The theoretical yield of 1,2-epoxycyclohexane is grams, when starting with 3.0 grams of trans-2-bromocyclohexanol. (Enter the number using 3 significant figures, i. e. 1.22) Given: 3.0 g of trans-2-bromocyclohexanol FW: 179.05 25 mL of 10% NaOH FW: 40 and density: 1.11 g/mL 1,2-epoxycyclohexane FW: 98.15

Explanation:

1. Gather all the information in one place.

Mᵣ:                179.05          40           98.15

                BrC₆H₁₀OH + NaOH ⟶ C₆H₁₀O + NaBr + H₂O

m/g:                3.0

V/mL:                                  25

C/%:                                     10

ρ/g·mL⁻¹                              1.11

Let's call BrC₆H₁₀OH "R" (reactant) and C₆H₁₀O "P" (product).

They have given us the amounts of two reactants and asked us to calculate the amount of product. This is a limiting reactant problem.

1. Calculate the moles of each reactant

(a) BrC₆H₁₀OH

(b) NaOH

(i) Mass of solution

(ii) Mass of NaOH

(iii) Moles of NaOH

3. Calculate the moles of C₆H₁₀O you can obtain from each reactant

The molar ratios are all 1:1, so

(a) 0.0168 mol R          ⟶ 0.0168 mol P

(b) 0.0694 mol NaOH ⟶ 0.0694 mol P

R is the limiting reactant, because it gives fewer moles of P.

3. Calculate the theoretical yield.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o