Question

Percent yield of alkyl halide (Show all calculations for limiting reactant, theoretical yield and percent yield.) Note I have 1.746g of product obatined of pure alkyl halide Also my unkown that i chose for this experiment was 3-methyl-1-butanol---- I started with 18.996g---15.281g after reflux-- 14.993g after IR spectroscopy

Answer 1

  

You need to begin with a balanced chemical equation and define the limiting reactant.

you will have to measure the amount of that reactant that you will be using, in order to calculate product yield.  

#of moles of the starting material (3-methyl-1-butanol) = mass/molecular weight

Mass of your reactant = 18.996 g

MW of 3 methylv1 butanol = 88.148 g/mol

So # of moles of reactant = 0.2155 mol

You haven't mentioned the procedure / all the reactants involved in the reaction. To calculate the theoretical yield, a completely balanced chemical equation is necessary. Check if the ratio of reactant required to form product is 1: 1, then, # of moles of product = #of moles of reactant 0.2155 mol = 0.2155 mol.

The limiting reagent/reactant in this experiment is 3-methyl-1-butanol.  

Now, the theoretical yield (in grams) = #of moles of product( mol) * MW of product(g/mol)

(To obtain the above values, you need to know the chemical equation for the reaction)  

To calculate percent yield = theoretical yield/actual yield × 100 %

So % yield = theoretical yield/ 1.746 * 100 %