GENERAL CHEMISTRY · STOICHIOMETRY

Convert with units.
Conserve with ledgers.

Learn all seven official Stoichiometry topics through mass, mole, entity, density, and reaction ledgers—then retrieve them with 28 randomized five-choice questions.

Start the lessons ↓Build a balanced mix →

7guided lessons

28practice questions

7official topics

21study objectives

A repeatable calculation routine

Name. Convert. Solve. Check.

01Name
Write the requested quantity and its unit before touching numbers.
02Convert
Move masses to moles or align units with explicit factors.
03Solve
Apply one relationship while preserving chemical identities.
04Check
Check units, magnitude, significant figures, atoms, and charge.

Current exam policy: both SI/metric and U.S. customary units may appear. A periodic table is provided for General Chemistry. The digital calculator shown in the current guide is for Quantitative Reasoning. Read the official guide ↗

Seven integrated topic lessons

Keep separate ledgers.

Mass, moles, entities, volume, and coefficients are related, but they are not interchangeable. Each lesson makes the conversion boundary visible.

LESSON 1 · 14 MIN

Study + retrieve

Molar mass as a conversion bridge

Calculate a formula’s molar mass, orient mass–mole conversions by units, and use molar mass as a bounded identity constraint.

ESSENTIAL QUESTIONWhich formula inventory produces the grams-per-mole bridge?

Ca(OH)₂ molar-mass inventory

Atomic-mass contributions using Ca = 40.0, O = 16.0, and H = 1.0
Element	Full count	Atomic mass	Contribution
Ca	1	40.0	40.0
O	2	16.0	32.0
H	2	1.0	2.0
Molar mass			74.0 g/mol

Parentheses auditThe outer 2 multiplies both O and H

STUDY TABLE · CAPTION INCLUDED

Inventory every atom

Expand parentheses and multiply each atomic mass by the full atom count before adding contributions.

Outer subscripts distribute
Sum element contributions

Orient the bridge

Molar mass is grams per mole. Divide grams by g/mol to obtain moles; multiply moles by g/mol to obtain grams.

Cancel the starting unit
Write the requested unit first

Limit identification claims

A molar-mass mismatch can reject a candidate formula, but agreement alone may not identify a unique formula or structure.

Constraint, not fingerprint
Isomers share molar mass

Worked example

Using Ca = 40.0, O = 16.0, and H = 1.0, find the molar mass of Ca(OH)₂.

1
Expand the parentheses: Ca₁O₂H₂.
2
Add contributions: 40.0 + 2(16.0) + 2(1.0).
3
The total is 74.0 grams for one mole of formula units.

ConclusionThe molar mass of Ca(OH)₂ is 74.0 g/mol using the supplied atomic masses.

Close the notes first

Retrieve the ledger.

01What does the 2 outside (OH)₂ multiply?

Both O and H inside the parentheses.

The outer subscript applies to the entire grouped unit.

02Which operation converts grams to moles?

Divide by molar mass in g/mol.

The gram units cancel, leaving moles.

03Does one matching molar mass prove a unique structure?

No.

Different formulas can share a nominal mass, and isomers share both formula and molar mass.

LESSON 2 · 14 MIN

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Moles, entities, and formula inventories

Convert moles to a labeled entity count, scale atom inventories by subscripts, and distinguish molecular from formula-unit language.

ESSENTIAL QUESTIONWhat entity is being counted, and how many atoms belong to each one?

0.250 mol CO₂ entity inventory

One sample described at molecule and atom scales
Requested inventory	Starting amount	Multiplier	Result
CO₂ molecules	0.250 mol	6.022 × 10²³/mol	1.51 × 10²³
C atom amount	0.250 mol CO₂	1 mol C/mol CO₂	0.250 mol C
O atom amount	0.250 mol CO₂	2 mol O/mol CO₂	0.500 mol O

Entity labelMolecules for CO₂ · formula units for ionic solids

STUDY TABLE · CAPTION INCLUDED

Label the entity

A mole counts specified entities: atoms, molecules, ions, or formula units. The label is part of the answer.

