How do you calculate the percent yield of triphenylmethanol?

To calculate the of triphenylmethanol, you divide the actual yield by the theoretical yield and multiply by 100.

EXAMPLE

Assume that you prepared phenylmagnesium bromide by reacting 2.1 mL of bromobenzene (density 1.50 g/mL) with 0.50 g of magnesium in anhydrous ether. To this, you slowly added a solution of 2.4 g benzophenone in anhydrous ether. You obtained 2.6 g of triphenylmethanol. What was your percent yield?

Solution

The equations are (##"Ph = C"_6"H"_5##):

##"PhBr" + "Mg" → "PhMgBr"##

##"Ph"_2"C=O" + "PhMgBr" → "Ph"_3"COMg Br"##

##"Ph"_3"COMg Br" + "H"^+ → "Ph"_3"COH" + "Mg"^(2+) + "Br"^-##

Calculate the limiting reactant (all molar ratios are 1:1).

##2.1 cancel("mL PhBr") × (1.50 cancel("g PhBr"))/(1 cancel("mL PhBr")) × "1 mol PhBr"/(157.0 cancel("g PhBr")) = "0.020 mol PhBr"##

##0.50 cancel("g Mg") × "1 mol Mg"/(24.30 cancel("g Mg")) = "0.021 mol Mg"##

##2.4 cancel("g Ph₂CO") × ("1 mol Ph"_2"CO")/(182.2cancel("g Ph₂CO")) = "0.013 mol Ph"_2"CO"##

Since benzophenone has the fewest moles, it is the limiting reactant. It will form 0.013 mol of triphenylmethanol.

Calculate the theoretical yield.

The molar mass of ##"Ph"_3"COH"## is 260.3 g/mol. So

##"Theoretical yield" = 0.013 cancel("mol Ph₃COH") ×("260.3 g Ph"_3"COH")/(1 cancel("mol Ph₃COH")) = "3.4 g Ph"_3"COH"##

Calculate the percent yield.

##"% yield" = "actual yield"/"theoretical yield" × 100 % = (2.6 cancel("g"))/(3.4 cancel("g")) × 100 % = 76 %##