Mole ratios · printable packet

Using a balanced equation's coefficients as a ratio of amounts, with a molar mass at each end.

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.

1. What you will learn

By the end of this lesson you will be able to take a mass of one substance through a balanced equation to the mass of another: divide by the first molar mass, multiply by the coefficient of what you want over the coefficient of what you have, and multiply by the second molar mass. You will be able to run the chain in either direction, and to say why the coefficients can never be applied to masses directly.

2. What you already have

You can balance an equation, and you can convert between mass and moles in either direction. This lesson adds the one step between them, and then the whole calculation is three steps you already know joined in a line.

3. Words for this lesson

4. Grams out, moles across, grams back

Term	What it means
Mole ratio	The ratio of two substances' coefficients in a balanced equation.
Stoichiometry	Working out how much of one substance a reaction uses or makes from another.
Theoretical yield	The mass of product the equation allows, before anything is lost.
Reactant	A substance used up in a reaction.
Product	A substance made by a reaction.

A balanced equation is a statement about counts, so its coefficients are a ratio of amounts. To use them on a mass you have to be in moles first, and to report a mass you have to come back out.

$$\text{g of A} \xrightarrow{\ \div M_\mathrm{A}\ } \text{mol of A} \xrightarrow{\ \times \frac{c_\mathrm{B}}{c_\mathrm{A}}\ } \text{mol of B} \xrightarrow{\ \times M_\mathrm{B}\ } \text{g of B}$$

Three arrows, three numbers, and each number comes from somewhere different. The first molar mass comes from A's formula, the ratio from the balanced equation, and the second molar mass from B's formula. Nothing else is involved.

The middle arrow is the only place the chemistry enters. The two ends are unit conversions; they would be the same if the reaction were something else entirely. That is worth knowing because it tells you where to look when an answer is wrong: a mistake in the ratio gives an answer out by a small whole-number factor, and a mistake at either end gives one out by a molar mass.

The ratio is the coefficient of what you want over the coefficient of what you have. Written that way round it is hard to invert by accident.

Another way: picture

Think of a recipe. Two eggs make one omelet, whatever an egg weighs. To go from a pound of eggs to a weight of omelets you first turn the pound into a number of eggs, use the recipe on the number, and then turn the omelets back into a weight. The recipe never speaks in ounces.

Another way: steps

For $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, from 28 g of nitrogen:

$28 \div 28 = 1$ mol of $\mathrm{N_2}$.
Ratio of ammonia to nitrogen is $2 : 1$, so $1 \times 2 = 2$ mol of $\mathrm{NH_3}$.
$2 \times 17 = 34$ g of ammonia.

5. The method, step by step, and how to check it

Balance the equation. Every number that follows depends on the coefficients, so check them first.

Identify have and want. Which substance did the question give you a quantity of, and which does it ask about? Write both down with their coefficients.

Convert the given quantity to moles. A mass divided by its own molar mass. If you were given moles, skip this.

Apply the ratio. Multiply by the coefficient of what you want and divide by the coefficient of what you have.

Convert to the unit asked for. Moles of the wanted substance times its own molar mass gives grams.

Check the work. Two checks are worth the seconds they take. First, the size of the ratio: if the wanted substance has the larger coefficient, its moles should come out larger than the moles you started with, and smaller if its coefficient is smaller. Second, conservation: work out the mass of every reactant and every product your answer implies, and confirm the two totals match. A ratio used upside down fails both checks at once.

6. Why each step is allowed

Converting to moles first is required because the coefficients count particles, and moles are a count. Grams are not, so the coefficients cannot act on them.

Multiplying by want over have is allowed because the equation says the two amounts are always in that ratio, however large or small the batch. If 1 mol of nitrogen makes 2 mol of ammonia, 0.3 mol makes 0.6 mol, by proportion.

Using a different molar mass at the end is required because the substance changed. A mole of ammonia does not weigh what a mole of nitrogen does.

