Lab Report Percent Yield
| Percent Yield Lab | | | | 4/20/2012 |
Mrs.Sardella
Per4
Matt , Kait
Mrs.Sardella
Per4
Matt , Kait
|
Introduction
*Limiting Reactant: A reactant that is completely consumed during a chemical reaction, limiting the amount of product that is produced.
*Excess Reactant: A reactant that remains after a reaction is over.
*Theoretical Yield: The amount of product that is predicted by stoichiometric calculations
*Actual Yield: The amount of product that is recovered after a reaction is complete.
*Percentage Yield: The actual yield of a reaction, expressed as a percentage of the theoretical yield.
The equation for calculating the percent yield of a reaction is:
Percentage Yield = Actual Yield X100% Theoretical …show more content…
Yield
Purposes
The purpose of this lab was to determine the limiting and excess reactants for the reaction between potassium iodide and lead (II) nitrate. The second purpose of this lab was to determine the theoretical yield of the reaction. The third purpose of this lab was to determine the percentage yield of the reaction.
Pre-Lab Calculations 1. Potassium Iodide (aq) + Lead (II) Nitrate (aq) Potassium Nitrate (aq) + Lead(II) Iodide (s) 2KI(aq) + Pb(NO3)2 (aq ) 2KNO3 (aq) +PbI2 (s)
2.Given/Unkn: 1.25g 1.25g /Ay=?
m=?
Mm (39.1)+(126.9) (317.2) (39.1) +(14) + (16x3) (207.2)+(126.9x2) =166g/mol 331.2g/mol 69.4g/mol 461.0g/mol
#n (EQUN) 2 mol 1 mol 2mol 1mol
M of PbI2 w KI = 1.25g KI X 1mol KI X 1 mol PbI2 X 461.0g PbI2 = 1.735 g of PbI2 LR 166.0g KI 2 mol KI 1 mol PbI2
M of PbI2 W Pb(NO3)2 = 1.25 g Pb(NO3)2 X 1 mol Pb(NO3)2 X 1 mol PbI 2 X 461.0 g PbI2 331.2 g Pb(NO3) 2 1 mol Pb(NO3)2 1 mol PbI2 =1.739 g od PbI2
3.The theoretical yield of the precipitate in the reaction is 1.735 g of potassium iodide.
Observation Table Material | Mass with no substance | Mass with substance | Physical property of Reactants | Beaker A | 66.97g | 68.22g including KI | White Powder | Beaker B | 65.91 g | 67.16 g including Pb(NO3)2 | Solid white crystals | Filter paper | 0.56g | 2.16g including PbI2 | Yellow Precipitate |
Post Lab Calculations 1. Ay=?
=Mass of the filter paper including substance – mass of filter paper with no substance = mass of …show more content…
PbI2
= 2.16g – 0.56g = 1.6g of PbI2 2. % yield =?
%yield = Actual Mass Theoretical Mass X 100% =1.6g 1.735g X 100% = 92.2 %
Discussion In this lab potassium iodide and lead (II) nitrate are mixed, and a reaction takes place. The lab begins with two separate beakers. The first beaker held the Potassium Iodide and the second held the Lead nitrate. When deionized water was added and the substances were combined together immediately a precipitate was formed, as well as a bright yellow, cloudy liquid appeared. After the filteration processes the product that our group was left with was a yellow powder consistency. If no sources of error occurred, and the lab took place under perfect conditions the percentage yield of this double displacement reaction should be 100%. However, the percentage yield that my group received 92.2%, which means that somehow throughout the lab process 8%, was lost. There could have been many sources of error. For example, some of them could have been left on the stirring rod, in the beakers, on the funnel or it could have even been spilt. We put a hole in the filter; however we used a new filter and avoided a large error. When measuring the filter paper and its contents the contents of the contents spilled or not placed in the filter paper will produce lower actual yield because there will be less product obtained. All these possible sources of error affected the actual yield thus disturbing the % yield outcome, preventing a perfect 100%.
Conclusion
In conclusion, all of the purposes of this lab were met and the various calculations were performed to achieve percent yield, theoretical yield, actual yield, limiting and excess reactant. Through this lab a greater understanding of reaction yields and stoichiometry were gained. My group and I were able to gain a much greater knowledge of how much of a reactant is required for a reaction to take place, and the amount of the left over reactants, how much product was produced and the % yield. Our final yield was 92.2%, which was not a perfect 100% but it was still a successful experiment.