One mole = Avogadro’s number
State molecules or formula units

Scale the formula inventory

Multiply formula-unit moles by a subscript to obtain moles of that atom; multiply again only when a coefficient or sample amount requires it.

Subscript scales one entity
Sample moles scale every entity

Respect formula meaning

Molecular formulas describe discrete molecules; ionic formulas normally describe simplest ratios in an extended lattice.

CO₂ molecules
NaCl formula units

Worked example

How many CO₂ molecules and moles of oxygen atoms are present in 0.250 mol CO₂?

1
Multiply 0.250 mol by 6.022 × 10²³ molecules/mol to obtain 1.51 × 10²³ molecules.
2
Each CO₂ molecule contains two oxygen atoms.
3
Multiply 0.250 mol CO₂ by 2 mol O atoms per mol CO₂ to obtain 0.500 mol O atoms.

ConclusionThe same sample contains 1.51 × 10²³ CO₂ molecules and 0.500 mol of oxygen atoms.

Close the notes first

Retrieve the ledger.

01Does one mole always have the same mass?

No; it always has the same entity count.

Mass depends on the substance’s molar mass.

02How many moles of Cl atoms are in 2.00 mol CaCl₂?

4.00 mol Cl atoms.

Each formula unit contains two chloride ions.

03What term should describe counted NaCl units?

Formula units.

NaCl is represented as an ionic ratio rather than discrete NaCl molecules.

LESSON 3 · 13 MIN

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Percent composition with a mass ledger

Calculate elemental mass percent from a formula or measured sample and state what a mismatch can—and cannot—prove.

ESSENTIAL QUESTIONWhich mass belongs in the numerator, and what belongs in the denominator?

H₂O mass ledger

Formula-mass contributions using H = 1.0 and O = 16.0
Element	Count	Atomic mass	Contribution
H	2	1.0	2.0
O	1	16.0	16.0
Formula mass			18.0

O mass %16.0 ÷ 18.0 × 100% = 88.9%

STUDY TABLE · CAPTION INCLUDED

Build the formula mass

Multiply each atomic mass by its subscript, including any multiplier outside parentheses, then sum every contribution.

Subscript × atomic mass
Parentheses multiply enclosed atoms

Choose the total denominator

For formula composition, divide one element’s formula-mass contribution by the total formula mass. For a measured sample, divide component mass by total sample mass.

Part ÷ whole
Use compatible mass units

Bound the conclusion

A mismatch with a predicted pure composition can reject that simple model, but composition alone may not identify the contaminant or error source.

Compare within stated precision
Mismatch is not unique identification

Worked example

Using H = 1.0 and O = 16.0, calculate the oxygen mass percent in H₂O.

1
Compute the formula mass: 2(1.0) + 16.0 = 18.0.
2
Oxygen contributes 16.0 of the 18.0 mass units.
3
Calculate (16.0 ÷ 18.0) × 100% = 88.9% to the shown precision.

ConclusionOxygen is 88.9% of the formula mass; hydrogen is the remaining 11.1%.

Close the notes first

Retrieve the ledger.

01Why is the largest subscript not automatically the largest mass contribution?

Atomic mass and subscript both determine contribution.

Two light atoms can contribute less mass than one much heavier atom.

02What denominator is used for a measured component’s mass percent?

The total sample mass.

Percent composition is part divided by the whole sample.

03What can an observed composition mismatch establish?

It can show inconsistency with the proposed pure composition under the stated assumptions.

It cannot uniquely identify impurity, loss, or measurement error without more evidence.

LESSON 4 · 15 MIN

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Empirical and molecular formula ladders

Convert composition to a simplest mole ratio, resolve small fractional ratios, and scale an empirical formula to a molecular formula.

ESSENTIAL QUESTIONHow do masses become whole-number atom ratios?

Composition-to-formula ladder

100 g basis for 40.0% C, 6.7% H, and 53.3% O
Element	Mass	Moles	÷ smallest
C	40.0 g	3.33	1.00
H	6.7 g	6.7	2.01
O	53.3 g	3.33	1.00

Empirical formulaCH₂O

STUDY TABLE · CAPTION INCLUDED

Turn every mass into moles

Treat percentages as grams in a 100 g basis when appropriate, then divide each elemental mass by its atomic mass.

