Exp 4 - Hydrated Salt Formula

SKU3013 : Chemistry I

D20152072027

HYDRATED SALT FORMULA Objective : 1. To determine the volume of water content in hydrated salt. 2. To determine the formula of hydrated salt. Material / apparatus : Crucible with cover Hot plate Analytical balance Glass rod Hydrated salt :

CuSO 4 • x H 2 O

Method : 1. The crucible is heated for a while. 2. The crucible is weighed. 3. Approximately 2 gram of hydrated salt is added, weighed and the data is recorded. 4. The salt is heated gradually in crucible without cover for around 3 – 5 minutes. 5. The color changes are observed. 6. Heating is stopped and waited until the crucible cool down. 7. Observed. 8. Weighed. 9. The salt is grinded using glass rod and heated again. 10. The salt is left cool to room temperature. 11. Weighed again. The heating is continued until the weight constant. 12. The salt temperature is recorded after the final weight. 13. Few drops of water is added at room temperature and the temperature is recorded. 14. Observed.

Data : Mass CuSO4 before heating After heating (mass crucible) After heating (mass crucible) + CuSO4

2.0152 g 32.0568 g 34.0720 g 1

SKU3013 : Chemistry I

D20152072027

After heating (mass crucible + CuSO4) – first reading After heating (mass crucible + CuSO4) – second reading After heating (mass crucible + CuSO4) – third reading Average reading

33.3412 g 33.3333 g 33.3312 g 33.3352 g

Calculation : 1. Calculate the mole number of hydrated salt, results of unhydrated/anhydrous salt and weight of leave water from this reaction. Mass of empty crucible Mass of crucible with hydrated salt Mass of hydrated salt Mass of crucible with anhydrous salt Mass of anhydrous salt Mass of water lost Moles of hydrated salt Moles of anhydrous salt Moles of water Number of water molecules per formula

32.0568 g 34.072 g 2.0152 g 33.335 g 1.278 g 0.737 g 0.0126 mol 0.008 mol 0.0409 mol 5 mol

Mass of hydrated salt : 34.072 g – 32.0568 g = 2.0152 g Mass of water lost : 34.072 g – 33.335 g = 0.737 g Mass of anhydrous salt : 33.335 g – 32.0568 g = 1.278 g Moles of hydrated salt :

( 2.0152 g ) ×

2. If the formula of hydrated salt is

1 mol =0.0126 mol 159.608 g

CuSO 4 • x H 2 O

, verify x. Compare with the real

molecular formula. Compound

Copper (II) Sulfate, CuSO4 2

Water, H2O

SKU3013 : Chemistry I

Mass

Number of mole

Ratio number of mole

∴

D20152072027

= 33.335 g – 32.0568 g

= 34.072 g – 33.335 g

= 1.278 g

= 0.737 g

( 1.278 g ) ×

1 mol 159.608 g

( 0.7368 g ) ×

1mol 18.015 g

¿ 0.008 mol

¿ 0.0409 mol

0.008 =1 0.008

0.0409 =5.11 ≈5 0.008

Formula from this experiment is

CuSO 4 • 5 H 2 O

it is similar with the real molecular formula.

3. List any error sources. Comment the x value.

3

(copper (II) sulfate pentahydrate) and

SKU3013 : Chemistry I

D20152072027

(a) Not heating the crucible before adding the sample. If this step is skipped, any water in the crucible prior to adding the sample will be accounted as being part of the compound. This will give results of a higher mass of water being lost, which will analyzed as a greater number of water molecules per formula unit. (b) Starting to heat the sample on high flame instead of gradually. This could cause the compound to foam and spill over the crucible. If the lost mass is not accounted for the calculations, the molar mass will be messed up. (c) Taking the mass of the anhydrous salt too soon before having two consecutive masses being the same. This will yield a lower number of water molecules per formula unit. 4. What is the name of process when water release from hydrated salt, endothermic or exothermic process? For this experiment the process when water release from hydrated salt is exothermic process because the temperature after adding a few drop of water is higher than the temperature after adding a few drop of water. This process will release the energy. Exothermic (outside heating describes a process or reaction that releases energy usually in the form of heat.

