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Allen 1

Tamoy Allen Bryant Muniz March 23, 2020 Experiment 8: Molecules and Reactions “Foiled again”

Introduction: In chemistry, reactions between elements and molecules form new compounds. Several factors, like the stoichiometry of a reaction, play an essential role when reactants, are combined to form new products. Stoichiometry plays such an important role because it follows the law of conservation, where mass cannot be created or destroyed. In other words, the total mass of the reactants is equal to the total mass of the products. One factor that falls under the stoichiometry of a reaction is the ratio of reactants to products in the reactions (helps with the law of conservation). Another factor is the number of atoms that are in one mole of the reactant. However, a known number of atoms cannot be measured, so the mass of the materials is measured instead. In part A of the experiment, different masses of Aluminum react with different volumes of Copper (II) sulfate. After the reaction was completed, the product was washed dried and weighed. In part B of the experiment, a reaction between Iron and Sulfur to form Iron (II) sulfide was analyzed to see if it followed the law of conservation and the law of definite proportions.

Materials:          

Two 150 mL Beaker Graduated cylinder 0.250 M Copper (II) Sulfate 6M HCl Heating plate Ethanol Aluminum foil Electronic weight Scale Distilled water Lab Manuel

Observations and Experimental:

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Part A: Chemistry of aluminum and copper (II) Sulfate 2 A l ( S ) +3 CuS O 4 ( aq ) → A l2 ( S O 4 )3 (aq ) +3 C u( s ) The reaction of Aluminum and Copper (II) Sulfate Trial

m(Al), g

mL 0.250 M CuSO4

m(Cu), g

n(Al)

n(Cu)

Ratio n(cu)/n(Al)

Drops 6M HCl

% yield

1

0.1

25

0.51

0.0037

0.008

2.16

5

144

2

0.12

30

0.62

0.0044

0.0098

2.23

6

145

3

0.14

35

0.42

0.0052

0.0066

1.16

7

84.8

4

0.16

40

0.42

0.006

0.007

1.17

8

129

5

0.18

45

0.8

0.0067 0.01258

1.88

9

126

6

0.2

50

1.16

0.007

2.57

10

171

0.018

***This table shows the class data and results of the experiment where Aluminum and Copper (II) Sulfate reacts to make Aluminum (II) sulfate and Copper) ***

Sample Calculation for trial 1: Molar mass of Cu: 63.55 g

Molar Mass of Al: 26.98 g

Moles of Cu→ 0.51 g ⋅

1 mol =.008 mol of Cu 63.55 g

Moles of Al →0.10 g ⋅

1 moI =.0037 mol of Al 26.98 g

Ratio=

.008 mols Cu cu =2.162 mol Al .0037 mols Al

Perecent Yield=

actual × 100 actual=0.51 g Cu theoretical

Theoretical :0.0037 mols Al x

3 mol Cu ( SO 4 ) =0.00555 mol Cu 2 Al

0.00555 mol Cu x 63.55 g Cu=0.353 g Cu Percent Yield=

0.51 g Cu x 100=144 % 0.353 g Cu

a) Moles of Cu formed vs. Moles of Al reacted

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moles of Cu Formed (n)

Moles of Cu formed vs. Moles of Al reacted 0.02 0.02 0.02 0.01 0.01 0.01 0.01 0.01 0 0 0

f(x) = 2.17 x − 0

0

0

0

0.01

0.01

0.01

0.01

0.01

0.01

Moles of Al reacted (n) n(Cu)

Linear (n(Cu))

*** This graph shows the moles of Cu that were formed in the experiment when it reacted with Al***

b) Ratio of moles of Cu/moles of Al vs. Moles of Aluminum reacted

Ratio of moles (Cu/Al) vs Moles of Al reacted 3.00

Ratio moles (Cu/Al) (n)

2.50 2.00 f(x) = 0.51 x + 1.86 1.50 1.00 0.50 0.00

0

0

0

0.01

0.01

0.01

0.01

0.01

0.01

Moles of Al reacted (n) Ratio n(cu)/n(Al)

Linear (Ratio n(cu)/n(Al))

