Final exam chemistry study guide: from zero to hero

This guide is designed as a full A-to-Z revision set for final exam preparation, starting from absolute basics and building toward confident problem solving across the four source topics. The imported notes build a clear path: first, learn how chemical reactions are written and interpreted; next, learn how the mole connects particles, mass, and gas volume; then use that quantitative foundation for percent composition, empirical formulas, and molecular formulas; finally, move into solutions, molarity, and dilution. Each topic depends on the one before it, so this guide is structured to make the logic accumulate rather than feel like a pile of separate rules.

The imported 6.1 Modeling Chemical Reactions Notes starts with the language of reactions: reactants on the left, products on the right, and an arrow showing what is formed. The imported 5.2 Molar Relationships Notes then frames the mole as the central bridge between what cannot be counted directly and what can be measured in the lab. The imported 5.3 Percent Composition and Empirical Formulas Notes extends that bridge into formula reasoning, showing how masses and mole ratios reveal a compound's composition. The imported 5.4 Concentration of a solution Notes finishes the sequence by defining concentration and molarity as ways to describe how much solute is present in a solution.

A strong exam answer in this unit usually does four things well:

1. Names the idea correctly. Example: coefficient, molar mass, empirical formula, molarity.

2. Chooses the right conversion path. Example: grams → moles → particles.

3. Shows units at every step. This is how errors get caught.

4. Checks whether the answer makes sense. A final value should match the size of the input.

Use this guide in two ways. Read it once from top to bottom to build the whole picture. Then come back by topic and drill the worked patterns until the steps feel automatic.

How chemists represent reactions on paper

Chemistry uses equations as models. They are not random symbol strings. They are compact descriptions of real particles being rearranged. In the imported 6.1 Modeling Chemical Reactions Notes, a reaction is represented by placing the reactants on the left side of the arrow and the products on the right side. The arrow means yields or reacts to produce.

There are stages of representation, and exam questions often move between them. A word equation uses full names. A skeleton equation uses chemical formulas. A balanced chemical equation goes one step further and shows the relative amounts with coefficients. That progression matters because it mirrors how chemistry becomes more precise.

The basic vocabulary

A small set of terms appears again and again:

• Reactants: substances present before the reaction

• Products: substances formed after the reaction

• Word equation: uses names only

• Skeleton equation: uses formulas but is not yet balanced

• Chemical equation: formula-based representation of a reaction

• Coefficient: number in front of a formula showing how many units are involved

• Subscript: small number inside a formula showing how many atoms of an element are in one unit

• State symbols: (s), (l), (g), (aq)

The state symbols matter because the imported notes explicitly include them as part of correct notation:

• (s) means solid

• (l) means liquid

• (g) means gas

• (aq) means aqueous, dissolved in water

From words to formulas

A typical exam task begins in words and asks for a skeleton equation. For example:

sodium hydrogen carbonate reacts with hydrochloric acid to form sodium chloride, water, and carbon dioxide

The correct skeleton equation, matching the imported 6.1 Modeling Chemical Reactions Notes, is:

NaHCO₃(s) + HCl(aq) → NaCl(aq) + H₂O(l) + CO₂(g)

That answer is only possible if the names are translated into correct formulas first. This is why naming and formulas are not separate skills. They feed each other.

Coefficients and subscripts do different jobs

This distinction is one of the most tested ideas in the unit.

If the formula is CO₂, the subscript 2 means one carbon dioxide molecule contains two oxygen atoms. If the equation is 3CO₂, the coefficient 3 means there are three carbon dioxide molecules.

The imported 6.1 Modeling Chemical Reactions Notes is especially clear on a crucial rule: subscripts cannot be changed when balancing. Changing a subscript changes the identity of the substance. H₂O and H₂O₂ are not two versions of the same thing. They are different compounds with different properties.

Catalysts and arrow notation

Some equations show extra information around the arrow. A catalyst may be written above or below the arrow. In the imported reaction for the decomposition of hydrogen peroxide, potassium iodide KI appears above the arrow. That means it helps the reaction happen faster but is not consumed as a reactant.

