Chemical Formula Writing Tutorial Number your paper from 1 -25 and write the answers as you go through the tutorial. Thank you!!

When we write formulas for ionic compounds, it is always important to balance charges on the ions. +

Let’s see what compound would form when Na ions (which have a charge of 1+) combine with F ions (which have a charge of 1-). F 1- F 1- Na 1+ F 1- Na 1+ F 1- F 1- Na 1+ Na 1+ F 1- F 1- 1+ Na Na 1+ F 1- F 1- Na 1+ Na 1+

+ = (1+) (1-) 1- 1- 1+ 1- 1+ 1- 1- 1+ 1+ 1- 1- 1+ Na 1+ 1- 1- 1+ 1+ Let’s assume we have a big pile of Na1+ ions and big pile of F1- ions. We’ll move one of the Na ions into the box… We’ll move one of the Na ions into the box… and one F ion in there too… Notice how the charges of 1+ and 1- cancel out (they add up to 0) That means that we have a balanced chemical formula: one Na for every one F So the formula is simply NaF + = (1+) (1-) F 1- F 1- Na 1+ F 1- Na 1+ F 1- F 1- Na 1+ Na 1+ F 1- F 1- 1+ Na Na 1+ F 1- F 1- Na 1+ Na 1+

+ = 1- (1+) (2-) 2- 1+ 2- 1+ 2- 2- 1+ 2- 1+ 2- 1+ Na 2- 1+ 2- 2- 1+ 1+ Now let’s see what compound forms between Na1+ ions and S2- ions. We’ll move one of the Na ions into the box… We’ll move one of the Na ions into the box… and one S ion in there too… Notice how the charges of 1+ and 2- do NOT cancel out this time. But if we add in one more Na ion, then they do cancel out. (they all add up to 0) But if we add in one more Na ion So the formula this time is Na2S Because it takes two Na ions to balance the charge on one S ion. + = 1- (1+) (2-) S 2- Na 1+ (1+) + (1+) + (2-) = 0 S 2- Na 1+ S 2- S 2- Na 1+ S 2- Na 1+ S 2- 1+ Na S 2- Na 1+ S 2- S 2- Na 1+ Na 1+

1- 3+ 1- 3+ 1- 3+ 1- 3+ 1- 3+ 3+ 1- 3+ 1- 3+ 1- 1- 3+ (3+)+(1-) = 2+ Now let’s see what compound forms between Al3+ ions and F1- ions. We’ll move one of the Al ions into the box… We’ll move one of the Al ions into the box… and one F ion in there too… Notice how the charges of 3+ and 1- do NOT cancel. If we add in one more F ion, they still do not cancel out. If we add in one more F ion But when we add a third F ion But when we add a third F ion, now they do cancel out. So the formula this time is AlF3 Because it takes three F ions to balance the charge on one Al ion. F 1- Al 3+ F 1- Al 3+ (3+)+(1-) = 2+ F 1- Al 3+ F 1- (3+)+(1-)+(1-) = 1+ Al 3+ F 1- (3+)+(1-)+(1-)+(1-) = 0 Al 3+ Al 3+ F 1- Al 3+ F 1- Al 3+ F 1- F 1- Al 3+

3+ 3+ 2- 2- 3+ 3+ 2- 2- 2- 3+ 3+ 3+ 2- 2- 3+ 2- 2- 3+ + = 0 (6+) (6-) Now let’s see what compound forms between Al3+ ions and S2- ions. This one is not so easy… Can you figure out how many Al ions and how many S ions it takes so that they all cancel out? Write down your prediction and show it to the instructor 1: _________________ Is this what you predicted? Notice how the two Al ions add up to a charge of 6+ and how the three S ions add up to a charge of 6- and that’s how they cancel out + = 0 So the formula for this compound is Al2S3 Al 3+ Al 3+ S 2- S 2- Al 3+ Al 3+ S 2- S 2- S 2- Al 3+ Al 3+ Al 3+ S 2- S 2- Al 3+ (6+) (6-) S 2- S 2- Al 3+

Now hopefully you understand the idea of balancing charges Now hopefully you understand the idea of balancing charges. If not, go back and review it some more because you will need this skill when writing correct names and formulas for ionic compounds.

