Title:

The Use of Analytical Balance in Determining the Water Content of a Given Hydrated Salt.

Objectives:

To determine weight of water in a hydrated salt.

Introduction:

Most experiment requires measuring substances as starting point. Balances are used to

measure weight. There are two types of balances: two-decimal place balances and analytical

balances. High accuracy is needed in certain experimental work such as material analysis or

those involving small change in material mass. Unknown masses of materials are normally

estimated with the use of two-decimal place balances, before they are determined accurately

with analytical balances. As the analytical balance is very expensive and sensitive, adequate

training has to be acquired by users so that they can use it correctly.

In this experiment we will learn to use the balance properly and to evaluate the magnitude of

various common errors encountered in weighing and find out the weight of water in a

selected hydrate.

Apparatus and Materials:

Sand, CuSO4 · xH2O salt, Evaporating dish, Analytical balance, Small test tubes,

thermometer, electrical heater.

Procedure:

1. An electrical evaporating dish was filled with sand to about half-full. The electric heater

was turned on and sand was heated slowly until the temperature reached 120 – 125ºC.

2. Small test tubes were heated in the hot sand for 10 minutes to dry them and then cooled

down to room temperature.

3. When the test tube were cooled, they were weighed first with two-decimal balance and

then with an analytical balance. The weights were recorded.

4. The test tube was filled with CuSO4· xH2O salt to about 0.5 cm depth. The weight was

determined.

5. The test tube with CuSO4· xH2O salt was heated in the sand at 120–125 for about ℃

30minutes.When it is cooled, the test tube was weighed.

6. Reheated for another 10 minutes cooled it and reweighed when cooled.

7. All the weights were recorded.

Results:

Materials Weight

Test Tube 15.0394g

Test Tube + CuSO4· xH2O 16.8222g

Test Tube + CuSO4· xH2O after heat for 30 minutes 16.1725g

Discussion:

In chemistry, weighing the analytical sample is often the very first step of any quantitative

analytical chemical method. In the laboratory, there are 2 types of balances: two-decimal

place balances and analytical balance. Two-decimal place balances digitally display a mass

reading, in grams, to 2 decimal places while analytical balances digitally display a mass

reading, in grams, to 4 decimal places. By comparing them, analytical balance is much more

accurate as it is sensitive to ± 0.0001 g. (Brewer, 2006 )

Water is present in many compounds especially inorganic salts. Hydrates are compounds that

incorporate water molecules into their fundamental solid structure. In a hydrate (which

usually has a specific crystalline form), a defined number of water molecules are associated

with each formula unit of the primary material. The moles of water present are significantly

contributive in terms of mass. However the water molecules are often held loosely and can be

easily removed by heating the hydrate. The material is said to be anhydrous and is referred to

as an anhydrate when all hydrating water is removed. (Eddy, 2001 )

Hydrate salt + Heat → Anhydrous Salt + Water

An anhydrous form of the compound will be yielded when the compound are dehydrated

through heating. In this experiment the hydrates used is CuSO4· xH2O. CuSO4· xH2O is dark

blue in its hydrated form, upon heating it slowly converts to bluish white CuSO4.

CuSO4· xH2O(s) + Heat → CuSO4(s) + xH2O(g)

Blue Bluish white

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 )

In this experiment, the reaction of removing water molecules is reversible, thus, in order to

get rid of water completely; water vapour formed during heating must be removed by either

using a tissue paper, cotton wool or using a desiccator. The desiccator is a metal container

with a tight fitting lid and small amount of drying agent or desiccant inside to remove traces

of water from an almost-dry sample. A desiccant is a compound that rapidly absorbs water

from surrounding. (Eddy, 2001 )

There are several precautions that needs to be taken care of throughout the experiment to

prevent errors and injuries. While measuring using the analytic balance, be sure to do it on a

draft-free location on a solid bench that is free of vibrations as it will affect its accuracy. Do

not place the chemical directly on the pan, use a weighing bottle, beaker, watch glass, etc.

Corrosive liquids and solids are always placed in a vapor tight, pre-weighed container before

weighing on an analytical balance. Lastly, this experiment involves heating, thus pay

attention to prevent burning. (Sibert, 2013 )

Conclusion:

Analytical balance is very sensitive and able to measure up to ± 0.0001 g sensitivity. The

weight of dried of CuSO4 is 1.1331 and the weight of water in CuSO4· xH2O is 0.6497g. The

ratio of mole of H2O to CuSO4, X is 5, hence the empirical formula of the salt is CuSO4·

5H2O

Reference:

Brewer, W. E. (2006 , September 6). Introduction to Analytical Measurements. Retrieved from The University of South Carolina Department of Chemistry & Biochemistry: http://www.chem.sc.edu/analytical/chem321L/labs/Expt1.pdf

Eddy, D. (2001 , June 2). CHEMISTRY 103: PERCENT WATER IN A HYDRATE. Retrieved from College of Engineering & Science - Louisiana Tech University: http://www.chem.latech.edu/~deddy/chem103/103Hydrate.htm

Sibert, G. (2013 , June 10 ). ANALYTICAL BALANCE. Retrieved from http://www.files.chem.vt.edu/RVGS/ACT/lab/Analytical_Balance.html

Questions and Answers:

1. Determine the weight of water in the salt and calculate the value x for the salt.

(CuSO4· xH2O).

Weight of CuSO4· xH2O = 16.8222g – 15.0394g

= 1.7828g

Weight of CuSO4 = 16.1725g – 15.0394g

=1.1331g

Weight of water = 1.7828g – 1.1331g

= 0.6497g

Number of mole CuSO4 = 1,1891

64.0+32.07+16.0 ( 4 ) g/mol

= 1,1891

160.07g /mol

= 7.4286 X 10-3 mol

Number of mole H2O = 0.64 97

1.0 (2 )+16.0 g /mol

= 0.64 97

18.0g /mol

= 0.03298 mol

Mole ratio of CuSO4 = 7.4286 X 10−3mol7.4286 X 10−3mol

= 1

Mole ratio of H2O = 0.03609mol

7.4286 X 10−3mol

= 4.8588

≈ 5

Value of X = 5

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.

Elements present (X) Na2SO4 H2O

Mass (x) 7.95g 7.05

Molar mass of (X) 142.07g 18.0g

Number of mole of (X)

n (X )= m(X)M (X)

¿ 7.95 g142.07

= 0.05596 mol

¿ 7.05g18.0

= 0.3917 mol

Simplest mole ratio

n (X )0.0559mol

0.0559mol0.0559mol

= 1

0.3917mol0.0559mol

= 6.999

≈ 7

The value of x = 7

Empirical formula = Na2SO4 · 7H2O