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The order that makes balancing quick: forced elements first, groups as blocks, oxygen last, and double to clear a half.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to balance a chemical equation without guessing: list the elements, settle the ones that appear in a single formula on each side, count any group that survives the reaction as one block, leave hydrogen and oxygen until last, clear a half by doubling every coefficient, and finish by checking that the numbers share no common factor. You will be able to do it on a decomposition, a synthesis and a reaction with two substances on each side.
You know what balancing means and what it may change: the numbers in front, and nothing else. You have balanced a decomposition and a two-in, two-out reaction by counting. This lesson is about not having to guess.
| Term | What it means |
|---|---|
| Polyatomic ion | A group of atoms that acts as one unit, such as sulfate or nitrate. |
| Diatomic element | One that exists as pairs: hydrogen, nitrogen, oxygen and the halogens. |
| Forced element | One that appears in exactly one formula on each side. |
| Lowest terms | Coefficients that share no whole-number factor above 1. |
Every element in an equation is in one of two situations, and which one decides when you should deal with it.
Forced. The element appears in exactly one formula on the left and one on the right. Then its two coefficients are locked to each other: choose one and the other follows. There is no search here, only arithmetic, and it costs nothing to do immediately.
Free until later. The element appears in two or more formulas on a side — oxygen usually, hydrogen often. Nothing pins it down yet, because several coefficients feed into its total. Balancing it early means choosing a number that some later step will have to undo.
So the method is:
The one extra move: if a step gives you a half — and oxygen is where it happens, because oxygen comes in pairs — finish the balancing with the half in place and then double every coefficient. That clears it in one go, and it is always in lowest terms afterwards.
Another way: picture
Picture the elements as a row of dials, each showing left-count and right-count. A forced element's two dials are geared together: turn one and the other moves. A spread-out element's dial is driven by three or four gears at once, so it is the last one you can set.
Another way: steps
On $\mathrm{Al + CuSO_4 \rightarrow Al_2(SO_4)_3 + Cu}$:
List the elements, and mark any polyatomic group that appears unchanged on both sides as a single block.
Find the forced elements. Those in exactly one formula on each side.
Balance them one at a time. Adjust only a coefficient that nothing else has fixed yet.
Balance the blocks. Count sulfates, nitrates and hydroxides as items.
Balance the spread-out elements last. Usually oxygen, often hydrogen.
Clear any half. Double every coefficient.
Check the work. Count every element on both sides one final time, not just the last one you adjusted — a late change can undo an early one. Look for a common factor among all the coefficients. And confirm that every formula is still exactly as given: a balanced equation with an edited subscript is not an answer.
Starting with forced elements is allowed because their coefficients are linked by a single equation. Two atoms of iron on the left and one per formula on the right means the iron coefficient must be 2; there is nothing to choose.
Counting a surviving group as a block is allowed because the group's atoms always move together. If sulfates balance, then the sulfur and the four oxygens in each sulfate balance with them, automatically.
Leaving spread-out elements last is allowed because their total depends on several coefficients. Once those are fixed by the forced elements, only one coefficient affecting the spread-out element is left free, and it is forced too.
Doubling to clear a half is allowed because multiplying every coefficient by the same number keeps every count equal on both sides. A balanced equation stays balanced under any common multiple, which is also why the final check for a common factor is needed.
Once an equation is balanced, its coefficients are a recipe. $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$ says that one nitrogen molecule and three hydrogen molecules make two ammonia molecules — and, because the ratio holds for any number of them, that a hundred nitrogens take three hundred hydrogens and make two hundred ammonias. A million take three million.
That is the bridge to the rest of the course. The mole, two units from now, is a way of counting molecules by the billions of billions, and every calculation in that unit begins by reading a ratio like this one off a balanced equation. A wrong coefficient here is a wrong ratio there, and every mass worked out from it is wrong by the same factor.
It is also why lowest terms matter beyond tidiness. $\mathrm{2N_2 + 6H_2 \rightarrow 4NH_3}$ gives exactly the same recipe, so it would never give a wrong answer — but two people comparing their working could each think the other had made a mistake. One agreed form saves that argument.
