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Carbon, then hydrogen, then oxygen — and what to do when the oxygen comes out as 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 the complete combustion of any hydrocarbon by taking carbon first, hydrogen second and oxygen last, and to finish the job by doubling every coefficient when the oxygen comes out as a half. You will be able to handle a fuel that already contains oxygen, and to balance an incomplete combustion, where the carbon leaves as carbon monoxide and the oxygen count changes with it.
You can balance an equation in order: forced elements first, groups as blocks, the spread-out ones last. Combustion is that method applied to one family of reactions, and the order comes out the same every time — which is what makes it worth learning as a routine.
| Term | What it means |
|---|---|
| Hydrocarbon | A compound of carbon and hydrogen only. |
| Complete combustion | Burning in plenty of air, giving carbon dioxide and water. |
| Incomplete combustion | Burning with too little oxygen, giving carbon monoxide or soot. |
| Fuel | Whatever is being burned. |
| Carbon monoxide | CO: colorless, odorless and poisonous. |
Complete combustion always has the same shape:
$$\text{fuel} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}.$$
That makes the order to balance in obvious rather than a judgment call.
For a hydrocarbon $\mathrm{C}_a\mathrm{H}_b$ with one unit of fuel: $a$ carbon dioxides, $b/2$ waters, and $a + b/4$ oxygen molecules. If either of those last two is a half, double every coefficient and the equation is finished.
That is the whole procedure, and it works for methane and for octane without changing.
Another way: picture
Picture the fuel molecule being taken apart: each carbon atom is handed a pair of oxygen atoms and leaves as carbon dioxide, each pair of hydrogen atoms is handed one oxygen atom and leaves as water. Count what has been handed out, and that is the oxygen you needed.
Another way: steps
On $\mathrm{C_4H_{10}}$:
Write the skeleton. Fuel plus oxygen gives carbon dioxide plus water — or carbon monoxide, if the air is short.
Balance the carbon. One carbon dioxide (or monoxide) per carbon atom in the fuel.
Balance the hydrogen. One water per two hydrogen atoms.
Count the oxygen on the right. Two for each carbon dioxide, one for each monoxide, one for each water.
Subtract the fuel's own oxygen, if it has any.
Halve for oxygen molecules, and if that leaves a half, double every coefficient.
Check the work. Count carbon, hydrogen and oxygen on both sides once more. Are all four coefficients whole numbers with no common factor? And does the oxygen coefficient make sense — for a hydrocarbon, it is always larger than the carbon dioxide's, because the water needs oxygen too.
Taking carbon first is allowed because carbon appears in exactly one formula on each side, so its count is forced by the fuel's formula alone.
Taking hydrogen next is allowed for the same reason: hydrogen is only in the fuel and the water.
Leaving oxygen last is allowed, and necessary, because oxygen appears in the oxygen gas, both products, and sometimes the fuel. Until carbon and hydrogen have fixed the products, nothing fixes oxygen; afterwards, only the oxygen gas coefficient is still free, so oxygen is forced too.
Subtracting the fuel's own oxygen is allowed because conservation counts every oxygen atom, wherever it started. An ethanol molecule's oxygen ends up in a product just as surely as the oxygen gas's does.
Doubling to clear a half is allowed because multiplying every coefficient by two keeps every element's count equal on both sides, while turning the half into a whole number.
Ethanol is $\mathrm{C_2H_6O}$, and the method does not change — but the last step has one extra subtraction in it.
Carbon: 2, so 2 carbon dioxide. Hydrogen: 6, so 3 water. The right-hand side then holds $2 \times 2 + 3 = 7$ oxygen atoms. One of those came in with the ethanol itself, so the oxygen gas only has to supply 6, which is 3 molecules:
$$\mathrm{C_2H_6O + 3O_2 \rightarrow 2CO_2 + 3H_2O}.$$
Forgetting the fuel's own oxygen is the commonest slip in this lesson, and it always makes the oxygen coefficient too large by exactly half the number of oxygens in the fuel.
The same fuel burned in a poor supply of air gives carbon monoxide instead of carbon dioxide, and with even less air, carbon itself — the soot on a chimney and the yellow, smoky flame of a burner whose air hole is shut.