*** This graph shows the product of moles that were made when reacted with Al were almost the same***

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

Based on the theoretical coefficients determine the limiting reagent and the excess reagent o The limiting reagent in this reaction is Aluminum, and the excess reagent is Copper. There is a 2:3 ratio of Aluminum to Copper based on the balanced chemical formula.  What do your graphs show? o The graphs show the law of mass conservation and the law definite proportions. This is especially evident in the second graph, where the ratio of the moles of Copper and Aluminum is the same as the number of moles of aluminum reacted.  Calculate average percent yield and discuss possible sources of error o The average percent yield was 133.3%. Some possible sources of error would be the drying of the Copper. One source of error was decanting the solution after all the Aluminum was dissolved. When decanting, some of the product did not settle and remained on top. The product that did not settle went into the beaker the solution was decanted in. In addition, some groups may not have dried the Copper enough on the heating plate before weighing products, which could have added weight to the final product.

Part B: Whistling Test Tube Analyzing another reaction Fe+S → FeS

M (Fe+ S), g

M (Fee s pro duc ed), g

Theor. m(Fes) ,g

rati o m(F e/S)

Ratio n(Fe/ S)

n(S)

Theor. excess Fe, g

m (Fe excess ), g

m (Fe, react ed), g

2

.062

1.52

1.49E+ 00

3.51

.0629

2

.0624

5.51

5.43

5.48

1.76

1.01

.0537

1.5

.0468

0.39

3.84E01

2.61

.0468

1.5

.0468

4.11

3.78

4.11

1.74

1

2

.0358

1

.0312

0.26

2.51E01

1.74

.0312

1

.0312

2.74

2.55

2.74

1.74

1

4

3.25

.0582

1.75

.0546

0.2

1.94E01

3.05

.0546

1.75

.0546

4.8

4.61

4.8

1.74

1

5

2.25

.0403

1.25

.039

0.073

7.08E02

2.18

.039

1.25

.039

3.43

3.22

3.43

1.74

1

6

3

.0537

1

.0312

1.26

1.23E+ 00

1.77

.0317

1

.0312

2.77

2.69

2.74

1.77

1.02

E m (Fe X used) ,g P

n(Fe)

m (S used) ,g

1

5

.0895

2

3

3

n (Fe react ed)

m (S react ed), g

n (S react ed)

The reaction of Iron and Sulfur ***This table shows the results of the experiment where Iron and Sulfur reacts to make Iron (II) Sulfide) ***

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a) Mass of Iron reacted vs. Mass of Sulfur reacted

Mass of Iron Reacted vs. Mass of Sulfur Reacted 4 Mass of Iron Reacted (g)

3.5

f(x) = 1.75 x + 0

3 2.5 2 1.5 1 0.5 0 0.8

1

1.2

1.4

1.6

1.8

2

2.2

Mass of sulfur Reacted (g) m(Fe, reacted), g

Linear (m(Fe, reacted), g)

This graph shows that more grams of Iron reacted in the reaction than Sulfur.

b) Mass of Iron reacted vs. Mass of Iron (II) sulfide formed

Mass of Iron reacted vs. Mass of Iron (II) Sulfide Formed 4

Mass of Fe reacted (g)

3.5

f(x) = 0.63 x + 0.14

3 2.5 2 1.5 1 0.5 0

2

2.5

3

3.5

4

4.5

5

5.5

6

Mass of FeS formed (g) m(FeS produced),g

Linear (m(FeS produced),g)

This graph shows that more grams of Iron (II) Sulfide was formed when more grams of Iron reacted

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c) Mass of Sulfur reacted vs. Mass of Iron (II) Sulfide formed

Mass of Sulfur reacted vs. Mass of Iron (II) Sulfide produced Mass of S reacted (g)

2.5 2

f(x) = 0.36 x + 0.08

1.5 1 0.5 0

2

2.5

3

3.5

4

4.5

5

5.5

6

Mass of FeS Produced (g)

m(FeS produced),g

Linear (m(FeS produced),g)

This graph shows more grams of Iron (II) Sulfide was formed as the mass of Sulfur increased

d) Moles of Sulfur Reacted vs. Moles of Iron Reacted

Moles of Sulfur reacted vs. Moles of Iron reacted moles of Sulfur reacted (n)

