I need help with molarity calculation The starting material [(n-C_4H_9)_4N]R_eO_

Question

I need help with molarity calculation

The starting material [(n-C_4H_9)_4N]R_eO_4* (3.0 g, 6.1 mmol) is placed in a 250-mL round-bottomed flask fitted with a reflux condenser. The condenser is connected to a mercury bubbler system, which consists of a cylindrical glass reservoir (22-mm internal diameter) containing 40 mL of mercury and fitted with a 4-mm-diameter gas inlet tube extending 95 mm into the mercury pool. The reaction vessel is purged with N_2 and benzoyl chloride (30 mL, 26 mmol) is then syringed into the reaction flask. The resulting mixture is refluxed for 1.5 h under a positive pressure of N_2. Under these conditions, the boiling point of the benzoyl chloride dagger should be very close to 209 degree. Dagger The resulting dark green solution is allowed to cool and is then treated with a solution of [(n-C_4H_9)_4N]Br (5.0g, 16 mmol) in ethanol (75 mL) that has been saturated with hydrogen chloride gas. This mixture is then refluxed for an additional hour, and the resulting solution is evaporated to approximately one half its original volume under a stream of N_2. The resulting blue crystals are filtered off, washed with three 10-mL portions of ethanol, then with diethyl ether, and dried under vacuum. Yield: 94% (3.28 g). Anal.. for C_32H_72Cl_8N_2Re_2: C, 33.69; H, 6.36. Found: C, 33.97; H, 6.12.

Explanation / Answer

The procedure is given for the synthesis of a Re-complex (C32H72Cl8N2Re2 and molar mass = 1140 g/mol; Let's call it 'P' afterwards) using a Re-precursor salt ([(n-C4H9)4N)]ReO4) and molar mass = 490 g/mol; Let's call it 'R' afterwards). Molar mass (M) is calculated using the formula: M = m/n, where 'm' = mass of the compound and 'n' = no of moles of the compound. Both values are provided in the 'Procedure'.

Clearly, the product is almost 'two times' of that of the reactant (as can been seen from the ratio of 'C', 'H', 'N' and 'Re' of R and P. Keep in mind that this is a two step reaction. This means that this reaction is going in 1:1 molar ratio in both the steps. In the first step, R and benzoyl chloride must be in almost 1:1 molar ratio. In the second step, the product formed in the first step and ([(n-C4H9)4N)]Br must be in 1:1 molar ratio.

However, the question asks for the calculation of molarities of various reactants. First of all, we do not need to know the molarity of Hg as it is being used just for checking if N2 gas is coming into the reaction set-up or not (N2 will form bubbles while going through Hg; larger the frequency of the bubbles, more is the pressure of N2 in the reaction vessel). Hg is not taking part in the reaction and neither N2 gas. N2 gas is just providing the inert atmosphere required for the reaction to go forward (Le Chatelier's principle's) as well to avoid formation of side products that can form if O2 is there.

Secondly, in the first step, when 30 mL of benzoyl chloride is added to the R, the reaction volume is 30 mL only. So, molarities of the reactants, viz., R and benzoyl chloride can be deduced using the relation, Molarity = n/V, where 'n' is no of moles and 'V' is volume in Litres. As 'n' is provided for both reactants and V = 30 mL, we can straightaway calculate their molarities using the just mentoned relation. Molarity for R = ~0.203 mol/L and for benzoyl chloride = 0.184 mol/L.

Since the reaction is under reflux conditions (to avoid loss of benzoyl chloride upon evaporation). However, benzoyl chloride must be consumed as it has reacted with R. Now, [(n-C4H9)4N)]Br is taken in 75 mL HCl gas saturated ethanol. As we discussed above, [(n-C4H9)4N)]Br will be in same molar ratio as of R, therefore, its molarity would also be equal to that of R. Note that molarity of HCl is of no importance here as it is taken in high excess.

so the molarity of ([(n-C4H9)4N)]Br = (16 x 10-3)/(75 x 1000) mol/L = 0.213 mol/L.

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