Question

Given 7.15g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield?

Answer 1

Given the mass of butanoic acid(CH3CH2CH2COOH) = 7.15 g

Molecular mass of butanoic acid(CH3CH2CH2COOH) = 88.1 g/mol

Hence moles of butanoic acid(CH3CH2CH2COOH) = mass/molecular mass = 7.15 g / 88.1 g/mol = 0.08116 mol

When butanoic acid reacts with excess ethanol they form ethyl butyrate(CH3CH2CH2COOCH2CH3 ).

The balanced chemical reaction is

CH3CH2CH2COOH + CH3CH2OH ------- > CH3CH2CH2COOCH2CH3

1 mol, --------------------- 1 mol, -------------------- 1 mol

From the above balanced chemical reaction it is clear that 1 mole of butanoic acid produces 1 mole of ethyl butyrate.

Hence 0.08116 mol of butanoic acid will also form 0.08116 mol of ethyl butyrate.

Hence moles of ethyl butyrate formed if we consider 100% yield = 0.08116 mol

molecular mass of ethyl butyrate(CH3CH2CH2COOCH2CH3) =

Hence mass of ethyl butyrate(CH3CH2CH2COOCH2CH3) formed = 0.08116 mol x 116.16 g/mol

= 9.43 g (answer)