Question #c5bdc

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 %#