This is the pattern:

• left side: what starts the reaction

• arrow: what happens

• right side: what is formed

• above/below arrow: conditions such as catalyst or heat

For example:

H₂O₂ → H₂O + O₂ with KI above the arrow

The catalyst is part of the reaction conditions, not part of the balanced reactant count.

Why equations are models

Chemical equations represent particle rearrangement. The particles before and after are not the same substances, but the atoms are the same atoms in new groupings. That idea connects directly to conservation of mass and to later mole calculations. When the imported modeling notes describe Fe and O₂ forming Fe₂O₃, the point is not just the symbols. The point is that iron atoms and oxygen molecules are reorganized into iron(III) oxide units.

A chemical equation is a compact map of what happens to atoms during a reaction.

Common question types

• Define reactant, product, coefficient, subscript

• Write a word equation

• Convert a word equation into a skeleton equation

• Add state symbols

• Identify the catalyst

• Explain why equations are models of particles rather than just symbols

Fast exam checks

Before moving on, check every reaction representation with this list:

1. Are the correct formulas used?

2. Are reactants on the left and products on the right?

3. Are state symbols included if required?

4. Are coefficients used correctly?

5. Were subscripts left unchanged?

Balancing equations and the law of conservation of mass

A chemical equation must be balanced because matter is not created or destroyed in an ordinary chemical reaction. The imported 6.1 Modeling Chemical Reactions Notes states the law of conservation of mass clearly: the mass of the products equals the mass of the reactants. In particle terms, the same atoms are present before and after the reaction. They are just rearranged.

Balancing is therefore not a formatting trick. It is the equation version of saying: the atom count must match on both sides.

What “balanced” actually means

An equation is balanced when every element has the same number of atoms on both sides of the arrow.

Take this skeleton equation:

Fe + O₂ → Fe₂O₃

Count atoms:

• Left: Fe = 1, O = 2

• Right: Fe = 2, O = 3

Not balanced.

The balanced version is:

4Fe + 3O₂ → 2Fe₂O₃

Now count again:

• Left: Fe = 4, O = 6

• Right: Fe = 4, O = 6

Balanced.

The rule students break most often

Never change subscripts to balance an equation. Only change coefficients.

Wrong: H₂ + O₂ → H₂O₂

This does balance atoms, but it changes water into hydrogen peroxide. The imported 6.1 Modeling Chemical Reactions Notes explicitly warns against this mistake because it changes the substance itself.

Right: 2H₂ + O₂ → 2H₂O

A reliable balancing procedure

Use the same method every time.

1. Write the correct skeleton equation.

2. Count atoms of each element on both sides.

3. Add coefficients to balance one element at a time.

4. Treat unchanged polyatomic ions as units if they appear on both sides.

5. Recount everything.

6. Reduce coefficients to the smallest whole-number ratio if possible.

Example 1: simple diatomic balancing

H₂ + O₂ → H₂O

Start with oxygen because it is odd on the right.

Put 2 in front of water:

H₂ + O₂ → 2H₂O

Now H is 4 on the right, so put 2 in front of hydrogen:

2H₂ + O₂ → 2H₂O

Done.

Example 2: a reaction from the imported notes

The imported 6.1 Modeling Chemical Reactions Notes gives this reaction:

AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + Ag(s)

Treat the nitrate ion NO₃ as a unit because it appears intact on both sides.

• Right side has 2 nitrates

• Left side has 1 nitrate

Put 2 in front of AgNO₃:

2AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + Ag(s)

Now silver is 2 on the left, so put 2 in front of Ag:

2AgNO₃(aq) + Cu(s) → Cu(NO₃)₂(aq) + 2Ag(s)

Balanced.

Common traps

Trap 1: balancing one element and forgetting to recheck the others

After every change, recount all elements. A fix for oxygen can break hydrogen. A fix for nitrate can change silver.

Trap 2: ignoring diatomic elements

Certain elements appear naturally as two-atom molecules in equations, especially:

• H₂

• N₂

• O₂

• F₂

• Cl₂

• Br₂

• I₂

If oxygen is a reactant by itself, it is usually O₂, not O.

Trap 3: using fractions and forgetting to clear them

Sometimes a fractional coefficient appears in the middle of balancing. That is acceptable temporarily, but final coefficients should be whole numbers.