These are the group number designations and charges for each column Before we go any further, let’s have a quick review of the representative groups in the periodic table and their charges: These are the group number designations and charges for each column I II III IV V VI VII VIII 1+ 2+ 3+ 4- 3- 2- 1-

Now… say you want to determine the correct chemical formula for the compound barium bromide. Where do you start? When ionic compounds are named, the positive ion is always listed first, then the negative ion. So look at the ion chart given to you by your instructor. Find barium on the left hand-side of the chart – that’s where all the positive ions are. You found that barium is “Ba” and that it is in the group of ions whose charge can be determined by their positions in the periodic table. So let’s find Ba in the table. Ba is right here And because it is in group II, you know the Ba ion must have a charge of 2+.

So in the compound barium bromide, we know the barium ion is Ba2+ So in the compound barium bromide, we know the barium ion is Ba2+. What about the bromide ion? Bromide must be the negative ion, since it is listed last. So look at the ion chart again, and find bromide on the right hand-side of the chart – that’s where all the negative ions are. You found that bromide is “Br” and that it is also in the group of ions whose charge can be determined by their positions in the periodic table. So let’s find Br in the table. Br is right here And because it is in group VII, you know the Br ion must have a charge of 1-.

So if barium is Ba2+ and bromide is Br1- then apply what you learned earlier to determine the correct, charge-balanced formula for barium bromide. 2: _____________ Show your answer to the instructor. Did you have BaBr2 ? If so, good job. You realized that it requires two Br1- ions to balance the charge on one Ba2+ ion.

Now let’s try another one: potassium nitride. 3: _________________ First we look up potassium on the chart and find that its symbol is “K” and that its charge can be determined by its position in the per. table: And K is right here so its charge is 1+. And nitride is on the right side of the chart, its symbol is “N” And N is right here so its charge is 3-. For the compound to be charge balanced, we must have three K1+ ions for every N3-. So its formula must be K3N.

Now one more: zinc sulfide. 4: ________________ First we look up zinc on the chart and find that its symbol is “Zn” and that its charge is given to us on the chart: 2+. (That’s easy.) And sulfide is on the right side of the chart, its symbol is “S” And S is right here so its charge is 2-. Since 2+ and 2- already add up to zero, the formula is an easy one: It’s ZnS.

Now what about going the other way: Given a compound’s chemical formula, how do you write its correct name? For example: what is the correct name for the compound Ag2O? That’s actually quite easy: since it’s already charge-balanced, all you do is look up Ag on the left-side of the chart. Its name is “silver.” Now look up O on the right side of the chart. Its name is “oxide.” So the correct name for Ag2O is silver oxide.

Now you try one: What is the correct name for AlF3 Now you try one: What is the correct name for AlF3? 5:___________ Show your answer to the instructor. Did you get aluminum fluoride? And did you spell fluoride correctly?

Now, you might have noticed that when we were showing charges for the elements of the periodic table, we kind of skipped over all the elements in the d-block? That’s because many of these elements form two or more different ions Take iron for example. It can form ions with charges of 2+ or 3+. And then there’s copper. It can form ions with charges of 1+ or 2+. Likewise for gold. It can form ions with charges of 1+ or 3+. So, how do we keep all these variable charge ions straight? With Roman numerals! Fe 2+ Fe 3+ Cu 1+ Cu 2+ Au 1+ Au 3+

If you are talking about a copper ion with a charge of 1+ (Cu1+), you say “copper (I).” Copper with a charge of 2+ (Cu2+), is “copper(II).” Iron ions come in two forms: Fe2+ is “iron(II),” and Fe3+ is “iron(III).” That makes writing formulas for the compounds easier: you don’t have to bother looking them up in the periodic table to tell what charge they have… Their charge is always given to you by the Roman numeral! But it makes naming compounds a little trickier: you can’t just say “copper bromide.” People won’t know what compound you are talking about. Instead you have to specify: “copper(I) bromide” or “copper(II) bromide,” since they are two different compounds with different formulas and difference properties.