Discussion : A number of ionic compounds contain one or more waters of hydration in their formulas. A good example of this is copper (II) sulfate which exists in an anhydrous form, CuSO 4(s), as well as a 4

SKU3013 : Chemistry I

D20152072027

pentahydrate form, CuSO4•5H2O. Many anhydrous compounds have a strong tendency to absorb water vapor from the air, thus becoming hydrated compounds. On the other hand, some hydrated compounds tend to spontaneously lose their water of hydration when they are placed in a dry environment. In this experiment, the number of water molecules associated with each formula unit of a salt will be determined. Some compounds have water molecules in their structure when they form a solid. These substances are called hydrates. The amount of water present is in a definite mole ratio of water to compound. The amount of water in the hydrated compound will be determined by heating the sample of the compound in order to drive off the water. An anhydrous form of the compound will be yielded when the compound are dehydrated through heating. In this experiment the hydrates used is CuSO 4 • xH2O. CuSO4 • xH2O is blue in its hydrated form, upon heating it slowly converts to white CuSO4. CuSO4 • xH2O (s) + Heat → CuSO4 (s) + xH2O (g) Hydrate is the chemical compound that contains water. A common hydrate is the familiar form of Copper (II) Sulfate. Chemically, it is cupric sulfate pentahydrate, CuSO 4•5H2O. When a crystal of the substance is formed, five molecules of water (H 2O) are combined in the crystal with each molecule of Copper (II) Sulfate (CuSO4). This water is called water of crystallization. When cupric sulfate pentahydrate is heated, the water of crystallization is driven off and anhydrous cupric sulfate is formed. It has several properties different from the pentahydrate, e.g., color, density, and crystal structure. Hydrate salt + Heat → Anhydrous Salt + Water

5

SKU3013 : Chemistry I

D20152072027

Hydrated salt

Anhydrous salt

When anhydrous salt is dissolved in water, a blue solution will be formed due to the formation of a crystalline structure, upon hydration give CuSO4 the blue color. Like at the end of the experiment, we add two drops of water to the dish and it will turn blue again. The blue color is due to the water of crystallization (CuSO4•5H2O). When this is removed by heating, basic copper sulphate is left (CuSO4•5H2O) and this can be further broken to simple CuSO4. It is essential you realize the water is not there as free liquid, but as a molecule in its own right and it is bound to the copper atom. These molecules fulfill a structural role in as much as they fill voids in the crystal structure and let the salt ions take up a regular and therefore crystalline, shape of minimal energy content. In the case of all transition metals, the water of hydration is linked to the d (or higher) orbital and these distort.

After few drops of water is added 6

SKU3013 : Chemistry I

D20152072027

In this experiment we need to find the number of moles of water molecules in the hydrate used. Firstly, the hydrated salt was weighed at first before heating, and then the sample was heated to allow the water to volatilize. After heating, the dried sample was reweighed. The loss in weight corresponds to the water content. Dividing the mass of the water lost by the original mass of hydrate used is equal to the fraction of water in the compound. (Eddy, 2001) Conclusion : From this experiment, we can know the weight of the dried CuSO 4 is 1.278 g and the weight of water in CuSO4· xH2O is 0.737 g. The ratio of mole of H2O to CuSO4 is 1 : 5 where x is 5. Therefore, the empirical formula of the salt is CuSO4· 5H2O. Reference : Brown, T. L. (2014). Chemistry : The Central Science (3rd ed.). Pearson Australia Group Pty Ltd Eddy, D. (2001, June 2). CHEMISTRY 103: PERCENT WATER IN A HYDRATE. Retrieved from College

of

Engineering

&

Science

-

Louisiana

http://www.chem.latech.edu/~deddy/chem103/103Hydrate.htm SKU3013 Chemistry I, Laboratory Manual

7

Tech

University

: