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Fall 2009 |
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| Exercise: Nomenclature |
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Supplemental Study Modules |
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1. K+ & F- - First we need to write the formula such that the final compound has a charge of zero, that is the charges of the ions must cancel each other.
In this first case the charges are equal, so they have a 1:1 ratio, and the formula is simply KF. Note that NO CHARGES are shown, and the positive ion is listed first!
To name the compound we first name the positive ion with its element name (potassium), and then replace the ending of the negative elemental ion name (fluorine) with -ide, thus:
KF = Potassium fluoride
2. Al3+ & S2- This ion pair does not have equal charges so we must multiply the two ions by common factors to give them the same TOTAL charge for each ion type. Thus 2 x 3+ = 6+ is countered by 3 x 2- = 6-, and the formula is Al2S3.
To name the compound we first name the positive ion with its element name (aluminum), and then replace the ending of the negative elemental ion name (sulfur) with -ide, thus:
Al2S3 = Aluminum sulfide
Since both of these elements have only one possible charge as elemental ions, we do not need to indicate the ratio - we know it from our vast chemical knowledge!
3. NH4+ & CN- - In this case the positive ion, or cation, is a molecular ion, that is NH4+ is a single entity or species, and will be combined as an entity NOT as individual atoms.
Once again the charges are equal, and the formula reflects this 1:1 ratio: NH4CN.
NH4CN = Ammonium cyanide
Since both of these ions have only one possible charge as ions, we do not need to indicate the ratio - we know it from our vast chemical knowledge!
4. Hg22+ & C2H3O2- - The mercury +1 ion is a molecular ion, that is there are always two mercury atoms combined as a single species - we never see the monoatomic ion Hg+. As a result mercury +1 acts as a 2+ ion, and we need two acetate molecular ions to combine with it, Hg2(C2H3O2)2.
Mercury can occur as both +1 and +2 ions, so we must indicate its charge (also known as "oxidation state") as well, Mercury(I), thus:
Hg2(C2H3O2)2 = Mercury(I) acetate
5. Pb2+ & SO42- - Lead has two oxidation states, +2 and +4 so Pb2+is named as the Lead(II) ion, and SO42- is the sulfate molecular ion, again a single species. These ions have the same charge, so we see a 1:1 ratio:
PbSO4 = Lead(II) sulfate
6. Sn4+ & Cr2O72- - Tin has two oxidation states, +2 and +4 so Sn 4+is named as the Tin(IV) ion, and Cr2O72- is the dichroamate molecular ion, again a single species. These ions differ in charge by a factor of two, so we see a 1:2 ratio:
Sn(Cr2O7)2 = Tin(IV) dichromate
7. H+ & SO42- - These ions differ in charge by a factor of two, so we see a 2:1 ratio: H2SO4.
Compounds containing the hydrogen ion, H+, such as this one, are named as acids with the -ate ending of negatively charged ion, the anion, replaced with the -ic ending,thus:
H2SO4 = Sulfuric acid
(Remember, for anions ending in -ite, the acid will end in -ous etc., e.g. H2SO3 = Sulfurous acid.)
8. H+ & ClO4- - This ion pair is matched in charge so a 1:1 ratio results: HClO4
ClO4- is one member of a family of four related ions having the same charge but with differing numbers of oxygens, 1, 2, 3 & 4: ClO-, ClO2-, ClO3-, & ClO4-. To name them the first two are given the -ite ending while the next two are given the -ate inding. The first, having a lower number of oxygens is also assigned the prefix hypo- meaning below, while the last, having a higher number is given the prefix per- meaning beyond. Thus:
HClO4 = Perchloric acid
The entire family is named as follws: ClO- Hypochlorite ion (HClO is then Hypochlorous acid), ClO2- is Chlorite ion (HClO2 is then Chlorous acid), ClO3- is Chlorate ion (HClO3 is then Chloric acid), and ClO4- is Perchlorate ion (HClO4 is then Perchloric acid).
9. Cs+ & NO2- - This ion pair is matched in charge so a 1:1 ratio results: CsNO2
Note that the anion here, NO2- has one fewer oxygen than the familiar nitrate ion, NO3-, while sharing the same -1 charge. By the rules above it is then named as nitrite.
CsNO2 = Cesium nitrite
10. Ba2+& O22- - This ion pair is matched in charge so a 1:1 ratio results: BaO2
Note that this molecular ion has two oxygens and a -2 charge, so each oxygen has a charge of -1, making this the peroxide ion.
BaO2 = Barium peroxide
Return to Chemical Nomenclature Exercises
© R A Paselk
Last modified 23 September 2008