0.07 0.06

f(x) = x − 0

0.05 0.04 0.03 0.02 0.01 0 0.03

0.03

0.04

0.04

0.05

0.05

0.06

0.06

0.07

Moles of Fe reacted (n) n(Fe reacted)

Linear ( n(Fe reacted))

This graph shows the moles of Sulfur reacted was essentially the same as the moles of Fe reacted

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e) Ratio of moles Fe/ Moles S vs. Moles of Iron

Ratio of Moles (Fe/S) vs Moles of Iron

Ratio of moles (Fe/S) vs. Moles of Fe 1.02

1.01 f(x) = 0.18 x + 0.99 1.00

0.99

0.98 0.03

0.04

0.05

0.06

0.08

0.09

0.1

Moles of Fe (n) Linear (n(Fe))

n(Fe)

This graph shows that the ratio of moles of Fe/S and the moles of Fe that was started with in the reaction is almost the same

f) Mass of sulfur plus mass of Iron Reacted vs. Mass of Iron (II) Sulfide formed

Mass of Sulfur plus Mass of Iron reacted vs. Mass of Iron (II) Sulfide formed 6 f(x) = 0.99 x + 0.22

Mass of (Fe+S) reacted(g)

5 4 3 2 1 0

2

2.5

3

3.5

4

4.5

5

5.5

6

Mass of FeS produced (g) m(FeS produced),g

Linear (m(FeS produced),g)

This graph shows the mass of the Iron and sulfur together is essentially the same of the Iron (II) Sulfide that was formed in the reaction.

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Results Questions 1. Analyze and draw a conclusion for each graph. Five out of the six graphs have a positive slope indicative of a positive correlation between the moles the reaction started with and the moles that were formed. In graph E (which shows the relationship between the ratio of Iron and Sulfur vs. the moles of Iron), the line is more horizontal, indicating the ratio of moles of Iron and Sulfur at the end of the reaction was the same with how much Iron the reaction started with. 2. By looking at your data and graphs, describe the relationship between the reacted moles of iron and reacted moles of sulfur. The graphs show a positive linear slope. The relationship between the reacted moles of iron and the reacted moles of sulfur is directly proportional (1:1 ratio). 3. What is the law that this reaction is following? The reaction is showing the law of mass conservation and the law of definite proportions. After looking at the mass of the reactants and the mass of the products, it is evident that the mass was conserved. 4. Using the graph, predict the mass of iron necessary to react with 1.3 moles of sulfur. Since there is a one to one ratio between Iron and Sulfur 1.3 moles of Iron would react with 1.3 moles of sulfur, which is 72.6 g. 1.3 mol S ⋅

1 mol Fe 55.845 g Fe ⋅ =72.6 g 1mol S 1 mol Fe

Discussion/Conclusion: The experiments in this lab were used to help students analyze and conclude the stoichiometry of reactions concerning the law of conservation and the law of definite proportions. The law of conservation states that mass cannot be created or destroyed. The law of definite proportions states that when elements combine with another to form another compound contains a fixed and constant proportion concerning mass. In part A of the lab, aluminum foil was dissolved in Copper (II) sulfate to form solid Copper (reddish/brown precipitate) and Aluminum (II) sulfate. As the reaction went into completion, the Aluminum was slowly being replaced by Copper. In this reaction, Aluminum was the limiting reagent, and Copper was the excess reagent. Aluminum was determined to be the limiting reagent because it was the reactant that we had the least of. The limiting reagent in a reaction also determines how much product has formed the reaction. After the reaction was completed, the substance was decanted, and the solid Copper was washed, dried, and weighed. After being washed, dried, and weighed, the mass and moles of the product were determined. These values were then compared to the values of the reactants. It is evident from the graphs and the tables in part A that the mass was conserved. In graph B, in part A, the ratio of the moles of Copper and Aluminum was the same as the number of moles of Aluminum reacted. During part A of the experiment, most of the percent yield percentages calculated came out to be over 100%.

