Aim: How to Name compounds with polyatomic ions.
Materials: Index Cards, markers
Background information: REVIEW: (These compounds to follow ARE NOT binary compounds. They contain three or more elements, as opposed to only two in a binary compound. Don’t use the Greek method. That naming technique is used only for binary compounds of two nonmetals. That means, if you see a formula like BaSO4, the name is not barium monosulfur tetraoxide. Many unaware students over the years have made this error and suffered for it.)
Goals: students should be able to learn to recognize the presence of a polyatomic ion in a formula.
Activity:
students will 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 use them. The cations used will be a mix of fixed charges AND variable charges. You must know which are which.
Students will associate the charges with each polyatomic ion. For example, NO3¯ is called nitrate and it has a minus one charge.
Notes: glencoe chemistry, zumdahl & zumdahl
1. 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 very first formula used is Fe(NO3)2. This means that two NO3¯ are involved in the compound. Without the parenthesis, the formula would be FeNO32, a far cry from the correct formula.
2. How to say a formula. When you speak a formula involving parenthesis out loud, you use the word “taken” as in the formula for ammonium sulfide, which is (NH4)2S. Out loud, you say “N H four taken twice S.” OR with the formula for copper(II) chlorate, which is Cu(ClO3)2. You say ” Cu Cl O three taken twice.”
Example #1 – write the name for Fe(NO3)2
Step #1 – decide if the cation is one showing variable charge. If so, a Roman numeral will be needed. In this case, iron does show variable charge.
If a variable charge cation is involved, you must determine the Roman numeral involved. You do this by computing the total charge contributed by the polyatomic ion. In this case, NO3¯ has a minus one charge and there are two of them, making a total of minus 2.
Therefore, the iron must be a positive two, in order to keep the total charge of the formula at zero.
Step #2 – determine the name of the polyatomic ion. Nitrate is the name of NO3¯.
The correct name is iron(II) nitrate. The common name would be ferrous nitrate.
Example #2 – write the name for NaOH
Step #1 – the cation, Na+, does not show a variable charge, so no Roman numeral is needed. The name is sodium.
Step#2 – OH¯ is recognized as the hydroxide ion.
The name of this compound is sodium hydroxide.
There are three things you must memorize: the name (hydroxide), the symbol (OH) and the charge (minus one).
Example #3 – write the name for KMnO4
Step #1 – the cation, K+, does not show a variable charge, so no Roman numeral is needed. The name is potassium.
Step#2 – MnO4¯ is recognized as the permanganate ion.
The name of this compound is potassium permanganate.
Example #4 – write the name for Cu2SO4
Step #1 – decide if the cation is one showing variable charge. If so, a Roman numeral will be needed. In this case, copper does show variable charge.
If a variable charge cation is involved, you must determine the Roman numeral involved. You do this by computing the total charge contributed by the polyatomic ion. In this case, SO42¯ has a minus two charge and there is only one, making a total of minus 2.
Therefore, the copper must be a positive one. Why? Well, there must be a positive two to go with the negative two in order to make zero. Since the formula shows two copper atoms involved, each must be a plus one charge.
Step #2 – determine the name of the polyatomic ion. Sulfate is the name of SO42¯.
The correct name is copper(I) sulfate. The common name would be cuprous sulfate.
Example #5 – write the name for Ca(ClO3)2
The first part of the name comes from the first element’s name: calcium. You also determine that it is not a cation of variable charge.
The second part of the name comes from the name of the polyatomic ion: chlorate.
This compound is named calcium chlorate.
Example #6 – write the name for Fe(OH)3
Iron is an element with two possible oxidation states. The iron is a +3 charge because (1) there are three hydroxides, (2) hydroxide is a minus one charge, (3) this gives a total charge of negative three and (40 there is only one iron, so it must be a +3.
Therefore the first part of the name is iron(III).
The second part of the name is hydroxide, the name of the polyatomic ion.
The name of this compound is iron(III) hydroxide (or ferric hydroxide when using the common method).
Home work:
The cations in this first set are all of fixed oxidation state, so no Roman numerals are needed.
Write the correct name for:
1) AlPO4
2) KNO2
3) NaHCO3
4) CaCO3
5) Mg(OH)2
6) Na2CrO4
7) Ba(CN)2
8) K2SO4
9) NaH2PO4
10) NH4NO3
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 name for:
11) Sn(NO3)2
12) FePO4
13) Cu2SO4
14) Ni(C2H3O2)2
15) HgCO3
16) Pb(OH)4
17) Cu2Cr2O7
18) Cu(ClO3)2
19) FeSO4
20) Hg2(ClO4)2
These formulas mix the use of the two types of cations.
Write the correct name for:
21) KClO3
22) SnSO4
23) Al(MnO4)3
24) Pb(NO3)2
25) Mg3(PO4)2
26) CuH2PO4
27) CaHPO4
28) Fe(HCO3)3
29) Na2CO3
30) MnSO4
Homework: handout completion of chemical formulas or (p224 #19-23) ion pairs
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