Trap 4: thinking equal total atoms is enough

The counts must match by element, not just in total. Four atoms on one side and four on the other means nothing if the element types do not match.

Quick worked examples

1. Carbon monoxide formation

C + O₂ → CO

Balanced: 2C + O₂ → 2CO

1. Glucose combustion from the imported notes

C₆H₁₂O₆ + O₂ → CO₂ + H₂O

Balance C first: C₆H₁₂O₆ + O₂ → 6CO₂ + H₂O

Balance H: C₆H₁₂O₆ + O₂ → 6CO₂ + 6H₂O

Now oxygen:

• Right side: 6CO₂ gives 12 O

• 6H₂O gives 6 O

• total = 18 O

• Left side already has 6 O in glucose, so need 12 more from oxygen gas = 6O₂

Balanced: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Exam-style prompts

• Balance the equation.

• Explain why the equation is not balanced.

• State the law that requires balancing.

• Identify the mistake in a student's attempt.

• Explain why coefficients can change but subscripts cannot.

If the formulas are correct and the atom counts match for every element, the balancing is correct.

The mole: the bridge between the microscopic and the measurable

The mole is the central counting unit of chemistry. The imported 5.2 Molar Relationships Notes presents it as a roadmap because it links things that seem completely different: mass in grams, number of particles, and gas volume at STP. Without the mole, chemistry would be stuck between two worlds: atoms are too small to count directly, but lab measurements are made in grams and liters.

The mole solves that problem. It is the bridge between the invisible particle scale and the measurable lab scale.

What a mole means

One mole is a fixed number of representative particles:

1 mol = 6.02 × 10^23 particles

That number is Avogadro's number.

The word particles changes depending on the substance:

• atoms for elements like He, Fe, Cu

• molecules for covalent compounds like H₂O, CO₂, NH₃

• formula units for ionic compounds like NaCl, MgSO₄

This is why chemistry problems ask not just “how many particles?” but also “what kind of particles?”

The mole roadmap idea

The imported 5.2 Molar Relationships Notes stresses that when converting between mass, volume, and representative particles, you usually move through moles as the intermediate step.

That gives a stable pattern:

• grams ↔ moles using molar mass

• particles ↔ moles using Avogadro's number

• gas volume at STP ↔ moles using 22.4 L/mol

This is the core exam logic. Many multi-step questions are really just combinations of these three conversions.

Why units matter so much

The notes emphasize using units to guide the setup. This is one of the best exam survival strategies. If units cancel correctly, the setup is usually correct.

For example, to convert grams to moles:

g × (mol / g) = mol

To convert moles to particles:

mol × (particles / mol) = particles

To convert gas volume at STP to moles:

L × (mol / L) = mol

If the unwanted unit cancels and the target unit remains, the path makes sense.

The three anchor relationships

1. Mole-mass relationship

Use molar mass.

1 mol substance = molar mass in g

2. Mole-particle relationship

Use Avogadro's number.

1 mol = 6.02 × 10^23 representative particles

3. Mole-volume relationship for gases at STP

Use molar volume.

1 mol gas = 22.4 L gas at STP

Why the mole appears everywhere else in the unit

The mole is not a chapter that ends. It keeps coming back.

• In percent composition, you use mole-based masses.

• In empirical formulas, you turn grams into moles.

• In molecular formulas, you compare molar mass to empirical formula mass.

• In molarity, you calculate moles of solute.

• In dilution, the key idea is that moles of solute stay constant.

If a chemistry problem involves quantity, the mole is usually somewhere in the middle.

A simple overview table

The habit that makes mole problems easier

Always ask:

1. What do I have?

2. What do I want?

3. What conversion connects them?

4. Do I need moles in the middle?

In these imported notes, the answer is very often yes.

Molar mass and gram-to-mole calculations

Molar mass is the mass of one mole of a substance, expressed in g/mol. It comes from adding the atomic masses of every atom in the chemical formula. This is the number that lets you move between grams and moles.

The imported 5.2 Molar Relationships Notes uses this relationship constantly: to convert moles to mass, multiply by molar mass; to convert mass to moles, divide by molar mass.

How to calculate molar mass

The full formula matters. Every subscript counts. Parentheses count too.

