Please find the balanced equation, theoretical yield, atom economy and estimated e factor )

Question

Please find the balanced equation, theoretical yield, atom economy and estimated e factor )

based on the procedure

To prepare the catalyst, sodium molybdate dihydrate (0.30 g; 1.2 mmol), and 4 M HCl (0.5 mL; 2.0 mmol) were added to a vial. About 1 mL of water was added to complete dissolution. Into a second vial benzyl tri-ethyl ammonium chloride (BTEAC) (0.525 g; 2.30 mmol) and ca.3 mL water were stirred until dissolved. The BTEAC solution was heated at 70 ° C with stirring.To the stirred solution was added the molybdate solution dropwise. After addition was complete, the solution was stirred for an additional five minutes, then removed from heat and vacuum filtered.

Answer 1

One thing first, because it changes everything downstream: your procedure never names the product that was filtered off. Balanced equation, theoretical yield, atom economy and E factor all depend on knowing what you made, so check your lab manual and swap in the right formula if it differs from what I use below.

That said, your own numbers point strongly at one answer. Work out the mole ratios:

Na2MoO4.2H2O: 0.30 g / 241.95 g/mol = 1.24 mmol
BTEAC (C13H22ClN): 0.525 g / 227.78 g/mol = 2.30 mmol
ratio BTEAC : Mo = 2.30 / 1.24 = 1.9, essentially 2:1

A 2:1 ratio of a singly charged cation to a doubly charged anion is exactly what you need for benzyltriethylammonium molybdate, so I will use that. The HCl is there to acidify and drive the salt out of solution rather than to appear in the product.

1. Balanced equation

2 C13H22ClN + Na2MoO4 -> (C13H22N)2MoO4 + 2 NaCl

2. Theoretical yield

Find the limiting reagent first. 1.24 mmol Mo would need 2.48 mmol BTEAC, but only 2.30 mmol is present, so BTEAC is limiting.

Product molar mass: 2(192.33) + 159.94 = 544.6 g/mol

moles product = 2.30 mmol / 2 = 1.15 mmol

theoretical yield = 1.15 mmol x 544.6 g/mol = 0.628 g

Divide your actual isolated mass by this to get percent yield.

3. Atom economy

Atom economy = (molar mass of desired product / sum of molar masses of all reactants) x 100

= 544.6 / (2 x 227.78 + 241.95) x 100 = 544.6 / 697.5 x 100 = 78.1%

The 21.9% lost is the two NaCl plus the two waters of hydration. Note atom economy is a property of the equation only - it does not care how well the reaction actually worked. A reaction can have 100% atom economy and 5% yield.

(If your course defines it using anhydrous Na2MoO4 at 205.91 g/mol, the figure rises to about 82.6%. State which convention you used.)

4. Estimated E factor

E factor = mass of waste / mass of product, and the interesting part is what you count as waste.

Reagents only: total input 0.30 + 0.525 + 0.5 g of HCl solution = 1.33 g. Waste = 1.33 - 0.628 = 0.70 g.

E = 0.70 / 0.628 = 1.1

Including the roughly 4 mL of water used as solvent: total input about 5.33 g, waste 4.70 g.

E = 4.70 / 0.628 = 7.5

What those two numbers are meant to teach you

The E factor jumps by a factor of seven purely from counting the solvent, and that is the whole point of the metric. Atom economy looked at the equation and gave a respectable 78%, but the equation cannot see the 4 mL of water, the workup, or anything you rinse the filter with. In real process chemistry solvent is typically 80 to 90% of all waste, which is why bulk pharmaceuticals run E factors of 25 to 100 while oil refining sits near 0.1.

Here at least the solvent is water, which is about the most benign choice available. Say that in your write-up - it is the sort of judgement the green chemistry metrics exist to prompt.