You have three mixtures consisting of powdered inorganic solids on your desk. They could contain the — Organic Chemistry Chemistry Question
PREPARATORY PROBLEM 31 (PRACTICAL)
You have three mixtures consisting of powdered inorganic solids on your desk. They could contain the following compounds:
a) (NH4)2CO3, AgNO3, BaCl2 · 2 H2O, NH4NO3, NiCl2 · 2 H2O
b) ZnO, KI, Pb(NO3)2, BaSO4, MnO2, Mg
c) CaCO3, NH4I, FeSO4 · 7 H2O, TiO2, CuCl2 · 2 H2O
You can use distilled water, HCl, HNO3, NH3, and NaOH solutions, all with the concentration of 2 mol dm–3 and, moreover, pH paper, test tubes and a Bunsen burner. (Not all of the compounds listed are present in the unknown samples.)
Determine which compounds are present in the mixture and which are not.
Model Answer
Example of one of the solutions:
The following compositions were handed out:
a) AgNO3, BaCl2·2 H2O, NH4NO3; absent: (NH4)2CO3, NiCl2·2 H2O
Colorless ⇒ no nickel.
No gas evolution observed with acids ⇒ no carbonate.
Sample does not dissolve in water ⇒ the precipitate must be AgCl. Indeed, the solution clears up on adding ammonia.
Heating the solution with NaOH produces ammonia, as seen on the indicator paper at the mouth of the test tube (or smelt).
Ag+ + Cl– = AgCl
AgCl + 2 NH3 = Ag(NH3)2+ + Cl–
NH4+ + OH– = NH3↑ + H2O
b) ZnO, Pb(NO3)2, Mg; absent: BaSO4, KI, MnO2
Grey powder consisting of a dark and white component ⇒ contains Mg or MnO2.
After adding water a grey, heterogeneous substance remains ⇒ ZnO or BaSO4 might also be present.
Slowly dissolves without residue in nitric acid with gas evolution ⇒ Mg and ZnO is present.
A white precipitate remains in HCl ⇒ Pb(NO3)2 is present. KI can be excluded as yellow PbI2 was not seen.
Mg + 2 H+ = Mg2+ + H2↑
Pb2+ + 2 Cl- = PbCl2
c) CaCO3, CuCl2·2 H2O; absent: NH4I, FeSO4·7H2O, TiO2
Sample dissolves completely in acids with gas evolution ⇒ TiO2 absent, CaCO3 present. FeSO4 must also be absent because CaSO4 is not formed.
The green powder dissolves in acid leaving a bluish solution. Ammonia gives a dark blue discoloration ⇒ CuCl2
NH4I cannot be present together with Cu2+ as iodine would be produced.
Cu2+ + 4 NH3 = Cu(NH3)4^2+
Note your experimental findings in detail. Explain every conclusion (positive or negative).
Model Answer
Example of one of the solutions:
The following compositions were handed out:
a) AgNO3, BaCl2·2 H2O, NH4NO3; absent: (NH4)2CO3, NiCl2·2 H2O
Colorless ⇒ no nickel.
No gas evolution observed with acids ⇒ no carbonate.
Sample does not dissolve in water ⇒ the precipitate must be AgCl. Indeed, the solution clears up on adding ammonia.
Heating the solution with NaOH produces ammonia, as seen on the indicator paper at the mouth of the test tube (or smelt).
Ag+ + Cl– = AgCl
AgCl + 2 NH3 = Ag(NH3)2+ + Cl–
NH4+ + OH– = NH3↑ + H2O
b) ZnO, Pb(NO3)2, Mg; absent: BaSO4, KI, MnO2
Grey powder consisting of a dark and white component ⇒ contains Mg or MnO2.
After adding water a grey, heterogeneous substance remains ⇒ ZnO or BaSO4 might also be present.
Slowly dissolves without residue in nitric acid with gas evolution ⇒ Mg and ZnO is present.
A white precipitate remains in HCl ⇒ Pb(NO3)2 is present. KI can be excluded as yellow PbI2 was not seen.
Mg + 2 H+ = Mg2+ + H2↑
Pb2+ + 2 Cl- = PbCl2
c) CaCO3, CuCl2·2 H2O; absent: NH4I, FeSO4·7H2O, TiO2
Sample dissolves completely in acids with gas evolution ⇒ TiO2 absent, CaCO3 present. FeSO4 must also be absent because CaSO4 is not formed.
The green powder dissolves in acid leaving a bluish solution. Ammonia gives a dark blue discoloration ⇒ CuCl2
NH4I cannot be present together with Cu2+ as iodine would be produced.
Cu2+ + 4 NH3 = Cu(NH3)4^2+
Include reaction equations where possible.
Model Answer
Example of one of the solutions:
The following compositions were handed out:
a) AgNO3, BaCl2·2 H2O, NH4NO3; absent: (NH4)2CO3, NiCl2·2 H2O
Colorless ⇒ no nickel.
No gas evolution observed with acids ⇒ no carbonate.
Sample does not dissolve in water ⇒ the precipitate must be AgCl. Indeed, the solution clears up on adding ammonia.
Heating the solution with NaOH produces ammonia, as seen on the indicator paper at the mouth of the test tube (or smelt).
Ag+ + Cl– = AgCl
AgCl + 2 NH3 = Ag(NH3)2+ + Cl–
NH4+ + OH– = NH3↑ + H2O
b) ZnO, Pb(NO3)2, Mg; absent: BaSO4, KI, MnO2
Grey powder consisting of a dark and white component ⇒ contains Mg or MnO2.
After adding water a grey, heterogeneous substance remains ⇒ ZnO or BaSO4 might also be present.
Slowly dissolves without residue in nitric acid with gas evolution ⇒ Mg and ZnO is present.
A white precipitate remains in HCl ⇒ Pb(NO3)2 is present. KI can be excluded as yellow PbI2 was not seen.
Mg + 2 H+ = Mg2+ + H2↑
Pb2+ + 2 Cl- = PbCl2
c) CaCO3, CuCl2·2 H2O; absent: NH4I, FeSO4·7H2O, TiO2
Sample dissolves completely in acids with gas evolution ⇒ TiO2 absent, CaCO3 present. FeSO4 must also be absent because CaSO4 is not formed.
The green powder dissolves in acid leaving a bluish solution. Ammonia gives a dark blue discoloration ⇒ CuCl2
NH4I cannot be present together with Cu2+ as iodine would be produced.
Cu2+ + 4 NH3 = Cu(NH3)4^2+