Example 1: Al₂O₃

• Al: 2 × 27.0 = 54.0

• O: 3 × 16.0 = 48.0

Total: 54.0 + 48.0 = 102.0 g/mol

This matches the imported example where 1 mol Al₂O₃ = 102.0 g.

Example 2: Fe(OH)₂

• Fe: 1 × 55.9 = 55.9

• O: 2 × 16.0 = 32.0

• H: 2 × 1.0 = 2.0

Total: 55.9 + 32.0 + 2.0 = 89.9 g/mol

This is the same value used in the imported notes.

Example 3: Ca(NO₃)₂

Be careful with parentheses.

• Ca: 1 × 40.1 = 40.1

• N: 2 × 14.0 = 28.0

• O: 6 × 16.0 = 96.0

Total: 40.1 + 28.0 + 96.0 = 164.1 g/mol

Converting moles to grams

Use: moles × (g / mol) = grams

From the imported 5.2 Molar Relationships Notes:

Find the mass of 9.45 mol Al₂O₃.

9.45 mol × 102.0 g/mol = 964 g Al₂O₃

The value is reasonable because nearly 10 moles of a substance with a molar mass near 100 g/mol should have a mass near 1000 g.

Converting grams to moles

Use: grams × (mol / g) = moles

From the imported notes:

Find the moles in 92.2 g Fe₂O₃.

First calculate molar mass:

• Fe: 2 × 55.8 = 111.6

• O: 3 × 16.0 = 48.0

• total = 159.6 g/mol

Then convert:

92.2 g × (1 mol / 159.6 g) = 0.578 mol Fe₂O₃

Multi-step thinking

Some questions begin with grams and end with something else, like particles or liters. In those cases, grams are not the destination. They are the starting point.

Pattern:

grams → moles → next quantity

Example: How many moles are in 75.0 g N₂O₃?

Molar mass:

• N: 2 × 14.0 = 28.0

• O: 3 × 16.0 = 48.0

• total = 76.0 g/mol

Then: 75.0 g × (1 mol / 76.0 g) = 0.987 mol

Common mistakes

• Forgetting to multiply atoms inside parentheses

• Using the mass of one element instead of the whole compound

• Reversing the conversion factor

• Forgetting units

• Copying the formula incorrectly

Exam question types

• Calculate the molar mass of a compound

• Convert grams to moles

• Convert moles to grams

• Explain how subscripts affect molar mass

• Solve a multi-step conversion that begins with mass

Quick self-checks

• If the sample mass is less than the molar mass, the answer should be less than 1 mol

• If the sample mass is about twice the molar mass, the answer should be about 2 mol

• If the number of moles is large, the mass should also be large if molar mass is moderate or high

Particles, Avogadro's number, and counting chemistry

Chemistry counts impossibly small things by grouping them into moles. The key conversion is Avogadro's number:

1 mol = 6.02 × 10^23 representative particles

The imported 5.2 Molar Relationships Notes treats this as one branch of the mole roadmap. Once moles are known, particle counts follow directly.

What counts as a representative particle

The correct particle word depends on the substance:

Using the right word matters in exam responses. 1 mol H₂O means 6.02 × 10^23 molecules of H₂O, not atoms of H₂O.

Moles to particles

Use: mol × (6.02 × 10^23 particles / 1 mol)

Example: How many molecules are in 2.0 mol CO₂?

2.0 mol × 6.02 × 10^23 molecules/mol = 1.20 × 10^24 molecules

Particles to moles

Use: particles × (1 mol / 6.02 × 10^23 particles)

Example: How many moles are in 3.01 × 10^23 molecules H₂O?

3.01 × 10^23 × (1 mol / 6.02 × 10^23) = 0.500 mol

Molecules versus atoms inside a compound

This is a classic exam trap. A question may ask for atoms of an element, not molecules of the compound.

Example: How many hydrogen atoms are in 2.0 mol H₂O?

Step 1: convert to molecules of water

2.0 mol H₂O × 6.02 × 10^23 molecules/mol = 1.20 × 10^24 molecules H₂O

Step 2: each water molecule has 2 H atoms

1.20 × 10^24 molecules × 2 = 2.40 × 10^24 H atoms

The same logic works for oxygen atoms in CO₂, nitrogen atoms in NH₄NO₃, and so on.

