Download Experiment 6 - Analysis of an Unknown Chloride

experiment date(s):
Experiment 5 – Compound Identity by Mass Relationships
Goal: To illustrate the utility of chemical formula and reaction stoichiometry.
Discussion: You will have one of the following 4 compounds as an unknown, and must
determine its identity. The first two are referred to as hydrogen carbonate salts because they
contain the hydrogen carbonate ion, HCO3-. The second two are carbonate salts because they
contain the carbonate ion, CO32-.
NaHCO3
KHCO3
Na2CO3
K2CO3
There are two experimental procedures used to distinguish between the four compounds.
1. Distinguish between carbonate and hydrogen carbonate
Experiment: Heat the solid. If the compound is a carbonate, nothing much happens. At
most, the sample will lose between 0.01 and 0.04 grams of mass. If you have a hydrogen
carbonate, it will lose a significant amount of mass, usually greater than 0.1 grams.
In the equations below, M = metal = Na or K.
Carbonate: M2CO3 (s)
__heat___
>
Hydrogen carbonate: 2 MHCO3 (s)
M2CO3 (s)
__heat___
>
(no change)
M2CO3 (s) + H2O (g) + CO2 (g)
(mass is lost
when gases escape)
Once it is determined whether the compound is a carbonate or hydrogen carbonate, then it is time
to determine whether the metal is sodium or potassium.
2. Distinguish between sodium and potassium salts
Experiment: Convert the solid to a chloride salt by reaction with HCl. The mass ratio of
chloride salt to original compound is different for Na vs. K.
(a) If the compound is a carbonate, then the equation for conversion to the chloride salt is as
follows:
M2CO3 (s) + 2 HCl (aq)
_____
>
2 MCl (s) + H2O (g) + CO2 (g)
Notice that for every one mole of carbonate salt reacting, two moles of chloride salt are
produced.
To determine the theoretical mass ratio, start with any mass of the original compound (the
example starts with 1.00 g) and calculate the mass of product expected. The example below is for
sodium carbonate. You may need to perform this same calculation for potassium carbonate.
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experiment date(s):
1.00 g Na2CO3∣ 1 mol Na2CO3 ∣ 2 mol NaCl ∣ 58.5 g NaCl∣ = 1.10 g NaCl
∣ 106 g Na2CO3 ∣ 1 mol Na2CO3∣ 1 mol NaCl ∣
ratio of chloride : original is 1.10 : 1.00
(b) If the compound is a hydrogen carbonate, then the equation for conversion to the chloride
salt looks as follows (this is the sum of equations 1 and 2 mentioned in the lab book):
MHCO3 (s) + HCl (aq)
_____
>
MCl (s) + H2O (g) + CO2 (g)
In this case, notice the mole ratio of hydrogen carbonate salt to chloride salt is one to one. Again,
calculate the theoretical mass ratio for these two sodium salts.
1.00 g NaHCO3∣ 1 mol NaHCO3 ∣ 1 mol NaCl ∣ 58.5 g NaCl∣ = 0.696 g NaCl
∣ 84.0 g NaHCO3 ∣ 1 mol NaHCO3∣ 1 mol NaCl∣
ratio of chloride : original is 0.696 : 1.00
Perform the same calculation for the potassium salt, as needed, to complete the advanced study
assignment or data page.
Calculate the actual mass ratio of chloride salt to original compound to determine which
theoretical ratio it matches. This determines the identity of the original compound.
Procedure:
(1) Due to residual moisture, you will most likely observe mass loss with both the carbonate and
hydrogen carbonate. Here’s how you tell the difference.
carbonates will lose around 0.01 to 0.04 g
hydrogen carbonates will lose ≥ 0.1 g
(2) Add 30-35 drops of HCl to your sample, rather than the 25 drops described in the lab book.
Waste disposal:
The contents of the crucible can be rinsed down the sink with plenty of water.
ASA:
(1) Complete all of the ASA.
(2) You can also fill out some of the data page ahead, to save time in lab:
(a) atomic and molar masses at the top of the data page
(b) theoretical ratios at the end of the data page
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