Sometimes the order runs out: every element you try seems to depend on another. Three checks usually unstick it.
Is there a block you have taken apart? A sulfate or a nitrate counted as separate atoms adds two or three extra elements to juggle. Put it back together.
Is there a half waiting to be written? If oxygen needs an odd number of atoms and comes in pairs, write the half, finish, then double.
Is one formula wrong? If nothing at all will balance, check each formula against its name before trying more coefficients. An equation with a wrong formula in it may have no balanced form at all, or only a form that describes a different reaction. Balancing cannot repair a formula; it can only reveal that something is off, and the fix is always to the formula, never to a subscript chosen to make the counts work.
Balancing the combustion of ethane by counting gives $\mathrm{C_2H_6 + \tfrac{7}{2}O_2 \rightarrow 2CO_2 + 3H_2O}$, and the fraction is not a mistake. Two carbons need two carbon dioxides, six hydrogens need three waters, and those together need seven oxygen atoms — which is three and a half molecules.
A coefficient counts units and has to be a whole number, so the last move is to double everything:
$$\mathrm{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O}.$$
It is worth letting the half appear rather than trying to avoid it. Learners who refuse to write it end up guessing at the fuel's coefficient, and the guess is nearly always wrong for the same reason the half was right.
For a reaction with a single sensible route, the balanced equation in lowest terms is the only one. That is worth knowing because it means there is nothing to argue about: two people who balance the same equation correctly write the same four numbers, in the same order, every time.
It also means the lowest-terms rule is not fussiness. $\mathrm{4H_2 + 2O_2 \rightarrow 4H_2O}$ is a true statement about four hydrogen molecules. It is not the equation for the reaction, because the equation is a statement about the ratio, and $4 : 2 : 4$ is the same ratio as $2 : 1 : 2$ written with a factor stuck to it.
Every gasoline engine sold in the United States has a computer that holds the mixture of air and fuel at one particular ratio, and that ratio comes straight from a balanced equation. Gasoline is mostly octane-like molecules; for octane itself the combustion balances as $\mathrm{2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O}$.
Balancing it is this lesson's method exactly. Carbon is forced: eight per octane, so eight carbon dioxides. Hydrogen is forced: eighteen per octane, so nine waters. Oxygen goes last: $8 \times 2 + 9 = 25$ atoms, which is twelve and a half molecules — a half, cleared by doubling everything.
Those coefficients set how much oxygen the engine must draw in for each gram of fuel. Worked through with the masses this course reaches in its mole unit, they give an air-to-fuel ratio of about 14.7 to 1 by mass. The oxygen sensor in the exhaust pipe of a car in Detroit or Denver reports whether there is leftover oxygen, and the computer adjusts the fuel injectors to hold the mixture right at that balanced point, where the catalytic converter works best and the least fuel is wasted.
The steel mills along Lake Michigan in Gary, Indiana reduce iron ore with carbon monoxide: $\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}$. The balanced coefficients tell the mill how much coke to burn for each ton of ore, and the oxygen, spread over three formulas, was the last element balanced.
Balancing oxygen first. It is the element most learners reach for, and it is almost always the worst one to start with, because it is usually in three or four of the formulas. Every other element will move it, so anything you do to it early is wasted.
Taking a group apart. Counting one sulfur and four oxygens when the sulfate went in whole and came out whole. Not wrong, just four times the work and four times the chances to slip.
Refusing to write the half. A learner who will not write $\tfrac{7}{2}$ starts guessing coefficients for the fuel. Write it, finish, then double.
Stopping at the first set that works. A doubled answer balances. The habit that catches it is one final look: do these numbers share a factor?
And the one from the last lesson, which does not go away: if you find yourself editing a subscript, you have stopped balancing and started inventing a substance.
Spot the surviving group.
$\mathrm{Pb(NO_3)_2 + KI \rightarrow PbI_2 + KNO_3}$
Nitrate goes in whole and comes out whole.
Balance the nitrate blocks.
$2 \ \mathrm{KNO_3}$
Two on the left, so two on the right.
Balance the potassium.
$2 \ \mathrm{KI}$
Forced by the potassium nitrate.
