Before the chemical formula of a compound can be known, the mass (and therefore the mols) of each element in a sample of that compound must be experimentally determined. Although this can be a lengthy procedure involving dozens of chemical reactions, the ultimate goal is very simple: find out the mass of each element present in the compound. Once you know the mass of each element, you can calculate the mols of each element.
Let's review the "mol" concept for a moment. The mass of a single atom of an element is expressed in atomic mass units (1 amu = 1/12 the mass of an atom of 12C). Since it is impossible to weigh atoms, it is necessary to perform chemical experiments with many more than just a few atoms. There is no difficulty in using large numbers of atoms so long as the numbers of atoms are maintained in the proper ratio. For example, a single cake might contain three eggs and one cup of sugar. A thousand cakes, however, would necessitate multiplying all the ingredients by a thousand; however, the ratio of three eggs to one cup of sugar would be maintained.
One "mol" of anything is 6.02 x 1023 items (Avogadro's number). This number is of particular importance because it allows the chemist to use atomic masses from standard tables simply by changing the unit from amu to grams. For example, one atom of hydrogen has an average mass of 1.008 amu, whereas one "mol" of hydrogen atoms has an average mass of 1.008 g ( NOTE THAT ONLY THE UNIT CHANGES). Using mols of substances, experiments may be performed on a large enough scale that initial and final quantities may actually be weighed. So, no matter how large or small the quantities are, the number of atoms and "mols" of atoms required to perform a certain reaction will be in the same ratio.
The empirical formula (simplest whole number ratio of atoms in a compound) represents not only the composition of a single unit of the compound, but also the overall composition of a much larger sample as well. It is for this reason that composition data are always treated by first converting the masses to "mols."
The whole number ratio of mols of the different elements is then obtained simply by dividing each number of mols by some number which will produce a whole number. This is usually accomplished by dividing the smallest number of mols (or 1/2, or 1/3, or 1/4 of this number) into all the others. The method used to get the whole numbers is unimportant so long as the proper ratio is maintained. The ratio of mols is exactly the same as the ratio of atoms in the compound.
The particular compound you will be
working with today is called a
"Hydrates" contain a fixed amount of water which is
to as "water of crystallization". The water in the formula
may be represented in several ways, for example, both MgCl2·6H2O
correct formula for hydrated magnesium chloride (a hexahydrate). Water
although it is incorporated into the crystal structure of the solid
like the ions, can be easily removed by heating the solid.
while heating breaks down the crystal structure and allows the water to
escape, it often leaves the sample in the form of a powder.
it is a relatively simple matter to decompose a sample by heating to
the number of molecules of water in each formula unit of the compound.
The purpose of today's lab is to
determine the empirical formula for
the compound: CuxCly·(H2O)z.
You will accomplish this by first heating a sample of the hydrate to
the water of crystallization. Then you will determine the amount
of copper present by using zinc to 'displace' the copper from
You will determine the weight of chlorine by difference. From
data, you will be able to determine the mols of copper, chlorine, and
water in the sample. Finally, you will use these mols determine
the simplest ratio of Cu (x), Cl (y), and H2O (z), i.e., the
Cleaning the Crucible and Lid:
Let's work through an EXAMPLE (your numbers will be different!!) and see how we can use your data to determine the empirical formula for this compound. Assume a 2.300 g sample of CuxCly·(H2O)z weighed only 1.950 g after heating in the crucible. This means that the original sample contained 0.350 g or 0.0192 mols of H2O:
mols of H2O = (wt. of H2O lost / 18.02 g per mol)
mols of H2O = (0.350 g / 18.2 g/mol) = 0.0192 mol
Let's further assume that after the zinc was added, you collected a total of 1.234 g of solid copper. This means that that the original sample contained 0.0194 mols of Cu:
mols of Cu = (wt. of Cu / 63.55 g per mol)
mols of Cu = (1.234 g / 63.55 g/mol) = 0.0194 mol
Since we know the amount of Cu and H2O the original sample contained, we can determine the amount of Cl and therefore the mols of Cl by difference:
grams of Cl = wt. of original sample - wt. of H2O lost - wt. of Cu
grams of Cl = 2.300 g - 0.350 g - 1.234 g
grams of Cl = 0.715 g
mols of Cl = (wt. of Cl / 35.45 g per mol)
mols of Cl = (0.715 g / 35.45 g/mol) = 0.0202 mol
We now know that your sample contains 0.0194 mols of Cu, 0.0202 mols of Cl, and 0.0192 mols of H2O. We can determine the empirical formula for this sample by dividing these numbers by the smallest number of mols:
0.0194 mols of Cu / 0.0192 mols = 1.010 Cu
0.0202 mols of Cl / 0.0192 mols = 1.052 Cl
0.0192 mols of H2O / 0.0192 mols = 1.000 H2O
Within experimental error, these ratios are all one. Therefore, this sample contains a ratio of one Cu to one Cl to one H2O and its empirical formula is: Cu1Cl1(H2O)1 or CuCl·H2O.
Important Note: the data for the
above example was chosen to
a clear result for the empirical formula, just like the examples in
text. However, as you know, in the lab, every measurement
errors, some large, some small. Because of this, your molar
for copper, chlorine, and water, may not turn out to be nice whole
For example, you may obtain data like this: 1.123 Cu, 2.592 Cl,
7.311 water. In this case, you would round the copper to 1, round
the chlorine up to 3, and round the water down to 7. In other
unless your data comes out exactly .500, use the rounding rules.
(Updated 2/14/13 by C.R. Snelling)