Question

3mL of cyclohexanol (density = 0.9624 g/mL, Molecular weight = 100.158 g/mol) reacts with excess sulfuric acid to produce cyclohexene (density = 0.811 g/mL, Molecular weight = 82.143 g/mol). What is the theoretical yield in Moles of cyclohexene? (hint: reaction scheme same as problem 1, 1:1 mole ratio) (10 pts)

Answer 1

Answer:

The theoretical yield of moles cyclohexene is 0.0288 moles

Explanation:

Step 1: Data given

Volume of cyclohexanol = 3 mL

Density cyclohexanol = 0.9624 g/mL

Molar mass of cyclohexanol = 100.158 g/mol

Sulfuric acid is in excess

Density of cyclohexene = 0.811 g/mL

Molar mass of cyclohexene = 82.143 g/mol

Step 2: The balanced equation

C6H12O + H2SO4 → C6H10 + H3O + HSO4

Step 3: Calculate mass cyclohexanol

Mass cyclohexanol = density cyclohexane * volume

Mass cyclohexanol = 0.9624 g/mL * 3mL

Mass cyclohexanol = 2.8872 grams

Step 4: Calculate moles cyclohexanol

Moles cyclohexanol = mass cyclohexanol / molar mass

Moles cyclohexanol = 2.8872 grams / 100.158 g/mol

Moles cyclohexanol = 0.0288 moles

Step 5: Calculate moles cyclohexene

For 1 mol cyclohexanol we need 1 mol H2SO4 to produce 1 mol cyclohexene

For 0.0288 moles cyclohexanol we'll have 0.0288 moles cyclohexene

The theoretical yield of moles cyclohexene is 0.0288 moles

Answer 2

Answer:

$n_{C_6H_{10}}=0.03molC_6H_{10}$

Explanation:

Hello,

In this case the undergoing chemical reaction is shown on the attached picture whereas cyclohexanol is converted into cyclohexene and water by the dehydrating effect of the sulfuric acid. Thus, for the starting 3 mL of cyclohexanol, the following stoichiometric proportional factor is applied in order to find the theoretical yield of cyclohexene in moles:

$n_{C_6H_{10}}=3mLC_6H_{12}O*\frac{0.9624gC_6H_{12}O}{1mLC_6H_{12}O}*\frac{1molC_6H_{12}O}{100.158gC_6H_{12}O}*\frac{1molC_6H_{10}}{1molC_6H_{12}O} \\n_{C_6H_{10}}=0.03molC_6H_{10}$

Besides, the mass could be computed as well by using the molar mass of cyclohexene:

$m_{C_6H_{10}}=0.03molC_6H_{10}*\frac{82.143gC_6H_{10}}{1molC_6H_{10}} \\\\m_{C_6H_{10}}=2.4gC_6H_{10}$

Even thought, the volume could be also computed by using its density:

$V_{C_6H_{10}}=2.4gC_6H_{10}*\frac{1mLC_6H_{10}}{0.811gC_6H_{10}} \\V_{C_6H_{10}}=3mLC_6H_{10}$

Best regards.