1. Calculate the molar masses of benzaldehyde, potassium permanganate, and benzo

Question

1. Calculate the molar masses of benzaldehyde, potassium permanganate, and benzoic acid. See page 2 for the formulas. How many moles of benzaldehyde are equal to 0.988g of that substance? Report your answer to the correct number of significant digits. 2. 3. I 0.0179 moles of benzaldehyde are used in the reaction done in this experiment, how many moles of KMn04 are required? 4. How many grams of KMnO are equal to 0.0141 moles of KMn0? 5. If you start the reaction with 0.100 moles of benzaldehyde and 0.105 moles KMn0 and you produce 0.857g of benzoic acid, what is your percent yield? Page 1/6 10/23/2017 CHM210 Lab: The Preparation &Purification; of Benzoic acid Purpose: The purpose of this lab is to perfor reaktion and purify the crystals and port a percent yield. m a reaction to prepare benzoic acid by an oxidation Introduction: The axidation of anorganic compound is a very common process, In this laboratory activiry benzaldehyde will be oxidized to benzoic acid. Potassum permanganate (KMn04) will be used as the oxidizing agen. The oxidation will occur in aqueous NaDH. Since the benzoic acid will react with the aqueous NaOH (base) a salt, sodium benzoate, will form in Step 1. This salt (lonic compound) is soluble in water. In Step 2. the aqueous sodium berzoate will then be reacted with aqueous HCI and

Explanation / Answer

1.

Benzaldehyde:
Molecular formula: C7H6O
Molar mass = (12 x 7) + (1 x 6) + (16 x 1)
                      = 84 + 6 + 16
                      = 106 g/mol

Potassium permanganate:
Molecular formula: KMnO4
Molar mass = (39 x 1) + (55 x 1) + (16 x 4)
                      = 39 + 55+ 64
                      = 158 g/mol

Benzoic acid:
Molecular formula: C7H6O2
Molar mass = (12 x 7) + (6 x 1) + (16 x 2)
                      = 84 + 6+ 32
                      = 122 g/mol

2.

Molecular formula of Benzaldehyde (C7H6O) = 106 g/mol
So, 106 g of Benzaldehyde (C7H6O) = 1 mol
or, 1 g of Benzaldehyde (C7H6O) = (1/106) mol
or, 0.988 g of Benzaldehyde (C7H6O) = (0.988/106) mol
                                                                   = 0.0093 mol

3.

1 mole of benzaldehyde is oxidized with 1 mole of potassium permanganate to produce 1 mole of benzoic acid.

So, for oxidizing 0.0179 mole of benzaldehyde 0.0179 moles of potassium permanganate is required.

4.

Molar mass of Potassium permanganate = 158 g/mol
So, 1 mole of Potassium permanganate = 158 g
or, 0.0141 moles of Potassium permanganate = 0.0141 x 158 g
                                                                                  = 2.2278 g

5.

Since, 1 mole of benzaldehyde produces 1 mole of benzoic acid.
So, 0.100 moles of benzaldehyde produces 0.100 moles of benzoic acid.

Molar mass of benzoic acid = 122 g/mol
So, 1 mole of benzoic acid = 122 g
or, 0.100 mole of benzoic acid = 0.100 x 122 g
                                                       = 12.2 g (Theoretical yield)

Actual yield = 0.857 g

So, percent yield = (Actual yield / Theoretical yield) x 100
                              = (0.857 / 12.2) x 100
                               = 7.02%

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