Experiment 2 CHM138: Determination Of Percent Composition In Hydrate Compounds PDF

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CHM 138LAB REPORTEXPERIMENT 2NAME : ALYA MAISARAH BINTI KAMARUZAMANMATRIX NUMBER : 2020877818GROUP : AS1141ATITLE : DETERMINATION OF PERCENT COMPOSITION IN HYDRATECOMPOUNDSLECTURER’S NAME : MADAM SITI NORIAH BINTI MOHD SHOTORINTRODUCTIONSeveral salts that occur naturally in nature or purchased from ...

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CHM 138 LAB REPORT EXPERIMENT 2 NAME

:

ALYA MAISARAH BINTI KAMARUZAMAN

MATRIX NUMBER

:

2020877818

GROUP

:

AS1141A1

TITLE

:

DETERMINATION OF PERCENT COMPOSITION IN HYDRATE COMPOUNDS

LECTURER’S NAME

:

MADAM SITI NORIAH BINTI MOHD SHOTOR

INTRODUCTION

Several salts that occur naturally in nature or purchased from local retailer are hydrated salts. When an ionic solid is crystallized from water solution, the crystal which forms often contains chemically-bound water molecules. These water molecules maybe referred to as the waters of crystallization or water hydration. The ratio of water molecules to the ions of the salt is a constant. The • in the formula indicates a kind of chemical bond that usually can be easily broken. For example, copper (II) sulphate pentahydrate can be converted to anhydrous copper (II) sulphate by heating: CuSO4. 5H2O(s)

CuSO4 (s) + 5H2O (g) Δ

The Δ sign means that heat is applied. Copper sulphate pentahydrate is a blue crystal, while anhydrous salt. The anhydrous (without water) form of the salt is produced when all the water of hydration is lost. Some salts have their water bound so tightly that make an anhydrous salt is nearly impossible. For example, iron(III) chloride hexahydrate. Before the anhydrous salt would be formed, the salt would be decompose. Total mass of hydrate salt is a sum of mass of the anhydrous salt and mass of water of hydration. This is the formula: Total mass of hydrate salts = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑛ℎ𝑦𝑑𝑟𝑜𝑢𝑠 𝑠𝑎𝑙𝑡 + 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟𝑠 𝑜𝑓 ℎ𝑦𝑑𝑟𝑎𝑡𝑖𝑜𝑛 The percentage composition of water in hydrate can also be calculated by using this formula: Percentage of water =

     

𝑥 100%

OBJECTIVES 1. To prove the chemical formula for Copper sulphate pentahydrate using percent composition concept and the atomic mass. 2. To identify molecular formula of compound A from the thermal decomposition.

APPARATUS

Crucible and lid

Bunsen burner

Clay triangle

White tile

Ring clamp retort stand

Crucible tongs

CHEMICALS    

Copper (II) sulphate pentahydrate Compound A (Compound A can be any of the following) 6 M HNO3 1 M HCl

Li2 SO 4.H2O

MgSO4.7H2O

FeSO4.7H2O

HNO3

Copper (II) Sulphate pentahydrate SrCl 2.6H2O

CaSO4.2H2O

PROCEDURE

a) Determination of Percent Water in Copper (II) Sulphate Pentahydrate

1. Wash and dry a porcelain crucible and lid. Heat both the crucible and cover for about 2-3 minutes to ensure that any water that is adsorbed to the walls of the crucible or cover is driven off. Allow the crucible and cover to cool on a piece of wire gauze 2. After that, weigh the crucible and its lid. Record the mass into the table. 3. Fill the crucible about ¾ full with the hydrate which is, Copper (II) Sulphate Pentahydrate (CuSO4 . 5H2O) 4. Weigh the crucible with the hydrate and then, record the mass in the table. 5. Place the crucible on the clay triangle on a ring stand. With the cover on the crucible slightly ajar, heat the crucible and its contents, gently at first and then strongly, for approximately 810 minutes. The hydrate will probably have a different appearance when the water has been driven off. 6. Let the crucible cool for a few minutes. Then, weigh the anhydrous sample and crucible.

