Lab 5 - Lab pertaining to percent water hydration. PDF

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Lab pertaining to percent water hydration....

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Experiment 5 Lab Report: Percent Water in a Hydrated Salt

Pre Lab Questions: 1. Calcium chloride, a deliquescent salt, is used as a desiccant in laboratory desiccators to maintain a dry environment. Explain. Calcium chloride, a deliquescent salt, is used as a desiccant in laboratory desiccators to maintain a dry environment because deliquescence is the property of having a water vapor pressure lower than the surrounding air. That means that a delinquent is used to absorb water. So, deliquescent substances are able to absorb water from the atmosphere, therefore, Calcium chloride, a deliquescent salt, is used as a desiccant in laboratory desiccators to maintain a dry environment. 2. Experimental Procedure ,PartA.1.What is the purpose of firing the crucible? The purpose of firing the crucible is to clean it. To explain, heating the crucible up helps rid and disinfect it of any moisture that can be contained in the inside of the crucible. 3. A 1.803-g sample of gypsum, a hydrated salt of calcium sulfate, CaSO₄, is heated at a temperature greater than 170°C in a crucible until a constant mass is reached. The mass of the anhydrous CaSO4 salt is 1.426 g. Calculate the percent by mass of water in the hydrated calcium sulfate salt. The percent by mass of water in the hydrated calcium sulfate salt, is as follows. (1.803g-1.426g)/1.803g=2.090g. 2.090g x 100% = 20.9%. Therefore, the percent mass of water in the hydrated calcium sulfate salt is 20.9%. 4. The gravimetric analysis of this experiment is meant to be quantitative; therefore, all precautions should be made to minimize errors in the analysis. a. The crucible and lid are handled exclusively with crucible tongs in the experiment. How does this technique maintain the integrity of the analysis? The crucible and lid are handled exclusively with crucible tongs in the experiment. This technique maintains the integrity of the analysis by touching the crucible with one’s hands will contaminate the experiment, by transferring oils from the hands onto the crucible. This in turn, changes the mass of the reading of the crucible. b. Mass measurements of the crucible, lid, and sample are performed only at room temperature. Why is this technique necessary for a gravimetric analysis?

Mass measurements of the crucible, lid, and sample are performed only at room temperature. This technique is necessary for a gravimetric analysis because scales are calibrated at room temperature. Therefore, taking a mass measurement of a hot object, can and will contaminate the experiment, by causing an error measurement of the weight in grams. 5. The following data were collected from the gravimetric analysis of a hydrated salt: Trial 1

Trial 2

Trial 3

19.437

20.687

18.431

21.626

25.111

22.167

21.441

24.701

21.762

1. Mass of fired crucible and lid (g)

2. Mass of fired crucible, lid, and hydrated salt (g)

3. Final mass of fired crucible, lid, and anhydrous salt (g)

Complete the following table. (See Report Sheet.) See Data Analysis C, D for determining the standard deviation and relative standard deviation for a set of data. Express all of the calculated data with the correct number of significant figures (see Data Analysis A).

Calculations 1. Mass of hydrated salt (g)

2. Mass of anhydrous salt (g)

3.Mass of water lost(g)

4. Percent by mass of

Trial 1

Trial 2

Trial 3

2.189g

4.424g

3.736g

2.004g

4.015g

3.331g

0.1850g

0.4090g

0.4050g

(0.1850g/2.189g)x 100=

(0.4090g/4.242g)x 100=

(0.4050g/3.736g)x 100=

volatile water in hydrated salt (%)

5. Average percent H2O in hydrated salt (%H2O)

6. Standard deviation of %H2O

7. Relative standard deviation of %H2O in hydrated salt (%RSD)

8.451%

→ → →

9.245% (8.451%+9.245%+ 10.84%)/3= 9.512% Sq rt of {[(8.451-9.152)²+(9 .245-9.512)²+(10.84 -9.512)²]/3-1}= 1.216 (1.216/9.512)x100= 13.29%

