To a mixture of 0.7 g of sodium acetate trihydrate in 3 mL of anhydrous DMSO in

Question

To a mixture of 0.7 g of sodium acetate trihydrate in 3 mL of anhydrous DMSO in a 10 mL round bottomed flask containing a stir bar is add 500 mL (0.5 mL) of 1-bromo-3- methylbutane, via syringe. the product is 0.40 g (37-93% yield). which is the limiting reactant and what is the percent yield if the experimental value for the product is 0.2545 g?

DMSO Br Or Na 100°c Isoamy acetate Isopentvl acetate 3-Methylbutyl acetate I-01 Isoamy bromide Isopentyl bromide sodium acetate NaOAc 1-Bro mo-3-methylbutane OVERVIEW

Explanation / Answer

Density of isopentyl bromide = 1.26g/mL

Then mass of isopentyl bromide = 0.5 mL × 1.26 = 0.63 g

No. Of moles of isopentyl bromide = mass/molar mass = 0.63/151.05 = 0.00417 moles .

No. Of moles of sod.acetate trihydrate = 0.7/136.079 = 0.00514 moles

So as moles of idoisopen bromide are less (one mole of each reactant reacts with each other ) , hence it is limiting reactant.

So as 1 mole of isopentyl bromide gives 1 mole of product .

Then 0.00417 moles of isopentyl bromide will give = 0.00417 moles of isopentyl acetate .

Mass of isopentyl acetate = moles × molar mass = 0.00417 × 130.19 g/mol = 0.54 g .

This is the theoretical yield .

Experimental yield = .2545 g

%yield = .2545/0.54 = 47.13 % .

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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