How to find theoretical yield of 1-bromobutane, formed from 1-butanol, sodium br

ID: 539063 • Letter: H

Question

How to find theoretical yield of
1-bromobutane, formed from 1-butanol, sodium bromide, and sulfuric acid.

The initial weights of reactants
Mass NaBr: 3.015 g
Mass butanol: 1.805 g

pre-weighed vial: 23.053 g
distillate vial: 25.957 g
diff. in weights 2.904g

I also need
Final bromotbutane: (% bromobutane GC/100)(final weight of mix. in gram)
I dont know how to find the final weight of mixture.

I aslo need the percent yield but first I need theoretical yeild

June 22 Male a, 6 acid r ea o2goc

Mass of NaBr = 3.015 g

Therefore moles of NaBr = mass/molar mass = 3.015 g / 102.89 g/mol = 0.029 moles.

Similarly moles of butanol = 1.805 g / 74.122 g/mol = 0.024 moles.

Now balanced equation ; NaBr + C4H9OH + catal. H2SO4 = C4H9Br.

So one mole of NaBr reacts with one mole of butanol to give one mole of butylbromide.

Since moles of butanol is less, it will determine the moles of product formed.

Therefore the theoretical yield of bromobutane in moles = 0.024 moles

Therefore in grams, theoretical yield = moles x molar mass = 0.024 moles x 137.02g/mol = 3.33 g.

Since the experimental yield is 2.904 g

Therefore percentage yield = (2.904g x 100)/3.33g = 87.2%

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