Question

How to calculate limiting reagent and percent yield?

4-methylcyclohexanol --> 4-methylcyclohexene

I used 23 mL 4-methylcyclohexanol and 5 mL of Phosphoric acid. Do we use phosphoric acid to calculate theoretical yield if it is a catalyst? If we do then my percent yield comes out to be like 160% which is wrong because I have to use the limiting reagent which would be phosphoric acid. What do I do?

23mL 4-methylcyclohexanol (.914g/1mL) = 21g

21g (1mol/114.2g) = .184 mol

.184 mol (96.1g/1mol) = 17.7g 4-methylcyclohexene

Excess reagent

Phosphoric acid

5 mL (1.685g/1mL) = 8.425g

8.425g (1mol/98g) = .073mol

.073mol (96.1g/1mol) = 7.0 g 4-methylcyclohexene

This is what I got so far. But when I go to calculate percent yield I get 11.7/7 * 100 = like 165%

Answer 1

Sol :-

Calculation of Limiting reagent is :

The reagent which is completely consumed in a chemical reaction that is the amount left of it after the completion of reaction is zero.

Given reaction is :

4-Methylcyclohexanol + Phosphoric acid -----------------> 4-Methycyclohexene

Given,

Volume of 4-methylcyclohexanol = 23 mL

Because, Density = Mass / Volume

So,

Mass of 4-methylcyclohexanol = Volume x density

= 23 mL x 0.914 g/mL

= 21.02 g

Also,

Number of moles = Given mass / Gram molar mass

So,

No. of Moles of 4-methylcyclohexanol = 21.02 g / 114.0 g/mol

= 0.184 mol

Similarly,

Given,

Volume of phosphoric acid = 5 mL

Mass of phosphoric acid = 5 mL x 1.685 g/mL

= 8.425 g

But in this reaction 85% of H₃PO₄ by mass Then

Mass of phosphoric acid = 0.85 x 8.425

= 7.16 g

Also,

No. of Moles of phosphoric acid = 7.16 / 98 g/mol

= 0.073 mol

Since, 1 mol of 4-Methylcyclohexanol reacts = 1 mol of H₃PO₄ as in the chemical reaction.

So,

0.184 mol of 4-Methylcyclohexanol reacts = 0.184 mol of H₃PO₄ as in the chemical reaction.

Because,

Given moles of H₃PO₄ are less that is only 0.073 mol

Hence, H₃PO₄ is the Limiting reagent.