[Chemistry Class Notes] on Theoretical Yield Formula Pdf for Exam

So after performing an experiment, we follow a protocol to report the yield of product in percentage and compare it with theoretical yield (desired yield) to approve the usage and efficiency of a reaction. 

In order to calculate the percentage yield of the product we use the following formula:

 

 

 

Let us understand the application of this formula in the following example.

 

Solved example: (i) For the balanced equation shown below, if 93.8 grams of PCl5 were reacted with 20.3 grams of H2O, how many grams of H3PO4 would be produced?

 

(ii) If the amount of H3PO4 was found to be 20 gm instead of the expected value, what will be the percentage yield?

 

Solution: 

(i) P = 31; Cl = 35.5; H = 1; O = 16; 

molar mass of PCl5 = 31* 1 + 35.5 * 5 = 208.5 grams/mol

number of moles of PCl5, n = 93.8/208.5 = 0.449 moles

molar mass of H2O = 18gm

number of moles = 20.3/18 = 1.127 moles

20.3/18 = 1.127 moles of H2O

So, we can see that the Limiting reagent is H2O

molar mass of H3PO4 = 1 * 3 + 31 * 1 + 16 * 4 = 98

number of moles of PCl5 used in the reaction would be 

1×1.1274

 

Options:

(a) 50 %

(b) 30 %

(c) 80 %

(d) 100 %

Answer: (a)

Solution: 

C = 12; H = 1; Cl  = 35.5; O = 16; 12 * 2 = 24; 

 

Molar Mass of C2H5Cl = 1 * 5 + 35.5 + 2 * 12 = 64.5 grams/mole

 

number of moles of C2H5Cl = 18.3/64.5 = 0.283

 

Molar mass of O2 = 32 grams/mole

 

number of moles of O2 = 37.3/32 = 1.165

As we can see the limiting reagent is C2H5Cl

 

So, number of moles of O2 used in the reaction will be 

 

13/4 * .283 = 0.919

From the balanced equation we can see that 4 moles of C2H5Cl gives 2 moles of Cl2

So, 0.283 moles of C2H5Cl will give 0.283 * 0.5 moles = 0.1415 mole

Amount of Cl2 produced in grams = 0.1415 * 71 = 10.04 gm.