Question

1. A student started with 20 grams of a fat (with a formula weight of 696 grams/mole). The student follows the procedure given in this experiment except that the student used 3.25 grams of NaOH dissolved in the ethanol:water mixture.

a) Which of these two reactants is the limiting reactant? (Assume that you have plenty of all other reactants.)

b) What is the theoretical yield of soap for this procedure if the soap has the formula C12H23O2Na?

c) The student obtained 13.78 grams of soap. What is the per cent yield?

Answer 1

mol of fat = mass/MW = 20/696 = 0.02873

mol of base = mass/MW = 3.25/40 = 0.08125

since this is a soap, then

1 mol of fat = 3 mol of base

then

0.02873 mol of fat need = 0.02873*3 = 0.08619 mol of NAOH needed

0.08125 - 0.08619 = -0.00494 mol of NaOH; then NaOH is LIMITING

b)

Theoretical --> 0.08125 mol of Na

since ratio is 1:1

then

0.08125 mol of C12H23O2Na

MW = 12*12+ 23 + 16*2 + 23 = 222 g/mol

mass yield = mol*MW = 0.08125 *222 = 18.0375 g

c)

% yield = real/theoretical * 100 = 13.78/18.0375*100 = 76.396 %