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This could be due to improper decantation techniques in addition to not letting the solid Copper dry enough, which could have added to the overall mass of the product formed. In part B of the lab, Iron and Sulfur combined to create Iron (II) Sulfide. In part B of the lab, the mass was given for each mini experiment to compute the moles and other things needed to complete the chart. During this part of the lab, it was easier to take note of the law of conservation and definite proportion. As seen in the graphs (a-f), there was a positive correlation between the reactants and the products formed. The amount (grams) of reactants the experiment started with was similar to the number of products the experiment ended with. This was depicted in graph F, where the mass of the Iron and sulfur combined was the same as Iron (II) Sulfide that was produced. Graph D shows moles of sulfur reacted vs. moles of Iron Reacted, where the moles of sulfur that reacted was the same for Iron reacted, also showing the law of definite proportions.

Focus Questions: Part A: 1. What is the balances equation for the reaction of aluminum and copper (II) Sulfate? 2 A l ( S ) +3 CuS O 4 ( aq) → A l2 ( S O 4 )3 ( aq ) +3 C u( s) 2. What is the average percent yield of the reaction? The average percent yield was 133.3% Part B: 1. Was the law of mass conservation proved in this reaction? Yes, the law of conservation was proved in this reaction. This was evident in the table and graphs where the amount (grams) of reactants was similar to the number of products produced. This was also shown in graph f, where the mass of the Iron and sulfur together was the same as the Iron (II) Sulfide that was formed in the reaction. 2. What is the percent yield of this reaction? The average percent yield is 99.92%

≅100 %

Sample calculation: Experiment 1: 0.0642 mol S ⋅

1 mol Fe =0.0642 mol Fe 1mol S

( 0.0642 mol fe ) (55.845 g )=3.59 g Fe 3.51 g Fe × 100=9 8 % 3.59 g Fe

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3. What graph will prove the law of definite proportions for this reaction? The graph that will prove the law of definite proportions for this reaction is graph d (Moles of Sulfur Reacted vs. Moles of Iron Reacted). The reaction has a one to one ratio between sulfur and Iron. The moles of sulfur that reacted were the same for Iron reacted.

References: The Editors of Encyclopedia Britannica, and Erik Gregersen. "Law of Definite Proportions.", https://www.britannica.com/science/law-of-definite-proportions. Post lab question: 1. Based on the appearance of your reaction (aluminum with copper (II) sulfate) in the beaker, which reagent do you think was consumed, and which reagent had some left over? Explain. The reagent that was consumed was Aluminum (the limiting reagent), and Copper was leftover (the excess reagent). As the solution was being stirred continuously, the Aluminum started to dissolve, and a reddish/brownish participate started to form at the bottom of the beaker, which is also indicative of Copper being left over. 2. If 5.0 g of iron metal is reacted with 15.0 g of Cl2 gas, how many grams of ferric chloride (FeCl3) will form? In this reaction Cl2 is the limiting reagent and Fe is the excess reagent. With Cl2 being the limiting reagent only 22.87 g of FeCl3 can be formed. 2 F e (s) +3 Cl 2 ( g ) → 2 FⅇC l3 ( S ) Molar mass of Fe : 55.845 g Molar mass of Cl2: 70.90 g 1 mol Fe =0.0895 mol Fe Moles of Fe:5.0 g x 55.845 g Moles of Cl2 : 15.0 g x

1 mol Cl2 =0.212 mol Cl 2 70.90 g

Mols FⅇC l3 :0.0895 mols Fe x Mols FⅇC l3 :0.212 mols Cl2 x 0.141 mols F ⅇC l 3 x

2 mols FⅇC l3 =0.0895 mol FⅇC l3 2 mol Fe

(excess reagent)

2 mols FⅇC l3 =0.141mol FⅇC l3 (limiting reagent) 3 mol Cl2

162.195 g FⅇC l 3 =22.87 g FⅇC l 3 1 mol FⅇC l3

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3. For the reaction above the amount if ferric chloride obtained in the lab was 9.15 g. Calculate the percent yield of the reaction. 9.15 g Fe Cl3 x 100 %=40 % 22.87 g Fe Cl3 4. In general, what are some reasons for obtaining a percent yield less than 100%. Some reasons for obtaining a percent yield less than 100% is that some of the reactants at the beginning of the experiment may have been lost due to evaporation, or there was not enough time allotted for the reaction to go into completion completely. This would result in the product not being produced. Another reason maybe some of the products could have been lost while decanting, drying, and washing process of the experiment....