General Chemistry 202-SN1-RE • CEGEP, Montreal

Revision sheet: stoichiometry, limiting reactant and yield (202-SN1-RE)

Stoichiometry is not hard, it is long: each question chains three or four conversions, and a single one done the wrong way sinks the whole answer. The discipline is to convert to moles on the very first line and to return to grams only on the last.

This sheet fixes the conversion map, gives the limiting reactant method in the only form that never misleads, and gathers the mistakes that come back from one exam to the next on yield, purity and combustion analysis. It is written for CEGEP students in General Chemistry 202-SN1-RE (formerly 202-NYA). Once you have read the sheet, the corrected exercise set on the same chapter puts every reflex to the test.

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The thread of the chapter

The coefficients of an equation read as MOLES, never as grams. All of stoichiometry therefore comes down to leaving grams as early as possible and returning to them only on the very last line.

This chapter is part of 202-SN1-RE, General Chemistry

The essentials

The conversion map

• $n=\dfrac{m}{M}$ : from grams to moles. It is the only entrance to the chapter.
• Number of entities $=n\times N_{A}$ , with $N_{A}=6.022\times 10^{23}$ /mol.
• Gas: $V=n\times V_{m}$ , with $V_{m}=22.4$ L/mol at STP, standard temperature and pressure ( $0$ °C), and $24.5$ L/mol at SATP, standard ambient temperature and pressure ( $25$ °C).
• Solution: $n=C\times V$ , with $V$ IN LITRES.
• The coefficients of the balanced equation relate only MOLES. There is no direct conversion between two masses.

Every conversion goes through the MOLE: no arrow connects a mass directly to a volume, or two masses to each other. The central box is the one that carries the coefficients of the equation.

No arrow connects a mass directly to a volume of gas, or one mass to another mass: everything goes through the mole. Drawing this map at the top of your answer saves time and avoids the costliest error.

Limiting reactant and yield, in the right order

• For EACH reactant, calculate $\dfrac{n}{\text{coefficient}}$ . The SMALLEST ratio identifies the limiting reactant.
• The whole rest of the calculation starts from the limiting reactant: amount of product, theoretical mass, volume released.
• Excess remaining $=n_{initial}-n_{consumed}$ , the amount consumed being calculated from the limiting reactant.
• Percent yield $=\dfrac{\text{actual amount}}{\text{theoretical amount}}\times 100$ , always less than or equal to $100\%$ .
• Purity $=\dfrac{\text{mass of the active species}}{\text{mass of the sample}}\times 100$ : it is NOT a yield, it is a property of the starting material.

There are more dihydrogen molecules, and yet it is the one that runs out: its ratio

n/3

2

against

3

for dinitrogen, of which one molecule is left over.

The word “excess” in a problem statement (“treated with an excess of hydrochloric acid”) saves you from looking for the limiting reactant: the other reactant is limiting by definition, and the statement has just earned you five minutes.

Formulas and combustion analysis

• Empirical formula: convert the mass percentages to moles, divide by the smallest, then multiply all the ratios by a whole number if decimals such as $1.5$ or $1.33$ remain.
• Molecular formula $=$ empirical formula multiplied by $\dfrac{M_{molar}}{M_{empirical}}$ , which must come out as a whole number.
• Combustion: all the carbon ends up in the $CO_{2}$ and all the hydrogen in the water, so $n_{C}=n_{CO_{2}}$ and $n_{H}=2\,n_{H_{2}O}$ .
• The oxygen of the compound is ALWAYS obtained by difference from the mass of the sample, never from the dioxygen consumed.
• Yields in series: $R_{global}=R_{1}\times R_{2}\times\ldots$ , as soon as each step processes all the product of the previous one.

Each absorber traps a single product: its gain in mass gives all the hydrogen or all the carbon. The oxygen is found by difference from the mass of the sample.

A ratio of $2.5$ in an empirical formula is not rounded to $3$ : it is multiplied by $2$ , and so are all the other ratios. Rounding changes the compound.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Naming the limiting reactant from its mass or from its raw amount of substance

the whole question, 8 points

What not to write

“I have $0.0300$ mol of carbonate against $0.0549$ mol of acid: the carbonate is the scarcer reagent, so it is limiting.”

