Question

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Mass of sodium bromide used = 11.09 Volume of butyl alcohol used = 10.0 ml 1a. Mass of butyl alcohol used (calculate; show work) = Mass of the pre-weighed E-flask or beaker =..... Mass of pre-weighed E-flask or beaker + crude product (n-butyl bromide) = 102.6-98.8 ....101.609... 1 b. Mass of the crude product (n-butyl bromide - show work) =................... Ic. Theoretical yield based on each of the starting materials (Please use two Dimensional Analysis (DA) equations, one for the maximum amount of product obtainable from n-butyl alcohol you used and the other from the mass of sodium bromide (ignore sulfuric acid which was used in excess), then use the lesser of the two to calculate the %Yield. • Identity of the Limiting reagent (LR) based on the above two DA equations = 7 Mex mass of the product obtainable on the basis ofta Yyteld of the distilled produtet obtained: Not

Answer 1

Answer 1C )

First we calculate mass of butyl alcohol = volume x density = 10 x 0.81 = 8.1 gm since

Moles of n butyl alcohol = mass / molecule weight = 8.1 / 74.12 = 0.109 g / mol.

Moles of sodium bromide = mass / molecular weight = 17 /102.9 = 0.165 g / mol.

Theoretical yield when n-butyl alcohol as maximum = Moles of n- butanol x molecular weight of n-butyl bromide = 0.109 x 137.02 = 14.93 gm.

Theoretical yield when sodium bromide as maximum = moles of Sodium bromide x molecular weight of n-butyl bromide = 0.165 × 137.02 = 22.60 gm .

% yield of n-butyl bromide =

=( practical yield /theoretical yield )x 100  

=( 3.8 / 14.93 ) x 100

=25.45 %

Answer 1 d )

From the above reaction n- butyl alcohol is limiting reagent because it having less moles, hence they decide the formation of amount of products ( n butyl bromide )