Why can't I just convert grams directly to grams using the coefficients?

Because coefficients count particles, not mass. In 2H_2 + O_2 arrow 2H_2O, two moles of hydrogen weigh only about 4 grams while one mole of oxygen weighs 32 grams, so the mass ratio is nothing like 2 to 1. The mole is the only unit the coefficients understand, so you must convert into moles, cross the arrow, and convert back out.

How do I know which way to write the mole ratio?

Put the substance you were given in the denominator and the substance you want in the numerator. Then the given unit cancels and the wanted unit survives. If you write the chain out and see the same unit twice in a numerator, the ratio is flipped. Labeling each mole value with its chemical formula makes this obvious immediately.

What if the problem gives me moles instead of grams, or asks for particles?

The middle of the pathway never changes. Skip the first molar mass step if you are already given moles, and skip the last one if the answer is wanted in moles. To finish in particles, multiply the moles of the wanted substance by Avogadro's number instead of by molar mass; to finish in liters of gas at standard temperature and pressure, multiply by 22.4 liters per mole.

Do coefficients affect significant figures in my answer?

No. Coefficients from a balanced equation are exact counts, like saying there are exactly 3 molecules, so they place no limit on significant figures. Your answer should carry the same number of significant figures as the measured mass you started with. Molar masses taken to two decimal places from the periodic table are usually precise enough not to be the limiting factor.

CHEM-6.4

Stoichiometry: Mole Ratios & Mass-Mass Problems

Learn how to use mole ratios from balanced equations to solve mass-mass stoichiometry problems with the grams to moles to moles to grams pathway.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Stoichiometry: Mole Ratios & Mass-Mass Problems, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

You already know how to turn grams into moles using molar mass. Stoichiometry takes that skill and pushes it across the arrow of a chemical equation: if you burn 25.0 grams of propane, exactly how much water forms? A balanced equation is a recipe written in particles, and its coefficients are the conversion factors that link one substance to another.

In this lesson you will build the four-step pathway that chemists use constantly: grams of the given substance, moles of the given, moles of the wanted, grams of the wanted. Every mass-mass problem in this course follows that same road, so once the pattern clicks, the only thing that changes is the equation and the molar masses. You will also see the two places students most often go off the rails: forgetting to balance first, and treating coefficients as if they counted grams.

Coefficients Count Particles, Not Grams

A balanced equation such as

2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}

says that two molecules of hydrogen react with one molecule of oxygen. Scale that up by Avogadro's number and it says that 2 moles of

\mathrm{H_2}

react with 1 mole of

\mathrm{O_2}

to form 2 moles of

\mathrm{H_2O}

. The coefficients are a ratio of counts, exactly like a recipe that calls for two eggs per cup of flour.

What the coefficients do not tell you is mass. Two moles of

\mathrm{H_2}

weigh about 4.03 grams while one mole of

\mathrm{O_2}

weighs 32.00 grams, so the mass ratio is nowhere near 2 to 1. Mass is still conserved overall (

4.03 + 32.00 = 36.03

grams of product), but the individual mass amounts follow no simple pattern. This is why you can never multiply grams by a coefficient ratio and call it done.

A mole ratio is a fraction built from two coefficients, written so that the unwanted unit cancels:

\frac{2\ \text{mol H}_2\text{O}}{1\ \text{mol O}_2} \qquad \text{or} \qquad \frac{1\ \text{mol O}_2}{2\ \text{mol H}_2\text{O}}

Both are correct; which one you use depends on what you are given and what you want. Always write the unit of the substance you are starting from in the denominator.

Before any of this works, the equation must be balanced. An unbalanced equation gives false coefficients, and a false coefficient poisons every step that follows. Balance first, every single time, and circle the coefficients you plan to use.

The Grams to Moles to Moles to Grams Pathway

Mass cannot be converted directly into mass across a reaction, because the mole ratio only speaks the language of moles. So every mass-mass problem detours through the mole world and back.