What to write

“Compare $\dfrac{n}{\text{coefficient}}$ : $\dfrac{0.0300}{1}=0.0300$ for the carbonate against $\dfrac{0.0549}{2}=0.0274$ for the acid. It is the ACID that is limiting.”

On the left, the raw amounts make the carbonate look like the scarcer reagent. On the right, once each amount is divided by its coefficient, it is the ACID that has the smaller ratio: it runs out first.

Why: Neither the mass nor the raw amount can be trusted: it is the equation that fixes how many moles of each are needed. Only division by the coefficient puts the two reactants on the same scale. Here the masses happen to name the right reagent, but only by luck: the raw amounts name the wrong one.

2. Changing the subscripts of a formula to balance an equation

the whole question, and often the whole problem that follows

What not to write

“To balance $H_{2}+O_{2}\rightarrow H_{2}O$ , I write $H_{2}+O_{2}\rightarrow H_{2}O_{2}$ .”

What to write

“Adjust ONLY the coefficients in front of the formulas: $2H_{2}+O_{2}\rightarrow 2H_{2}O$ .”

Doubling the water molecule still gives water; changing its subscript makes hydrogen peroxide. Only the coefficients are adjusted.

Why: Changing a subscript changes the SUBSTANCE: hydrogen peroxide is not water. Coefficients say how many entities react, subscripts say what they are made of.

3. Reading the coefficients of the equation as grams

the whole question, 6 points

What not to write

“The equation says $N_{2}+3H_{2}\rightarrow 2NH_{3}$ , so $1$ g of nitrogen gives $2$ g of ammonia.”

What to write

“The coefficients relate MOLES: $1$ mol of nitrogen gives $2$ mol of ammonia, that is $28.0$ g giving $34.1$ g.”

The coefficients count moles, which are not conserved:

1+3

give

2

. The grams, on the other hand, are conserved:

1

g of nitrogen cannot give

2

g of ammonia.

Why: Mass is conserved in a reaction, but the number of moles is not. Reading the coefficients as grams therefore violates the conservation of mass, which the mass balance of the problem always lets you spot.

4. Calculating the oxygen of a combustion analysis from the dioxygen consumed

the whole question, 6 points

What not to write

“The sample was burned with $0.250$ mol of dioxygen, so it contained $0.500$ mol of oxygen.”

What to write

“The oxygen of the COMPOUND is obtained by difference: $m_{O}=m_{sample}-m_{C}-m_{H}$ . The dioxygen consumed never enters this calculation.”

Why: The dioxygen from the air also supplies oxygen to the products: the two sources cannot be separated. Only the mass balance on the sample isolates the oxygen that came from it.

5. Calculating the yield from the excess reactant

3 points, and a yield that is always underestimated

What not to write

“The carbonate is in excess and could give $1.32$ g of gas; I get $1.10$ g, so the yield is $83\%$ .”

What to write

“The theoretical yield is ALWAYS calculated from the LIMITING reactant: here $1.21$ g, so the yield is $\dfrac{1.10}{1.21}\times 100=91\%$ .”

The maximum possible is calculated from the LIMITING reactant:

1.21

g, hence

91\%

. Calculated from the excess reactant, the maximum is unreachable and the yield wrongly drops to

83\%

Why: The excess reactant will never be entirely consumed, so it cannot fix the maximum amount of product. A yield calculated on it is systematically too low.

6. Adding or averaging successive yields

3 points, and an absurd order of magnitude

What not to write

“Two steps at $80\%$ each give an overall yield of $160\%$ , or else $80\%$ on average.”

What to write

“Yields MULTIPLY: $0.80\times 0.80=0.64$ , that is $64\%$ .”

The second step does not work on

100

g but on the

80

g that remain: it keeps

80\%

of them, that is

64

g. Yields multiply, they are never averaged.

Why: The second step works only on what came out of the first. Each step takes its share from a total that is already reduced, which is exactly a multiplication of fractions.

7. Rounding a ratio of an empirical formula instead of multiplying it

the whole question, 5 points

What not to write

“The ratios are $1$ and $1.5$ , so the empirical formula is $AB_{2}$ .”

