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The subscripts are not in the name, so they have to be worked out — from the one fact that a formula unit carries no overall charge.
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By the end of this lesson you will be able to write the formula of an ionic compound from its name: find the charge on each ion from the group number, from a roman numeral or from the polyatomic ion itself, then find the smallest counts that make the positive and negative charge totals equal. You will be able to say why crossing the charges over works, where it overshoots, and when a bracket is needed — and you will see what a wrong formula does to a balanced equation that never looks wrong.
You can name an ionic compound from its formula, and you know that a roman numeral states the charge on the metal ion. You can also read a group number as a count of outer electrons. This lesson runs the naming process backwards, and the one new idea is that the subscripts are not stated anywhere in the name — they have to be worked out, from the fact that the compound has no overall charge.
| Term | What it means |
|---|---|
| Formula unit | The smallest group of ions the formula describes. |
| Neutral | Carrying no overall charge: positive and negative totals are equal. |
| Subscript | The small number after a symbol or a bracket. |
| Crossing over | The shortcut of making each charge the other ion's subscript; it may need cancelling. |
| Cancelling number | The smallest number both charge sizes divide into. |
A compound sitting in a jar has no charge on it. So in one formula unit, the positive charge and the negative charge are the same number:
$$(\text{how many positive ions}) \times (\text{their charge}) = (\text{how many negative ions}) \times (\text{their charge})$$
That is the only rule, and everything else in this lesson is arithmetic with it.
Where the charges come from. A metal in group 1 has one outer electron to give away, so its ion is $1+$; group 2 gives two, so $2+$; group 3 gives three, so $3+$. A non-metal in group 7 is one electron short of a full shell, so it takes one and its ion is $1-$; group 6 takes two, so $2-$. A polyatomic ion has a charge of its own — hydroxide and nitrate are $1-$, carbonate and sulfate are $2-$, ammonium is $1+$. And where the name carries a roman numeral, the numeral is the charge and there is nothing to work out.
Then find the smallest counts. For calcium and chloride, $2+$ against $1-$: one calcium and two chlorides, because $1 \times 2 = 2 \times 1$. For aluminum and oxide, $3+$ against $2-$: two aluminum and three oxides, because $2 \times 3 = 3 \times 2$.
The shortcut, and its catch. Crossing the charges over — each charge becoming the other ion's subscript — gives the right counts most of the time, because $q_1 \times q_2 = q_2 \times q_1$ always. But it gives the smallest counts only when the two charges share no common factor. Magnesium $2+$ with oxide $2-$ crosses to $\mathrm{Mg_2O_2}$, and the answer is $\mathrm{MgO}$. Cross over if you like, then cancel.
Brackets. If more than one polyatomic ion is needed, the ion goes in a bracket and the count goes outside it: two nitrates is $(\mathrm{NO_3})_2$, never $\mathrm{NO_6}$. The bracket is what keeps the ion whole.
Another way: picture
Think of it as paying an exact amount with two kinds of coin. A $3+$ ion is a three-unit coin and a $2-$ ion is a two-unit coin, and you have to make the two piles equal with whole coins and no change. Six is the smallest total that works, so two of one and three of the other — and that is $\mathrm{Al_2O_3}$. Where the coins are both worth two, one of each is enough, and that is why magnesium oxide is $\mathrm{MgO}$.
Another way: steps
To write a formula from a name:
Split the name. The first word is the positive ion; the second is the negative ion.
Find the two charges. From the group, from a roman numeral, or from a polyatomic ion's own charge.
Find the cancelling number. The smallest number both charge sizes divide into.
Divide to get the counts. The cancelling number over each charge.
Write the formula. Positive ion first; counts as subscripts; a bracket round a polyatomic ion counted more than once; no subscript of 1.
Check the work. Multiply each count by its charge — are the two totals equal? Do the counts share a common factor? If so, cancel. Is every polyatomic ion still whole, with its own subscripts untouched inside its bracket? And read the formula back into a name: do you get the name you started from?
Reading a main-group charge from the group is allowed because an atom reaches a full outer shell by the shorter route, losing or gaining as few electrons as possible.
Taking a roman numeral as the charge is allowed because that is what the numeral was put in the name to say. Nothing else in the name could tell you a transition metal's charge.
Making the totals equal is allowed because matter we can handle is electrically neutral. Every formula unit, and so every sample, carries as much positive charge as negative.
Using the smallest counts is allowed because a formula unit is defined as the smallest neutral group. Larger counts describe the same lattice, so they are cancelled down.
Keeping a polyatomic ion whole is allowed because it really is one particle: its atoms are held together by covalent bonds and carry the charge between them. Changing its inside subscripts would make a different ion.
