Atoms, bonding, gases, balanced equations, moles, solutions, pH, buffers, equilibrium and an organic and biochemistry preview, with short fill-in-the-blank Python calculators.

Chemistry Fundamentals

## Why this matters

A balanced equation is the recipe of a reaction. It tells a chemist how much of each reagent to order, tells an engineer how much carbon dioxide a furnace releases per kilogram of fuel, and is the starting point for every yield and dosing calculation in the next lessons. If the equation is not balanced, every number calculated from it is wrong.

## The core idea: atoms are conserved

In a chemical reaction atoms are rearranged, not created or destroyed (the **law of conservation of mass**). So every element must appear the same number of times on the left (reactants) and the right (products).

```text
reactants  ->  products
CH4 + O2   ->  CO2 + H2O        unbalanced
```

Count the atoms: left has 1 C, 4 H, 2 O; right has 1 C, 2 H, 3 O. Hydrogen and oxygen do not match.

### Coefficients, not subscripts

You fix the count by putting **coefficients** (whole numbers) in front of formulas. You never change the **subscripts** inside a formula, because that changes the substance: H₂O is water, H₂O₂ is hydrogen peroxide.

```text
2 H2O   = 2 molecules of water = 4 H and 2 O
H2O2    = a different chemical
```

## A step-by-step method

1. Write correct formulas for every reactant and product.
2. Count each element on both sides (a small table helps).
3. Balance elements that appear in only one compound on each side first; save free elements (O₂, H₂, Fe) and hydrogen/oxygen for last.
4. Adjust one coefficient at a time and recount.
5. Reduce to the smallest whole-number set and check every element once more.

### Worked example 1: burning methane

```text
CH4 + O2 -> CO2 + H2O
```

- Carbon: 1 and 1. Already balanced.
- Hydrogen: 4 on the left, 2 on the right. Put a 2 in front of H₂O.

```text
CH4 + O2 -> CO2 + 2 H2O
```

- Oxygen: right side now has 2 (in CO₂) + 2 (in 2 H₂O) = 4. Left has 2. Put a 2 in front of O₂.

```text
CH4 + 2 O2 -> CO2 + 2 H2O
```

| element | left | right |
|---|---|---|
| C | 1 | 1 |
| H | 4 | 4 |
| O | 4 | 2 + 2 = 4 |

Balanced.

### Worked example 2: burning propane

```text
C3H8 + O2 -> CO2 + H2O
```

- Carbon: 3 on the left, so 3 CO₂.
- Hydrogen: 8 on the left, so 4 H₂O.
- Oxygen: right has 3 × 2 + 4 × 1 = 10, so 5 O₂.

```text
C3H8 + 5 O2 -> 3 CO2 + 4 H2O
```

### Worked example 3: making ammonia (the Haber process)

```text
N2 + H2 -> NH3
```

Nitrogen: 2 on the left, so 2 NH₃. Hydrogen: now 6 on the right, so 3 H₂.

```text
N2 + 3 H2 -> 2 NH3
```

### Worked example 4: rusting iron

```text
Fe + O2 -> Fe2O3
```

Oxygen comes in pairs on the left and threes on the right; the lowest common multiple is 6. So 3 O₂ and 2 Fe₂O₃, which needs 4 Fe:

```text
4 Fe + 3 O2 -> 2 Fe2O3
```

Check: Fe 4 = 4, O 6 = 6.

## Fractions are a useful trick

If you end with an odd count of oxygen, you can briefly use a fraction and then multiply everything by 2. For ethane, `C2H6 + 7/2 O2 -> 2 CO2 + 3 H2O` becomes `2 C2H6 + 7 O2 -> 4 CO2 + 6 H2O`. The final answer must use whole numbers.

## Reading what an equation tells you

`N2 + 3 H2 -> 2 NH3` means:

- 1 molecule of nitrogen reacts with 3 molecules of hydrogen to make 2 molecules of ammonia, **and**
- 1 **mole** of N₂ reacts with 3 moles of H₂ to make 2 moles of NH₃.

