Qualitative Analysis: Test for Gas
2019 O Level, Open-ended Question 8(a)(ii)
When nickel reacts with another chemical, a gas is formed. Describe how we can test the gas to find out if it is hydrogen gas or oxygen gas.
Describe the test for hydrogen.
Test the gas with a burning splint. Hydrogen gas will extinguish the flame with a ‘pop’ sound.
Theory: The reaction here is combustion, whereby the hydrogen gas produced reacts with the oxygen gas in the atmospheric air. The burning splint provides the activation energy, that is transformed to the reactants to initiate the reaction.
Describe the test for oxygen.
Test the gas with a glowing splint. Oxygen will relight the glowing splint.
Theory: The oxygen produced increases the rate of combustion, hence relighting the glowing splint.
Qualitative Analysis Flowchart
2018 O Level, Open-ended Question 7
Identify metal R.
Iron, Fe
How to infer: As the metal can be oxidised to form a coloured solution in T, it must be a transition metal. As solution T is green in colour, it must be the iron(II) ion that comes from the iron metal in R.
Identify gas S.
Hydrogen gas, H2
How to infer: Hydrogen gas gives a ‘pop’ sound with a burning splint. Also, acid-metal reaction produces hydrogen gas.
Identify solution T.
Iron(II) chloride, FeCl2
How to infer: Iron reacts with hydrochloric acid to form the chloride salt, iron(II) chloride. Its green colour suggests the presence of iron(II) ions, while the subsequent test for cation with sodium hydroxide confirms its presence.
Identify precipitate U.
Silver chloride, AgCl
How to infer: The silver ion in the silver nitrate reagent forms an insoluble salt with the chloride ions in the green solution T. This is the test to confirm the presence of chloride ions.
Identify precipitate V.
Iron(II) hydroxide, Fe(OH)2
How to infer: The hydroxide ions from the sodium hydroxide reagent forms insoluble iron(II) hydroxide with the iron(II) ions in solution T. The insoluble iron(II) hydroxide appears as the green precipitate. This is the test for iron(II) cation.
2017 O Level, Open-ended Question 8
Compound C has the general formula XCl2. X represents a metal in Group II of the Periodic Table.
The relative atomic masses, Ar of chlorine and Group II elements are: Be, 9; Mg, 24; Cl, 35.5; Ca, 40; Sr, 88; Ba, 137.
Identify and write either the name or chemical formula of solution E.
Calcium carbonate, CaCO3
Note: Limewater is calcium hydroxide, Ca(OH)2. It is an alkali that can react with carbon dioxide, an acidic oxide. The reaction is neutralisation, which forms water and the salt calcium carbonate. As calcium carbonate is an insoluble salt, it appears as white precipitate that does not dissolve in water.
Identify and write either the name or chemical formula of solution D.
Carbon dioxide, CO2
Note: The presence of carbon dioxide is confirmed by the subsequent test, whereby carbon dioxide is bubbled into limewater to yield white precipitate of calcium carbonate.
Identify and write either the name or chemical formula of solution C.
Strontium chloride, SrCl2
Mr = Ar of X + 35.5×2
159 = Ar of X + 71
Ar of X = 88, which is the Ar of Sr
Identify and write either the name or chemical formula of solution B.
Strontium carbonate, SrCO3
Note: B must be a metal carbonate, as the salt C and carbon dioxide are produced when acid is added to it, alongside water which is not mentioned. The cation is strontium ion, as inferred by the presence of strontium ion in C.
Identify and write either the name or chemical formula of solution A.
Strontium nitrate, Sr(NO3)2
Note: The addition of aluminium and sodium hydroxide to solution A is the test for the nitrate anion. Aluminium reduces nitrate to ammonia, which is an alkaline gas that turns damp red litmus paper blue. Therefore, A contains the nitrate anion, as well as the strontium cation that is found in B and C.
Write a balanced chemical equation for any one of the reactions in the flowchart above.
Test for nitrate anion: 8Al(s) + 3NO3– + 5OH– + 18H2O ⟶ 8[Al(OH)4]– + 3NH3
A to B: Sr(NO3)2 + Na2CO3 ⟶ SrCO3 + 2NaNO3
B to C and D: SrCO3 + 2HCl ⟶ SrCl2 + CO2 + H2O
D to E: Ca(OH)2 + CO2 ⟶ CaCO3 + H2O