Mass ÷ molar mass
Ratios must be molar

Reduce one shared ratio

Divide all mole values by the smallest. If values are near halves, thirds, or quarters, multiply every ratio by the same small integer.

One common divisor
One common multiplier

Scale to the molecular formula

Divide molecular molar mass by empirical-formula mass. The result must be a whole-number multiplier applied to every empirical subscript.

Molar mass ÷ empirical mass
Scale all subscripts together

Worked example

A compound is 40.0% C, 6.7% H, and 53.3% O. Find its empirical formula using C = 12.0, H = 1.0, O = 16.0.

1
Assume 100 g: 40.0 g C, 6.7 g H, and 53.3 g O.
2
Convert to moles: 3.33 C, 6.7 H, and 3.33 O.
3
Divide by 3.33 to obtain approximately 1:2:1.

ConclusionThe simplest whole-number ratio is CH₂O.

Close the notes first

Retrieve the ledger.

01Why can mass percentages not be used directly as subscripts?

Subscripts represent numbers of atoms, so the composition must first be converted to moles.

Elements have different atomic masses.

02A reduced ratio is 1.00:1.50. What should happen?

Multiply both ratios by 2 to obtain 2:3.

A shared multiplier preserves the measured ratio.

03How is a molecular formula multiplier found?

Divide molecular molar mass by empirical-formula mass.

A molecular formula is an integer multiple of the empirical formula.

LESSON 5 · 13 MIN

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Balance with independent ledgers

Select coefficients that conserve every element, preserve species identities, and reduce to a simplest whole-number ratio.

ESSENTIAL QUESTIONWhat may change when an equation is balanced—and what must not?

Independent atom ledgers

4 Al + 3 O₂ → 2 Al₂O₃

Atom totals after coefficients are applied
Element	Reactants	Products	Status
Al	4	4	Balanced
O	6	6	Balanced

Identity checkFormulas unchanged · coefficients 4:3:2

STUDY TABLE · CAPTION INCLUDED

Count each element independently

Create left and right atom totals for every element. A coefficient multiplies every atom in its formula.

Separate element ledgers
Coefficient scales the whole species

Never repair with subscripts

Subscripts define the substance. Changing O₂ to O₃ or H₂O to H₂O₂ changes chemical identity rather than balancing the stated reaction.

Coefficients may change
Formulas remain fixed

Audit the final ratio

Recount atoms, check net charge when ionic species are shown, and divide coefficients by any common factor when simplest form is requested.

Atoms and charge are separate checks
Reduce a common factor

Worked example

Balance Al + O₂ → Al₂O₃ using the smallest whole-number coefficients.

1
Use 2 Al₂O₃ to make six product oxygen atoms.
2
Use 3 O₂ to supply six reactant oxygen atoms.
3
Use 4 Al to match the four product aluminum atoms.

Conclusion4 Al + 3 O₂ → 2 Al₂O₃ conserves Al and O in the simplest ratio.

Close the notes first

Retrieve the ledger.

01Why may a coefficient change but not a subscript?

A coefficient changes amount; a subscript changes substance identity.

Balancing preserves the specified reactants and products.

02Must molecule count be the same on both sides?

No; each element’s atom count must be the same.

Reactions can combine or split particles.

03What is wrong with 4 H₂ + 2 O₂ → 4 H₂O as a final answer?

It is balanced but not in the simplest whole-number ratio.

Every coefficient can be divided by 2.

LESSON 6 · 12 MIN

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Density with a measurement ledger

Solve mass, volume, or density with compatible units, use displacement correctly, and bound identity conclusions from a measured density.

ESSENTIAL QUESTIONWhich mass belongs to which volume under which conditions?

