Percentage Yield theoretical yield actual yield Example ...

[Pages:4]SCH 3U

May 11th 2012

Percentage Yield

In a reaction, ideally all of our limiting reagent is converted into the desired product. This amount of product is call our theoretical yield. In practice, however, these theoretical yields are rarely achieved and the amount of product achieved is usually less. This amount of product is called an actual yield.

Percentage yield gives us a way to see how efficient a reaction is. The higher the percentage yield the more efficient the reaction because our actual yield is closer to our theoretical yield. We calculate it using the following equation:

Example Problem #1 Methanol, CH3OH, can be made in a synthesis reaction using carbon dioxide and hydrogen:

CO2 + 3 H2 CH3OH + H2O During an investigation, 20.0 g of hydrogen was reacted with excess carbon dioxide to produce 102.0 g of methanol. What is the percentage yield of this reaction?

Solution: Step 1: Write the balanced chemical equation, and find the molar masses that we require:

CO2 + 3 H2 CH3OH + H2O

molar massH2 = 2.02 g/mol molar massCH3OH = 32.05 g/mol

Step 2: Convert the mass of given substance to moles:

Step 3: Convert the amount of given substance to amount of required:

Step 4: Convert the moles of required substance to mass of required substance: Step 5: Calculate the percentage yield:

SCH 3U

May 11th 2012

Statement: The percentage yield in the synthesis of methanol reaction is 96.4%.

SCH 3U

May 11th 2012

Worksheet 3.6 ? Percentage Yield

1. How to actual yield and theoretical yield differ? The theoretical yield is what you expect to have given an amount of limiting reagent. The actual yield is the amount you receive in practice, and is often less.

2. A student carefully neutralized a sample of hydrochloric acid with sodium hydrogen carbonate in an open container:

HCl + NaHCO3 H2O + CO2 + NaCl When the reaction was complete, the student heated the remaining solution to remove the water. What effect, if any, would each of the following have on the percentage yield of NaCl? Justify your answers.

a) Not drying the mixture long enough. -This would mean that there is still some water left as a product, and therefore the mass of the product would be higher. We would mistakenly assume this added mass is NaCl, so our actual yield would increase and our percent yield would increase as well.

b) Adding insufficient sodium hydrogen carbonate to completely react all of the acid. -This would decrease the amount of NaCl formed, so our actual yield and theoretical yield are both lessened.

c) Using an impure form of sodium carbonate. -The impurities could react with HCl to form something else, so there would be less NaCl as a product. Percent yield will be lowered.

d) Splattering the reaction mixture out of the container during the reaction. -The mass of our product will be less, so our actual and percent yield will be lower.

3. Consider the following reaction: NaCl + H2O Cl2 + NaOH + H2

a) If the percentage yield of chlorine in this reaction is 95% and the theoretical yield is 142 g of chlorine, what is the actual yield of chlorine? -actual yield = (95%*142g)/100% = 134.9 g

b) What mass of sodium chloride is required to produce this yield of chlorine? molar massCl2 = 70.90 g/mol molar massNaCl = 58.44 g/mol molesCl2 = (134.9g)/(70.90g/mol) = 1.90 mol molesNaCl = 1.90*(1/1) = 1.90 mol massNaCl = 1.90 mol * 58.44 g/mol = 111.04 g

SCH 3U

May 11th 2012

4. A chemist adds 3.00 g of zinc to a solution containing an excess of silver nitrate. Only 7.2 g of silver metal is collected at the end of the experiment.

a) Write a balanced chemical equation for this reaction. Zn + 2 AgNO3 2 Ag + Zn(NO3)2

b) Determine the theoretical yield of silver metal. molar massZn = 65.41 g/mol molar massAg = 107.87 g/mol molesZn = (3.00 g)/(65.41 g/mol) = 0.046 mol molesAg = 0.046 mol * (2/1) = 0.092 mol massAg = (107.87 g/mol)*(0.092 mol)= 9.89g

c) c) Determine the percentage yield of this reaction. % yield = (7.2g/9.89g)x100% = 72.8%

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download