Step	Conversion used	What changes
1	Divide by molar mass of the given	grams of given to moles of given
2	Multiply by the mole ratio from the balanced equation	moles of given to moles of wanted
3	Multiply by molar mass of the wanted	moles of wanted to grams of wanted

Written as one dimensional-analysis string for a general reaction, it looks like this:

g_{\text{given}} \times \frac{1\ \text{mol given}}{\mathcal{M}_{\text{given}}} \times \frac{\text{coef. wanted}}{\text{coef. given}} \times \frac{\mathcal{M}_{\text{wanted}}}{1\ \text{mol wanted}} = g_{\text{wanted}}

Notice the shape: molar mass, mole ratio, molar mass. The middle factor is the only place the chemical equation enters. The two outside factors are pure bookkeeping from the periodic table.

A reliable habit is to set the whole calculation up as one chain before touching a calculator, then check that every unit except grams of the wanted substance cancels diagonally. If

\text{mol H}_2

appears in a numerator and nowhere in a denominator, you have flipped the mole ratio.

Finally, round only at the end. Carrying an intermediate value rounded to two digits and then multiplying twice more can shift your answer by several percent. Report the answer with the same number of significant figures as the measured mass you started from; coefficients are exact counts and never limit significant figures.

Where Students Actually Go Wrong

The most common error is using the coefficient ratio on grams. If a problem gives 50.0 grams of

\mathrm{N_2}

for

\mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}

, writing 100.0 grams of ammonia because the ratio is 1 to 2 is wrong. Grams must become moles first.

The second most common error is an upside-down mole ratio. The fix is mechanical: the substance you were given goes on the bottom of the ratio so its unit cancels. If you are given moles of

\mathrm{N_2}

and want

\mathrm{NH_3}

, the ratio is

\frac{2\ \text{mol NH}_3}{1\ \text{mol N}_2}

.

A third trap is molar mass of the wrong species, especially with diatomic elements. Oxygen gas is

\mathrm{O_2}

at 32.00 grams per mole, not 16.00. The same goes for

\mathrm{H_2}

\mathrm{N_2}

\mathrm{Cl_2}

\mathrm{Br_2}

, and

\mathrm{I_2}

.

A fourth is forgetting subscripts inside parentheses when computing molar mass. In

\mathrm{Ca(NO_3)_2}

there are two nitrogens and six oxygens, giving 164.10 grams per mole, not 102.09.

Mistake	Symptom	Repair
Equation not balanced	Ratio uses 1 to 1 for everything	Balance, then re-read coefficients
Ratio inverted	Units do not cancel	Given substance goes in the denominator
Monatomic molar mass for a diatomic gas	Answer off by a factor near 2	Check the formula in the equation
Rounding mid-problem	Answer differs in the second digit	Keep extra digits until the last step

When an answer looks absurd, compare it to conservation of mass: the total mass of products cannot exceed the total mass of reactants consumed. A calculated 300 grams of product from 20 grams of reactant is a signal to recheck the chain, not a discovery.

Reading the Problem and Choosing the Path

Stoichiometry problems hide their structure in words, so translate before you calculate. Ask three questions: what substance am I given and in what unit, what substance is wanted and in what unit, and what is the balanced equation?

Phrases like produced, formed, and yields point to products. Phrases like required, needed, consumed, and reacts with point to reactants. A problem can run product to reactant just as easily as reactant to product; the pathway is identical, only the ratio flips.

You will also meet shorter versions of the same road. If the problem gives moles instead of grams, you skip step one. If it asks for moles instead of grams, you skip step three. A mole-to-mole problem is a single multiplication by the mole ratio. Recognizing which steps you actually need saves time and prevents accidental extra conversions.

Units other than grams often bolt onto the ends of the chain, using conversions from earlier in this unit: particles through Avogadro's number, or gas volume at standard temperature and pressure through 22.4 liters per mole. The middle of the chain never changes. Whatever comes in must be reduced to moles, crossed over by the mole ratio, then rebuilt into whatever unit is requested.

One more habit worth building: label every number with both a unit and a chemical formula. Writing 2.27 mol is ambiguous, but 2.27 mol

\mathrm{H_2O}

is not. Half of all stoichiometry confusion comes from a number losing track of which substance it describes partway down the page. Labeled numbers also make it obvious at a glance whether your mole ratio is oriented correctly.

Assumptions Behind Mass-Mass Answers

The number you calculate from the pathway is a theoretical yield: the mass you would get if the reaction ran to completion, if the substance you were given were entirely consumed, and if no product were lost. Those are real assumptions, and this lesson quietly makes all of them.