Displacement-to-density ledger

Measurements for a 52.0 g fully submerged object
Quantity	Reading	Operation	Result
Initial volume	18.50 mL	Final − initial	6.50 mL object
Final volume	25.00 mL	Final − initial	6.50 mL object
Density	52.0 g	mass ÷ volume	8.00 g/mL

BoundaryAgreement supports identity; it does not prove uniqueness

STUDY TABLE · CAPTION INCLUDED

Preserve mass per volume

Density equals mass divided by volume. Rearrange only after naming the requested quantity and its units.

d = m/V
m = dV

Subtract displacement readings

For a fully submerged nonreactive object, object volume is the final cylinder reading minus the initial reading.

Final − initial
Do not use final alone

Treat density as evidence

Density is intensive under fixed conditions. Agreement with a reference supports a candidate but does not establish unique identity by itself.

Sample size does not change density
Check temperature and uncertainty

Worked example

A 52.0 g object raises a cylinder from 18.50 mL to 25.00 mL. Find its density.

1
Object volume = 25.00 − 18.50 = 6.50 mL.
2
Density = 52.0 g ÷ 6.50 mL.
3
Report three significant figures because both measured inputs support three.

ConclusionThe object’s measured density is 8.00 g/mL.

Close the notes first

Retrieve the ledger.

01What is the reciprocal-density error?

Dividing volume by mass instead of mass by volume.

Density is mass per unit volume.

02Why is 25.00 mL not the object volume in the example?

It includes the initial liquid volume.

Only the change in reading is displaced by the object.

03Does doubling a homogeneous sample double its density?

No.

Mass and volume scale together, so their ratio stays constant under the same conditions.

LESSON 7 · 17 MIN

Study + retrieve

Balanced-equation calculation chains

Use coefficients as mole ratios, identify the limiting reactant, and distinguish theoretical, actual, and percent yield.

ESSENTIAL QUESTIONWhich reactant sets the maximum product, and where do mass conversions belong?

Limiting-reactant product ceilings

N₂ + 3 H₂ → 2 NH₃

Theoretical NH₃ from each available reactant
Starting reactant	Available	Mole-ratio conversion	NH₃ ceiling
N₂	2.00 mol	× 2/1	4.00 mol
H₂	3.00 mol	× 2/3	2.00 mol
Smaller compatible ceiling			2.00 mol NH₃

Limiting reactantH₂ · it sets the theoretical maximum

STUDY TABLE · CAPTION INCLUDED

Enter and leave in moles

Convert a measured mass to moles before using coefficients, then convert product moles to the requested unit afterward.

Mass → moles → ratio
Coefficients are not gram ratios

Compare product ceilings

Calculate the product amount each reactant could form. The smaller compatible product amount identifies the limiting reactant.

Test every supplied reactant
Smallest mass is not automatically limiting

Keep yield roles separate

Theoretical yield is the model maximum from the limiting reactant; percent yield is actual divided by theoretical times 100%.

Theoretical in denominator
Above 100% needs investigation

Worked example

For N₂ + 3 H₂ → 2 NH₃, 2.00 mol N₂ reacts with 3.00 mol H₂. Find the limiting reactant and theoretical NH₃ amount.

1
From N₂: 2.00 mol × (2 mol NH₃ / 1 mol N₂) = 4.00 mol NH₃.
2
From H₂: 3.00 mol × (2 mol NH₃ / 3 mol H₂) = 2.00 mol NH₃.
3
The smaller product ceiling comes from H₂.

ConclusionH₂ is limiting and the theoretical amount is 2.00 mol NH₃.

Close the notes first

Retrieve the ledger.

01What kind of ratio comes directly from balanced coefficients?

A particle or mole ratio.

Mass ratios also require each species’ molar mass.

02How is a limiting reactant identified reliably?

Compare the product amount each reactant can form.

Starting mass alone ignores molar mass and reaction coefficients.

03What is the percent-yield denominator?

The theoretical yield.

It represents the calculated maximum under the reaction model.

All 28 Stoichiometry problems

Now choose the ledger yourself.

This full-bank set randomizes question order and all five answer options. Each numeric question is checked so exactly one choice matches the computed result.

28 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Score notice

All seven official Stoichiometry topics are included.

Use the mixed-practice route to retrieve across topics after completing the lessons.

Raw accuracy guides study; it does not predict an official General Chemistry score.