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Chemistry practical 

Paper type: Essay
Pages: 5 (1158 words)
Categories: Chemistry
Downloads: 16
Views: 2

Hydrated Iron(II) Sulphate crystals contains water of crystallization. Water of crystallization is the fixed amount of water that an hydrated salt contains.

The formula of an hydrated salt is generally written as M.xH2O, where the M is an hydrous salt formula, and x is the number of moles of water per mole of hydrated salt.

In this experimental investigation, we are going to determine the formula of hydrated Iron(II) Sulphate, which is given to us in the instruction as FeSO4.

xH2O. Our main objective is to determine the value of x.

Determining the value of x will involve the use of two different methods then comparing the results.

One of the methods involves heating the hydrated to decompose it into an hydrous salt of Fe+2 and water, the calculating the value of x, employing the method called ‘formula of calculating the empirical method’. This involves assuming the FeSO4 as an element, lets call it Y and H2O as another element and lets call it Z.

The other method involves making the hydrated salt into a solution then perform some titration with a suitable aqueous solution, preferably, potassium permanganate(VII).

The reaction between Acidified iron(II) Sulphate can be represented by

5 Fe+2 + MnO- 2 + 8H+ ? 5Fe3+ + Mn2+ + 4H2O

in this reaction, the Iron is oxidised from oxidation number +2 to +3, i.e. lost one electron, manganete lost five electrons (oxidised), and the hydrogen is reduced. The other un-represented species in the reaction did not change during the course of the reaction. Calculating the number of moles in the solution and the can allow as to calculate the formula.


Iron (II) Sulphate salt normally exists as a hydrated salt. Hydrated salt contains a fixed amount of water and in most case, the number of moles of water is discrete.

In some chemistry books, the value of x is given as 7, i.e. there are seven moles of water per mole of Iron (II) Sulphate hydrated salt.

Design of an experiment

This investigation involves performing at least two different experiments and comparing the results.

Two methods will be used to find the value if x.

Experiment 1

Decomposition methods.

Variables and there control

In this experiment, there are so many variables that needed to be controlled for the results to be varied. The following are some of them.

– Temperature: we had to wait for the crucible and the contents to cool before measuring other wise, the water was still evaporating whilst hot, until after two minutes when the water had stopped evaporating every time we did the experiment.

– Mass (controlled variable), the mass of the crucible needed to measured as accurately as possible. We had to use an electronic beam balance to measure it and the experiment was done away from the taps to prevent water plinking on to the crucible, which could influence our results.

Diagram of apparatus


The procedure is given in the instruction. It involves heating the salts, noting down the change in mass every 2 minutes, however, you have to wait after heating before measuring the mass, until the crucible is cool.

Control of Variables

Mass; using an electronic beam balance to measure the mass reduced the uncertainty in the measurement. Also the experiment is best done I a dry environment, keeping water and other liquids away to prevent the salt absorbing the water before weighing or else there could have been a more errors in the measurement.

Data collections

Mass of the crucible = 17.09g

Crucible + FeSO4.xH2O = 18.49

FeSO4.xH2O = 18.49 -17.09 = 1.40g

Time /s

Mass of the crucible /g

Mass of FeSO4 /g

Mass of H2O /g





















Method 2

Method II involved making the iron (II) sulphate into a solution and then titrating it with a solution of potassium permanganate with a concentration of 0.0100-mol dm-3. full instruction is given in the question paper which is attached at the end.


> Mass of anhydrous salt dissolved in sulphuric acid

> Volume of acid

> Volume of water

> Volume of Potassium sulphate used in titration

Method for collection of sufficient data

In order for us to collect sufficient data, the experiment is repeated until we get concordant results.

Control of variables

In this method, all the major variables needs to be taken into account if our method is to be varied, not only this for us to compare with the first method, all else we can not tell which is the varied formula of iron (II) Sulphate. The most important variables been: –

– The Volume of sulphuric acid needs to be measured as accurately as possible to reduce random errors.

– The uncertainty of the measuring devise (Measuring cylinder, pipette, Burette, etc.) needs to taken into account during calculations.

Data Collection

Mass of FeSO4.xH2O dissolved in water = 3.00g

Volume of FeSO4.xH2O / (?0.25) cm3

Burette Reading /cm3

Initial reading

(? 0.25) cm3

Final reading

(?0.25) cm3

Titre Value

(?0.25) cm3








Needs editing

In this experiment to find the formula of an hydrated salt, there are several factors that needs to be controlled if the experiment is to be varied. The following are the factors and their methods of control

1. Volume of water and acid: the quantity of water and acid used to add to the salt solution needs to be exactly 250cm3 or another value which can be determined on the graduations on the measuring apparatus i.e. volumetric flask and pipette.

2. Mass of the salt dissolved in the acid: the mass of the salt dissolved in the acid was measured as accurately as possible. We had to measure the same mass three times and we were getting the same 3grams.

The second method to find the formula of hydrated Iron(II) Sulphate salt involves dissolving the salt in sulphuric acid and water making up to 250cm3 solution. a method called titration is then employed, titrating it with potassium hydroxide. Full instructions are given in the question paper.


From method 1

Iron (II) sulphate decompose on heating giving anhydrous Iron(II) Sulphate salt and water. This can be represented by the equation

FeSO4.xH2O(s) ? FeSO4(s) + xH2O (g)

Mass of anhydrous salt that remained after decomposition is

18.49 -17.95 = 0.86g

Number of moles of anhydrous FeSO2 salt

= 0.85g/Mr(FeSO4)


= 0.00559 moles.

Mass of water produced

= 18.49g -17.95

= 0.55g

Number of moles of water produced

= 0.55./Mr(H2O)

= 0.54/18

= 0.0306 moles

Let H2O be X and FeSO4 be y

There fore, the ratio of number of moles of Y to X is

0.00559 moles : 0.0306 moles

0.00559 moles 0.00559 moles

1 : 5.5

1 : 6

Hence x is equal to six and the formula becomes


Calculation from method 2

Needs thorough editing before submitting,

– Organise work,

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Chemistry practical . (2020, Jun 01). Retrieved from https://studymoose.com/chemistry-practical-new-essay

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