Question

1. The next three questions are based on a synthesis performed by Doc Inmaking. He is doing a common organic chemistry test for C=C groups. He starts with 0.6734 g of hexene, C₆H₁₂, and reacts it with excess Br₂. The Br₂ adds to the C=C group to form the product C₆H₁₂Br₂ according to this chemical equation.

C₆H₁₂(l) + Br₂(l) → C₆H₁₂Br₂(l)

Predict Doc's theoretical yield of C₆H₁₂Br₂ product. (Answer with four significant digits, units of g)

2.   Doc collects 1.357 g of his C₆H₁₂Br₂ product. What is his percent yield? (Answer with four significant digits, and units of %)

3. Based on the answers to the previous two questions, what is Doc's percent error for his synthesis experiment? (Answer to four significant digits, and units of %)

Answer 1

1) Given,

The balanced chemical reaction,

C₆H₁₂(l) + Br₂(l) C₆H₁₂Br₂(l)

Also given,

Mass of hexene = 0.6734 g

Calculating the number of moles of hexene,

= 0.6734 g of hexene x (1 mol /84.1608 g)

= 0.008 mol hexene

Now, using the moles of hexene and the mole ratio from the balanced chemical reaction, calculating the number of moles of C₆H₁₂Br₂,

= 0.008 mol hexene x ( 1 mol C₆H₁₂Br₂ / 1 mol hexene)

= 0.008 mol C₆H₁₂Br₂

Converting the number of moles to grams,

= 0.008 mol C₆H₁₂Br₂ x (243.97 g /1 mol)

= 1.952 g of C₆H₁₂Br₂

2) Given,

The experimental yield of C₆H₁₂Br₂ = 1.357 g

We know, the formula for percent yield,

Percent yield = [ Experimental yield / Theoretical yield ] x100

Percent yield = [ 1.357 / 1.952 ] x100

Percent yield = 69.52 %

3) We have,

Experimental yield = 1.357 g

Theoretical yield = 1.952 g

Percent error = [ (Theoretical yield - Experimental yield) / Theoretical yield] x 100

Percent error = [ (T1.952 g -1.357 g) / 1.952 g] x 100

Percent error = 30.48 %