What to write

“Multiply ALL the ratios by $2$ : the formula is $A_{2}B_{3}$ .”

Why: Rounding $1.5$ to $2$ changes the compound. The decimals to watch for are $0.5$ (multiply by $2$ ), $0.33$ and $0.67$ (by $3$ ), $0.25$ and $0.75$ (by $4$ ).

8. Using the molar volume at STP at room temperature

2 points, and a systematic error of

9\%

What not to write

“The gas is collected at $25$ °C: $n=\dfrac{V}{22.4}$ .”

What to write

“At $25$ °C and $101.3$ kPa, the molar volume is $24.5$ L/mol, not $22.4$ : $n=\dfrac{V}{24.5}$ .”

The same mole of gas occupies

9\%

more volume at

25

°C than at

0

°C: at room temperature, divide by

24.5

, not by

22.4

Why: The molar volume depends on temperature and pressure. Going from $0$ to $25$ °C expands the gas by $9\%$ , which is enough to spoil a purity or a yield calculated to the tenth.

9. Confusing purity and yield

3 points on the reasoning question

What not to write

“The limestone is $77.6\%$ pure, so the reaction has a yield of $77.6\%$ .”

What to write

“Purity describes the STARTING sample, yield describes the reaction. Here the reaction is complete, its yield is $100\%$ , and it is the sample that contains $22.4\%$ inert impurities.”

Purity describes what you put in the beaker, yield what the reaction did with it. Here the sample is

77.6\%

pure and the reaction is complete.

Why: The two quantities can coexist in the same problem, and they then multiply. Confusing them amounts to counting the same loss twice, or forgetting one.

10. Forgetting to convert millilitres to litres in $n=CV$

2 points, and a factor of

1000

What not to write

“ $250$ mL of a $0.200$ mol/L solution: $n=0.200\times 250=50.0$ mol.”

What to write

“ $250$ mL $=0.250$ L, so $n=0.200\times 0.250=0.0500$ mol.”

Why: Concentration is defined PER LITRE. A quantity of $50$ moles in a $250$ mL beaker would represent several kilograms of solute, which you can spot without any calculation.

Which method to choose

Which conversion to start with

The quantity the problem gives for the starting species

If a mass in grams → $n=\dfrac{m}{M}$ , after calculating $M$ from the periodic table

Example: $3.00$ g of $CaCO_{3}$ give $0.0300$ mol
If a volume of GAS with its temperature and pressure → $n=\dfrac{V}{V_{m}}$ if the conditions are STP or SATP, otherwise $n=\dfrac{PV}{RT}$

Example: $0.950$ L at $25$ °C gives $0.0388$ mol
If a volume of SOLUTION and a concentration → $n=CV$ with $V$ in litres

Example: $250$ mL at $0.200$ mol/L gives $0.0500$ mol
If a number of atoms or molecules → $n=\dfrac{N}{N_{A}}$

Example: $3.01\times 10^{23}$ molecules give $0.500$ mol
If a mass percent and a sample mass → first the mass of the active species, only then the conversion to moles

this is the case for impure samples, where purity applies BEFORE any stoichiometry

Whatever the branch, the output is the same: moles. Only once you are in moles can the coefficients of the equation be used.

Does the problem ask for a limiting reactant, a yield or a purity

What the problem gives and what it asks for

If the amounts of TWO reactants are given → look for the limiting reactant by $\dfrac{n}{\text{coefficient}}$ before any other calculation

Example: $3.00$ g of carbonate and $2.00$ g of acid
If the statement says “an excess of” for one of the reactants → the other one is limiting by definition, no comparison calculation is needed

Example: “treated with an excess of hydrochloric acid”
If the statement gives an amount obtained that is LOWER than the calculation → percent yield $=\dfrac{\text{actual}}{\text{theoretical}}\times 100$

Example: $1.10$ g obtained for $1.21$ g expected, that is $91\%$
If the statement mentions an impure sample, a crude material or an ore → purity $=\dfrac{\text{active mass}}{\text{total mass}}\times 100$ , calculated at the END

Example: $3.88$ g of carbonate in $5.00$ g of limestone, that is $77.6\%$
If the statement describes several successive steps → multiply the yields of the individual steps

and check that each step really processes all the product of the previous one, otherwise the reasoning changes

A problem can combine purity AND yield: purity applies to the starting material, yield to the reaction. The two multiply and are never confused.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Solving a limiting reactant problem

When to use it: The amounts of at least two reactants are given, or the statement asks what is left after the reaction.

