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What is the percent water found in colbolt (II) chloride hexahydrate?

I especaiily don't understand what hexahydrate and nonahydrate is.

Would someone mind explaning how to do the problem and the definations?

Thanks.

2006-10-10 13:51:16 · 3 answers · asked by Anonymous in Science & Mathematics Chemistry

3 answers

The "hexahydrate means there are 6 water molecules attached to the CoCl2 molecule.

It is written as: CoCl2-10H2O (note: there is a dot, NOT A DASH, placed inbetween the CoCl2 and the 10H2O, but I can't do that on my computer, so imagine a dot in place of the dash).

To find percent of each element, you simply do this as a general % problem:

% = (part / whole) x 100%

ex: first you must determine the molecule weight of the ENTIRE molecule (including the water). So just add up all the atomic weights, including the 6H20, and you will get a sum total "whole".

Then to determine, say, % Co, look up atomic weight of Co. That is the "part".

Take that part number and divide it by the whole number and multiply by 100 % That will be your answer. Try it.

If you do this correctly you will find you have approximately 19.1% cobalt in this compound.

And by the way, it is spelled "cobalt".

2006-10-10 14:02:57 · answer #1 · answered by MrZ 6 · 0 1

the hydrate part is saying how many water molecules are "attached" to the ionic compound in a crystal shape. Cobalt(II) Chloride hexahydrate has the formula CoCl2-6H2O. So you can just add up the mass of the water in the compound and the mass of the whole compound. Divide the water by the whole and multiply by 100 for a percent

2006-10-10 20:56:23 · answer #2 · answered by Greg G 5 · 0 0

Hexa is the prefix for 6. So there are six hydrate molecules in the compound.
to solve, find the atomic mass for each, then add them up. Then divide the atomic mass of one element by the sum to find the percent.
Or, in this case, divide the atomic mass of H and O added together by the sum to find the percent of hydrate in the compound.

2006-10-10 20:55:57 · answer #3 · answered by consumingfire783 4 · 0 0

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