Check the iodine and lead.
$\text{I: } 2 = 2; \ \text{Pb: } 1 = 1$
Each forced element settled the next.
Write the equation.
$\mathrm{Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3}$
Nitrogen and oxygen never counted separately.
Balance the iron.
$\mathrm{Fe_2O_3 + CO \rightarrow 2Fe + CO_2}$
One formula on each side: settle it now.
Look at the carbon.
$\text{forced, but by two free coefficients}$
Leave it linked for now.
Count the oxygen sources.
$\text{three formulas}$
Oxygen is spread out, so it goes last.
Try three of each carbon oxide.
$3\mathrm{CO}, \ 3\mathrm{CO_2}$
Carbon three and three.
Check the oxygen.
$3 + 3 = 6; \ 3 \times 2 = 6$
It falls out once the rest is fixed.
Write and check the equation.
$\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}$
No shared factor.
Balance the carbon.
$\mathrm{C_2H_6} \to 2\mathrm{CO_2}$
Forced by the fuel.
Balance the hydrogen.
$6 \text{ H} \to 3\mathrm{H_2O}$
Two hydrogens per water.
Count the oxygen on the right.
$2 \times 2 + 3 = 7$
Oxygen last.
Find the oxygen molecules.
$7 \div 2 = 3.5$
A half, because oxygen comes in pairs.
Write it with the half.
$\mathrm{C_2H_6 + 3.5O_2 \rightarrow 2CO_2 + 3H_2O}$
Balanced, but not whole numbers.
Double every coefficient.
$\mathrm{2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O}$
The half is cleared.
Check for a factor.
$2, 7, 4, 6: \text{ none shared}$
Seven shares nothing with the others.
Balance carbon and hydrogen.
$3\mathrm{CO_2}; \ 4\mathrm{H_2O}$
Both forced by the fuel.
Count the oxygen on the right.
$3 \times 2 + 4 = 10$
Oxygen arrives in pairs.
Write the equation.
Put the steps of balancing an equation into the order that gets you there fastest — the order you would use on iron(III) oxide reduced by carbon monoxide in a blast furnace.
Number the steps in order (write the number in the box):
Complete the worked solution: butane, $\mathrm{C_4H_{10}}$, burns completely to carbon dioxide and water. Carbon forces four carbon dioxides and hydrogen forces five waters. Find the oxygen atoms on the right, the oxygen molecules that makes, and the fuel's coefficient after doubling to clear the half.
Count the oxygen atoms on the right.
$(\text{carbon dioxides}) \times \text{two} + (\text{waters}) =$ o
Two per carbon dioxide, one per water.
Find the oxygen molecules.
$(\text{oxygen atoms}) \div \text{two} =$ m
Oxygen arrives in pairs, so a half appears.
Double every coefficient.
$\text{the fuel's coefficient becomes}$ f
Whole numbers clear the half in one move.
Two substances react and there is only one product, which means the product's formula decides everything. Balance the equation for magnesium ribbon burning in air.
This task has no paper form; do it on a device.
Balance the equation for chlorine displacing bromine from potassium bromide solution. Look first for anything that goes in whole and comes out whole — a sulfate, a nitrate, a hydroxide — because counting it as one block is far less work than counting its atoms one at a time.
This task has no paper form; do it on a device.
A fertilizer plant in Louisiana makes ammonia by the balanced reaction $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$. In one batch $6$ million molecules of nitrogen react completely. How many million molecules of hydrogen does that use?
The answer: a.
This equation is from gasoline burning in a car engine, and its coefficients are much larger than anything in this lesson. Nothing about the method changes — the numbers are just bigger, which is exactly why guessing stops working and the order matters. Balance it.
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.
One substance, two products, and everything on the right came out of the one formula on the left. Balance the equation for the inflator of a car airbag firing.
This task has no paper form; do it on a device.
You can balance an equation by working in order rather than by trial. Say which element you would start with in the equation for burning propane, and why it is not oxygen. Next: combustion, where the order matters most because oxygen is always the element left over.
17. Your turn: balance $\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$., step 3
$\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$
No half to clear, and no shared factor.