Percentage Yield
The idea here is to determine how much product should be formed, given the amount of limiting
reactant and the balanced equation. Then you run the reaction and weigh how much was really
formed Divide the real by the ideal, multiply by 100, and you have the % yield.
So let's suppose you are making calcium oxide by decomposing calcium carbonate:
You have 20.0 g of Caco. You decompose it by heat, and weigh the calcium oxide that
remains. You have 10.3 grams. What is the % yield?
A) always write a balanced equation first:
B) then look at the mole relationship of reactant and product
C) then calculate the number of grams of product (Cao) you should get.
D) divide real by predicted amount and multiply by 100. Make sure it makes sense.
Sample problem: A student makes sodium chloride by buming 2.3 grams of sodium in chlorine
gas. If the yield is 90%, how much sodium chloride is made?
Problem 2: For the chemical reaction Mg(s) + 2 HCl(aq) - H, (9) + MgC1, (aq), calculate the
% vield if 100. Grams of magnesium react with excess HCl to produce 310 grams of MgCl,

Q.No. 1:

            You have 20.0 g of CaCO₃. You decompose it by heat, and weigh the calcium oxide that  remains. You have 10.3 grams. What is the % yield?

               %age Yield = 92.37 %

Solution:

               The balance chemical equation for given decomposition reaction is;

                                           CaCO₃ → CaO + CO₂

Step 1: Calculate Moles of CaCO₃:

                Moles  =  Mass / M.Mass

                Moles  =  20.0 g / 100.08 g/mol

                Moles  =  0.199 moles of CaCO₃

Step 2: Calculate theoretical amount of CaO produced;

According to equation,

              1 moles of CaCO₃ produced  =  1 mole of CaO

So,

           0.199 moles of CaCO₃ will produce  =  X moles of CaO

Solving for X,

                     X  =  0.199 mol × 1 mol / 1 mol

                     X  =  0.199 mol of CaO

Also,

         Mass  =  Moles × M.Mass

         Mass  =  0.199 mol × 56.07 g/mol

         Mass  =  11.15 g of CaO

Step 3: Calculate %age Yield as;

         %age Yield  =  Actual Yield / Theoretical Yield × 100

         %age Yield  =  10.3 g / 11.15 g × 100

         %age Yield = 92.37 %

Q.No. 2:

            A student makes sodium chloride by buming 2.3 grams of sodium in chlorine  gas. If the yield is 90%, how much sodium chloride is made?

               Actual Yield =  5.785 g of NaCl

Solution:

               The balance chemical equation for given decomposition reaction is;

                                           2 Na + Cl₂ → 2 NaCl

Step 1: Calculate Moles of Na:

                Moles  =  Mass / M.Mass

                Moles  =  2.3 g / 23 g/mol

                Moles  =  0.10 moles of Na

Step 2: Calculate theoretical amount of NaCl produced;

According to equation,

              2 moles of Na produced  =  2 moles of NaCl

So,

           0.10 moles of Na will produce  =  X moles of NaCl

Solving for X,

                     X  =  0.10 mol × 2 mol / 2 mol

                     X  =  0.10 mol of NaCl

Also,

         Mass  =  Moles × M.Mass

         Mass  =  0.10 mol × 58.44 g/mol

         Mass  =  5.844 g of NaCl

Step 3: Calculate Actual Yield as;

         %age Yield  =  Actual Yield / Theoretical Yield × 100

Or,

         Actual Yield  =  %age Yield × Theoretical Yield ÷ 100

         Actual Yield  =  99 × 5.844 g ÷ 100

         Actual Yield =  5.785 g of NaCl

Q.No. 3:

            For the chemical reaction Mg(s) + 2 HCl (aq) → H₂ (g) + MgCl₂ (aq) calculate the  % yield if 100 grams of magnesium react with excess HCl to produce 310 grams of MgCl₂.

               %age Yield = 79.13 %

Solution:

               The balance chemical equation for given decomposition reaction is;

                                           Mg + 2 HCl → MgCl₂ + H₂

Step 1: Calculate Moles of Mg:

                Moles  =  Mass / M.Mass

                Moles  =  100 g / 24.30 g/mol

                Moles  =  4.11 moles of Mg

Step 2: Calculate theoretical amount of MgCl₂ produced;

According to equation,

              1 mole of Mg produced  =  1 mole of MgCl₂

So,

           4.11 moles of Mg will produce  =  X moles of MgCl₂

Solving for X,

                     X  =  4.11 mol × 1 mol / 1 mol

                     X  =  4.11 mol of MgCl₂

Also,

         Mass  =  Moles × M.Mass

         Mass  =  4.11 mol × 95.21 g/mol

         Mass  =  391.73 g of MgCl₂

Step 3: Calculate %age Yield as;

         %age Yield  =  Actual Yield / Theoretical Yield × 100

         %age Yield  =  310.0 g / 391.73 g × 100

         %age Yield = 79.13 %

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally