Question

1. Find the amount used in moles of cyclohexanol. 3ml used. Density 0.948 g/mL.

2. Caluclate the percent yield of of cyclohexane produced from the experiment. 1.12 g of cyclohexane was obtained by distilling 4mL of H2PO4 and 3mL of cyclohexane. The product was dried with 0.25 g of CaCl2.

Answer 1

Q1. Volume of cyclohexanol used in experiment = 3 mL

mass of cyclohexanol = (volume of cyclohexanol) * (density of cyclohexanol)

mass of cyclohexanol = (3 mL) * (0.948 g/mL)

mass of cyclohexanol = 2.844 g

moles of cyclohexanol = (mass of cyclohexanol) / (molar mass cyclohexanol)

moles of cyclohexanol = (mass of cyclohexanol) / (molar mass C₆H₁₂O)

moles of cyclohexanol = (2.844 g) / (100.16 g/mol)

moles of cyclohexanol = 0.0284 mol

Q2.

Assuming cyclohexanol is the limiting reactant and all other reactants are in excess,

moles of cyclohexane formed = moles of cyclohexanol

moles of cyclohexane formed = 0.0284 mol

Theoretical yield cyclohexane = (moles cyclohexane formed) * (molar mass cyclohexane)

Theoretical yield cyclohexane = (moles cyclohexane formed) * (molar mass C₆H₁₂)

Theoretical yield cyclohexane = (0.0284 mol) * (84.16 g/mol)

Theoretical yield cyclohexane = 2.39 g

Percent yield cyclohexane = (Actual mass cyclohexane obtained / Theoretical yield cyclohexane) * 100

Percent yield cyclohexane = (1.12 g / 2.39 g) * 100

Percent yield cyclohexane = (0.469) * 100

Percent yield cyclohexane = 46.9 %

Answer 2

10 g and 10 ml