Chemical Nomenclature: The Stock System
Rev. 021712
Part one - Binary compounds of a metal and a nonmetal
What we know as the Stock system of naming simple chemical compounds was proposed by
the German chemist Alfred Stock in 1919. He proposed that the archaic system for naming
compounds then in use could be simplified by including the oxidation number of the metal in the
name. His original proposal called for the use of Arabic numerals, but that was later changed to
Roman numerals.
To use the Stock system ….
1. Write the name of the metal.
2. If the metal exhibits more than one oxidation number, include a Roman numeral which is
equal to the oxidation number of the metal.
3. Write the name of the nonmetal with the name modified to end in "-ide".
The Roman numeral is needed to distinguish between compounds which would otherwise have
the same name. For instance, FeCl2 and FeCl3 can't both be "iron chloride". The Roman
numeral tells us which "iron chloride" it is. The Roman numeral is placed inside parentheses
which follow the name of the metal. No space is left between the name of the metal and the
parentheses, as in "iron(III) chloride". No Roman numeral is needed for the elements in groups
IA, IIA and IIIB, and for Al, Ga, In, Zn, Cd, and Ag since these elements have only one oxidation
number. All other metals require a Roman numeral.
Nonmetals and their modified names
arsenic - arsenide
bromine - bromide
carbon - carbide (C4- or C22-)
chlorine - chloride
fluorine - fluoride
iodine - iodide
nitrogen - nitride
oxygen - oxide
phosphorous - phosphide
selenium - selenide
silicon - silicide
sulfur - sulfide
tellurium - telluride
Examples
Write the name, given the formula:
Na2O → sodium oxide … no Roman numeral, Na is in group IA
Al2O3 → aluminum oxide … Al exhibits only one common oxidation number
FeCl3 → iron(III) chloride … chlorine is -1, iron is +3
CuCl2 → copper(II) chloride … chlorine is -1, copper is +2
Mn2O3 → manganese(III) oxide … oxygen is -2, manganese is +3
TiO2 → titanium(IV) oxide … oxygen is -2, titanium is +4
Write the formula, given the name:
Zinc chloride → ZnCl2 … zinc has only one oxidation state, +2, chloride is -1
Gallium oxide → Ga2O3 … gallium has only one oxidation state, +3, oxide is -2
thallium(I) phosphide → Tl3P … the (I) tells us to use +1 for thallium, phosphide is -3
lead(II) chloride → PbCl2 … the (II) tells us to use +2 for lead, chloride is -1
gold(III) fluoride → AuF3 … the (III) tells us to use +3 for gold, fluoride is -1
titanium(IV) oxide → TiO2 … the (IV) tells us to use +4 for titanium, oxide is -2
Write the name of each compound.
1.
FeO
2.
CoCl3
3.
KBr
4.
Ga2O3
5.
SnCl2
6.
CdS
7.
HgCl2
8.
MgCl2
9.
CrO3
10. TcF7
Write the formula of each compound.
1.
iron(II) oxide
2.
lead(II) bromide
3.
silver sulfide
4.
chromium(III) oxide
5.
gold(III) phosphide
6.
tin(IV) chloride
7.
rubidium selenide
8.
vanadium(V) oxide
9.
indium fluoride
10. manganese(VII) oxide
Review: Balance the following chemical equations
+
F2(g)
1.
Tc(s)
2.
CdS(s) +
3.
Ga2O3(s) +
4.
Fe2O3 +
5.
CoCl3(s)
→
TcF7
HCl(aq) →
HCl(aq) →
H2C2O4 →
+
CdCl2(aq) +
H2S(aq)
GaCl3(aq) +
FeO +
Na2CO3(aq) →
CO2 +
H2O(l)
H2O
Co2(CO3)3(s) +
NaCl(aq)
Answers
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
FeO → iron(II) oxide
CoCl3 → cobalt(III) chloride
KBr → potassium bromide
Ga2O3 → gallium oxide
SnCl2 → tin(II) chloride
CdS → cadminum sulfide
HgCl2 → mercury(II) chloride
MgCl2 → magnesium chloride
CrO3 → chromium(VI) oxide
TcF7 → technicium(VII) fluoride
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
iron(II) oxide → FeO
lead(II) bromide → PbBr2
silver sulfide → Ag2S
chromium(III) oxide → Cr2O3
gold(III) phosphide → AuP
tin(IV) chloride → SnCl4
rubidium selenide → Rb2Se
vanadium(V) oxide → V2O5
indium fluoride → InF3
manganese(VII) oxide → Mn2O7
1.
2.
3.
4.
5.
2Tc(s) + 7F2(g) → 2TcF7
CdS(s) + 2HCl(aq) → CdCl2(aq) + H2S(aq)
Ga2O3(s) + 6HCl(aq) → 2GaCl3(aq) + 3H2O(l)
Fe2O3 + H2C2O4 → 2FeO + 2CO2 + H2O
2CoCl3(s) + 3Na2CO3(aq) → Co2(CO3)3(s) + 6NaCl(aq)