Question

Determine the mmol of both starting materials (factoring in that formic acid is not pure, but rather 88% weight/volume, or 88g/100 ml), showing your work. Determine the limiting reagent in this synthesis. Lastly, calculate the theoretical yield of benzimidazole that you could expect to form.

Answer 1

Molecular Molar Mass Volume Density Mass Moles nmoles

formula (g/mol) (mL) (g/mL) (g)

`$C_6H_8N_2$` 108.14 0.108 0.001 1

HCOOH 46.02 0.064 1.22 0.07808 0.0017 1.7

mmoles of o-phenylenediamine = 1 mmoles

mmoles of formic acid = 1.7
`\approx` 2 mmoles

From the reaction of o-phenylenediamine and formic acid, we see,

1 mmole of o-phenylenediamine reacts with 1 mmole of formic acid.

But here, 2 mmoles of the formic acid , this means that the formic acid is an excess reagent and the o-phenylenediamine is the limiting reagent here.

The amount of product depends on the limiting reagent that is o-phenylenediamine. So, 1mmole of o-phenylenediamine will give 1mmole of product.

molar mass of Benzimidazole =
`118.14` g/mol

mmoles of Benzimidazole formed =
`1` mmol

Mass of benzimidazole formed = molar mass x
`(nmoles)/(1000)`

`$=(118.14 * 1)/(1000)$`

= 0.11814 g

So the theoretical yield of Benzimidazole is = 0.118 g = 118mg