Determining the Water Content of a Given Hydrated Salt

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

=

= = 7.4286 X 10-3 mol

( )

Number of mole H2O =

( )

=

= 0.03298 mol Mole ratio of CuSO4

= =1

Mole ratio of H2O

= = 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) ( ) ( )

( )

= 0.05596 mol

= 0.3917 mol

Simplest mole ratio ( ) =1

= 6.999 ≈7

The value of x = 7 Empirical formula = Na2SO4 · 7H2O