Working in kilograms or kilomoles throughout is allowed because every step is either a ratio or a molar mass, and scaling both ends by a thousand leaves every ratio unchanged. That is why a factory and a test tube use exactly the same method.

The mass check works because a balanced equation conserves atoms, and conserving atoms conserves mass. If the masses fail to balance, something upstream was wrong.

7. Why the coefficients cannot be used on masses

Take $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ and read the coefficients as grams: 2 g of hydrogen and 1 g of oxygen making 2 g of water. The masses do not even add up, and they should — matter is conserved.

Read as moles they work perfectly. 2 mol of hydrogen is 4 g, 1 mol of oxygen is 32 g, and the 36 g that go in come out as 2 mol of water, which is 36 g. Conserved exactly.

So the check is free and worth doing on a new calculation: convert every coefficient to a mass and see whether the two sides balance. They always will, if the equation is balanced and the arithmetic is right, because conservation of atoms is conservation of mass. It is the quickest way to catch a ratio used upside down.

8. Which way up the ratio goes

This is the one step in the whole course where an answer can be exactly wrong rather than approximately wrong, and the fix is a phrase: want over have.

For $\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}$, going from the oxide to the iron: you want iron, coefficient 2; you have the oxide, coefficient 1. So multiply by $2/1$.

Going back the other way — iron to oxide — you want the oxide and have the iron, so multiply by $1/2$. Same equation, ratio the other way up, because the question changed.

If an answer comes out a whole-number factor away from what you expected, this is almost always where it went.

9. One starting amount, every other substance

An equation with four substances holds six different mole ratios, and a single starting amount gives you all of them. That is worth seeing once, because it shows the ratio is chosen by the question, not by the equation.

Take the combustion of propane, $\mathrm{C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O}$, and start from 2 mol of propane. The oxygen needed is $2 \times 5/1 = 10$ mol. The carbon dioxide made is $2 \times 3/1 = 6$ mol. The water made is $2 \times 4/1 = 8$ mol. Every one of those used the same starting number and a different ratio.

The ratios between the other substances follow too. How much water comes with 6 mol of carbon dioxide? Want water, have carbon dioxide: $6 \times 4/3 = 8$ mol, the same answer by a different route. When two routes agree, the arithmetic is right; when they disagree, one of the ratios was inverted.

This is also why a balanced equation is so powerful. One measurement of any substance in the reaction fixes the amount of every other substance, in either direction.

10. The same chain in factor-label form

The three steps can be written as one line of fractions, each chosen so the unwanted unit cancels:

$$12 \text{ g Mg} \times \frac{1 \text{ mol Mg}}{24 \text{ g Mg}} \times \frac{2 \text{ mol MgO}}{2 \text{ mol Mg}} \times \frac{40 \text{ g MgO}}{1 \text{ mol MgO}} = 20 \text{ g MgO}.$$

Writing the substance beside each unit — g Mg, mol MgO — is what makes this safe. Grams of magnesium cancel grams of magnesium; moles of magnesium cancel moles of magnesium. If you find yourself cancelling moles of one substance against moles of another, the ratio is missing. Many American chemistry classes teach stoichiometry exactly this way, and it is the same three arrows as above, with the labels doing the checking.

11. In the world: the fertilizer plants of the Gulf Coast

Much of the ammonia made in the United States comes from plants along the Gulf Coast of Louisiana and Texas, running the Haber–Bosch reaction, $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$, around the clock. A single large plant makes about 2,000 metric tons of ammonia a day, and its managers plan every input with this lesson's chain.

Start from the product: 2,000 tons is $2{,}000{,}000 \text{ kg} \div 17 \approx 117{,}600$ kilomoles of ammonia. Want hydrogen, have ammonia: $3/2$, so about $176{,}500$ kilomoles of hydrogen. Times hydrogen's molar mass of 2 gives about 353 tons of hydrogen a day. Nitrogen, by the ratio $1/2$, comes to about $58{,}800$ kilomoles, or 1,650 tons, drawn from the air for free.

