Paper Example on Water of Crystallization

Discussion

When a set-up of an aqueous solution is set up for specific compound or salt, there is an evaporation of water molecules, and the water that evaporates is referred to as water of hydration. However, there is some water that remains to be part of the crystals when all the liquid has already evaporated. The salts that behave in this manner are referred to as hydrated salts or in other words hydrate (Zhou et al 78). For every hydrate there must be a particular composition; a definite number of water molecules are taken and combined with an indeed given formula of a given unit of salt. For instance, given a formula, CuSO4 10H2O describes that ten water molecules are in chemical combination with every formula of CuSO4. Here are a few other formulas that are considered to be common hydrates:

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CuSO4 10H2O Copper (II) sulfate decahydrateBaCl2 2H2O barium chloride dihydrateNa2CO3 5H2O Sodium carbonate pentahydrate The stability of hydrates depends. Some of the hydrates lose their water spontaneously while others do not just lose their water they must be heated at very high temperatures so that they can lose their water of hydration. Substances that are prone to absorbing water from the atmosphere are considered to be hygroscopic (a good example would be CuSO4). Some elements that are hygroscopic absorb a lot of water such that they form solutions.

Some of the hydrates require to be heated so that they can completely dry off the water of hydration. The resultant water-free compound is referred to as an anhydrous salt. Some of the equations that form and reaction process when the compound react are listed below:

Hydrated salt (solid) heat anhydrous salt (solid) + water of hydration (gas)

CuSO4 10H2O (solid) heat CuSO4 (white) + 10H2O (colorless)

When the reaction has taken place, it can be reversed since it will absorb the water from the atmosphere. For instance:

CuSO4 + 10H2O CuSO4 10H2O

Determining how much amount of water is comprised in a specific hydrate can be achieved through deciding the percentage composition that is within that particular compound. Experimentally, heating a hydrate should be done to a certain constant mass. When the compound has been heated one will notice that there will be a decrease in mass of that sample the reason behind is because it has lost water, the resultant product is what is referred to as anhydrous salt. It is very possible for one to determine the amount of water lost from the hydrate while at the same time assess its percentage (Kotz et al 89).

%water = Mass of water lost divided by the mass of hydrate multiplied by a hundred percent

A formula of a given hydrated salt is given as some moles which are present in water for a crystalline structure. Deliquescent compounds are salts that may absorb water from the atmosphere thus eventually resulting in being hydrated. The compounds have a very high affinity for water. When the compound is absorbing water from the atmosphere the whole structure of the salt is altered to incorporate the change that is the water molecules. Therefore, a crystal will change in the manner in which it appears to a different color. For one to calculate the amount of water that is contained in a specific crystal, the weight is taken then the difference from before and after being used.

Determination of the amount of water that is contained or absorbed in a particular salt the differences in the mass was a useful factor in this process. Therefore, the mass of the hydrous salt minus the mass of the anhydrous salt one will get the mass of water absorbed. Using this method requires the percentage of water, with the specific number of moles and also the complete formulae of the compound were used.

Hypothesis

Magnesium Sulfate was used in this experiment, and if the water is driven out of the salt, then it will remain to be colorless the reason being is that it is made up of group two metals, and for a fact, they do not form any colors.

Objectives

There will be the use of 'weight by difference' which is determined with mass quantities. A mass of an anhydrous compound will be heat as a sample to a cert constant mass. Determination of grams and will be converted to moles, the water percentage in a given sample, lastly determination formula for a sample compound.

Procedure

It started by the cleaning of the crucible for the sole reason of making sure that there are no particles which are there with a tissue paper. The placing of the clay triangle was to be done strategically in the iron ring. Then the next thing was to light the Bunsen burner since it will be used to heat the hydrated magnesium Sulfate. The crucible was now to be placed on the clay triangle so that it would heat gently for about two minutes.