b) Determination of Water Composition and Molecular Formula of Compound A. 1. Wash and dry a porcelain crucible and lid. 2. Weigh the porcelain crucible and its lid on analytical balance. Record the mass of the crucible and its lid in the table. 3. Heat the crucible and its lid over a Bunsen burner for 5 minutes. Make sure it is completely dry. 4. Then, dry and cool the crucible and its lid at room temperature. 5. Weigh the crucible again and record the mass of the crucible and its lid. 6. Drop off a certain amount of the hydrated salt which is Compound A into the crucible dish. 7. Put a few of Compound A into the crucible. 8. Weigh the compound A and the crucible on the analytical balance. Record the mass in the result table. 9. Dry the crucible with the Compound A in it over the Bunsen burner for 5 minutes. Put the lids on so that there is enough of a gap to water can escape. 10. After 5 minutes, allow the crucible to cool for 10 minutes at a room temperature. 11. Weigh the Compound A and crucible again to get a final mass measurement. Record the mass in the result table.

RESULTS a) Determination of Percent Water in Copper (II) sulphate pentahydrate Mass of empty crucible + lid (g) 12.73 12.73

Before heating After heating

Mass of crucible + lid + hydrate (g) 24.11 20.01

Before heating After heating

b) Determination of Water Composition and Molecular Formula of Compound A. Mass of empty crucible + lid (g) Before heating After heating

Trial 1 49.2122 49.2066

Trial 2 49.2122 52.1945

Average 49.2122 50.7006

Mass of crucible + lid + hydrate (g) Before heating After heating

52.1833 50.6819

54.2194 53.2013

53.2014 51.9416

CALCULATIONS a) Determination of Percent Water in Copper (II) sulphate pentahydrate

1. Mass of copper (II) sulphate pentahydrate. Mass copper (II) sulphate pentahydrate = Mass hydrate and crucible – Mass empty crucible = 24.11 g – 12.73 g = 11.38 g

2. Mass of copper (II) sulphate anhydrous. Mass copper (II) sulphate anhydrous = Mass anhydrate and crucible – Mass empty crucible = 20.01 g – 12.73 g = 7.28 g

3. Mass of water in copper (II) sulphate pentahydrate Mass water in copper (II) sulphate pentahydrate = Mass hydrate – Mass anhydrate = 24.11 g – 20.01 g = 4.1 g

4. Percent composition of water in copper (II) sulphate pentahydrate. % composition of water =

  ()   ()

𝑥 100%

= (4.1 g/ 11.38 g) x 100% = 36% 5. Give correct formula of copper (II) sulphate pentahydrate. CuSO4 = 7.28 g H2O = 4.1 g Mol CuSO4 : 7.28 g / (63.5 + 32 + 4(16)) g/mol = 0.0456 mol CuSO4 Mol H2O : 4.1 g / (2(1.008) + 16) g/mol = 0.228 mol H2O For CuSO4 : 0.0456 mol / 0.0456mol = 1 CuSO4 For H2O : 0.228 mol / 0.0456 mol = 5 H2O Correct formula: CuSO4 . 5H2O

b) Determination of water composition and molecular formula of Compound A.

1. Mass of Compound A. Mass compound A = Mass hydrate and crucible – Mass crucible = 53.2014 g – 50.7006 g = 2.5008 g

2. Mass of Compound A anhydrous. = Mass anhydrate and crucible – Mass crucible = 51.9416 g – 50.7006 g = 1.241 g 3. Mass of water in Compound A. = Mass hydrate – Mass anhydrate = 2. 5008 g – 1.241 g = 1.2598 g 4. Percent composition of water in Compound A. = ( Mass of water / Mass of hydrate ) x 100% = ( 1.2598 g / 2.5008 g ) x 100% = 50.38%