10.84%

Data Analysis, B

Data Analysis, C

Data Analysis, D

6. a. What is the percent by mass of water in iron(II) sulfate heptahydrate, FeSO₄•7H₂O (or what percent of the molar mass of FeSO₄•7H₂O is due to the waters of crystallization)? The percent by mass of water in iron (II) sulfate heptahydrate, FeSO₄•7H₂O (or what percent of the molar mass of FeSO₄•7H₂O is due to the waters of crystallization) is: Chemical formula = FeSO4 • 7H2O Molar mass of FeSO4 • 7H2O = 277.90 g/mol Molar mass of 7H2O = 7 x 18.00 = 126.00 g/mol %-H2O = (126.00 / 277.90) x 100% = 45.34% Therefore; the percent by mass of water in iron (II) sulfate heptahydrate, FeSO₄•7H₂O is 45.34%. b. What mass due to waters of crystallization is present in a 3.38-g sample of FeSO₄•7H₂O? The mass due to waters of crystallization is present in a 3.38-g sample of FeSO₄•7H₂O is : Molar mass of: FeSO₄•7H₂O and molar mass of compound without water FeSO₄:Molar mass =(FeSO₄•7H₂O) = 54+32+(16x4)+7(2+16) = 276 g/mole. Molar mass =(FeSO₄) = 54+32+(16x4) = 150 g/mole. Now that we have both molar masses, we must subtract the two. So, we take the molar mass value of iron(II) sulfate heptahydrate island subtract that from the molar mass value of iron(II) sulfate, the final answer is the total molar mass of water in crystal: Molar mass of(H₂O) = Molar mass of(FeSO₄ • 7H₂O) - Molar mass of(FeSO₄) = 276 - 150 = 126 g mole. Finally, the amount of water that there is in 3.2 g of the crystal, can be determined. Molar mass(FeSO₄ • 7H₂O) : Molar mass(H20) = mass(FeSO₄ • 7H₂O) :

mass(H₂O), 276 : 126 = 3.38 : x; x = 425.88 / 276 = 1.54 g of H₂O. Therefore, the mass due to waters of crystallization is present in a 3.38-g sample of FeSO₄•7H₂O is 1.54g/mole.

Report Sheet: 3. Instructor’s approval of flame and apparatus: along with completion of the experiment Unknown no. A

1. Mass of fired crucible and lid (g)

2. Mass of fired crucible, lid, and hydrated salt (g)

4.Mass of crucible, lid, and anhydrous salt: 1st mass measurement (g)

2nd mass measurement (g)

5. Final mass of crucible, lid, and anhydrous salt (g)

Calculations

Trial 1

Trial 2

Trial 3

24.403g

24.404g

24.398g

27.012g

26.152g

27.014g

26.184g

25.542g

26.337g

26.186g

25.543g

26.336g

26.186g

25.543g

26.336g

2.6990g

1.7480g

2.6160g

1.7770g

1.1390g

1.9380g

1. Mass of hydrated salt(g)

2. Mass of anhydrous salt(g)

3.Mass of water lost(g)

4. Percent by mass of volatile water in hydrated salt (%)

5.Average percent H2O in hydrated salt (%H2O)

6. Standard deviation of %H2O

7. Relative standard deviation of %H2O in hydrated salt (%RSD)

0.9220g

0.6090g

*(0.9220g/2.6990g) x 100%= 34.160%

(0.6090g/1.7480g)x (0.6780g/2.6160gx 100%= 100%= 34.840% 25.920%

→ → →

(34.160%+34.840% +25.920%)/3= 31.640%

*Sq rt of {[(34.160-31.640)² +(34.840-31.640)²+ (25.920-31.640)²]/ 3-1}= 4.9653 *(4.9653/31.640)x 100= 15.693%

0.6780g

Data Analysis, B

Data Analysis, C

Data Analysis, D

*Required Calculations as directed from Lab Manual *Calculations for Trial 1. Show your work. (0.9220g/2.6990g)x 100%= 34.160% *Calculation of standard deviation and %RSD. Standard deviation= Sq rt. of {[(34.160-31.640)²+(34.840-31.640)²+(25.920-31.640)²]/3-1}= 4.9653 Relative standard deviation of %H2O in hydrated salt (%RSD)= (4.9653/31.640)x 100=15.693% Conclusion: In conclusion, the purpose of this lab was to discover the identity of an unknown sample of hydrated salt. We did exactly this, in the procedure of this lab, by discovering the identity of an unknown sample of hydrated salt by calculating the %H₂O in the ion. According to our

final average of %H₂O which was a result of 31.640%. The results as per each trial were consistent. This allowed me to finally conclude that my results also support the final conclusion that if we heat a sample of hydrated salt, then the mass of the salt will decrease since water molecules escape the ion. Once again, which is what I ultimately concluded as per the experiment: Percent Water in a Hydrated Salt.

Post Lab Questions: 1. PartA.1.During the cooling of the fired crucible,water vapor condensed on the crucible wall before its mass measurement. The condensation did not occur following thermal decomposition of the hydrated salt in Part B. Will the percent water in the hydrated salt be reported as being too high, too low, or unaffected? Explain. During the cooling of the fired crucible,water vapor condensed on the crucible wall before its mass measurement. The condensation did not occur following thermal decomposition of the hydrated salt in Part B. The percent water in the hydrated salt would be reported as being too high. To explain, if water vapor condensed on the wall of the crucible before its mass measurement, the recorded percentage of water will be too high because this is due to the fact of the matter that the crucible would have a greater weight. Also, when it later would get burned off, the perceived loss would be higher. 2. PartA.1.The fired crucible is handled with (oily) fingers before its mass measurement. Subsequently, in PartB.1, the oil from the fingers is burned off. Will