1 Write and BALANCE the equation, then underline the coefficients you will use.
2 Convert each given quantity to moles, writing the unit on every line.
3 Calculate $\dfrac{n}{\text{coefficient}}$ for each reactant and write the two numbers side by side to make the comparison readable.
4 Name the limiting reactant explicitly in one sentence, then use ONLY it for the rest.
5 Calculate the expected amount of product, convert it to the unit asked for, and, if the statement asks, the excess left of the second reactant.

Concluding sentence

“The ratios $\dfrac{n}{\text{coefficient}}$ are $0.0300$ for the carbonate and $0.0274$ for the acid: the ACID is the limiting reactant, so it produces $0.0274$ mol of carbon dioxide, that is $1.21$ g, and $0.254$ g of carbonate is left.”

The trap: Writing the comparison without the word “limiting”. The sentence that names the limiting reactant is worth almost a point on its own, and it justifies the whole rest of the calculation.

Marking: Typically 1 mark for the balanced equation, 2 for the conversions, 2 for the comparison of the ratios, 2 for the amount of product, 1 for the excess.

Determining an empirical formula, then a molecular formula

When to use it: The statement gives mass percentages, masses of elements, or the products of a combustion.

1 Take $100$ g of sample if the statement gives percentages: each percentage becomes a mass in grams directly.
2 Convert each mass to moles using the molar mass of the ELEMENT, not of the compound.
3 Divide all the amounts by the smallest, which gives ratios one of which is $1$ .
4 Multiply all the ratios by the smallest whole number that makes them whole, never rounding a value such as $1.5$ .
5 If the molar mass of the compound is given, calculate $\dfrac{M_{molar}}{M_{empirical}}$ and multiply the empirical formula by that whole number.

Concluding sentence

“The molar ratios are $1.00$ and $1.50$ ; multiplied by $2$ , they give the empirical formula $A_{2}B_{3}$ with mass $M_{emp}$ , and since $\dfrac{M_{molar}}{M_{emp}}=2$ , the molecular formula is $A_{4}B_{6}$ .”

The trap: Using the molar mass of the compound in step 2. Each ELEMENT is converted separately, with its own atomic mass.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Conservation of mass on the equation

Add up the masses of the reactants consumed and those of the products formed: the two totals must coincide. It is the most powerful check of the chapter.

$28.0$ g of nitrogen plus $6.05$ g of hydrogen must give exactly $34.1$ g of ammonia.
A yield below one hundred percent

A yield above $100\%$ is impossible: it signals either a limiting reactant chosen wrongly or a theoretical mass calculated from the wrong reactant.

A yield of $112\%$ : the theoretical calculation started from the excess reactant.
The sum of the percentages of a percent composition

The mass percentages of all the elements of a compound must total $100\%$ , up to rounding.

A sum of $94\%$ : an element was forgotten or a molar mass is wrong.
The order of magnitude of the amount of substance

A few grams of a common compound give a fraction of a mole, never tens. An amount above one mole for less than $50$ g needs checking.

An $n=50$ mol in $250$ mL of solution: the millilitres were not converted.
The final ratio of the molecular formula

The quotient of the molar mass by the mass of the empirical formula must come out as a WHOLE NUMBER. A result such as $2.3$ signals a wrong empirical formula.

A ratio of $1.5$ : the empirical formula was not multiplied by two in the previous step.

The typical problem, taken apart

Impure limestone, from purity to limiting reactant

A $5.00$ g sample of limestone, made of calcium carbonate $CaCO_{3}$ and inert impurities, is treated with an EXCESS of hydrochloric acid. $0.950$ L of carbon dioxide is collected, measured at $25$ °C and $101.3$ kPa.