The coefficients are a **mole ratio**, which is exactly what the next lesson uses to predict yields. State symbols are often added: (s) solid, (l) liquid, (g) gas, (aq) dissolved in water.

## Common types of reaction

- **Combustion:** fuel + O₂ → CO₂ + H₂O (for hydrocarbons).
- **Synthesis:** A + B → AB (N₂ + 3H₂ → 2NH₃).
- **Decomposition:** AB → A + B (2H₂O₂ → 2H₂O + O₂).
- **Single replacement:** Zn + 2HCl → ZnCl₂ + H₂.
- **Double replacement / precipitation:** AgNO₃ + NaCl → AgCl(s) + NaNO₃.

## Common mistakes

- **Changing subscripts** to force a balance. That changes the chemical.
- **Forgetting to multiply through.** In 3 CO₂ there are 6 oxygen atoms, not 2.
- **Leaving a common factor.** 2 CH₄ + 4 O₂ → 2 CO₂ + 4 H₂O balances, but reduce it to 1, 2, 1, 2.
- **Not rechecking every element** after the last change.

## Quick recap

- Atoms are conserved, so each element must match on both sides.
- Change coefficients only; never change subscripts.
- Balance single-appearance elements first, then free elements, and recheck.
- Coefficients give the mole ratio used in stoichiometry.

## Try it

Open the **Do** tab and take the quiz: you will choose the correct coefficients for several equations.


Balance a chemical equation by changing coefficients, never subscripts, so every element has the same atom count on both sides.

Balancing equations

Introduction to Balancing Chemical Equations

## Why this matters

Reactions happen atom by atom, but you weigh chemicals by the gram. The **mole** is the bridge between the two. With it, a lab technician knows how many grams of salt make a buffer, a process engineer knows how much product a batch should yield, and a chemist can tell whether a 60% yield is a problem or normal. This is the single most used calculation in introductory chemistry.

## The mole

A mole is a counting unit, like a dozen, but much bigger:

```text
1 mol = 6.022 × 10^23 particles   (Avogadro's number, N_A)
```

It is chosen so that the **molar mass** of an element in grams per mole equals its atomic mass from the periodic table. Carbon's atomic mass is 12.011 u, so one mole of carbon atoms weighs 12.011 g.

### Molar mass of a compound

Add up the atomic masses of every atom in the formula.

```text
H2O   = 2 × 1.008 + 15.999           = 18.015 g/mol
CO2   = 12.011 + 2 × 15.999          = 44.009 g/mol
NaCl  = 22.990 + 35.45               = 58.44 g/mol
C6H12O6 (glucose) = 6 × 12.011 + 12 × 1.008 + 6 × 15.999 = 180.156 g/mol
```

### The three conversions

```text
moles     = grams / molar mass
grams     = moles × molar mass
particles = moles × 6.022 × 10^23
```

Example: how many moles are in 9.0 g of water? 9.0 / 18.015 ≈ **0.50 mol**, which is about 3.0 × 10²³ molecules.

## Stoichiometry: the road map

Stoichiometry uses the balanced equation's coefficients as a **mole ratio**. You cannot go straight from grams of A to grams of B; you always go through moles:

```text
grams A  ->  moles A  ->  (mole ratio)  ->  moles B  ->  grams B
     ÷ molar mass A      × coef B / coef A        × molar mass B
```

### Worked example 1: water from hydrogen

```text
2 H2 + O2 -> 2 H2O
```

How much water can 4.00 g of hydrogen make (with plenty of oxygen)?

1. Moles of H₂: 4.00 g / 2.016 g/mol = 1.984 mol
2. Mole ratio H₂O : H₂ = 2 : 2 = 1, so 1.984 mol of H₂O
3. Grams of H₂O: 1.984 × 18.015 ≈ **35.7 g**

### Worked example 2: carbon dioxide from methane

```text
CH4 + 2 O2 -> CO2 + 2 H2O
```

A furnace burns 1.00 kg (1000 g) of methane. How much CO₂ is released?

