The Use of Analytical Balance in Determining

Objective

1. Utilize the analytical balance to weigh an offered hydrated salt as precise as possible in order to identify the weight of water in the salt

Introduction

Balances are important laboratory equipment as they are utilized to identify the mass of materials. Today, many balances used in the lab are analytical balances which can provide readings up to four decimal locations or greater. High precision is needed in certain speculative work such as product analysis or those including little change in material mass.

Unknown masses of materials are generally estimated with using twodecimal location balances, before they are identified properly with the analytical balances. As the analytical balance is extremely expensive and sensitive, appropriate training has to be obtained by users so that they can utilize it properly.

Device and Materials
1. Sand
2. CuSO4 · xH2O salt
3. Evaporating dish
4. Analytical balance
5. Little testtubes
6. Drying flask
7. Thermometer
8. Electrical heating unit

Experimental Procedure

Results and estimations

Mass of Test tube: 15.8945 g Mass of Test tube + CuSO4 ∙ xH2O salt: 17.

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6698 g Mass of Test tube + CuSO4 ∙ xH2O salt after heat for 30 minutes: 17.1844 g Mass of Test tube + CuSO4 ∙ xH2O salt after reheated for 10 minutes: 17.1537 g Mass of CuSO4 ∙ xH2O salt: 1.7753 g.

Mass of CuSO4:1.2592 g.

Discussion

Many salts readily liquify in water to form solutions. When water is permitted to evaporate from the services, crystals appear. Frequently, the crystals that appear to be dry actually hold a bargain of water within the crystalline structure. If the crystals are warmed, nevertheless, this water is repelled.

These kinds of salt crystals are called hydrates. Hydrated salts are salts which have a certain amount of water chemically combined. The physical homes of a hydrate might vary from the residential or commercial properties of the salt without the water called anhydrous salts.

For instance, the colour may differ. However, the water contained in the hydrate is bound physically to the crystal of the salt and the two can be easily separated by physical means heating. After a hydrate is heated, the remaining salt is called anhydrous which is without water. To differentiate between the hydrate and the anhydrous salt, different chemical names and formulas are used. The dot indicates an attractive force between the polar water molecules and the positively charged metal ion. On heating, the attractive forces are overcome and the water molecules are released leaving behind the anhydrous salt. The water released on heating is called the water of hydration.

A weighed sample of CuSO4.5H2O will be heated at temperatures slightly above 100oC until all of its water content will be driven off as steam. After the heating, the mass of the anhydrous CuSO4, will be determined. There will be a colour change and a decrease in mass after heating since the molar mass anhydrous CuSO4 is less than the molar mass of CuSO4.5H2O. The loss in mass will be the amount of water that is driven off as steam. Copper is a transition metal and is blue in colour.

This is because of the incomplete filled d orbitals. There is slightly energy difference between d orbitals so the electron only needs to absorb small amount of energy to move to 3d orbital. This small energy difference are in the same range as the visible light so under certain frequency of light are absorbed, copper will appears blue in colour. Copper (II) sulphate salt will change from blue to white when is dehydrated. This is because of the d electrons which undergo orbital transition. This happens when copper sulphate datively bonds with water. When there is no transition happens when there are no dative covalent bonds. So the colour will appear white.

Conclusion

The water content in the hydrated salt is 0.5161 g.

References

* Copper (II) sulphate . Retrieved on June 2013 from http://en.wikipedia.org/wiki/Copper(II)_sulfate * Smith, Janice G. (2006).Organic Chemistry.McGraw-Hill
* Bettelheim, Brown, Campbell, and Farrell (2007).Introduction to General, Organic, and Biochemistry. (8th ed.) Pearson

Questions:
1.Determine the weight of water in the salt and calculate the value of x for the salt (CuSO4∙xH2O). Weight of water in salt = mass of CuSO4 . xH2O – mass of CuSO4
= 1.7753g – 1.2592g
= 0.5161 g

The value of x:
1 CuSO4∙XH2O (s) → 1 CuSO4 (s) + X H2O (g)

Mass of CuSO4 =17.1537 g- 15.8945 g
= 1.2592g

Number of mole of CuSO4 = mass / molar mass
= 1.2592 g / 159.62 gmol-1
= 0.00789 mole

Number of mole of H2O = mass/ molar mass
= 0.5161g/ 18.01528
= 0.02865 mole
X = 0.02865 moles of water / 0.00789 moles anhydrous salt = 3.63
≈4.0

2.A 15.00 g sample of an unstable hydrated salt, Na2SO4∙xH2O, was found to contain 7.05 g of water. Determine the empirical formula of the salt.

Na2SO4 . xH2O Na2SO4 + H2O
Mass of Na2SO4 . H2O = 15.00 g
Mass of water= 7.05 g
Mass of Na2SO4 = 15.00 g- 7.05 g
= 7.95 g

Number of moles of Na2SO4 = Mass / molar mass
= 7.95 g / 142.04 gmol-1
= 0.06 mole
Number of moles of water = mass/ molar mass
= 7.05 g/ 18 gmol-1
= 0.39 mole
X = 0.39 moles of water / 0.06 moles anhydrous salt
= 6.5
≈7.0

Empirical formula of Na2SO4 . xH2O= Na2SO4 . 7H2O