Reverse questions

Sometimes the problem gives a huge particle number and asks for grams. That becomes:

particles → moles → grams

Example: What mass corresponds to 6.02 × 10^23 molecules CO₂?

• 6.02 × 10^23 molecules = 1.00 mol

• molar mass of CO₂ = 44.0 g/mol

Mass = 44.0 g

Common traps

• Using atoms when the question asks for molecules

• Forgetting that ionic compounds are counted in formula units

• Stopping after molecules when the question asks for atoms inside the molecules

• Not using scientific notation correctly

Good habits

When reading a problem, underline:

• the number given

• the unit given

• the unit wanted

• the particle type wanted

In counting chemistry, the number is huge, but the logic is simple: moles are the translator.

Gas volume at STP and Avogadro's hypothesis

The imported 5.2 Molar Relationships Notes states Avogadro's hypothesis like this: equal volumes of gases at the same temperature and pressure contain equal numbers of particles. This is a big idea because it means gas volume under fixed conditions depends mainly on how many particles are present, not on what the gas is.

At STP in these notes, the conditions are:

• 0°C

• 100 kPa

At STP, one mole of any ideal gas occupies:

22.4 L/mol

This is called the molar volume of a gas.

Why gas identity does not control the volume at STP

A mole of helium and a mole of sulfur dioxide have very different masses, but at STP both occupy about 22.4 L. The imported notes explain this by the kinetic view of gases: gas particles have negligible volume compared with the container, and the gas expands to fill the space available.

That means:

• mass can differ

• particle type can differ

• volume per mole at STP stays the same

Moles to liters

Use: mol × (22.4 L / 1 mol)

From the imported notes: Find the volume of 0.60 mol SO₂ at STP.

0.60 mol × 22.4 L/mol = 13 L SO₂

Liters to moles

Use: L × (1 mol / 22.4 L)

Imported-note style examples:

• 67.2 L SO₂ × (1 mol / 22.4 L) = 3.00 mol

• 0.880 L He × (1 mol / 22.4 L) = 0.0393 mol

Liters to particles

This is a two-step problem:

L → mol → particles

Example: How many molecules are in 44.8 L O₂ at STP?

Step 1: 44.8 L × (1 mol / 22.4 L) = 2.00 mol O₂

Step 2: 2.00 mol × 6.02 × 10^23 molecules/mol = 1.20 × 10^24 molecules O₂

Important limitation

The imported 5.2 Molar Relationships Notes also makes an important comparison with liquid water: 22.4 L/mol applies to gases at STP, not to liquids or solids. A basketball filled with liquid water does not contain the same number of molecules as a basketball filled with gas, because liquid particles are packed much more closely.

That means:

• use 22.4 L/mol only for gases

• only at the specified STP conditions

• do not apply it to liquids, solids, or gases not at STP unless the course specifically says to

Common exam questions

• Find gas volume from moles at STP

• Find moles from gas volume at STP

• Find number of particles from gas volume

• Explain Avogadro's hypothesis

• Explain why different gases can have the same molar volume at STP

Quick reasoning check

• If the volume is 22.4 L, the amount is 1 mol

• If the volume is 11.2 L, the amount is 0.5 mol

• If the volume is 44.8 L, the amount is 2 mol

These checkpoints make mental estimation fast.

Percent composition from formulas and from data

Percent composition tells how much of a compound's mass comes from each element. The imported 5.3 Percent Composition and Empirical Formulas Notes defines it as percent by mass. This idea is built on the law of definite proportions: a compound contains its elements in a fixed ratio by mass.

That means every pure sample of water has the same mass percentage of hydrogen and oxygen, and every pure sample of propane has the same mass percentage of carbon and hydrogen.

The core formula

Use:

% by mass of element = (mass of element / mass of compound) × 100

There are two common paths:

• from a chemical formula

• from experimental mass data

Percent composition from a formula

The imported notes show this with water.

For H₂O:

• H mass in 1 mol = 2 × 1.0 = 2.0 g

• O mass in 1 mol = 1 × 16.0 = 16.0 g

• total molar mass = 18.0 g/mol

Now calculate:

• %H = (2.0 / 18.0) × 100 = 11.1%

• %O = (16.0 / 18.0) × 100 = 88.9%

The percentages add to 100%, which is a key check.