So … what is the formula for chromium(III) phosphide? 6: _________ First, we look up chromium(III) on the left side of the ion chart and find that it is Cr3+. Then we look up phosphide on the right side of the chart and find that it is P. This we need to find in the periodic table. And we can tell that P has a charge of 3-. 3+ and 3- cancel quite nicely, so the formula is simply CrP.

And what if it had been chromium(II) phosphide? 7: ____________ And phosphide is still P3-. We know that chromium(II) is Cr2+. These two ions aren’t so simple to balance, but we’ve done something like it before. It will take three Cr2+ ions to balance out two P3- ions. And so the correct formula is Cr3P2.

But what about going the other way: what if we want to determine the correct name for the compound: CuO. 8: ____________________ Again we look up Cu on the left side of the ion chart and find that its either copper(I) or copper(II), and we look up O on the right side and see that it is oxide.

So now we have to look at the formula CuO and determine whether the Cu ion has a charge of 1+ or 2+. Let’s first look up the charge on the negative ion. O is right here so it always has a charge of 2-. And the formula tells us that it took just one Cu ion to balance the charge on one O ion. That means that the Cu must have had a charge of 2+. If the Cu has a charge of 2+, then we’re talking about copper(II)… And so the correct name is copper(II) oxide. CuO (2+) + (2-) = 0

Cu2O So what is the name for Cu2O. 9: __________________ We know that Cu is either copper(I) or copper(II) and again O is oxide and it always has a 2- charge. This time the formula tells us that it took two Cu’s to balance the charge on just one O. That means that the charge on each indidual Cu must be just 1+, which means we’re talking about copper(I). So correct name is copper(I) oxide. Cu2O (1+)+(1+)+(2-) = 0

CuO = copper(II) oxide Cu2O = copper(I) oxide This may seem very strange at first that CuO is called copper(II) oxide and Cu2O is copper(I) oxide. But remember: the Roman numeral indicates the charge on the positive ion not how many there are.

How about this one: what is the correct name of CoF3. 10: _______ Again we look up Co on the chart and find that it’s either cobalt(II) or cobalt(III), and we look up F and see that it’s fluoride.

So now we have to look at the formula CoF3 and determine whether the Co ion has a charge of 2+ or 3+. Let’s first look up the charge on the negative ion. F is right here… so it always has a charge of 1-. And the formula tells us that it took three F ions to balance the charge on one Co ion. That means that the Co must have had a charge of 3+ So the correct name for CoF3 is cobalt(III) fluoride. CoF3 (3+)+(1-)+(1-)+(1-) = 0

How about this one: what is the correct name of PbS2. 11: _______ We look up Pb on the chart and find that it’s either lead(II) or lead(IV), and we look up S and see that it’s sulfide.

So now we have to look at the formula PbS2 and determine whether the Pb ion has a charge of 2+ or 4+. Let’s first look up the charge on the negative ion. S is right here… so it always has a charge of 2-. And the formula tells us that it took two S ions to balance the charge on one Pb ion. That means that the Pb must have had a charge of 4+ So the correct name for PbS2 is lead(IV) sulfide. PbS2 (4+)+(2-)+(2-) = 0

Try this one: what is the correct name of Cr3N2. 12: ___________ We look up Cr on the chart and find that it’s either chromium(II) or chromium(III), and we look up N and see that it’s nitride.