the percent water in the hydrated salt be reported as being too high, too low, or unaffected? Explain. The fired crucible is handled with (oily) fingers before its mass measurement. Subsequently, in PartB.1, the oil from the fingers is burned off. The percent water in the hydrated salt would be reported as being too high. To explain, the oil from an individual’s fingers being on the crucible would cause an increase in the reported percent of water. To explain, this is due to the initial measurement of the crucible being too high. But when the oil is burned off, the result would contribute to the mass reading of the crucible's measurement in the experiment. 3. Part A.1. The crucible is handled with (oily) fingers after its mass measurement but before the ~3 g sample of the hydrated salt is measured (Part A.2). Subsequently, in Part B.1, the oil from the fingers is burned off. Will the percent water in the hydrated salt be reported as being too high, too low, or unaffected? Explain. The crucible is handled with (oily) fingers after its mass measurement but before the ~3 g sample of the hydrated salt is measured (Part A.2). Subsequently, in Part B.1, the oil from the fingers is burned off. The percent water in the hydrated salt would be reported as being too high. To explain, the crucible would be contaminated; and if the crucible is contaminated after the initial measurement, but before the sample, the percentage of water loss will still be too high. Thus, this would cause a larger gap between the initial measurement of hydrated salt and final measurement of the anhydrous salt. 4. Part A.2. Suppose the original sample is unknowingly contaminated with a second anhydrous salt. Will the reported percent water in the hydrated salt be too high, too low, or unaffected by its presence? Explain. Suppose the original sample is unknowingly contaminated with a second anhydrous salt. The reported percent water in the hydrated salt would be too low by its presence. To explain, an anhydrous salt is a salt that lacks water and is dry. Since we are adding an additional anhydrous salt, this means that we are adding more dry mass but not more water. So, since we are adding more of a substance lacking water we will have a lower percent of water to be present than there should be. *5. After heating in Part A.1, the crucible is set on the lab bench, where it is contaminated with the cleaning oil used to clean the lab bench, but before its mass is measured. The analysis continues through Part B.1, where the mass of the anhydrous salt is determined. While heating, the cleaning oil is burned off the bottom of the crucible. Describe the error that has occurred; that is, is the mass of the anhydrous salt remaining in the crucible reported as being too high or too low? Explain. After heating in Part A.1, the crucible is set on the lab bench, where it is contaminated with the cleaning oil used to clean the lab bench, but before its mass is measured. The analysis continues through Part B.1, where the mass of the anhydrous salt is determined. While heating, the cleaning oil is burned off the bottom of the crucible. The

error that has occurred; that is, is the mass of the anhydrous salt remaining in the crucible reported as being too high. To explain, since the crucible was set on the bench and contaminated with the cleaning oil, the mass will be greater than it should be causing an error. So, when heating the anhydrous salt the mass of the percent water and the oil will evaporate. But, it is important to note that it will not evaporate as fast due to the fact that the burning is still burning for the same amount of time. This will result in the mass of the remaining anhydrous salt to be too high due to the cleaning oil. 6. Part B.1. The hydrated salt is overheated and the anhydrous salt thermally decomposes, one product being a gas. Will the reported percent water in the hydrated salt be reported too high, too low, or be unaffected? Explain. The hydrated salt is overheated and the anhydrous salt thermally decomposes, one product being a gas. The reported percent water in the hydrated salt would be reported too high. To explain, the sample is being overheated. The reading would be too high because the gap between the initial measurement and final measurement which would be less due to loss of sample’s measurement will widen. 7. PartB.1.Some of the hydrated salt spatters out of the crucible because of a too rapid heating process.Will the reported percent water in the hydrated salt be reported too high, too low, or unaffected? Explain. Some of the hydrated salt spatters out of the crucible because of a too rapid heating process. The reported percent water in the hydrated salt would be reported too high. To explain, If the sample is overheated and some of the salt is lost due to it splattering out of the crucible, then the reported percentage of water loss will still be too high because the gap between the initial measurement and final measurement which would be lower due to sample leaving the crucible measurement will widen. 8. Part B.2. Because of a lack of time, Bill decided to skip this step in the Experimental Procedure. Will his haste in reporting the “percent H₂O in the hydrated salt” likely be too high, too low, or unaffected? Explain. Because of a lack of time, Bill decided to skip this step in the Experimental Procedure. His haste in reporting the “percent H₂O in the hydrated salt” this can alter his reading and the reading would likely be too low. To explain, if Bill skips this step, he does not ensure that he has removed all of the water content in the salt. This action could alter his result depending on how well he heated his sample. If there is water still present in his sample, then his percent of water will be lower than expected, giving him an inaccurate result and error....