For the main-group elements, the ion's charge is a consequence of the group number, and the reason is the one from the last two lessons: an atom reaches a full outer shell by whichever route is shorter.
| Group | Outer electrons | What it does | Ion |
|---|---|---|---|
| 1 | 1 | gives one away | $1+$ |
| 2 | 2 | gives two away | $2+$ |
| 3 | 3 | gives three away | $3+$ |
| 5 | 5 | takes three in | $3-$ |
| 6 | 6 | takes two in | $2-$ |
| 7 | 7 | takes one in | $1-$ |
| 8 | 8 | neither | none |
Group 4 is missing from the table on purpose: with four outer electrons there is no short route either way, so carbon and silicon share rather than forming ions, and that is the subject of the bonding unit.
The transition metals are missing too, and for the reason the naming lesson gave: most of them form more than one ion, so the group cannot tell you the charge and the name has to. That is what the roman numeral is for.
It is tempting to treat writing a formula as a tidy-up step before the real chemistry. It is the opposite: it is the step every later one inherits.
A molar mass is the sum of the atomic masses in the formula, so a wrong subscript gives a wrong molar mass, which gives a wrong number of moles, which gives a wrong mass of product — and every one of those calculations is performed correctly. Nothing in the arithmetic complains.
Balancing inherits it too. Take aluminum burning in oxygen. If the product is written $\mathrm{Al_2O_3}$, the equation balances as $4\mathrm{Al} + 3\mathrm{O_2} \rightarrow 2\mathrm{Al_2O_3}$. Write the product as $\mathrm{AlO}$ instead — which is what crossing the charges the wrong way round gives — and the equation still balances, as $2\mathrm{Al} + \mathrm{O_2} \rightarrow 2\mathrm{AlO}$. A perfectly balanced equation about a substance that does not exist, and no checker in the world would catch it from the equation alone.
That is why the transfer question in this lesson is a balancing question. The balancing is easy; the formula underneath it is the part that had to be right.
A county road commission in Michigan buys de-icing chemicals by name from a bid sheet: sodium chloride, calcium chloride, magnesium chloride. The purchasing officer never sees a formula, but the engineer deciding how much to spread needs one, because the formula decides how many dissolved particles each pound of salt produces.
The names are enough. Sodium is in group 1, so $\mathrm{Na^+}$; chloride is $\mathrm{Cl^-}$; one of each, $\mathrm{NaCl}$, two ions per formula unit. Calcium is in group 2, so $\mathrm{Ca^{2+}}$; it needs two chlorides to cancel, $\mathrm{CaCl_2}$, three ions per formula unit. Magnesium is also in group 2, so magnesium chloride is $\mathrm{MgCl_2}$, three ions as well.
Those counts are why calcium and magnesium chloride keep working at colder temperatures than rock salt: three ions per formula unit get in the way of freezing more than two do. A bid sheet that listed calcium chloride as $\mathrm{CaCl}$, as a careless spreadsheet might, would throw off every calculation of how much to apply per lane-mile, while the arithmetic itself looked perfectly sound. The formula, worked out from the name, is the step everything else depends on.
The antacid sold as milk of magnesia is magnesium hydroxide. Magnesium is $2+$ and hydroxide is $1-$, so it takes two hydroxides, in a bracket: $\mathrm{Mg(OH)_2}$.
The charges are not crossed. Aluminum $3+$ with oxide $2-$ written as $\mathrm{Al_3O_2}$. Check it: three lots of $3+$ is $9$, two lots of $2-$ is $4$, and the unit carries a charge of $5+$. A formula unit is never charged, so this cannot be a formula.
Crossing over is done and the cancelling is forgotten. Magnesium $2+$ with oxide $2-$ written as $\mathrm{Mg_2O_2}$. The charges do cancel, so it is neutral; it is just not the smallest unit, and a formula names the smallest unit.
A polyatomic ion is broken open. Two nitrates written as $\mathrm{NO_6}$ or $\mathrm{N_2O_6}$. Nitrate travels as a whole $\mathrm{NO_3^-}$ and stays whole in the formula, so two of them is $(\mathrm{NO_3})_2$ with the bracket doing the multiplying.
The roman numeral is used as a subscript. Iron(III) chloride written as $\mathrm{Fe_3Cl}$. The three is the charge on the iron ion, and it forces three chlorides for one iron: $\mathrm{FeCl_3}$.
A subscript of 1 is written. $\mathrm{Na_1Cl_1}$. A symbol with no subscript already means one, so the ones are left off.
The charge is read off the group number for a transition metal. It cannot be, which is exactly why those names carry a numeral.
Find aluminum's charge.
$\text{group } 3 \Rightarrow \mathrm{Al^{3+}}$
From the group number.
Find oxide's charge.