1. Moles of CH₄: 1000 / 16.043 ≈ 62.33 mol
2. Ratio CO₂ : CH₄ = 1 : 1, so 62.33 mol CO₂
3. Grams: 62.33 × 44.009 ≈ 2743 g ≈ **2.74 kg**

Burning a kilogram of methane releases more than two and a half kilograms of CO₂, because each carbon picks up two oxygen atoms from the air.

### Worked example 3: ammonia from nitrogen

```text
N2 + 3 H2 -> 2 NH3
```

From 28.0 g of N₂: 28.0 / 28.014 ≈ 0.9995 mol N₂; ratio 2 : 1 gives 1.999 mol NH₃; × 17.031 g/mol ≈ **34.0 g** of ammonia.

## Limiting reactant and percent yield

Real reactions rarely have the exact ratio of reactants. The **limiting reactant** is the one that runs out first; it sets the maximum product. To find it, calculate the product each reactant could make on its own; the smaller answer wins.

Example: 4.0 mol H₂ and 1.5 mol O₂. Hydrogen alone could make 4.0 mol H₂O; oxygen alone could make 3.0 mol (ratio 2 : 1). Oxygen is limiting: **3.0 mol** of water at most.

The amount you calculate is the **theoretical yield**. What you actually collect is usually less (side reactions, losses in transfer, incomplete reaction):

```text
percent yield = actual / theoretical × 100%
```

If the theoretical yield is 35.7 g and you collect 30.4 g, the percent yield is 30.4 / 35.7 × 100 ≈ **85%**.

## Common mistakes

- **Using grams in the mole ratio.** Coefficients compare moles, never grams.
- **Forgetting the subscript multiplier** in molar mass (two hydrogens in H₂O).
- **Using an unbalanced equation.** Balance first, always.
- **Ignoring the limiting reactant** when both amounts are given.
- **Percent yields over 100%.** Usually a wet or impure product, not a miracle.

## Quick recap

- 1 mol = 6.022 × 10²³ particles; molar mass in g/mol comes from the periodic table.
- Grams ↔ moles with molar mass; moles of A ↔ moles of B with the mole ratio.
- The limiting reactant sets the theoretical yield; percent yield compares reality to it.

## Try it

Open the **Do** tab: you will complete a Python function that predicts the theoretical yield of a reaction, then take the quiz.


Convert between grams, moles and particles, and use a balanced equation's mole ratio to predict how much product a reaction can make.

The mole and stoichiometry

Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6

Worked example: Calculating molar mass and number of moles | AP Chemistry | Khan Academy

## Why this matters

Almost every lab procedure starts with "prepare a 0.1 M solution" or "dilute the stock 1 in 10". Clinical labs dilute patient samples that read above an instrument's range, biotech technicians make buffers and media, and water-treatment engineers dose chemicals by concentration. Getting the concentration wrong silently ruins an experiment or a result, so these calculations are checked again and again on the job.

## Vocabulary

- **Solute**: the substance being dissolved (salt).
- **Solvent**: the substance doing the dissolving, usually water.
- **Solution**: the uniform mixture of the two. "Aqueous" (aq) means dissolved in water.
- **Concentration**: how much solute is in a given amount of solution.

## Molarity

The most common concentration unit in chemistry is **molarity** (M), moles of solute per litre of **solution**:

```text
molarity (M) = moles of solute / litres of solution
moles        = molarity × litres
```

Note "litres of solution", not "litres of water added". When you make a solution properly, you dissolve the solute and then add solvent **up to** the final volume mark.

### Worked example 1: how much salt?

You need 500 mL of 0.200 M sodium chloride (NaCl, 58.44 g/mol).

```text
moles = 0.200 mol/L × 0.500 L = 0.100 mol
grams = 0.100 mol × 58.44 g/mol = 5.84 g
```

In a lab you would weigh 5.84 g of NaCl, dissolve it in less than 500 mL of water in a 500 mL volumetric flask, then fill to the mark and mix.

