Tuesday

May 3, 2016
Posted by **Deana** on Sunday, October 12, 2008 at 11:25pm.

-I understand how to get the % H2O in the experiment but how do you determine the formula?

- Chemistry -
**DrBob222**, Sunday, October 12, 2008 at 11:56pmThere are two or three ways to do it but this is what I do.

0.6523 - 0.3423 = 0.3100 g = mass H2O.

mols H2O = 0.3100/18 = 0.01721

mols CoCl2 = 0.3423/129.838 = 0.00264.

Mow you want the ratio of mols H2O to 1 mol CoCl2; therefore, let's MAKE CoCl2 1. How do we do that. Divide the mols CoCl2 by itself; i.e., 0.00264/0.00264 = 1.000

Then H2O must be treated the same way so that is 0.01721/0.00264 = 6.53 which rounds to 6.5 and the formula would be CoCl2*6.5H2O or 2CoCl2*13H2O.**I think most of the CoCl2 I read about is CoCl2*6H2O (although there are several different hydrates). Is this an experiment and you have typed in your experimental data? That might explain the funny number. If this is a problem you have been given, please check the numbers to make sure they were typed in correctly. Check my arithmetic to make sure I didn't make an error.** - Chemistry -
**GK**, Sunday, October 12, 2008 at 11:57pmConvert the mass of dry residue, CoCl2, to moles by dividing it by the formula mass of CoCl2.

Determine the mass of water (loss in mass of sample).

Convert grams of H2O to moles by dividing by its formula mass.

Determine the mole ratio of H2O/CoCl2:

(moles H2O) / moles (CoCl2)

Your ratio is somewhat higher than the correct ratio of 6. - Chemistry -
**Deana**, Monday, October 13, 2008 at 12:53amWell what I did to get the % H2O is just subtract the end product(0.3423)from the original sample (0.6523) to get the 0.3100 H20 then found the percent by dividing and got 47.5% H20. Is this the completely wrong method of going about this problem and should do the moles way?

- Chemistry -
**DrBob222**, Monday, October 13, 2008 at 11:37amNo, your work is ok for the percent water. Since you told us in the problem that you already knew how to do that, neither GK nor I addressed the percent part of the problem but went straight to the determination of the formula. By the way, I calculate 47.5% H2O also. You can avoid all of the small numbers (like I used) in the determination of the formula if you take a 100 g sample. That gives you 47.5 g H2O and 52.5 g CoCl2. Then convert those grams to mols and obtain the formula that way. You'll still end up with CoCl2*6.5H2O which is slightly high. I assume these are experimental data.

- Chemistry -
**Sarah**, Sunday, April 29, 2012 at 5:52pmA sample of CoCl2 XH2O with a mass of 1.62g was heated. After heating only 0.88 of CoCl2 remained. Determine the name and formula for the hydrate.