Example: propane from the imported notes

For C₃H₈:

• C mass = 3 × 12.0 = 36.0 g

• H mass = 8 × 1.0 = 8.0 g

• total = 44.0 g/mol

Then:

• %C = (36.0 / 44.0) × 100 = 81.8%

• %H = (8.0 / 44.0) × 100 = 18.2%

Rounded in the notes, this may appear as 82% and 18%.

Percent composition from measured data

Sometimes the formula is not given, but the masses are.

Imported example: A 13.60 g compound contains 5.40 g oxygen.

First find magnesium mass: 13.60 - 5.40 = 8.20 g Mg

Then:

• %Mg = (8.20 / 13.60) × 100 = 60.3%

• %O = (5.40 / 13.60) × 100 = 39.7%

Again, they add to 100%.

Percent composition as a conversion factor

The imported 5.3 Percent Composition and Empirical Formulas Notes takes this one step further. Once a percent is known, it can be used like a conversion factor.

Example from the notes: Hydrogen is 11.1% by mass in water.

In 20 g H₂O, mass of H is:

20 g H₂O × (11.1 g H / 100 g H₂O) = 2.2 g H

This is powerful because it turns a percentage into a mass-conversion tool.

Fertilizer-style questions

The notes use fertilizers like NH₃ and NH₄NO₃.

For NH₃:

• N = 14.0

• total = 17.0

• %N = (14.0 / 17.0) × 100 = 82.4%

For NH₄NO₃:

• total N = 2 × 14.0 = 28.0

• molar mass = 80.0

• %N = (28.0 / 80.0) × 100 = 35.0%

This explains why different compounds containing the same element can have very different percent compositions.

Hydrates and decomposition-style questions

A hydrate question may ask for the percent by mass of water or another component. The method stays the same:

• find part mass

• divide by whole sample mass

• multiply by 100

Common mistakes

• Dividing by the wrong total mass

• Forgetting to find the missing mass first

• Using atomic mass instead of the full compound mass

• Not checking whether percentages add to 100%

Exam question types

• Find percent composition from a formula

• Find percent composition from experimental mass data

• Find grams of one element in a given mass of compound

• Compare compounds by percent of a specific element

Percent composition answers the question: “Out of the total mass, how much belongs to this element?”

Empirical formulas from percent composition

An empirical formula gives the lowest whole-number ratio of atoms in a compound. It does not necessarily give the actual number of atoms in one molecule. The imported 5.3 Percent Composition and Empirical Formulas Notes is built around a standard workflow, and this workflow is one of the most important exam procedures in the unit.

What the empirical formula tells you

If a compound has molecular formula H₂O₂, the empirical formula is HO. That means the simplest ratio of H to O is 1:1, even though the actual molecule contains 2 H and 2 O atoms.

This is why empirical formulas are about ratio, not exact particle structure.

The standard workflow

When percentages are given, assume a 100 g sample. That turns percentages directly into grams.

Then follow these steps:

1. Assume 100 g if percentages are given

2. Convert grams to moles for each element

3. Divide all mole values by the smallest

4. Inspect for fractions

5. Multiply to clear fractions if needed

6. Write the simplest whole-number formula

Example from the imported notes: nitrogen and oxygen

Given:

• 25.9% N

• 74.1% O

Assume 100 g:

• 25.9 g N

• 74.1 g O

Convert to moles:

• 25.9 g N × (1 mol / 14.0 g) = 1.85 mol N

• 74.1 g O × (1 mol / 16.0 g) = 4.63 mol O

Divide by the smaller value, 1.85:

• N: 1.85 / 1.85 = 1.00

• O: 4.63 / 1.85 = 2.5

Now the ratio is 1 : 2.5. That is not whole-number form, so multiply both by 2:

• N: 1 × 2 = 2

• O: 2.5 × 2 = 5

Empirical formula: N₂O₅

Fraction traps

These are the common fractional patterns after dividing by the smallest mole value:

• 1.5 → multiply all by 2

• 2.5 → multiply all by 2

• 1.33 or 2.33 → multiply all by 3

• 1.25 or 2.25 → multiply all by 4

Do not round 2.5 down to 2. That destroys the real ratio.