The balancing is identical; only the product changes, and with it the oxygen count, because each carbon monoxide takes one oxygen atom where a carbon dioxide took two:
$$\mathrm{2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O}.$$
This is worth knowing about rather than only calculating. Carbon monoxide has no color and no smell, and it binds to the hemoglobin in blood far more tightly than oxygen does, so it stops blood carrying oxygen at concentrations far too low to notice. That is the reason a fuel-burning appliance needs a clear flue and a working alarm, and the reason a burner is never run in a sealed room.
A balanced combustion equation is more than a bookkeeping exercise; it is a recipe that says how much air a flame needs and how much exhaust it makes. Read $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$ aloud and it says: for every propane molecule, five oxygen molecules go in, and three carbon dioxides and four waters come out.
That ratio scales. Ten propane molecules need fifty oxygens; a million need five million. In the next lessons the same ratio will be read in moles, and then in grams, which is how an engineer sizes the air intake on a furnace or works out the carbon dioxide a car puts out per gallon. None of that is possible until the equation is balanced, which is why this lesson insists on the routine.
The recipe also explains a pattern worth noticing. The longer the carbon chain, the more oxygen each molecule needs, and the more carbon dioxide it makes: methane needs two oxygens, propane five, octane twelve and a half. A heavier fuel carries more energy per molecule and asks more of the air around it for the same reason.
For a hydrocarbon $\mathrm{C}_a\mathrm{H}_b$, the oxygen per fuel molecule is $a + b/4$. That gives a fast check before any careful counting. For butane, $\mathrm{C_4H_{10}}$, it is $4 + 2.5 = 6.5$, so the doubled equation must have 13 in front of the oxygen and 2 in front of the butane. If your careful balancing gives a different number, one of the two is wrong, and the estimate tells you where to look. The estimate only applies to fuels with no oxygen of their own; for an alcohol, subtract half the fuel's oxygen atoms from it.
Most states now require carbon monoxide alarms in homes with gas appliances or attached garages, and the reason is in the two equations this lesson balances. A natural gas furnace burns methane. With a clear flue and plenty of air, the reaction is $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$: two oxygen molecules for each methane, and harmless products that go up the chimney.
If a bird's nest blocks the flue, or a cracked heat exchanger lets the flame starve, the same methane burns as $\mathrm{2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O}$. Per methane molecule that is only one and a half oxygen molecules — the flame is making do with less — and the carbon leaves as carbon monoxide.
Carbon monoxide has no color and no smell, and it binds to the hemoglobin in red blood cells about 200 times more tightly than oxygen does. At a few hundred parts per million it causes headaches and confusion; at a few thousand it can kill within an hour. The Centers for Disease Control and Prevention estimates that hundreds of Americans die from unintentional carbon monoxide poisoning every year, most of them in winter. An alarm detects what no person can, and a yearly furnace inspection keeps the equation on the complete side.
A backyard grill burns propane as $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$. The burner's air holes are sized to let in five oxygen molecules for each propane, so the flame burns blue and clean instead of yellow and sooty. A standard 20-pound tank holds about 9 kilograms of propane, and the equation says that burning all of it uses five oxygen molecules per propane molecule — roughly 33 kilograms of oxygen, drawn from well over 100 kilograms of air. That is why a grill is only ever lit outdoors: indoors, the same flame would quickly use up the oxygen in a closed space, slide toward incomplete combustion, and fill the room with carbon monoxide. The grill's instructions, the fire code and the balanced equation all say the same thing in different words. Respect all three.
Balancing oxygen first. It is in three of the four formulas and every other decision moves it. It is the last element, every time.
Refusing to write the half. $\tfrac{7}{2}$ is the correct number of oxygen molecules for one unit of ethane. Write it, finish the equation, then double. Learners who will not write it start guessing at the fuel's coefficient and lose far more time.
Forgetting the oxygen inside the fuel. Ethanol brings one oxygen atom of its own to the reaction. If you count only the oxygen gas, the equation comes out with too much oxygen on the left and nothing obvious to point at.
Doubling and then not checking. After doubling, look once at the four numbers. If the doubling was forced by a genuine half they are in lowest terms; if you doubled out of habit they are not.
Writing carbon dioxide when the air was short. Incomplete combustion gives carbon monoxide, which needs less oxygen and is the reason a blocked flue is dangerous.
Write the skeleton.
$\mathrm{C_3H_8 + O_2 \rightarrow CO_2 + H_2O}$
Complete combustion.
Balance the carbon.
$3 \ \mathrm{CO_2}$
Forced by the fuel's formula.
Balance the hydrogen.
$8 \div 2 = 4 \ \mathrm{H_2O}$
Also forced.