The hydrogen is the expensive part, made from natural gas, which is why these plants sit near Gulf gas fields. A planner who used the coefficients on masses — three tons of hydrogen for every two of ammonia — would order 3,000 tons of hydrogen a day, more than eight times what the plant needs. The mass check catches it instantly: 1,650 tons of nitrogen plus 353 tons of hydrogen is 2,003 tons, matching the 2,000 tons of ammonia within rounding. Conservation of mass is the plant's own audit.

12. In the world: an airbag in a fraction of a second

Older airbag inflators decomposed sodium azide, $\mathrm{2NaN_3 \rightarrow 2Na + 3N_2}$. About 130 g of azide gives 3 mol of nitrogen, roughly 70 liters of gas, filling the bag in about 30 milliseconds.

13. Where this goes wrong

Using the coefficients on masses. The error the lesson exists for. A balanced equation counts particles and knows nothing about grams; the numbers in front only apply once a mass has become an amount.

The ratio upside down. Want over have. An answer out by exactly a factor of 2 or 3 is this, every time.

Using the same molar mass at both ends. The substance changed in the middle of the calculation, so the molar mass has to change with it. Dividing by A's molar mass and multiplying by it again is a calculation that has forgotten what it is about.

Skipping the balancing. The whole method rests on the coefficients, and unbalanced coefficients are not a ratio of anything. If the equation is not balanced, nothing after it means anything.

Skipping a ratio of 1. When the coefficients are equal it is tempting to leave the step out. Write it anyway; the habit is what catches the next equation, where the ratio is not 1.

14. Lime from limestone

15. Where the ratio is not 1

16. Iron from its ore, in kilograms

17. Your turn: what mass of ammonia comes from 14 g of nitrogen in $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$?

18. Guided practice

The equation $\mathrm{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ has a $1$ in front of the methane and a $1$ in front of the carbon dioxide. A student concludes that $1$ g of methane makes $1$ g of carbon dioxide. Is that right?

19. Guided practice

Complete the worked solution: magnesium ribbon burns as $\mathrm{2Mg + O_2 \rightarrow 2MgO}$. A strip has a mass of twelve grams; magnesium is twenty-four grams per mole and oxygen gas thirty-two. Find the moles of magnesium, the moles of oxygen it uses, and the grams of oxygen.

20. Guided practice

The equation is $\mathrm{Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2}$. If $3.9$ mol of iron(III) oxide reacts, how many moles of iron are formed?

21. Practice

In the reaction $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$, what mass of water can be made from $4$ g of hydrogen? Answer in grams, and assume nothing is lost.

22. Practice

Older car airbags inflate by decomposing sodium azide: $\mathrm{2NaN_3 \rightarrow 2Na + 3N_2}$. A design calls for $3$ mol of nitrogen gas. Sodium azide's molar mass is $65$ g/mol. What mass of sodium azide, in grams, must the inflator hold?

23. Somewhere new

A plant manager has to place an order. The plant runs $\mathrm{2H_2 + O_2 \rightarrow 2H_2O}$ and is contracted to deliver $9$ kg of water. How much hydrogen must be bought? Answer in kilograms, and assume the reaction goes to completion.

24. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

25. Test question

The reaction is $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$. Starting from $103.6$ g of nitrogen, work through to the mass of ammonia it can make, one stage at a time.

26. What you can do now

You can go from a mass of one substance to a mass of another through a balanced equation. Say what mass of ammonia 28 g of nitrogen can make, and which step in that calculation used the equation. Next: what happens when one of the two reactants runs out before the other.

Working for the steps left to you

	value
mass of the first substance, in g	103.6
amount of it, in mol
amount of the second substance, in mol
mass of the second substance, in g

17. Your turn: what mass of ammonia comes from 14 g of nitrogen in $\mathrm{N_2 + 3H_2 \rightarrow 2NH_3}$?, step 3