- #1

TheShapeOfTime

For example: CoCl2 * xH2O (The * is meant to be a dot). I need to find "x".

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- #1

TheShapeOfTime

For example: CoCl2 * xH2O (The * is meant to be a dot). I need to find "x".

- #2

chem_tr

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Yes, the hydrate formulae are found in this way, .by successive evaporations and subsequent weight measurements until a stable reading is achieved.

Suppose that you started with [itex]a[/itex] grams of [itex]CoCl_2\cdot xH_2O[/itex] and after several steps, you get [itex]b[/itex] grams of [itex]CoCl_2[/itex]. Then it means that [itex]b-a[/itex] grams of water was evaporated in the process, just divide it by 18 (molar mass of water) to find the overall x. Note that you may not be very precise in this calculation, so if you find [itex]x=2,5[/itex], you may conclude that it can be regarded as [itex]x=3[/itex], etc.

Suppose that you started with [itex]a[/itex] grams of [itex]CoCl_2\cdot xH_2O[/itex] and after several steps, you get [itex]b[/itex] grams of [itex]CoCl_2[/itex]. Then it means that [itex]b-a[/itex] grams of water was evaporated in the process, just divide it by 18 (molar mass of water) to find the overall x. Note that you may not be very precise in this calculation, so if you find [itex]x=2,5[/itex], you may conclude that it can be regarded as [itex]x=3[/itex], etc.

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- #3

TheShapeOfTime

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Of course, we suppose the hydrate compound is pure...

- #5

TheShapeOfTime

[tex]CoCl_2 \cdot xH_2O[/tex]Mertas said:

Of course, we suppose the hydrate compound is pure...

1.62 is the mass before evaportation

0.88 is the mass after evaportation

[tex]1.62 - 0.88 = 0.74[/tex]

[tex]0.74 / 18.02 = 0.041[/tex]

How can this be right?

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- #6

chem_tr

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Oops, I should have written [itex]a-b[/itex] of course. The positive difference between these two measurements gives the amount of water evaporated.

About the difference, [itex]0.041[/itex] moles of water is present in this compound, namely [itex]CoCl_2 \cdot xH_2O[/itex]. The molar amount of the initial compound is not known, but we may consider that [itex]0,88[/itex] grams of [itex]CoCl_2[/itex] is present, you can find the molar mass from [itex]Co:58.93[/itex] and [itex]Cl:35.45[/itex] grams/mol. You then set up a proportion equation to find how many moles of water are present in one mole of [itex]CoCl_2[/itex]. This will give [itex]\displaystyle x[/itex] you're looking for.

I recommend that you use greater amounts of salt and multiple determinations to minimize errors. For example, do the analysis triplicate at one time and use at least 5 or 10 grams of sample, then average the findings you obtained. This will be better.

About the difference, [itex]0.041[/itex] moles of water is present in this compound, namely [itex]CoCl_2 \cdot xH_2O[/itex]. The molar amount of the initial compound is not known, but we may consider that [itex]0,88[/itex] grams of [itex]CoCl_2[/itex] is present, you can find the molar mass from [itex]Co:58.93[/itex] and [itex]Cl:35.45[/itex] grams/mol. You then set up a proportion equation to find how many moles of water are present in one mole of [itex]CoCl_2[/itex]. This will give [itex]\displaystyle x[/itex] you're looking for.

I recommend that you use greater amounts of salt and multiple determinations to minimize errors. For example, do the analysis triplicate at one time and use at least 5 or 10 grams of sample, then average the findings you obtained. This will be better.

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- #7

TheShapeOfTime

Could you tell me a bit more about this proportion equation?chem_tr said:Oops, I should have written [itex]a-b[/itex] of course. The positive difference between these two measurements gives the amount of water evaporated.

About the difference, [itex]0.041[/itex] moles of water is present in this compound, namely [itex]CoCl_2 \cdot xH_2O[/itex]. The molar amount of the initial compound is not known, but we may consider that [itex]0,88[/itex] grams of [itex]CoCl_2[/itex] is present, you can find the molar mass from [itex]Co:58.93[/itex] and [itex]Cl:35.45[/itex] grams/mol. You then set up a proportion equation to find how many moles of water are present in one mole of [itex]CoCl_2[/itex]. This will give [itex]\displaystyle x[/itex] you're looking for.

I recommend that you use greater amounts of salt and multiple determinations to minimize errors. For example, do the analysis triplicate at one time and use at least 5 or 10 grams of sample, then average the findings you obtained. This will be better.

- #8

chem_tr

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I think you'll be able to do these.

- #9

TheShapeOfTime

[tex]

CoCl_2 = 129.83 g/mol

[/tex]

[tex]

\frac{0.88}{129.83} = 0.0068 mol

[/tex]

Second Calculation:

[tex]

\frac{0.041}{0.0068} = 6.0

[/tex]

I'm not sure what you wanted for `X', and I don't think I did the second calculation right.

- #10

chem_tr

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