Application Questions 1. C12H22O11 (s) + H 2O(l) 4C 2 H 6 O(l) + 4CO 2
Givens/unkn: 1000g
Mm: 342g/mol 18g/mol 46g/mol 44g/mol
#N (equn) 1mol 1mol 4mol 4mol
C12H22O11withC2H6O= 1000g X 1mol X 4mol X 46gmol = 538g 342g/mol 1 mol 1 mol
In my personal opinion I do not believe that ethanol is a worthwhile alternative to fuel.
I have come to this conclusion based on three main facts that I found while researching. The first reason I found was that due of the amounts of ethanol that humans would need in order to keep up with the demand of our lifestyle. North America would need all crops to be specifically corn for ethanol production. This would lead to food prices sky rocketing. The second reason why this is not an efficient alternative is because its costs more than petroleum to produce. And the third reason why ethanol is not a useful alternative is because a new engine would need to be created so that it could run on pure ethanol. This is because as of now the majority of engines would not be able to run on the product of pure ethanol. There are also many negative effects that impact the environment, in regards to the production of ethanol. The first effect is that the reaction produces carbon dioxide which is a greenhouse gas that the world is trying to get rid of. The second negative effect is that shipping the corn around the world to be produced would require gas, which emits more bad chemicals into the world. It is for these reasons that I personally do not believe that ethanol is a worthwhile alternative to
fuel. 2. Copper(I) oxide + copper(I) sulfide copper + sulphur dioxide 2Cu2O + Cu2S 6Cu2 + SO2
Givens: 50.0g 25.8g 58.0g Ay=? Ty=? %Y=? Mm: 143g/mol 159.1g/mol 63.5g/mol 64.1g/mol
#n (EQUN): 2 mol 1 mol 6 mol 1 mol
Mass of 6Cu2w2Cu2O = 50.0g X 1 mol X 6 mol X63.5 = 66.6g ER 143g/mol 1 mol 1 mol
Mass of of 6Cu2wCu2S = 25.8g X 1mol X 6mol X 63.5 = 61.8g TY LR 159.1g/mol 1 mol 1 mol
%yield = Actual yield X100% Theoretical Yield = 58.0g X100% 61.8g = 0.938 X100% =93.85%
Mrs.Sardella
Per4
Matt , Kait
Mrs.Sardella
Per4
Matt , Kait
|
Introduction
*Limiting Reactant: A reactant that is completely consumed during a chemical reaction, limiting the amount of product that is produced.
*Excess Reactant: A reactant that remains after a reaction is over.
*Theoretical Yield: The amount of product that is predicted by stoichiometric calculations
*Actual Yield: The amount of product that is recovered after a reaction is complete.
*Percentage Yield: The actual yield of a reaction, expressed as a percentage of the theoretical yield.
The equation for calculating the percent yield of a reaction is:
Percentage Yield = Actual Yield X100% Theoretical …show more content…
Yield
Purposes
The purpose of this lab was to determine the limiting and excess reactants for the reaction between potassium iodide and lead (II) nitrate. The second purpose of this lab was to determine the theoretical yield of the reaction. The third purpose of this lab was to determine the percentage yield of the reaction.
Pre-Lab Calculations 1. Potassium Iodide (aq) + Lead (II) Nitrate (aq) Potassium Nitrate (aq) + Lead(II) Iodide (s) 2KI(aq) + Pb(NO3)2 (aq ) 2KNO3 (aq) +PbI2 (s)
2.Given/Unkn: 1.25g 1.25g /Ay=?
m=?
Mm (39.1)+(126.9) (317.2) (39.1) +(14) + (16x3) (207.2)+(126.9x2) =166g/mol 331.2g/mol 69.4g/mol 461.0g/mol
#n (EQUN) 2 mol 1 mol 2mol 1mol
M of PbI2 w KI = 1.25g KI X 1mol KI X 1 mol PbI2 X 461.0g PbI2 = 1.735 g of PbI2 LR 166.0g KI 2 mol KI 1 mol PbI2
M of PbI2 W Pb(NO3)2 = 1.25 g Pb(NO3)2 X 1 mol Pb(NO3)2 X 1 mol PbI 2 X 461.0 g PbI2 331.2 g Pb(NO3) 2 1 mol Pb(NO3)2 1 mol PbI2 =1.739 g od PbI2
3.The theoretical yield of the precipitate in the reaction is 1.735 g of potassium iodide.
