Question

Calculate the theoretical yield:

Rx: isoamyl alcohol with mixture of sodium bromide and sulfuric acid to make isoamyl bromide

Isoamyl alcohol used: 8.2 mL, density: .810 g/cm3, molecular mass: 88.148 g/mol

sodium bromide used: 2.80 mL, density: 3.21 g/cm3, molecular mass: 102.894 g/mol

sulfuric acid used: 15.0 mL, density: 1.84 g/cm3, molecular mass: 98.079 g/mol

Answer 1

Balanced chemical reaction is

C₅H₁₂O + NaBr + H₂SO₄    C₅H₁₁Br + NaHSO₄ + H₂O

First calculate mole of each

mass = density X volume

mass of isoamyl alcohol = 8.2 X 0.810 = 6.642 gm

mole of isoamyl alcohol = 6.642 / 88.148 = 0.07535 mole

mass of NaBr = 2.80 X 3.21 = 8.988 gm

mole of NaBr = 8.988 / 102.079 = 0.088 mole

mass of H₂SO₄ = 15 X 1.84 = 27.6 gm

mole of H₂SO₄ = 27.6 / 98.079 = 0.2814mole

According to above given reaction all reactant react in equimolar proportion but isoamyl alcohol given only 0.07535 mole it is limiting reactant react completly and other two are excess reactant some amount reamain unreacted.

According to reaction 1 mole of isoamyl alcohol produce 1 mole of isoamyl bromide therefore 0.07535 mole of isoamyl alcohol produce 0.07535 mole of isoamyl bromide

molar mass of isoamyl bromide = 151.047 gm/mole

then 0.07535 mole of isoamyl bromide = 151.047 X 0.07535 = 11.38 gm

11.38 gm isoamyl bromide is the theoretical yield