So now we have to look at the formula Cr3N2 and determine whether the Cr ion has a charge of 2+ or 3+. Let’s first look up the charge on the negative ion. N is right here… so it always has a charge of 3-. And the formula tells us that it took three Cr ions to balance the charge on two N ions. That means each Cr ion must have had a charge of 2+ So the correct name for Cr3N2 is chromium(II) nitride. Cr3N2 (2+)+(2+)+(2+) + (3-)+(3-) = 0

The third thing you need to know about ionic naming is that some ions are not made from individual atoms, but from groups of atoms bonded together into a cluster. These are known as polyatomic ions (“poly” means many, so a polyatomic ion is simply an ion made from many atoms). These are almost always negative ions, and they almost always involve oxygen (O) atoms bonded to other nonmetals.

For example find the sulfate ion on the chart For example find the sulfate ion on the chart. It is a negative ion so it will be on the right side of the chart. If you have found it, you will see it is SO42-. That is, it is made up of one S atom and four O atoms all bonded together and with a 2- charge on the whole bundle.

2+ 2- So … what is the formula for magnesium sulfate? 13: ____________ First, we look up magnesium on the left side of the ion chart. We see its symbol is Mg and then we look it up in the periodic table and see that it has a charge of 2+. We look up sulfate, find it on the right side: SO4 with a 2- charge. Since 2+ and 2- already balance, the formula is simply MgSO4. (That is, it took one SO4 ion to balance the charge on one Mg ion.) Mg 2+ SO4 2-

And what is the formula for copper(I) phosphate? 14: ____________ First, we look up copper(I) on the left side of the ion chart. We see its symbol is Cu and the Roman numeral tells us its charge is 1+. We look up phosphate, find it on the right side: PO4 with a 3- charge. Clearly, it will take three Cu1+ ions to balance the charge on one PO43- ion. So the formula for copper(I) phosphate is Cu3PO4. Cu 1+ Cu 1+ PO4 3- Cu 1+

Try another one: what is the formula for tin(II) nitrate First, we look up tin(II) on the left side of the ion chart. We see its symbol is Sn and the Roman numeral tells us its charge is 2+. We look up nitrate, find it on the right side: NO3 with a 1- charge.

Clearly, it will take two NO31- ions to balance the charge on one Sn2+ ion. But how do we show two NO3 ions in the formula? If we write SnNO32, then it looks like the compound is made up of one Sn atom, one N atom and 32 O atoms!! That’s not what we want. We want to show one Sn atom and two NO3 bundles. Whenever a polyatomic ion is doubled or tripled (or whatever), parentheses are placed around it and then the subscript is placed outside the parentheses. So the correct formula for tin(II) nitrate is Sn(NO3)2. NO3 1- Sn 2+ NO3 1-

So … what is the formula for aluminum chlorate? 16: ____________ First, we look up aluminum on the left side of the ion chart. We see its symbol is Al and then we look it up in the periodic table and see that it has a charge of 3+. We look up chlorate, find it on the right side: ClO3 with a 1- charge. It will take three ClO31- ions to balance the charge on one Al3+ ion. Again, we use parentheses here because we need to show three ClO3’s So the correct formula for aluminum chlorate is Al(ClO3)3. ClO3 1- Al 3+ ClO3 1- ClO3 1-

Now, what about the formula for zinc hydroxide? 17: ____________ First, we look up zinc on the left side of the ion chart. We see its symbol is Zn and the chart tells us its charge is 2+. We look up hydroxide, find it on the right side: OH with a 1- charge. Clearly, it will take two OH1- ions to balance the charge on one Zn2+ ion. So the formula for zinc hydroxide is Zn(OH)2. OH 1- Zn 2+ OH 1-

Now, what about the formula for zinc hydroxide? 17: ____________ First, we look up zinc on the left side of the ion chart. We see its symbol is Zn and the chart tells us its charge is 2+. We look up hydroxide, find it on the right side: OH with a 1- charge. Clearly, it will take two OH1- ions to balance the charge on one Zn2+ ion. So the formula for zinc hydroxide is Zn(OH)2. Note that ZnOH2 at first glance does not look wrong... but it is. OH 1- The formula ZnOH2 means that the compound contains one Zn, one O and two H atoms. But we want it to show one Zn atom and two OH bundles. Zn 2+ OH 1- That’s why the parentheses are needed: Zn(OH)2