$\text{group } 6 \Rightarrow \mathrm{O^{2-}}$
From the group number.
Find the cancelling number.
$3 \text{ and } 2 \Rightarrow 6$
The smallest total both divide into.
Divide by each charge.
$6 \div 3 = 2; \ 6 \div 2 = 3$
Two aluminum ions, three oxide ions.
Write the formula.
$\mathrm{Al_2O_3}$
Three and two share no factor, so nothing cancels.
Find calcium's charge.
$\text{group } 2 \Rightarrow \mathrm{Ca^{2+}}$
From the group number.
Find carbonate's charge.
$\mathrm{CO_3^{2-}}$
A polyatomic ion's own charge.
Cross the charges over.
$\mathrm{Ca_2(CO_3)_2}$
Neutral, but not the smallest unit.
Look for a common factor.
$2 \text{ and } 2 \text{ share } 2$
Both counts divide by two.
Cancel the counts down.
$1 : 1$
One of each.
Write the formula.
$\mathrm{CaCO_3}$
No bracket: only one carbonate.
Read iron's charge.
$\text{iron(III)} \Rightarrow \mathrm{Fe^{3+}}$
The numeral states it.
Find sulfate's charge.
$\mathrm{SO_4^{2-}}$
A polyatomic ion's own charge.
Find the cancelling number.
$3 \text{ and } 2 \Rightarrow 6$
Six of each kind of charge.
Divide by each charge.
$6 \div 3 = 2; \ 6 \div 2 = 3$
Two irons, three sulfates.
Bracket the sulfate.
$(\mathrm{SO_4})_3$
More than one polyatomic ion needs a bracket.
Write the formula.
$\mathrm{Fe_2(SO_4)_3}$
Positive ion first.
Check the charges.
$2 \times 3 = 6; \ 3 \times 2 = 6$
Neutral, and in lowest terms.
Find both charges.
$\mathrm{Mg^{2+}}; \ \mathrm{NO_3^{-}}$
Group 2, and nitrate's own charge.
Find the counts.
$1 \times 2 = 2 \times 1$
One magnesium, two nitrates.
Write the formula.
Match each name to the formula it describes.
| $\mathrm{Al_2O_3}$ | $\mathrm{CaCO_3}$ | $\mathrm{Na_2CO_3}$ | $\mathrm{(NH_4)_2SO_4}$ | |
|---|---|---|---|---|
| aluminum oxide | ||||
| calcium carbonate | ||||
| sodium carbonate | ||||
| ammonium sulfate |
Complete the worked solution: write the formula of lead(IV) oxide. Each oxide ion carries a charge of two minus. Find the metal ion's charge, the smallest number both charges divide into, and how many of each ion one formula unit holds.
Read the metal ion's charge.
$\text{the roman numeral} =$ q
A numeral states the charge outright.
Find the cancelling number.
$\text{smallest number both charges divide into} =$ l
The two totals must reach it together.
Find the metal ions.
$(\text{cancelling number}) \div (\text{metal's charge}) =$ c
Positive ions needed.
Find the negative ions.
$(\text{cancelling number}) \div (\text{anion's charge}) =$ a
Negative ions needed.
Write the formula of copper(II) sulfate, and say how many sulfate ions one formula unit of it contains. Type formulas plainly, as in MgCl2.
The formula of copper(II) sulfate is a, and the number of sulfate ions in one formula unit is b.
Give both charges in magnesium sulfate and then its formula. Type formulas plainly, as in MgCl2.
In magnesium sulfate the magnesium ion carries a charge of a+ and the sulfate ion carries a charge of b-, so the formula is c.
aluminum sulfate is the compound a Chicago water plant doses into its settling tanks. Working only from the name, how many sulfate ions are in one formula unit of it?
The answer: a.
In a works, aluminum is burned in oxygen to make aluminum oxide. The product's formula is $\mathrm{Al_2O_3}$ — which is the formula the charges force, and nothing about the equation can be settled until it is right. Balance the equation.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
The two charges are given for each of these three compounds. Work out how many of each ion one formula unit contains.
| charge on the positive ion | charge on the negative ion | positive ions in one formula unit | negative ions in one formula unit | |
|---|---|---|---|---|
| zinc chloride | 2 | 1 | ||
| calcium hydroxide | 2 | 1 | ||
| potassium nitrate | 1 | 1 |
You can write a formula from a name by making the charges cancel, and you know why the smallest counts are the ones to write. Say out loud why magnesium oxide is not written as two magnesium and two oxide when crossing the charges over says it should be. Next: how the mole turns these formulas into masses you can actually weigh out.
15. Your turn: write the formula of magnesium nitrate., step 3
$\mathrm{Mg(NO_3)_2}$
The bracket keeps each nitrate whole.