Determine the purity of the limestone. The experiment is then repeated by reacting $3.00$ g of pure carbonate with $2.00$ g of hydrochloric acid: determine the limiting reactant, the mass of carbon dioxide formed and the mass of reactant left.

Data: $M(CaCO_{3})=100.09$ g/mol, $M(HCl)=36.46$ g/mol, $M(CO_{2})=44.01$ g/mol, $V_{m}=24.5$ L/mol at SATP.

The impure limestone is attacked by an excess of acid; the carbon dioxide released pushes the piston of the syringe out to

0.950

Step 1

Equation: $CaCO_{3}+2\,HCl\rightarrow CaCl_{2}+H_{2}O+CO_{2}$ . The coefficients you need are $1$ for the carbonate, $2$ for the acid and $1$ for the gas.

Why

Without a balanced equation, no ratio is available and the coefficient $2$ of the acid, which will decide the limiting reactant in the second part, would stay invisible.

Step 2

$n(CO_{2})=\dfrac{0.950}{24.5}=0.0388$ mol. We use $24.5$ and not $22.4$ because the measurement is made at $25$ °C.

Why

The choice of molar volume is justified in half a sentence, and this is where half of the answers lose $9\%$ on the whole rest of the problem.

Step 3

The ratio of the equation being $1:1$ between the carbonate and the gas, $n(CaCO_{3})=0.0388$ mol, which gives $m=0.0388\times 100.09=3.88$ g.

Why

The acid being in EXCESS according to the statement, the carbonate is limiting without any calculation: the amount of gas measured therefore tells you about it directly.

Step 4

Purity $=\dfrac{3.88}{5.00}\times 100=77.6\%$ .

Why

Purity refers to the STARTING sample. It says nothing about the yield of the reaction, which is complete here since the acid was in excess.

Step 5

Second part. $n(CaCO_{3})=\dfrac{3.00}{100.09}=0.0300$ mol and $n(HCl)=\dfrac{2.00}{36.46}=0.0549$ mol.

Why

Convert first, compare afterwards. Comparing the masses $3.00$ and $2.00$ or even the moles $0.0300$ and $0.0549$ still says nothing about the limiting reactant.

Step 6

Ratios: $\dfrac{0.0300}{1}=0.0300$ for the carbonate against $\dfrac{0.0549}{2}=0.0274$ for the acid. The smallest ratio is that of the ACID: it is the limiting reactant.

Why

Division by the coefficient is the only operation that puts the two reactants on the same scale. Here it reverses the conclusion that the raw amounts suggested.

Step 7

$n(CO_{2})=\dfrac{n(HCl)}{2}=0.0274$ mol, so $m(CO_{2})=0.0274\times 44.01=1.21$ g.

Why

Everything starts from the limiting reactant, and from it alone. Using the carbonate here would give $1.32$ g, a value the reaction cannot reach for lack of acid.

Step 8

Carbonate consumed: $0.0274$ mol. What is left is $0.0300-0.0274=0.00254$ mol, that is $0.254$ g.

Why

The excess is always calculated by difference between the initial and the consumed, and the consumed is read from the limiting reactant. This is the question that checks that the limiting reactant was identified correctly.

Step 9

Check by mass: $3.00+2.00=5.00$ g at the start; at the end, $1.21$ g of gas, $0.254$ g of carbonate left, and the rest as calcium chloride and water.

Why

The overall mass balance is the check that catches nearly every conversion error, without redoing any stoichiometric calculation.

The conclusion, written out

“The limestone is $77.6\%$ pure. In the second experiment, hydrochloric acid is the limiting reactant: only $1.21$ g of carbon dioxide forms and $0.254$ g of calcium carbonate is left unattacked.”

The classic mistake on this problem: Naming the carbonate as limiting because there are fewer moles of it, and announcing $1.32$ g of gas. The difference from the right answer is only $0.11$ g, which makes it undetectable without the check of the ratio to the coefficient.