Another imported example pattern

The notes also show a composition leading to HgSO₄. After converting to moles and dividing by the smallest, the ratios came out close to:

• Hg = 1.00

• S = 1.00

• O = 4.02

That is effectively 1 : 1 : 4, so the empirical formula is HgSO₄.

This illustrates another exam point: small decimal drift from rounding is normal. A value like 4.02 is treated as 4.

When masses, not percentages, are given

The workflow is almost identical. Skip the 100 g assumption and start directly with the given masses.

Why the answer must be simplest whole numbers

The empirical formula is defined as the lowest whole-number ratio. So if you obtain C₂H₄, reduce it to CH₂ for the empirical formula.

Common mistakes

• Forgetting to convert grams to moles before making ratios

• Dividing by the wrong mole value

• Rounding fractions too early

• Forgetting to multiply all elements when clearing a fraction

• Giving a formula that is not in simplest whole-number form

Exam question types

• Determine the empirical formula from percent composition

• Determine the empirical formula from mass data

• Explain why a ratio must be in whole numbers

• Identify and correct a student's ratio mistake

In empirical formula problems, the answer is hidden in the mole ratio, not in the original percentages.

Solutions, molarity, and what concentration means

A solution is a homogeneous mixture. The imported 5.4 Concentration of a solution Notes defines concentration as a measure of the amount of solute dissolved in a given quantity of solvent. That basic language is essential before the math starts.

Core solution vocabulary

• Solute: the substance being dissolved

• Solvent: the substance doing the dissolving

• Solution: the uniform mixture of solute and solvent

• Dilute: low amount of solute relative to solution amount

• Concentrated: high amount of solute relative to solution amount

The imported notes point out that dilute and concentrated are useful but qualitative. They do not give exact numbers. For exam calculations, chemistry uses molarity.

Definition of molarity

Molarity is the number of moles of solute dissolved in 1 liter of solution.

Formula:

M = moles of solute / liters of solution

The unit is mol/L, often written as M.

A very important wording detail

The denominator is liters of solution, not liters of solvent.

The imported 5.4 Concentration of a solution Notes directly warns about this. If you add 1 L of water to a solute, the final solution volume may be greater than 1 L. That means the result will not automatically be 1 M.

This is one of the most common exam mistakes.

Why qualitative words are not enough

A solution might look dark or pale, but without numbers it is hard to compare precisely. The imported notes use copper sulfate examples to show why a solution may seem concentrated or dilute only in comparison with another solution. Molarity solves that by giving an exact numerical concentration.

Connecting the idea to particles

A more concentrated solution has more solute particles in a given volume. A more dilute solution has fewer. The imported notes even describe a particle model: if solute particles stay the same but more solvent is added, the solution becomes less concentrated.

Molarity in plain language

• 1.0 M NaCl means 1.0 mol NaCl in every 1 L of solution

• 0.50 M glucose means 0.50 mol glucose in every 1 L of solution

Other percent-based concentrations

The imported notes also discuss:

• percent by volume % (v/v)

• percent by mass % (m/m)

These are different from molarity. Percent by mass uses masses. Percent by volume uses volumes. Molarity uses moles per liter.

Common exam questions

• Define solute, solvent, solution

• Distinguish dilute from concentrated

• Define molarity

• Explain why liters of solution are used, not liters of solvent

• Identify which quantity in a problem is the solute and which is the solvent

Molarity turns a vague description like “strong” or “weak” into a measurable concentration.

Molarity calculations: grams, moles, and liters

Once the definition is known, most solution questions reduce to one relationship:

M = n / V

where:

• M = molarity

• n = moles of solute

• V = liters of solution

The imported 5.4 Concentration of a solution Notes uses this equation in every direction. A strong exam student can rearrange it automatically.

The three main forms

• M = n / V

• n = M × V

• V = n / M

Finding molarity

Example from the imported notes: A solution contains 0.90 g NaCl in 100 mL of solution.