In particular, a mass-mass problem assumes that the substance you were handed is the one that runs out, meaning every other reactant is present in excess. When a problem gives you two reactant masses at once, that assumption breaks and you need the limiting-reactant reasoning that comes next in this unit. For now, a single given mass means that substance controls the outcome.

The pathway also assumes the reaction happens exactly as the balanced equation describes, with no competing side reactions. Real laboratory work rarely cooperates, which is why measured yields usually fall below the calculated value.

That said, the calculation is genuinely useful, not just an exercise. An engineer sizing a scrubber needs to know how many kilograms of limestone will neutralize the sulfur dioxide from a day of operation. A pharmacist scaling a synthesis needs the mass of each starting material for a target amount of product. In both cases the arithmetic is exactly what you are doing with 25.0 grams of propane, just with bigger numbers.

One useful check for the assumption of conservation: add the masses of all reactants consumed and all products formed in a fully worked problem. They should agree to within rounding. If they do not, a molar mass or a mole ratio is wrong somewhere in the chain.

Key terms

Stoichiometry.: The study of quantitative relationships between amounts of reactants and products in a chemical reaction, based on the coefficients of a balanced equation.
Mole ratio.: A conversion factor formed from the coefficients of two substances in a balanced equation, such as 4 mol water per 1 mol propane, used to convert moles of one substance to moles of another.
Molar mass.: The mass in grams of one mole of a substance, numerically equal to the sum of the atomic masses in its formula, with units of grams per mole.
Coefficient.: The whole number written in front of a formula in a balanced equation; it counts particles or moles, never grams, and is treated as an exact number for significant figures.
Dimensional analysis.: A problem-solving method in which quantities are multiplied by conversion factors arranged so unwanted units cancel, leaving only the desired unit.
Theoretical yield.: The mass of product predicted by stoichiometry assuming the reaction goes to completion and nothing is lost.
Excess reactant.: A reactant present in more than the amount needed; in a single-given mass-mass problem, all reactants other than the given one are assumed to be in excess.

Worked example

Propane burns in oxygen according to

\mathrm{C_3H_8} + 5\mathrm{O_2} \rightarrow 3\mathrm{CO_2} + 4\mathrm{H_2O}

. What mass of water is produced when 25.0 grams of propane burns completely in excess oxygen?

Start by confirming the equation is balanced: 3 carbons, 8 hydrogens, and 10 oxygens on each side. It is, so the coefficients are trustworthy.

Identify the given and the wanted. Given: 25.0 grams of

\mathrm{C_3H_8}

. Wanted: grams of

\mathrm{H_2O}

. That means the full three-step pathway.

Step 1, grams of given to moles of given. Molar mass of

\mathrm{C_3H_8}

3(12.01) + 8(1.008) = 36.03 + 8.064 = 44.09

grams per mole.

25.0\ \text{g C}_3\text{H}_8 \times \frac{1\ \text{mol C}_3\text{H}_8}{44.09\ \text{g C}_3\text{H}_8} = 0.5670\ \text{mol C}_3\text{H}_8

Step 2, mole ratio. The equation gives 4 moles of water per 1 mole of propane, and propane goes in the denominator so it cancels.

0.5670\ \text{mol C}_3\text{H}_8 \times \frac{4\ \text{mol H}_2\text{O}}{1\ \text{mol C}_3\text{H}_8} = 2.268\ \text{mol H}_2\text{O}

Step 3, moles of wanted to grams of wanted. Molar mass of

\mathrm{H_2O}

2(1.008) + 16.00 = 18.02

grams per mole.

2.268\ \text{mol H}_2\text{O} \times \frac{18.02\ \text{g H}_2\text{O}}{1\ \text{mol H}_2\text{O}} = 40.87\ \text{g H}_2\text{O}

The given mass had three significant figures, so report 40.9 grams of water.

Sanity check: 40.9 grams of water is larger than the 25.0 grams of propane, which is fine because oxygen mass was added to the products. It is far smaller than the total reactant mass, as conservation of mass requires.

Practice questions

For the reaction

2\mathrm{Al} + 3\mathrm{Cl_2} \rightarrow 2\mathrm{AlCl_3}

, how many moles of

\mathrm{Cl_2}

are required to react completely with 0.60 mol of Al?