Now … what is the formula for ammonium sulfide? 18: ____________ First, we look up ammonium on the left side of the ion chart. We see its symbol is NH4 (a positive polyatomic ion!) and its charge is 1+ We look up sulfide, find it on the right side: S and a quick glance at the periodic table shows us that its charge is 2- Clearly it will take two NH41+ ions to balance the charge on one S2- ion. So the correct formula for ammonium sulfide is (NH4)2S. NH4 1+ S 2- NH4 1+

Try this one: what is the formula for iron(III) carbonate? 19: ________ First, we look up iron(III) on the left side of the ion chart. We see its symbol is Fe and the Roman numeral tells us its charge is 3+. We look up carbonate, find it on the right side: CO3 with a 2- charge. We’ve seen this before… it will take two Fe3+ ions [2x(3+) = 6+] to balance the charge on three CO32- ions [3x(2-) = 6-]. So the correct formula for iron(III) carbonate is Fe2(CO3)3. CO3 2- Fe 3+ CO3 2- Fe 3+ CO3 2-

Now, finally, naming ionic compounds with polyatomic ions. What is the name of KNO3? 20: __________________ The trick here is knowing how to break it up into its two ions… Based on everything we’ve seen so far, it should make sense to break it up like: K NO3 and not like: KN O3 We look up K on the left side of the chart: “potassium” and NO3 on the right side: “nitrate”. So the name is simply potassium nitrate.

Try this one: What is the name of FeCO3? 21: _____________ Again, we need to know how to break the formula up into its two ions. It should make sense to break it up like: Fe CO3 and not: FeC O3 We look up Fe on the left side of the chart and see it is either “iron(II)” or “iron(III)” and CO3 on the right side and see that it is “carbonate”. and that it took just one CO32-bundle to balance the charge on one Fe ion. We notice that carbonate has a 2- charge That would mean that the Fe would have to have a charge of 2+. So the name must be iron(II) carbonate.

What is the name of NH4F? 22: _____________ Again, we need to know how to break the formula up into its two ions. It should make sense to break it up like: NH4 F and not: N H4F We look up NH4 on the left side of the chart and see it is “ammonium” and F on the right side and see that it is “fluoride”. No need for any Roman numerals this time… The name would simply be ammonium fluoride.

What is the name of Co(C2H3O2)3? 23: _____________ The parentheses help us know to break it up like: Co C2H3O2 We look up Co on the left side of the chart and see it is either “cobalt(II)” or “cobalt(III)” and C2H3O2 on the right side “acetate”. and that it took three of them to balance the charge on just one Co ion. We notice that C2H3O2 has a charge of 1- That would mean that the Co would have to have a charge of 3+. So the name must be cobalt(III) acetate.

Here’s a tricky one: NH4C2H3O2? 24: _____________ Actually, the only thing tricky about it is figuring out how to break it up into two recognizable ions… It should make sense to break it up like: NH4 C2H3O2 And not like: N H4C2H3O2 or NH4C2 H3O2 or NH4C2H3 O2 We look up NH4 and see it is “ammonium” We look up C2H3O2 and see it is “acetate” So the name must be ammonium acetate.

2+ 3- 2+ 3- 2+ One last question: Name Cr3(PO4)2 25: _______________ The parentheses help us know to break it up like: Cr PO4 We find Cr on the left side of the chart: it is either “chromium(II)” or “chromium(III)” and PO4 on the right side is “phosphate”. We notice that PO4 has a charge of 3- and that it took two of them to balance the charge on three Cr ions. Cr 2+ That would mean that the Cr would have to have a charge of 2+. PO4 3- So the name is chromium(II) phosphate. Cr 2+ PO4 3- Cr 2+