Step 1: convert grams to moles

Molar mass of NaCl = 58.5 g/mol

0.90 g × (1 mol / 58.5 g) = 0.0154 mol

Step 2: convert volume to liters

100 mL = 0.100 L

Step 3: apply formula

M = 0.0154 / 0.100 = 0.154 M

Rounded: 0.15 M

Finding molarity when moles are already given

Imported example: 0.70 mol NaCl in 250 mL

Convert volume: 250 mL = 0.250 L

Then: M = 0.70 / 0.250 = 2.8 M

Finding moles of solute

Use: n = M × V

Imported example: How many moles are in 1.5 L of 0.70 M NaClO?

n = 0.70 × 1.5 = 1.05 mol

Rounded to two significant figures: 1.1 mol

Another imported style: 335 mL of 0.425 M NH₄NO₃

Convert volume first: 335 mL = 0.335 L

Then: n = 0.425 × 0.335 = 0.142 mol

Finding volume

If the question gives moles and molarity:

V = n / M

Example: What volume is needed for 0.50 mol of a 2.0 M solution?

V = 0.50 / 2.0 = 0.25 L

When grams are given instead of moles

This is another classic two-step pattern:

grams → moles → molarity

Imported example: A 2.0 L solution contains 36.0 g glucose, molar mass 180 g/mol.

First: 36.0 g × (1 mol / 180 g) = 0.200 mol

Then: M = 0.200 / 2.0 = 0.10 M

Preparing solutions

The imported notes describe making a solution in a volumetric flask:

1. add some solvent first

2. add solute

3. dissolve

4. fill to the final volume mark

That reinforces the key definition: the final quantity is volume of solution, not initial solvent volume.

Common mistakes

• Leaving volume in mL instead of converting to L

• Using grams directly in M = n/V without converting to moles

• Using solvent volume instead of solution volume

• Rearranging the equation incorrectly

• Losing track of significant figures

Exam question types

• Find molarity from grams and volume

• Find molarity from moles and volume

• Find moles from molarity and volume

• Find volume from moles and molarity

• Explain the steps needed before substitution

Fast problem-sorting guide

Dilution and conservation of solute

A dilution happens when more solvent is added to a solution. The imported 5.4 Concentration of a solution Notes makes the key idea explicit: during dilution, the number of moles of solute does not change. What changes is the total volume. As volume increases, concentration decreases.

That is why dilution problems are built on conservation of solute.

The core idea in plain language

Before dilution:

• certain moles of solute are present

After dilution:

• the same moles of solute are spread through a larger volume

So:

• concentration goes down

• moles of solute stay the same

The imported particle model uses red balls for solute and green balls for solvent. If the number of red balls stays constant while more green balls are added, the fraction of red balls drops. That is dilution.

The dilution equation

Because moles stay constant:

M₁V₁ = M₂V₂

where:

• M₁ = initial concentration

• V₁ = initial volume used

• M₂ = final concentration

• V₂ = final total volume after dilution

Example from the imported notes

Prepare 100.0 mL of 0.400 M MgSO₄ from a 2.00 M stock solution.

Use: M₁V₁ = M₂V₂

(2.00)(V₁) = (0.400)(100.0 mL)

V₁ = 20.0 mL

So measure 20.0 mL of the stock solution and dilute with water to a final volume of 100.0 mL.

Another imported example

Prepare 250 mL of 0.20 M NaCl from 1.0 M NaCl.

(1.0)(V₁) = (0.20)(250)

V₁ = 50 mL

So:

1. measure 50 mL of the stock

2. transfer to a 250 mL volumetric flask

3. add water to the mark

Dilution versus adding more solute

These are not the same process.

• Dilution: add solvent, moles of solute stay the same

• Making a more concentrated solution by adding solute: moles of solute increase

If moles change, M₁V₁ = M₂V₂ does not apply directly.

How to recognize a dilution question

Look for phrases like:

• stock solution

• prepare a dilute solution

• dilute to

• make up to volume

• what volume of stock is needed

These are almost automatic signals to use M₁V₁ = M₂V₂.

Common mistakes

• Using the final water added instead of the final solution volume

• Forgetting that V₂ is the total final volume

• Mixing up stock concentration and final concentration

• Applying the equation when solute moles are not actually conserved

Exam question types

• Find the volume of stock solution needed

• Find the final concentration after dilution

• Explain why dilution lowers molarity

• Describe how to prepare a weaker solution from a stronger stock

In a dilution, the solute is not lost. It is just spread out through more solution.