Water Of Hydration Essay Example
Water Of Hydration Essay Example

Water Of Hydration Essay Example

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Objective: To determine experimentally the water of hydration for the hydrate Borax. Theory: In this experiment, we examined and found the water of hydration of the hydrate borax. Borax is derived from the element Boron, which is a metalloid chemical element with properties between those of carbon and aluminum(95CI). Boron is relatively rare, constituting only 3 PPM of the Earths crust(95CI). It is most commonly found in the borate minerals borax, and kernite(95CI). Borax is also otherwise known as a hydrated salt. First, let me describe exactly what a salt is. Salts are compounds composed of a metal ion plus a nonmetal ion(86BC85). Some examples of this would be NaCl, AgBr, and FePO4 . Salts may also be composed of a polyatomic ion which has a positive oxidation number(94CP159). Now I w

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ill define hydrated salts. They are salts that combine with water to form crystalline compounds called hydrates(93CL357). In other words, water becomes part of the salt crystal. Water + Certain compounds ---- > Hydrates In hydrates, water appears to be held in loose combination with the associated compound(93CL357).

This is shown by the ease in which hydrates give up water on heating or in a dry atmosphere. The water in combination with a hydrate is always in a definite proportion by weight(86BC85). This is shown to us in the formulas for hydrates. An example of this would be the formula for Borax, which is: Na2 B4 O7 . 10H2O. The centered dot between the formula for the salt and the formula for the water is not a decimal point, nor does it mean to multiply. It is simply a way of

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showing that there is a bond between the salt and the water(93CL357). When such salts are heated, this bond is usually broken, and the water is driven off(93CL357). For example: Na2 B4 O7 . 10H2O ---- > Na2 B4 O7 + 10H2O In each of these compounds, there is a definite amount of water attached to the salt crystal(93CL357). To simplify matters, we may say that in Na2 B4 O7 . 10H2O, for example, there are 10 water molecules for every 1 molecule of hydrated salt. Using this information, we will now try to find the water of hydration. This process however, contains many thought provoking steps.

The first thing you must do in order to find the water of hydration is find the molecular weight of the anhydrous form of the salt you will be working with. The anhydrous form of the salt contains no water. Molecular weight is found by adding up the atomic weights of the elements in the compound. The following shows an example of the molecular weight of borax: Na2 B4 O7 . 10H2O. Elements # Atoms Atomic Weight Molecular Weight Na 2 23 46 amu B 4 11 44 amu O 17 16 272 amu H 20 1 + 20 amu 382 amu In this example, you can see that weve multiplied the number of atoms of each element by its coefficients. In water, for example, you can see that there are 2 hydrogen atoms and 1 oxygen atom. However, there is a coefficient of 10 before it. When you multiply this coefficient of 10 to the number of atoms of each element you will get 10 for

oxygen and 20 for hydrogen. We got a total of 17 for the oxygen only because we had to add the number of atoms from the salt, which was 7.

After repeating this same procedure for each element, we got a total of 46 atomic mass units(amu) for the sodium, 44 amus for boron, 272 amus for oxygen, and 20 amus for Hydrogen. After adding all the elements atomic weights together, we got a total of 382 amus. This number is the atomic weight of the Borax. You now must begin the weighing of the hydrate and other materials involved such as papers and test tubes your working with. After weighing, you may start heating the hydrate to begin the process of finding the water of hydration. However, it is important to make accurate measurements and to record them correctly. In measuring, you must compare what your measuring to some calibrated device. Whether it is a meter stick, a scale, or a graduated cylinder, you must apply the same rules. By comparing the object your working with, to the different graduation marks on your calibrated device, you can come up with the correct measurements with the correct amount of significant figures.

The number of significant figures in a quantity is the number of digits that are known accurately, plus the first uncertain digit or the estimated number. Once youve found the molecular weight and weighed the chemicals properly, you may proceed with the heating process as directed. When your heati! ng, it is very important to make sure that youve driven all the water of hydration from the compound(93CL357). This is called heating

to constant weight(93CL357). When finished heating, we represented the data as follows: HEAT MX . nH2O ---- > MX + nH2O Hydrate anhydrous water salt In this equation, MX . nH20 represents the weight of the hydrated salt in grams. After heating, the bond breaks between the salt and the water, leaving MX + nH20(59CA168). Now, MX represents the weight of the anhydrous salt and, n represents the number of molecules per water molecule of the hydrated salt(59CA168). This leaves the H20, which represents the grams of water the hydrated salt lost during the heating, otherwise known as the water of hydration(59CA168).

In this lab, you must abide by many safety precautions in order to avoid harming yourself or someone else around you. Goggles and Aprons must be worn at all times to avoid damage to your eyes or skin. Borax is also a very harmful substance, so you must never handle it with your hands or any other part of the body. You also must be sure that while heating, your workspace is free from papers or other flammable substances. Acting like a professional in the lab is mandatory in order to ensure a safe environment for you and your classmates. Raw Data: See sheet labeled: Raw Data #2 Sample Calculations: Does not apply Results: Several results were found during this lab that I felt were relevant in finding the water of hydration for the hydrate, Borax. One such result was the weight of amt. of borax we were using, which we found to be .44 gms. This answer came about by subtracting the weight of the test tube which was 12.76 gms.

From the weight of the test tube and the hydrated salt which was 13.20 gms. Another result found was the weight of the anhydrous salt( .20 gms.).

We found this answer by subtracting the weight of the test tube which weighed 12.76 gms. From the weight of the test tube and the anhydrous salt which weighed 12.96 gms. We then found the weight driven from the borax, which was .24 gms. This answer which was stated the weight driven from borax is also known as the water of hydration. Conclusion: Finally, this brings us to our conclusion. We found that the water of hydration of the hydrate borax was .24 gms. This answer was obtained by subtracting the weight of the anhydrous salt(contains no water) which was .20 gms. from the weight of the hydrated salt(contains water) which was .44 gms. The water of hydration was found this way because after the hydrated salt (which contains water) was heated, all the water was driven out, giving us an anhydrous salt, which contains no water. By subtracting the two, you can see how much water escaped the salt, which indeed is the water of hydration.

Although a conclusion was made, there were still errors that may have greatly faltered it. During the procedure, a series of heatings were performed on the hydrate Borax. It was to be heated for 10 min. then left to cool for 15 min. It would then be weighed. You would then repeat this same procedure directly after one another. However, because the end of the period was near, we werent able to reheat directly after the cooling period. This

caused the hydrate to be exposed to the air for over one week before we could work with it again. Therefore, because the hydrate was left to cool for such long periods, water may have entered or escaped, not letting us reach its constant weight. This is a great error indeed, assuming that the whole point was to get the hydrate to its constant weight. If it wasnt at its constant weight, this could have greatly faltered the weight of the anhydrous salt and more importantly the water of hydration.

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