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Betsämning of crystal water in barium chloride

Subject: Chemistry
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Introduction

The purpose of the laboratory experiment 4 was to remove the water of crystallization from a variety of barium chloride and then work out how much that has been removed, ie, find out what x is in the formula BaCl xH O.

In order to arrive at the crystal water content was weighed bariumkloriden before it was heated and then after.

Background Theory about the lab

After reading the related chapter of chemistry book, we know that by measuring the mass of a substance loses when heated, one can calculate how much water it attaches.
Crystal water, it is called because it is bound to the substance in the crystal structures.

Fixtures

Material: Crucible, crucible tongs, wave, small spoon, burner, tripod, gauze, triangular and barium chloride.

First, it weighed only crucible to later know exactly how much bariumkloriden weighed. The crucible was then filled one third full of barium chloride, which also weighed before. With the aid of crucible tongs, tripod and triangle crucible was heated for about 5 minutes, stirring for not bariumkloriden would burn-in bottom. While warming themselves lumped bariumkloriden part but it was also the only thing you could see happen with the naked eye.
After the crucible and its contents to cool for a while so everything was weighed again.

Results, conclusions

Crucible mass was 12,131 g, together with bariumkloriden mass was 23.321 g
After 5 minutes of heating when the crystal water delivered as water vapor measured mass of 21.629 g

Using the periodic table are counted molar mass of BaCl and HO:
M (BaCl) = (137.3 + 2 * 35.45) g / mol = 208.2 g / mol
M (HO) = (2 * 1.008 + 16.00) g / mol = 18.016 g / mol

The mass of bariumkloriden without water is calculated by removing the crucible mass:
m (BaCl) = 21.629 g - 12.131 g = 9.498 g

Then the mass and molar mass of BaCl is complete, the amount of substance of the anhydrous bariumklodiden calculated:
n (BaCl) = m (BaCl) / M (BaCl) = 9.498 g / 208.2 g / mol 0.0456 * 10 mol

The amount of water lost during heating (weight before - weight after):
23.321 g - 21.629 g = 1.692 g

Substansmängdes of crystal water:
n (HO) = m (HO) / ​​M (HO) = 1.692 g / 18.016 g / mol 0.0939 * 10 moles

After counting out the amount of substance (s) of both barium chloride and water, one can see that the relationship between them is almost exactly 1:2.
Answer: x = 2, then the formula BaCl 2H O.

Sources of error

One source of error in this lab may be that all water is not boiled away or to bariumkloriden possibly has collected moisture while it cooled.

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In comparison to table collection where the answer also is 2 then our calculation quite accurate.

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The lab was to calculate how many units crystal water content of a unit containing barium. The result we got up in good agreement with table collection, ie two.

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