The removing the crucible using a crucible tong is advised, then placed somewhere right and allowed to cool for some time than measuring the mass. The crucible was weighed using an analytical balance. Magnesium Sulfate was later on put in bits of small amounts into the crucible then the mass was to be measured. Then the results were to be recorded down. Now the next process was to heat the crucible that contains the magnesium sulfate which took about 15 minutes. After 15 minutes were up, the crucible was removed from a clay triangle by a crucible tong. Then it was put down to cool down a little bit then the mass of the anhydrate was measured by the same analytical balance (Hein, Judith and Robert 105).

Then there was a final measure which was recorded. The anhydrate after its job was over it was to be disposed of and the other materials cleaned up to take them to their respective places where they were first gotten from.

Individual Data

The mass of an empty crucible was 16.749grams

The crucible after magnesium sulfate was put in it resulted in 18.293 grams

Hydrated Magnesium sulfate 1.544 grams

The crucible after it was filled with anhydrate magnesium sulfate 17.516 grams

Anhydrate magnesium sulfate gave a result of 0.78grams which was the amount of water that remained. The mass of hydrated Magnesium sulfate minus the Mass of anhydrate magnesium sulfate gives us 0.77 grams which were the amount of water (Moore and Conrad 90)

Individual Calculations and Accuracy

Here there is the calculation of the percentage of water that is contained in the hydrate.

The answer that will be calculated will also have an explanation as to why the sample size of the tested will not change the answer. The formula that will be employed here will be that the percentage of water equals to the mass of water divided by the mass of the hydrate multiplied by one hundred therefor from our calculation 0.7670 grams of water divided by 1.544 grams of hydrate multiplied by one hundred will give us 49.68% of water. And the formula of the hydrated salt and also the moles that are contained in the water that were found from the sample above 7670 grams of water multiplied by one mole of water will give us 0.04261 moles of water and 18 grams of water. The numbers of moles that presumably will be remaining from the crucible of anhydrous salt are calculated as 0.78 anhydrate Magnesium sulfate multiplied by one mole of magnesium sulfate will give us 0.006453 moles of the remaining magnesium sulfate and a whole 120 grams of Magnesium sulfate.

Now we will calculate the ration some moles to the nearest integer that is contained in the anhydrous salt. Here we found out 0.043 moles of water to a ratio of 0.0065 moles of magnesium sulfate; therefore it would be correct to say that 1 mole of magnesium is equaled to 7 moles of water. And the complete formula of the salt that we have just calculated will be given as MgSO4 7H2O and make a percentage error will be provided as (measured value subtract the accepted amount) then divided by accepted value and multiplied by one hundred here our measure value will be equals to 49. And the accepted value will be 67% which equals to 51. Calculating it from our formula above we will get (67%-51)/51. On the other hand, 12% multiplied by one hundred we will get an error of negative two point eight one seven.

Conclusion and Summary of the Essay

If a student were to determine a percent of water hydration in a crystal and conclude to be thirty-one percent. Then a correct value would be fifteen percent. The likely source of the error would be the two. That is the stable hydrate which is estimated to be weighing two 2.32 grams which are heated and given space to cool down for the sole reason of making sure that there is no water of hydration. The mass measured after the heating process respectively gives us 2.21 grams, 1.79 grams, and lastly 1.79 grams. For instance, if we were to get a residue with a molar mass of 178 then the moles of water that would be contained in the salt would be 2.32 minus 2.21 grams which will give us 0.11 grams of water then we will subtract 1.78 grams from 2.32 grams and get 0.54 grams of water. On the other hand, 2.32 grams subtract 1.9 grams we get an average mass of 1.93 grams of a stable anhydrous mole of anhydrous.

Works Cited

Kotz, John C, Paul Treichel, John R. Townsend, and David A. Treichel. Chemistry & Chemical Reactivity. , 2015. Print.

Moore, John W, and Conrad L. Stanitski. Chemistry: The Molecular Science. , 2014. Print.

Hein, Morris, Judith N. Peisen, and Robert L. Miner. Foundations of Chemistry in the Laboratory. Hoboken, NJ: J. Wiley & Sons, 2011. Print.

Zhou, Siyu, Yan Zhou, Ziye Ling, Zhengguo Zhang, and Xiaoming Fang. Modification of Expanded Graphite and Its Adsorption for Hydrated Salt to Prepare Composite Pcms. , 2018. Internet resource.