Question

B. Determine the Formula of an Unknown Hydrate Unknown sample number: 20 Nickel II Molecular formula of the unknown: 1. Mass of crucible + unknown hydrate Nisou "X"H60 (sulfate) 37.40g 36-3lg 1:09g 2. Mass of crucible 3. Mass of unknown hydrate( 1 - 2) 4. Mass of crucible + anhydrous salt a. after first heating b. after second heating 31.3919 37.02 g 0.719 0.389 5. Mass of unknown anhydrous salt (4b - 2) 6. Mass of water lost (1 - 4b)

use the information given to calculate the formula for mass/molar mass of the anhydrous salt of your unknown sample(show calculations)

Answer 1

You know you have a total of 1.09 g of the hydrated salt, whereas you also have 0.71 g of the anhydrous salt. This gives you that the total mass of water that you had in the sample is 0.38 g. What you need to do is get the mass of the anhydrous salt as follows:

Ni - 58.6934 g/mol

S - 32.065 g/mol

O - 15.9994 g/mol (x4) = 63.9976 g/mol

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NiSO4 - 154.756 g/mol

You can calculate the number of moles of the anhydrous salt as follows:

, where m is the mass of the salt and MW is its molecular weight

n = MW

0.719 154.756q/mol = 0.0046moles

You can also calculate the amount of moles of water, knowing that the molecular weight of water is:

H - 1.00794 g/mol (x2) = 2.01588 g/mol

O - 15.9994 g/mol

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H2O - 18.01528 g/mol

Hence the number of moles of water is:

" 0.389 18.01528g/mol = 0.0211moles

You can then get the ratio of moles of the substances as follows:

ratio = 0.0211moles = 4.5976 - 5 0.0046moles

You could assume your formula is NiSO4 x 5H2O

**It is either 4 or 5 molecules of water, but you also have to consider possible errors in the procedure, such as the fact that it is possible not all molecules of water were evaporated or even possible measurement errors.