Learn by heart

• $n=\dfrac{m}{M}$ , $n=CV$ with $V$ in LITRES, $n=\dfrac{V}{V_{m}}$ for a gas, $n=\dfrac{N}{N_{A}}$ .
• $V_{m}=22.4$ L/mol at STP ( $0$ °C) and $24.5$ L/mol at SATP ( $25$ °C).
• The coefficients relate MOLES, never grams.
• Limiting reactant: the smallest $\dfrac{n}{\text{coefficient}}$ , never the smallest mass.
• The whole product calculation starts from the LIMITING reactant; the excess left is calculated by difference.
• Percent yield $=\dfrac{\text{actual}}{\text{theoretical}}\times 100$ , at most $100\%$ . Yields in series MULTIPLY.
• Combustion: $n_{C}=n_{CO_{2}}$ , $n_{H}=2n_{H_{2}O}$ , and the oxygen of the compound by mass DIFFERENCE.
• Empirical formula: a ratio of $1.5$ is multiplied by $2$ , it is never rounded.

Frequently asked questions

How do you find the limiting reactant every time?

Convert each reactant to moles, then divide that amount by its coefficient in the balanced equation. The smallest result identifies the limiting reactant. Comparing the masses, or even the raw amounts of substance, does not work, because the equation does not consume the reactants in the same proportions.

What is the difference between purity and percent yield?

Purity describes the starting sample: it is the fraction of active material it contains, the rest being impurities. Percent yield describes the reaction: it is the fraction of the theoretical product actually obtained. Both can appear in the same problem, and they then multiply, but they are never confused.

Why do the yields of several steps multiply?

Because each step works only on what the previous one left it. Two steps at 80 percent leave 80 percent of 80 percent, which is 64 percent of the starting amount. Adding or averaging would give a result with no physical meaning at all.

How do you calculate the oxygen in a combustion analysis?

Always by mass difference. Find the mass of carbon from the carbon dioxide formed, the mass of hydrogen from the water formed, then subtract these two masses from the mass of the sample. What is left is the oxygen of the compound. The dioxygen consumed is never used in this calculation, because it also supplies oxygen to the products.

Which molar volume should you use for a gas?

Use 22.4 litres per mole at standard temperature and pressure, that is 0 degrees Celsius. Use 24.5 litres per mole at standard ambient temperature and pressure, that is 25 degrees. In any other case, go through the ideal gas law. The gap between the two values reaches 9 percent.

What do you do when a ratio in an empirical formula is not a whole number?

Multiply all the ratios by the smallest whole number that makes them whole; never round. A decimal of 0.5 calls for a factor of 2, a decimal close to one third or two thirds calls for a factor of 3, and a quarter or three quarters call for 4. Rounding would simply change the compound.

Practise it

Corrected exercises: Stoichiometry, limiting reactant and percent yield, 202-SN1-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-chem-sn1-stoichiometry. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: stoichiometry, limiting reactant and yield (202-SN1-RE)

The essentials

The conversion map

Limiting reactant and yield, in the right order

Formulas and combustion analysis

The mistakes that cost marks

1. Naming the limiting reactant from its mass or from its raw amount of substance

2. Changing the subscripts of a formula to balance an equation

3. Reading the coefficients of the equation as grams

4. Calculating the oxygen of a combustion analysis from the dioxygen consumed

5. Calculating the yield from the excess reactant

6. Adding or averaging successive yields

7. Rounding a ratio of an empirical formula instead of multiplying it

8. Using the molar volume at STP at room temperature

9. Confusing purity and yield

10. Forgetting to convert millilitres to litres in n=CVn=CVn=CV

Which method to choose

Which conversion to start with

Does the problem ask for a limiting reactant, a yield or a purity

How the answer is expected to be written

Solving a limiting reactant problem

Determining an empirical formula, then a molecular formula

Check before you hand in

The typical problem, taken apart

Impure limestone, from purity to limiting reactant

Learn by heart

Frequently asked questions

How do you find the limiting reactant every time?

What is the difference between purity and percent yield?

Why do the yields of several steps multiply?

How do you calculate the oxygen in a combustion analysis?

Which molar volume should you use for a gas?

What do you do when a ratio in an empirical formula is not a whole number?

Corrected exercises: Stoichiometry, limiting reactant and percent yield, 202-SN1-RE

See also

Looking for a 202-SN1-RE chemistry tutor in Montreal?

10. Forgetting to convert millilitres to litres in $n=CV$