Compounds Involving a Polyatomic Ion

Given Name, Write the Formula

These compounds to follow ARE NOT binary compounds. The contain three or more elements, as opposed to only two in a binary compound.

The Greek method WILL NOT be used. That naming technique is used only for binary compounds of two nonmetals. That means, if you see a formula like BaSO_{4}, the name is not barium monosulfur tetraoxide. Another example might be Na_{2}CO_{3}, the student incorrectly naming it as disodium monocarbon trioxide.

Consequently, a warning: it is important that you learn to recognize the presence of a polyatomic ion in a name. The ChemTeam recommends you make a set of flashcards with the name on one side and the ion __and its charge__ on the other. Then, carry them everywhere and actually __use__ them.

The cations used will be a mix of fixed charges AND variable charges. You must know which are which.

Another warning: you must also know the formula and charge associated with each polyatomic ion's name. For example, NO_{3}¯ is called nitrate and it has a minus one charge. The formula and charge are not inherent in the name.

When more than one polyatomic ion is required, parenthesis are used to enclose the ion with the subscript going outside the parenthesis. For example, the first example used below is copper(II) chlorate. The correct formula will require the use of parenthesis.

When you say a formula involving parenthesis out loud, you use the word "taken" as in the formula for ammonium sulfide, which is (NH_{4})_{2}S. Out loud, you say "N H four taken twice S." OR with the formula for copper(II) nitrate, which is Cu(NO_{3})_{2}. You say " Cu N O three taken twice."

On writing the formula from the ions, you may want to review Charge-Crossing.html or Least-Common-Multiple.html for more information.

Example #1 - write the formula for **copper(II) chlorate**

Step #1 - the first word tells you the symbol of the cation. In this case it is Cu.

Step #2 - the Roman numeral **WILL** tell you the charge on the cation. In this case it is a positive two.

Step #3 - the polyatomic formula and charge comes from the second name. In this case, chlorate means ClO_{3}¯.

Step #4 - remembering the rule that a formula must have zero total charge, you write the formula Cu(ClO_{3})_{2}.

This graphic summarizes example #1:

By the way, out load, this is said C U C L O three taken twice.

Example #2 - write the formula for **silver cyanide**

Step #1 - the first word tells you the symbol of the cation. In this case it is Ag^{+}.

Step #2 - silver has a constant charge of +1, it is not a cation with variable charge.

Step #3 - the polyatomic formula and charge comes from the second name. In this case, cyanide means CN¯.

Step #4 - remembering the rule that a formula must have zero total charge, you write the formula AgCN.

This graphic summarizes example #2:

Usually, at this point is the plaintive wail, "But how do you know that cyanide is CN¯?" The stock ChemTeam answer is "Well, how do you know anything? How do you know your phone number? How do you know your best friend's name? In fact, how do you know your name?" There are three things you must memorize: the name (cyanide), the symbol (CN, not Cn, by the way) and the charge (minus one). You must put in the time to learn this nomenclature stuff. It does not come easy and the ChemTeam realizes you'd rather be spending the time doing more important things: going cool places with friends, spending time with members of the opposite sex, spending your parents' money, eating, etc. Maybe some other time. Right now, let's move on to another example.

Example #3 - write the formula for **plumbic hydroxide**

Step #1 - the cation, Pb^{4+}, does show a variable charge. The "-ic" ending means the higher of the two, for this cation that means +4.

Step#2 - hydroxide is recognized as OH¯.

The formula of this compound is Pb(OH)_{4}. Notice that it is not PbOH_{4}.

This graphic summarizes example #3:

You might want to read about a problem with hydroxide that many students suffer from.

Example #4 - write the formula for **sodium phosphate**

Step #1 - the cation, sodium, is Na^{+}, and it does not show a variable charge.

Step#2 - phosphate is PO_{4}^{3}¯.

The formula of this compound is Na_{3}PO_{4}. Notice that no parenthesis are required since only one polyatomis is used.

This graphic summarizes example #4:

Example #5 - write the formula for **mercurous nitrate**

Step #1 - the cation, mercurous, does show a variable charge and its formula is unusual. It is Hg_{2}^{2+}. The "-ous" ending indicates the lower of the two charges mercury shows and that is the +1 charge. Remember that, in this particular case, Hg^{+} is __wrong__.

Step#2 - nitrate is NO_{3}¯.

The formula of this compound is Hg_{2}(NO_{3})_{2}. This formula is not reduced.

This graphic summarizes example #5:

Example #6 - write the name for **barium carbonate**

Step #1 - the cation, barium, does not show a variable charge and its symbol is Ba^{2+}.

Step#2 - carbonate is CO_{3}^{2}¯.

The formula of this compound is BaCO_{3}.

This graphic summarizes example #6:

Example #6 - write the name for **mercury(I) phosphate**

Step #1 - the cation, mercury(I), does show a variable charge and its symbol is Hg_{2}^{2+}. Notice that there is a subscript of two and the charge is +2. Therefore, EACH Hg atom is +1, leading to the name mercury(I).

Step#2 - phosphate is PO_{4}^{3}¯.

The formula of this compound is (Hg_{2})_{3}(PO_{4})_{2}.

Note the use of parenthesis around both parts (positive and negative) of the formula. It would be incorrect to write this formula: Hg_{6}(PO_{4})_{2}. We want the formula to show that mercury(I) comes in groups of two and that there are three of them.

The cations in this first set are all of fixed oxidation state, so no Roman numerals are needed.

Write the correct formula for:

1) silver carbonate

2) potassium hydrogen phosphate

3) aluminum hydroxide

4) sodium hydrogen carbonate

5) calcium acetate

6) potassium permanaganate

7) calcium perchlorate

8) lithium carbonate

9) magnesium hydrogen sulfite

10) sodium hypochlorite

These formulas involve the use of a polyatomic ion. The cations are all of variable oxidation state, so Roman numerals are needed.

Write the correct formula for:

11) tin(IV) chlorite

12) mercury(II) phosphate

13) tin(II) carbonate

14) mercurous acetate

15) lead(II) chromate

16) copper(I) sulfite

17) stannous dichromate

18) iron(III) nitrate

19) ferric sulfate

20) ferrous hydroxide

These formulas mix the use of the two types of cations.

Write the correct formula for:

21) potassium perchlorate

22) lead(IV) hydrogen phosphate

23) aluminum sulfate

24) iron(II) bicarbonate

25) barium iodate

26) tin(II) hydrogen sulfide

27) magnesium dihydrogen phosphate

28) plumbous cyanide

29) silver phosphate

30) cobalt(III) nitrite

and two special additions:

31) ammonium sulfate

32) ammonium nitrate