Introduction:

The smallest drop of water that can be seen by the naked eye is made up of billions and billions of water molecules. The "mol concept" is a tool that we can use to better grasp such astronomical numbers. A mol is a unit that is used to represent a very large number of atoms or molecules. One mol of any substance is 6.02 x 1023 (Avogadro's number) particles of that substance. Just as you always assume that there are 12 eggs in a dozen, there will always be 6.02 x1023 particles in 1 mol of any substance. To give you an idea of how large of a number that really is, if you had a mol of baseballs, it would cover the entire surface of our planet to a height of 100 miles (that's were the space shuttle orbits)!

The molar mass of an element is its atomic weight on the periodic table expressed in grams per mol. For example, the molar mass of carbon is defined as 12.0000 g/mol. The molar mass of a compound is the formula weight in grams for one mol of that substance. Some examples are shown below: 

Molecular Formula
Formula Weight
Molar Mass
NaCl
    Na              Cl
(1 x 23.0) + (1 x 35.5) = 58.5 amu

58.5 g/mol
CaCl2
    Ca               Cl 
(1 x 40.0) + (2 x 35.5) = 111 amu

111 g/mol
Na3(PO4)2
     Na              P                 O 
(3 x 23.0) + (2 x 31.0) + (8 x 16.0) = 259 amu

259 g/mol

The molar mass of an element or compound can be used as a conversion factor between grams and mols. For example, how many grams are in 3 mols of CaCl2?

Or, how many mols of CaCl2 are in 55.5 g of CaCl2?

Mol relationships in a chemical reaction can be determined by looking at the balanced reaction equation as shown below for the reaction of aluminum (Al) with hydrochloric acid (HCl) to produce aluminum chloride (AlCl3) and hydrogen gas (H2):

A balanced reaction equation has numbers in front of each substance called coefficients. If there is no number in front of a substance, assume the coefficient to be 1. These coefficients tell us the ratio of how many atoms or molecules of each substance will be consumed and produced in that chemical reaction. From the reaction equation above, we can see that for every 2 mols of Al, we will produce 2 mols of AlCl3. This mol relationship can also be used as a conversion factor. There are two conversion factors that we can derive for each reactant and product in this balanced reaction equation:  

mol Al
2 mol AlCl3
2 mol Al
3 mol H2
6 mol HCl
2 mol AlCl3
6 mol HCl
3 mol H2
2 mol AlCl3
2 mol Al
3 mol H2
2 mol Al
2 mol AlCl3
6 mol HCl
3 mol H2
6 mol HCl

The mol to mol relationships or equalities can be used as conversion factors between mols of one substance to mols of another substance in the same chemical reaction. For example, if you started with 1.0 mol of Al in the reaction above, how many mols of H2 gas would be produced?

Or, if you want to produce 4.0 mols of AlCl3, how many mols of Al would you need to start with?

Purpose:

In this experiment, you will be conducting two experiments to convince yourself that the molar relationships discussed above do indeed work.  In the first experiment, you will be reacting sodium bicarbonate (NaHCO3) with hydrochloric acid (HCl) to produce sodium chloride, water, and carbon dioxide:

In the second experiment, you will be reacting sodium carbonate (Na2CO3) with hydrochloric acid (HCl) to produce sodium chloride, water, and carbon dioxide:

Note that although the products are the same in both reactions, the molar ratios are different.  In the first experiment, one mol of NaCl is produced for each mol of NaHCO3.  However, in the second experiment, 2 mols of NaCl are produced for each mol of Na2CO3.

At the beginning of the experiment, you will obtain the mass in grams of the sodium bicarbonate or the sodium carbonate. Using the conversion factors discussed above, you will be able to carry out the following conversions:

The grams of NaCl that you determine in these calculations is called the theoretical yield. At the end of the experiment, you will determine the mass in grams of the sodium chloride product and this is called the actual yield. According to the law of conservation of mass, the actual yield should be equal to the theoretical yield. However, due to human and experimental errors, you very rarely obtain the theoretical yield. The percent yield is calculated using the following equation:

Percent yield = (Actual yield/Theoretical yield) x 100%

Procedure:

  1. Thoroughly clean a large Pyrex test tube with soap and water.  Then rinse it with distilled water.  Fold a paper towel into a long thin strip and use this to remove as much of the water as possible.  It is VERY important that you remove all of the water!
  2. Measure its mass to the nearest 0.001g. 
  3. Then add a single boiling chip and measure the combined mass to the nearest 0.001 g.
  4. With a scoopula, add approximately 1.5 g of sodium bicarbonate, NaHCO3, to the test tube and measure the mass to the nearest 0.001 g.  Note: Do not try to measure exactly 1.500 g. Your measurement should be about 1.5 g but recored exactly.  For example, 1.441 g or 1.582 g are fine.
  5. Obtain about 10 mL of 6M hydrochloric acid in your 10 mL graduated cylinder. CAUTION:  HCl causes acid burns - avoid contact with your skin.  Use a pasteur (disposable) pipette to slowly add the acid to the NaHCO3.  Did the test tube get hotter or colder (exothermic or endothermic)?
  6. When you add the acid to the NaHCO3, you should observe the evolution of gas (bubbles). Continue to add the acid slowly until the reaction is complete  (bubbling has stopped).  Do not add more acid than is needed.  The more acid you add, the longer it is going to take to remove it by boiling.
  7. Swirl the contents of the test tube to make sure the HCl has come in contact with all of the NaHCO3.  If any unreacted NaHCO3 remains (bubbles), add a few more drops of HCl to complete the reaction.
  8. Clamp the test tube at a 45 angle and gently heat the liquid over the Bunsen burner (outer blue cone) until it boils. Take care to avoid loss of liquid from boiling over. If the liquid begins to splatter, remove the heat immediately.  Remove the flame and then continue to heat. Continue to dry the solid slowly until all moisture appears to have evaporated.
  9. Allow the test tube to cool to room temperature and then measure its mass to the nearest 0.001 g.
  10. Reheat the sample strongly (inner blue cone) for 3-4 minutes. Allow it to cool to room temperature and reweigh.  If this weight does not agree to within 0.01 g with the weight in Step 9, reheat and remeasure the mass until two consecutive weights are within 0.01 g of each other.  This is known as weighing to constant dryness and 'proves' that all of the water is gone.
  11. Repeat Steps 1-10 using sodium carbonate, Na2CO3.

Calculation of Results:

Let's work through an EXAMPLE (remember, your numbers will be different!!) and see how you will use your data to determine the molar ratios and theoretical yields.  Assume you started the experiment with a 1.456 g sample of sodium bicarbonate (NaHCO3) and at the end of the experiment, you were left with 1.005g of salt (NaCl).  Since everything in Chemistry is about moles, the first thing we will do is calculated the moles of sodium bicarbonate and salt:

mols of NaHCO3  =  (mass of NaHCO3 used / molecular weight of NaHCO3)  =  1.456g / 84.007 g/mol  =  0.0173 mol

mols of NaCl  =  (mass of NaCl obtained / molecular weight of NaCl)  =  1.005g / 58.44 g/mol  =  0.0172 mol

Now, from the balanced chemical equation in the Purpose, we can see that one mole of sodium bicarbonate, should produce one mole of salt.  Within experimental error, your data confirms this, 0.0174 moles of sodium bicarbonate reacted to produce 0.0173 moles of salt.

The next thing we need to do is calculate the theoretical yield.  This is how many grams of salt we would expect to obtain from the sodium bicarbonate if we carried the experiment out perfectly:

Theoretical yield of NaCl  = grams of NaHCO3 x (1mol / 84.007g) x (1mol NaCl / 1mol NaHCO3) x (58.44g / 1mol NaCl)  =

= 1.456g NaHCO3 x (1mol / 84.007g) x (1mol NaCl / 1mol NaHCO3) x (58.44g / 1mol NaCl)  = 1.013g

So, if everything went perfectly, we would expect to obtain 1.013g of NaCl at the end of the experiment.  Now, the Percent Yield (% yield) is calculated by dividing your actual yield of NaCl (in grams) by the theoretical yield (in grams) and then multiplying by 100:

% Yield  =  (actual grams of NaCl / theoretical grams of NaCl)  x 100 =

=  (1.005g / 1.013g)  x  100  =  99.21%

So, the % Yield for this experiment was 99.21% which is excellent!  You now have all of the results for the sodium bicarbonate portion of the experiment.  All that is left to do is carry out the same set of calculations for the sodium carbonate portion of the experiment.  The one 'hitch' is to remember that the molar ratio of sodium carbonate to salt is 1:2, not 1:1 as it was for the calculations we just finished.

Calculations and Conclusions:

(Updated 9/30/13 by C.R. Snelling)