Observation Table Material | Mass with no substance | Mass with substance | Physical property of Reactants | Beaker A | 66.97g | 68.22g including KI | White Powder | Beaker B | 65.91 g | 67.16 g including Pb(NO3)2 | Solid white crystals | Filter paper | 0.56g | 2.16g including PbI2 | Yellow Precipitate |
Post Lab Calculations 1. Ay=?
=Mass of the filter paper including substance – mass of filter paper with no substance = mass of …show more content…
PbI2
= 2.16g – 0.56g = 1.6g of PbI2 2. % yield =?
%yield = Actual Mass Theoretical Mass X 100% =1.6g 1.735g X 100% = 92.2 %
Discussion In this lab potassium iodide and lead (II) nitrate are mixed, and a reaction takes place. The lab begins with two separate beakers. The first beaker held the Potassium Iodide and the second held the Lead nitrate. When deionized water was added and the substances were combined together immediately a precipitate was formed, as well as a bright yellow, cloudy liquid appeared. After the filteration processes the product that our group was left with was a yellow powder consistency. If no sources of error occurred, and the lab took place under perfect conditions the percentage yield of this double displacement reaction should be 100%. However, the percentage yield that my group received 92.2%, which means that somehow throughout the lab process 8%, was lost. There could have been many sources of error. For example, some of them could have been left on the stirring rod, in the beakers, on the funnel or it could have even been spilt. We put a hole in the filter; however we used a new filter and avoided a large error. When measuring the filter paper and its contents the contents of the contents spilled or not placed in the filter paper will produce lower actual yield because there will be less product obtained. All these possible sources of error affected the actual yield thus disturbing the % yield outcome, preventing a perfect 100%.
Conclusion
In conclusion, all of the purposes of this lab were met and the various calculations were performed to achieve percent yield, theoretical yield, actual yield, limiting and excess reactant. Through this lab a greater understanding of reaction yields and stoichiometry were gained. My group and I were able to gain a much greater knowledge of how much of a reactant is required for a reaction to take place, and the amount of the left over reactants, how much product was produced and the % yield. Our final yield was 92.2%, which was not a perfect 100% but it was still a successful experiment.
Application Questions 1. C12H22O11 (s) + H 2O(l) 4C 2 H 6 O(l) + 4CO 2
Givens/unkn: 1000g
Mm: 342g/mol 18g/mol 46g/mol 44g/mol
#N (equn) 1mol 1mol 4mol 4mol
C12H22O11withC2H6O= 1000g X 1mol X 4mol X 46gmol = 538g 342g/mol 1 mol 1 mol
In my personal opinion I do not believe that ethanol is a worthwhile alternative to fuel.
I have come to this conclusion based on three main facts that I found while researching. The first reason I found was that due of the amounts of ethanol that humans would need in order to keep up with the demand of our lifestyle. North America would need all crops to be specifically corn for ethanol production. This would lead to food prices sky rocketing. The second reason why this is not an efficient alternative is because its costs more than petroleum to produce. And the third reason why ethanol is not a useful alternative is because a new engine would need to be created so that it could run on pure ethanol. This is because as of now the majority of engines would not be able to run on the product of pure ethanol. There are also many negative effects that impact the environment, in regards to the production of ethanol. The first effect is that the reaction produces carbon dioxide which is a greenhouse gas that the world is trying to get rid of. The second negative effect is that shipping the corn around the world to be produced would require gas, which emits more bad chemicals into the world. It is for these reasons that I personally do not believe that ethanol is a worthwhile alternative to
fuel. 2. Copper(I) oxide + copper(I) sulfide copper + sulphur dioxide 2Cu2O + Cu2S 6Cu2 + SO2
Givens: 50.0g 25.8g 58.0g Ay=? Ty=? %Y=? Mm: 143g/mol 159.1g/mol 63.5g/mol 64.1g/mol
#n (EQUN): 2 mol 1 mol 6 mol 1 mol
Mass of 6Cu2w2Cu2O = 50.0g X 1 mol X 6 mol X63.5 = 66.6g ER 143g/mol 1 mol 1 mol
Mass of of 6Cu2wCu2S = 25.8g X 1mol X 6mol X 63.5 = 61.8g TY LR 159.1g/mol 1 mol 1 mol
%yield = Actual yield X100% Theoretical Yield = 58.0g X100% 61.8g = 0.938 X100% =93.85%