Question

Given the following atomic weights, that the density of acetic acid is 1.05, and that 8.7 g of isoamyl alcohol, 5.8 mL of acetic acid, and 0.5 mL of sulfuric acid make 4.2 g of isoamyl acetate, what is the % yield of isoamyl acetate from its limiting reagent? Give at least two significant figures.

C = 12, H = 1, O =16

Answer 1

Given data,

Density of acetic acid = 1.05

Volume of acetic acid = 5.8 mL

We know molar mass of isoamyl alcohol = 88.14 g / mol

The mass of acetic acid = Volume x density

= 5.8 x 1.05

= 6.05 g

Number of moles of acid = Mass of acetic acid / molar mass of acetic acid

= 6.05 / 60

= 0.100 moles

Number of moles of isoamyl alcohol = Mass of isoamyl alcohol / molar mass of isoamyl alcohol

= 8.7 / 88.14

= 0.098 moles

Here alcohol is the limiting reagent. So, the number of moles of isoamyl acetate formed will be 0.098 moles.

The mass of  isoamyl acetate = 0.098 x 130.19

= 12.75 g

Pecent yield  = ( 4.2 / 12.75) x 100

= 0.3294 x 100

Percent yield = 32.94%