Count the oxygen on the right.
$3 \times 2 + 4 = 10$
An even count.
Write the equation.
$\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$
No half, nothing to double.
Write the skeleton.
$\mathrm{C_2H_2 + O_2 \rightarrow CO_2 + H_2O}$
The fuel of a welding torch.
Balance the carbon.
$2 \ \mathrm{CO_2}$
Forced.
Balance the hydrogen.
$2 \div 2 = 1 \ \mathrm{H_2O}$
Forced.
Count the oxygen on the right.
$2 \times 2 + 1 = 5$
An odd number, and oxygen comes in pairs.
Write it with the half.
$2.5 \ \mathrm{O_2}$
The honest answer before the last step.
Double every coefficient.
$\mathrm{2C_2H_2 + 5O_2 \rightarrow 4CO_2 + 2H_2O}$
No shared factor.
Write the skeleton.
$\mathrm{C_2H_6O + O_2 \rightarrow CO_2 + H_2O}$
The fuel already holds oxygen.
Balance the carbon.
$2 \ \mathrm{CO_2}$
Forced.
Balance the hydrogen.
$6 \div 2 = 3 \ \mathrm{H_2O}$
Forced.
Count the oxygen on the right.
$2 \times 2 + 3 = 7$
Both products.
Subtract the fuel's oxygen.
$7 - 1 = 6$
One came in with the ethanol.
Halve for oxygen molecules.
$6 \div 2 = 3$
No half this time.
Write the equation.
$\mathrm{C_2H_6O + 3O_2 \rightarrow 2CO_2 + 3H_2O}$
Forgetting the fuel's oxygen would have given 3.5.
Balance carbon and hydrogen.
$6 \ \mathrm{CO_2}; \ 7 \ \mathrm{H_2O}$
Both come straight off the formula.
Count the oxygen on the right.
$6 \times 2 + 7 = 19$
An odd number.
Clear the half.
Balancing the combustion of ethyne carbon-first, with a single unit of fuel, gives $\tfrac{5}{2}$ molecules of oxygen on the left. Everything else is a whole number. What is the right next move?
Complete the worked solution: one molecule of an alkane, $\mathrm{C_{3}H_{8}}$, burns completely to carbon dioxide and water. Carbon forces one carbon dioxide for each carbon atom. Find the waters, the oxygen atoms on the right, and the oxygen molecules needed.
Find the waters.
$(\text{hydrogen atoms}) \div \text{two} =$ w
Two hydrogens per water.
Count the oxygen atoms on the right.
$\text{two} \times (\text{carbon dioxides}) + (\text{waters}) =$ o
Oxygen goes last.
Find the oxygen molecules.
$(\text{oxygen atoms}) \div \text{two} =$ m
A half here means doubling everything.
The fuel is propane, burning in a camping stove. With plenty of air, the only products are carbon dioxide and water. Balance the equation.
This task has no paper form; do it on a device.
The products of burning ethane have already been counted. Work backwards and fill in the two coefficients on the left.
a $\mathrm{C_2H_6} \ + \ $ b $\mathrm{O_2} \ \rightarrow \ 4\mathrm{CO_2} \ + \ 6\mathrm{H_2O}$
A backyard grill in Texas burns propane: $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$. If $8$ million propane molecules burn completely, how many million oxygen molecules must the grill draw in?
The answer: a.
In a gas boiler with a blocked flue there is not enough air, and the carbon comes out as carbon monoxide instead of carbon dioxide. That is why a blocked flue is dangerous: carbon monoxide has no smell and stops blood carrying oxygen. Balance the equation for methane burning this way.
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.
Burning ethanol completely gives carbon dioxide and water. Write the four coefficients of the balanced equation into the table.
| in front of the fuel | in front of $\mathrm{O_2}$ | in front of $\mathrm{CO_2}$ | in front of $\mathrm{H_2O}$ | |
|---|---|---|---|---|
| Coefficient |
You can balance a combustion equation in one pass, and you know why oxygen is always the element left until last. Say what the coefficient of oxygen would be for burning one molecule of ethane, and what you would then do about it. Next: equations where charge has to balance as well as atoms.
17. Your turn: balance the combustion of hexane, $\mathrm{C_6H_{14}}$., step 3
$\mathrm{2C_6H_{14} + 19O_2 \rightarrow 12CO_2 + 14H_2O}$
Nineteen halves, doubled.