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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) _ _h eat ___> M2CO3 (s) (no change)
Hydrogen carbonate: 2 MHCO3 (s ) __h eat ___> 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 (a q ) _____ > 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.