5. Identify Compound A Percent composition of water in: a. LiSO4 . H2O = [(2(1.008) + 16) g/mol / (6.94 + 32 + 4(16) + 2(1.008) + 16) g/mol] x 100% = 14.89% b. MgSO4 . 7H2O = [(14(1.008) + 7(16)) g/mol / (24.3 + 32 + 4(16) + 14(1.008) + 7(16)) g/mol] x 100% = 51.18% c. FeSO4 . 7H2O = [(14(1.008) + 7(16)) g/mol / (55.8 + 32 + 4(16) + 14(1.008) + 7(16)) g/mol] x 100% = 45.38% d. SrCl2.6H2O = [(12(1.008) + 6(16)) g/mol / (87.62 + 2(35.45) + 12(1.008) + 6(16)) g/mol] x 100% = 40.54% e. CaSO4.2H2O = [(4(1.008) + 2(16)) g/mol / (40 + 32 + 4(16) + 4(1.008) + 2(16)) g/mol] x 100% = 20.94% Therefore, compound A is MgSO4 . 7H2O. This is because its percent composition of water, 51.18% is the nearest to the result in question 4 which is 50.38%.

DISCUSSION In this experiment, we used copper sulphate pentahydrate as a hydrate. To find the correct formula for this hydrate is by finding the percentage composition of water. From the calculations, we obtained that the stoichiometric ratio between copper (II) sulphate and water was found to be 1:5. Therefore, the correct formula for copper (II) sulphate pentahydrate is CuSO4 . 5H2O. While for compound A, the percent composition of water is 50.38%. So that, to find the substance for compound A, we must use the formula to find the percentage composition of water which is: Percentage composition of water =

       

𝑥 100%

After using this formula, we compare the percentage composition of water for all compounds and MgSO4 . 7H2O obtained the nearest percent which is 51.18% compared to the others. Therefore, compound A is MgSO4 . 7H2O. After that, there are a few safety precaution which is be careful when handling the hot crucible after heating. Severe burns can result if you are not very careful when handling a Burner, tripod, clay triangle, or crucible when these are hot. Next, many hydrates dissolve in their water of hydration on initial heating forming a concentrated solution which easily spatters and sometime violently (like popping pop corn). Therefore, heat gently initially and then strongly only after most of the water has been driven off.

CONCLUSION

At the end of the experiment, we can prove the chemical formula for Copper (II) sulphate pentahydrate using percent composition concept and the atomic mass. So, the chemical formula of Copper (II) sulphate pentahydrate is CuSO4 . 5H2O. From this experiment, we also can identify the molecular formula of compound A from the thermal decomposition. Therefore, compound A is MgSO4 . 7H2O. This is because the percentage composition of water MgSO4 . 7H2O, 51.18% is the nearest to the data obtained which is 50.38%.

QUESTIONS

1. Give the reason why the empty crucible should be heated before starting the experiment. Heating the empty crucible before starting the experiment is to drive off contaminants (moisture) that would measure as added error. 2. Why the process of heating hydrate compound should be started slowly first? If the compound is heated strongly, the fully hydrated salt may "spit" the water out, carrying some of the salt with it and reducing the yield. This explosive spitting out of water is called decrepitation. This will give inaccurate mass at the calculation. 3. What is the important of percent composition of water in the hydrate compounds? Percent composition is important because it helps one to know the chemical composition of certain substances. Percent composition is used to calculate the percentage of an element in a mixture. One can also derive an empirical formula from percent composition. The empirical formula derived from percent composition can help one find the actual molecular weight. REFERENCE 1. https://view.officeapps.live.com/op/view.aspx?src=https%3A%2F%2Fwww.northallegheny.o rg%2Fcms%2Flib9%2FPA01001119%2FCentricity%2FDomain%2F1255%2FAP_Chem_Lab__Hydrate_Lab.doc 2. https://www.npsd.k12.nj.us/cms/lib04/NJ01001216/Centricity/Domain/472/Hydrate%20La b.pdf 3. https://www.quora.com/Why-was-it-important-to-heat-the-empty-crucible-to-a-constantmass-before-adding-the-hydrated-salt 4. https://www.studocu.com/my/document/universiti-teknologi-mara/generalchemistry/other/lab-report-limiting-reagent/7773725/view 5. https://www.reference.com/science/percent-composition-important-aba9d47f0376c892 6. https://salve.digication.com/alexanderantonopoulos/Lab_2_Determine_the_Percentage_of _Water_in_a_Hydra...