[Lilly]

We reacted 1g of C6H5NH3+Cl- in 30cm3 of water with 6grams of sodium ethanoate in 25cm3 of water and 2cm3 of (CH3CO)2O. To get antifebrin CH3CONHC6H5. The equation is
C6H5NH3+Cl- + (CH3CO)2O -------> CH3CONHC6H5 + CH3COOOH + HCL

I ended up with 0.21grams of antifebrin, what was the percentage yield?

This was my working out:
Mr of reactant C6H5NH3+Cl- = 129.5
Moles= Mass /Mr
1.0g/129.5=0.0077mol of reactant

Mr of anitifibrin=135
Moles=Mass/Mr
0.21 gained from experiment / 135=0.00156

0.00156/ 0.0077 x100= 20% Is this Correct according to experiment results? Thanx in advance!

[Astrokel]

hey lily, it looks ok but how much exactly did you use for acetic anhydride?

[Lilly]

2cm3 of ethanoic anhydride.

Thanx

[Astrokel]

and the concentration or density?

[Lilly]

That was not given.

[Borek]

My bet is that there was an excess of anhdyride. You may safely assume that its density was not lower that 0.5 g/mL, not many liquids get that low. That means at least 1 g of anhdyride used. Compare numbers of moles of reactants calculated assuming 1g+1g.

Besides, wiki lists its density as 1.08 g/mL.