Question

1) How many ul of a 2% Br2 solution are required to completely react with .90 mmol of chalcone? Then with 1.10 mmol of chalcone? Show work.

2) What is the theoretical yield of dibromchalcone if you react 195 mg chalcone with excess Br2? Show work

Answer 1

Solution :-

1) Chalcone molar mass = 208.26 g / mol

0.90 mmol chalcone needs 0.90 mmol Br2 because mole ratio is 1 : 1

2% solution means 2 g Br2 in 100 ml solution

So lets calculate the mass of 0.90 mol Br2

(0.90 mmol * 1 mol / 1000 mmol)*(159.808 g / 1 mol Br2) = 0.1438 g Br2

Now lets calculate the volume of the solution needed

0.1438 g * 100 ml / 2 g = 7.19 ml

So lets convert 7.19 ml to ul

7.19 ml * 1*10^3 ul / 1 ml = 7.19*10^3 ul solution needed

Now lets calculate for the 1.10 mmol of chalcone

0.90 mmol needed 7.19*10^3 ul

Therefore 1.10 mmol * 7.19*10^3 ul / 0.90 mmol = 8.78*10^3 ul 2% Br2 solution

Part 2 )

(195 mg chlcone * 1 g /1000 mg) *(1 mol / 208.26 g) *(1 mol dibromochalcone/1 mol chalcone ) = 0.00093633 mol

Now lets convert moles to mass of the dibromochalcone

Mass = moles * molar mass

Mass of dibromochalcone = 0.00093633 mol * 368.68 g per mol = 0.3446 g

0.3446 g * 1000 mg / 1 g = 344.6 mg dibromochalcone

So the theoretical yield is 344.6 mg dibromochalcone