Lab Experiment #517
Determining the Water Content of an Ionic Hydrate
Anhydrous Compound: a compound that does not contain any water.
Ionic Hydrate: an ionic solid that contains water molecules within its structure.
Waters of Hydration: water molecules that combine with an anhydrous compound to form an ionic hydrate.
Waters of hydration are not just physically trapped within the solid, but are chemically bound to specific sites within the hydrate's lattice structure, therefore they are included in the ionic hydrate's chemical formula.
CuSO4(s) + 5H2O(l) ----------------> CuSO4 . 5H2O(s)
(please read the above period on the right as a raised dot, as this is how hydrates are written)
The above chemical formula shows the initial state of CuSO4 as a solid anhydrous compound. When 5 waters of hydration are added, they become part of its lattice structure and the formula is the written with the raised dot. Note the subscripts denoting liquids and solids.
To name an ionic hydrate, begin with the initial anhydrous compound, in the above case, CuSO4. Because Cu has a charge of +2 and SO4 has a charge of -2, this compound would be copper (II) sulfate.
Next, determine the ratio of components, including the water. We have 1 Cu++, 1 SO4 -- molecule, and 5 molecules of water, so the ratio is 1:1:5. We have five waters of hydration, which we indicate using prefixes. The prefixes are mono, di, tri, tetra, penta, hexa, hepta, octa, nona, and deca. Because we have five, we use the prefix "penta" and the word "hydrate". Thus the full name for the ionic hydrate is:
copper(II)sulfate pentahydrate
To analyze the water content of an ionic hydrate, the ionic hydrate is decomposed:
CuSO4 . 5H2O(s) ---------heat---------> CuSO4(s) + 5H2O(g)
Note that subscripts denoting solids and gases.
By measuring the mass of the ionic hydrate before heating and then again after heating, we can determine how much water was lost, and therefore the percentage of water lost and therefore, the number of waters of hydration the initial ionic hydrate contained.
Mass lost upon heating = mass of water,
Mass % of water = mass of water, g/mass of original sample, g (100%)
So if we had 1.00g as our mass of water and an initial mass of 5.00g ionic hydrate before heating:
1.00g/5.00g= 0.20, 0.20x100% = 20%
Next we determine the chemical formula of the ionic hydrate to find the number of moles of anhydrous compound in the original sample. To do this, take the mass of sample after heating (as in, the mass of the compound without any water left in it), and then find the molar mass of the anhydrous compound, in the above case, CuSO4 in a 1:1 ratio.
Recap on calculating molar mass:
Using the periodic table of elements, find the the mass of each individual atom and multiply it by the number of atoms present. In this case, CuSO4, there is one copper, one sulfur, and 4 oxygens:
Cu = 63.546
S= 32.065
O= 15.9994x4= 63.9976
Total: 159.60g/mol
So the number of moles in our sample after heating is the weight of the sample divided by the molar mass of the compound, 159.60, hence if we had 3.00g of anhydrous compound CuSO4 left after heating away the waters of hydration, to find out how many moles of CuSO4 we have: 3.00g CuSO4/159.60 g/mol CuSO4, which gives us 1.9 x 10^-2 moles CuSO4.
We then use the same technique for determining molar mass to find the molar mass for H2O, which is 18.02g/mol. We divide the amount we calculated for our water mass by the molar mass of water to determine the number of moles of water in our original sample (before heating).
1.00g/18.02g/mol = 5.5 x 10^-2 moles H2O
To determine the number of waters of hydration in the ionic hydrate, we use the mole calculations for both CuSO4 and H2O in the following equation:
number of waters of hydration = #moles of water/# moles CuSO4
number waters of hydration = 5.5 x 10^-2 mol/1.9 x 10^-2 mol = 2.8, which is approx 3, and thus:
CuSO4 . 3H2O, or "copper(II)sulfate trihydrate".
Yaaaaaay!
Up Next: Standardizing Hydrochloric acid solutions- DOUBLE YAY!! Due 10/11/11