0.40 mol
0.60 mol
0.90 mol
1.8 mol

Answer: 0.90 mol

The mole ratio comes straight from the coefficients: 3 mol

\mathrm{Cl_2}

per 2 mol Al. Multiply with aluminum in the denominator so its unit cancels:

0.60\ \text{mol Al} \times \frac{3\ \text{mol Cl}_2}{2\ \text{mol Al}} = 0.90\ \text{mol Cl}_2

. The value 0.40 mol comes from flipping the ratio upside down, and 1.8 mol comes from multiplying by 3 while ignoring the 2. Because both quantities are already in moles, no molar masses are needed here.

Methane burns according to

\mathrm{CH_4} + 2\mathrm{O_2} \rightarrow \mathrm{CO_2} + 2\mathrm{H_2O}

. What mass of oxygen gas is required to burn 10.0 grams of methane completely?

20.0 g
31.9 g
39.9 g
62.4 g

Answer: 39.9 g

Molar mass of

\mathrm{CH_4}

12.01 + 4(1.008) = 16.04

grams per mole, so

10.0 \div 16.04 = 0.6234

mol

\mathrm{CH_4}

. The mole ratio is 2 mol

\mathrm{O_2}

per 1 mol

\mathrm{CH_4}

, giving 1.247 mol

\mathrm{O_2}

. Multiplying by the molar mass of

\mathrm{O_2}

, which is 32.00 grams per mole, gives 39.9 grams. The answer 20.0 grams comes from doubling the grams of methane directly, the classic error of applying a coefficient to mass instead of moles.

Iron(III) oxide is reduced by carbon monoxide:

\mathrm{Fe_2O_3} + 3\mathrm{CO} \rightarrow 2\mathrm{Fe} + 3\mathrm{CO_2}

. A student needs 100.0 grams of iron. Explain the pathway you would use and calculate the mass of

\mathrm{Fe_2O_3}

required, assuming excess carbon monoxide.

Answer: About 143 grams of $\mathrm{Fe_2O_3}$ .

This problem runs backward, from product to reactant, but the pathway is identical. Convert grams of iron to moles using the molar mass of Fe, 55.85 grams per mole:

100.0 \div 55.85 = 1.7905

mol Fe. Apply the mole ratio with iron in the denominator:

1.7905 \times \frac{1\ \text{mol Fe}_2\text{O}_3}{2\ \text{mol Fe}} = 0.8953

mol

\mathrm{Fe_2O_3}

. Then multiply by the molar mass of

\mathrm{Fe_2O_3}

, which is

2(55.85) + 3(16.00) = 159.70

grams per mole, giving 143.0 grams. Reporting four significant figures matches the 100.0 grams given. Notice that the required oxide mass is greater than the iron produced, which makes sense because the oxygen atoms leave as carbon dioxide.

FAQ

Why can't I just convert grams directly to grams using the coefficients?: Because coefficients count particles, not mass. In $2\mathrm{H_2} + \mathrm{O_2} \rightarrow 2\mathrm{H_2O}$ , two moles of hydrogen weigh only about 4 grams while one mole of oxygen weighs 32 grams, so the mass ratio is nothing like 2 to 1. The mole is the only unit the coefficients understand, so you must convert into moles, cross the arrow, and convert back out.
How do I know which way to write the mole ratio?: Put the substance you were given in the denominator and the substance you want in the numerator. Then the given unit cancels and the wanted unit survives. If you write the chain out and see the same unit twice in a numerator, the ratio is flipped. Labeling each mole value with its chemical formula makes this obvious immediately.
What if the problem gives me moles instead of grams, or asks for particles?: The middle of the pathway never changes. Skip the first molar mass step if you are already given moles, and skip the last one if the answer is wanted in moles. To finish in particles, multiply the moles of the wanted substance by Avogadro's number instead of by molar mass; to finish in liters of gas at standard temperature and pressure, multiply by 22.4 liters per mole.
Do coefficients affect significant figures in my answer?: No. Coefficients from a balanced equation are exact counts, like saying there are exactly 3 molecules, so they place no limit on significant figures. Your answer should carry the same number of significant figures as the measured mass you started with. Molar masses taken to two decimal places from the periodic table are usually precise enough not to be the limiting factor.

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