1.6 mL

Since the mole ratio of Toluic acid: Oxalyl Chloride is 1: 1.1, Toluic acid is the limiting reagent and only 1.0 mole equivalent of toulylchloride can form.

Toluoylchloride:

Molar mass of Toluoylchloride = 154.6 g/mol

Expected Moles of Toluoylchloride produced = 2.9 x 10-3 moles (1:1 mole ratio with toulic acid)

Expected mass of Toluoylchloride produced = expected moles x molar mass = 2.9x10-3mol * 154.6 g/mol

= 0.50g.

Note: We would use this expected mass for further caclulations. If you did this experiment, it is practically impossible to obtain 100% yield. So you may have a different mass here (less than 0.50 g if its completely dried). So please use that mass instead.

  Toluolyl Chloride + Diethylamine ---------> Final product

Toluolyl Chloride?:

Mass of Toluolylchloride? Used = 0.50 g

Molar mass of Toluolylchloride?= 154.6 g/mol

Moles of Toluolylchloride? used = mass/molar mass = 0.50 g/154.6 g/mol = 3.2 x 10-3 moles

Diethylamine?:

Molarity of Diethylamine? = 2M = 2 moles/Litre

Mole ratio = 2.4 of moles of Toluolylchloride? = 2.4 x 3.2 x 10-3 = 7.8 x 10-3 moles

Molar mass of Diethylamine? = 73.14 g/mol

Gms of Diethylamine? = moles x molar mass = 7.8 x 10-3 moles x 73.14 g/mol = 0.57 g

Density of diethylamine = 0.7074 g/mL

Volume of Diethylamine? = mass/density = 0.57 g/ 0.707 g/mL = 0.81 mL

Since the mole ratio of toulylchloride: diethylamine is 1:2.4, toulylchloride is the limiting reagent and only 1 mole equivalent of the product can form.

Final product (C12H17ON):

Molar mass of Final product = 191.2 g/mol

Expected Moles of Final product produced = 3.2 x 10-3 moles (1:1 mole ratio with touloylchloride)

Expected mass of Final product produced = expected molxmolar mass = 3.2x10-3mol * 191.2g/mol

= 0.61g.

Note: Again this is expected mass of product. Also known as theoretical yield. If you did this experiment, it is practically impossible to obtain 100% yield. So you may have a different mass obtained (less than 0.61 g if its completely dried). The mass you obtained from the experiment would be your actual yield. Actual yield/ theoretical yield * 100% will give you theoretical yield.

Toluic acid Oxalylchloride Toluolyl Chloride M.W/Conc 136.2 g/mol 2M 154.6 g/mol Mole ratio 1 1.1 1.0 Moles 2.9 x 10-3 3.2 x 10-3 2.9 x 10-3 Amount 0.4 g 1.6 mL 0.50 g

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.