### Worked example 2: what is the concentration?

9.0 g of glucose (180.16 g/mol) is dissolved to make 250 mL of solution.

```text
moles    = 9.0 / 180.16 ≈ 0.0500 mol
molarity = 0.0500 mol / 0.250 L = 0.200 M
```

## Other units you will meet

| unit | meaning | where it shows up |
|---|---|---|
| % w/v | grams of solute per 100 mL of solution | saline: 0.9% w/v NaCl = 0.9 g per 100 mL |
| mg/mL or mg/L | mass per volume | clinical results, drug doses, water quality |
| ppm | parts per million (about mg/L in dilute water) | environmental testing |
| mM, µM | millimolar (10⁻³ M), micromolar (10⁻⁶ M) | biology and biochemistry |

Converting between them always goes through molar mass and unit prefixes. For example, 0.9% w/v NaCl is 9 g/L, and 9 / 58.44 ≈ 0.154 M.

## Dilution: C₁V₁ = C₂V₂

When you add solvent, the amount of solute does not change; it is just spread through more volume. Moles before = moles after:

```text
C1 × V1 = C2 × V2
```

`C1, V1` are the stock concentration and the volume of stock you take; `C2, V2` are the final concentration and final volume. Any volume unit works as long as both sides use the same one.

### Worked example 3: a 1-in-10 dilution

Make 250 mL of 0.10 M from a 1.0 M stock.

```text
V1 = C2 × V2 / C1 = 0.10 × 250 / 1.0 = 25 mL
```

Take 25 mL of stock and add solvent **to a final volume of** 250 mL (that is about 225 mL of solvent, but you fill to the mark, you do not add exactly 225 mL).

### Worked example 4: from a concentrated stock

Make 100 mL of 0.50 M hydrochloric acid from 6.0 M stock.

```text
V1 = 0.50 × 100 / 6.0 ≈ 8.33 mL
```

### Dilution factor

The **dilution factor** is V₂ / V₁ (or C₁ / C₂). A 1-in-10 dilution has a factor of 10. Serial dilutions repeat a step: three 1-in-10 dilutions in a row give 1-in-1000. Clinical labs multiply a diluted sample's reading by the dilution factor to report the original value.

## Safety note

This lesson is paper calculation only. Preparing solutions, especially from concentrated acids and bases, must be learned in a teaching or workplace lab with your instructor or supervisor and the lab's own procedures; employer or institution training is required before you do it for real. One rule you will hear on day one: when diluting concentrated acid, **add acid to water**, never water to acid, because the heat released can make the liquid boil and spatter. Do not experiment with household chemicals at home. See OSHA's [laboratory safety guidance](https://www.osha.gov/laboratories). Checked on 2026-10-23.

## Common mistakes

- **Dividing by the volume of solvent added** instead of the final solution volume.
- **mL vs L.** Molarity needs litres: 250 mL = 0.250 L.
- **Swapping C₁ and C₂.** The stock is always the more concentrated one; if V₁ comes out larger than V₂, you swapped them.
- **Adding V₁ + V₂.** V₂ is already the final volume, stock included.

## Quick recap

- Molarity = moles of solute per litre of solution.
- grams = molarity × litres × molar mass.
- C₁V₁ = C₂V₂ because dilution keeps the moles of solute the same.
- Fill to the final volume; add acid to water; learn the hands-on part under supervision.

## Try it

Open the **Do** tab: you will finish a Python dilution calculator, then take the quiz.


Calculate molarity, work out how much solute a solution needs, and plan a dilution with C1V1 = C2V2.

Solutions and concentration

Solutions: Crash Course Chemistry #27

Dilutions | Chemistry | Khan Academy

Balancing equations, the mole and stoichiometry, and solution concentration and dilution.