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 hydrate.  "Hydrates" contain a fixed amount of water which is referred to as "water of crystallization".  The water in the formula may be represented in several ways, for example, both MgCl2·6H2O and MgCl2·(H2O)6 are correct formula for hydrated magnesium chloride (a hexahydrate). Water of crystallization, although it is incorporated into the crystal structure of the solid just like the ions, can be easily removed by heating the solid.  However, while heating breaks down the crystal structure and allows the water to escape, it often leaves the sample in the form of a powder.  Therefore, it is a relatively simple matter to decompose a sample by heating to determine the number of molecules of water in each formula unit of the compound.

I have added a link to a short video that goes into more detail about hydrates and explains the process we will be using in today's lab.  It also has several examples of how you would calculate the empirical formula of different hydrates based on your experimental results.  I HIGHLY advise you to watch this video as part of your preparation for your quiz and lab.


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 remove the water of crystallization.  Then you will determine the amount of copper present by using zinc to 'displace' the copper from solution.  You will determine the weight of chlorine by difference.  From this 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 empirical formula:

CuxCly·(H2O)z(s)  +  heat    CuxCly(s)  +  zH2O(g)

CuxCly(aq)  +  Zn(s)    xCu(s)  +  yCl-(aq)  +  Zn+2(aq)


Cleaning the Crucible and Lid:

  1. Thoroughly wash your crucible with soap and water.  Be sure to rinse with distilled water and dry with a paper towel.
  2. Place the crucible on a clay triangle which is supported by an iron ring.
  3. Adjust your Bunsen burner to produce a large inner blue cone (it should sound like a jet engine).  Now heat the crucible for 5 minutes or until it glows yellow-orange.  Then allow to cool to room temperature.
  4. Weigh the crucible on the balance to the nearest 0.001 g.

Determining the Amount of Water:
  1. Accurately weigh between 1.000 - 1.500 g of the hydrate into the crucible.  Weigh the crucible and sample to the nearest 0.001 g.
  2. Place the crucible and sample on the clay triangle and heat with a low flame (small blue flame, but no inner blue cone).
  3. Use your scoopula to continually stir your sample.
  4. Hold the burner in your hand and wave it back and forth across the crucible so no part gets too hot.  DO NOT OVERHEAT THE SAMPLE!  IF IT MELTS YOU MUST START OVER!
  5. You should begin to notice that the hydrated green crystals will start to turn to the brown anhydrous color.  Continue heating until all of the green crystals have turned brown.  If there are any green crystals in the crucible or on your scoopula, you need to continue heating.
  6. After all of the water has been driven off (no green crystals left), continue to gently heat and stir the sample for an additional minute or two.
  7. Remove the crucible and sample from the clay triangle and allow to cool to room temperature.
  8. Weigh the crucible, and product on the balance to the nearest 0.001 g..  The difference between this weight and your initial weight of the sample is the amount of water in your sample.
Determining the Amount of Copper:
  1. Using your scoopula, transfer the brown crystals from the crucible to a 150 mL beaker.
  2. Rinse the crucible with 5-10 mLs of distilled water and add this to the beaker.  Repeat a couple of more times and add these washings to the beaker as well.
  3. Add approximately 10 mL of 3M sulfuric acid to your beaker.  You should now have about 50-60 mL of solution in the beaker.  If not, add enough distilled water to bring it up to this volume.
  4. Swirl the contents of the beaker to make sure all of the brown crystals dissolve.  Note any color changes you see.
  5. Obtain a piece of pure zinc plate from your instructor and add it to the solution in the beaker (it will be approximately 1" x 3").
  6. You will notice a series of physical and chemical changes (copper precipitating and the color of the solution will begin to fade).  As the copper comes out of solution, it will coat the zinc plate, so you need to periodically shake the plate to remove any copper that has built up.  Allow these reactions to continue for 20 - 30 minutes or until the solution is completely colorless..
  7. After all of the reactions have ceased, use your scoopula to GENTLY scrape off any copper still on the zinc plate.  Rinse the zinc plate with water, dry it with paper towels and return to your instructor.
  8. Make sure you view and understand How to Set up a Vacuum Filtration before coming to lab!  (I modified it from the procedure used at Dartmouth Univ.)
  9. You will now use a Buchner funnel to separate the solid copper from the solution.  Weigh a piece of filter paper to 0.001 g.  Add this filter paper to the Buchner funnel and wet the filter paper with distilled water before filtering.
  10. Use a distilled water bottle to quantitatively transfer all of the contents of the beaker onto the filter paper.  NOTE:  IT IS VERY IMPORTANT THAT ALL OF THE COPPER BE TRANSFERRED TO THE FILTER.
  11. Wash the copper on the filter paper with several 10 mL portions of distilled water to remove any of the solution.  Use your glass stirring rod to breakup any clumps (be careful that none of the copper sticks to the stirring rod).
  12. Allow the copper and the filter paper to air dry for an additional 5 minutes after the last water drop has passed through the filter.
  13. Thoroughly clean your watch glass and use your grease pencil to put your initials on the edge. NOTE:  DO NO FLAME DRY YOUR WATCH GLASS IT WILL SHATTER!
  14. Weigh the clean watch glass.
  15. Use a folder paper towel to carefully 'wick' away any water droplets that may be sticking to the sides of the Buchner funnel.  Be carefully not to touch the copper!
  16. Quantitatively transfer the copper and the filter paper to the watch glass.
  17. Dry the watch glass/filter/copper in an oven for 10 minutes to remove any residual water.
  18. Let the watch glass cool to room temperature and weigh.
  19. Repeat Steps 16 and 17 until you obtain a constant weight (within 0.01 g).  Now you are sure that all of the water has been removed.
  20. Subtracting the original weight of the watch glass and the filter paper from this weight gives you the amount of the copper in your sample.
Determining the Amount of Chlorine:
  1. There are a number of chemical means to independently determine the amount of chlorine in your sample.  However, due to safety and time constraints, you will determine the amount of chlorine simply by difference.
  2. From the above procedures, you know the weight of the original sample.  You also know the weight of the water in the sample, and the weight of the copper in the sample.  Therefore, the amount of chlorine is simply the weight of the sample minus the weight of the water and the weight of the copper.
Calculation of Results:

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 give a clear result for the empirical formula, just like the examples in your text.  However, as you know, in the lab, every measurement contains errors, some large, some small.  Because of this, your molar results for copper, chlorine, and water, may not turn out to be nice whole numbers.  For example, you may obtain data like this:  1.123 Cu, 2.592 Cl, and 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 words, unless your data comes out exactly .500, use the rounding rules.

(Updated 9/26/13 by C.R. Snelling)