Chemistry Calculations Guide

The Mole Concept

The mole is the chemist's counting unit—a bridge between the atomic world and the macroscopic world we can measure.

Avogadro's Number

1 mole = 6.022 × 10²³ particles

This number of atoms, molecules, or ions is called Avogadro's number (N�?. One mole of any substance contains exactly this many particles.

Key Mole Relationships

Moles from Mass

n = m / M

n = moles, m = mass (g), M = molar mass (g/mol)

Number of Particles

N = n × N�?/div> N = number of particles, N�?= 6.022 × 10²³

Gas at STP

V = n × 22.4 L

Volume at standard temperature and pressure

The Mole Triangle

Moles (n)

Mass (g) ÷ M or × M

Particles × N�?or ÷ N�?/span>

Volume (L) × 22.4 or ÷ 22.4

Balancing Chemical Equations

A balanced equation has equal numbers of each type of atom on both sides, obeying the Law of Conservation of Mass.

Step-by-Step Method

1

Write Unbalanced Equation

List all reactants and products with their chemical formulas.

2

Count Atoms

Tally each element on both sides of the equation.

3

Balance One Element at a Time

Start with the most complex molecule. Add coefficients (not subscripts!).

4

Balance Hydrogen and Oxygen Last

These often appear in multiple compounds.

5

Verify

Recount all atoms on both sides to confirm balance.

Example: Balance Fe + O�?�?Fe₂O�?/h4>

Unbalanced: Fe + O�?�?Fe₂O�?br>Fe: 1�?, O: 2�? (not balanced)

Balance Fe: 2Fe + O�?�?Fe₂O�?br>Fe: 2�? �? O: 2�? �?/span>

Balance O: Need 3 O on left. LCM of 2 and 3 is 6.
4Fe + 3O�?�? 2Fe₂O�?/span>

Verify: Fe: 4�? �? O: 6�? �?Balanced!

Stoichiometry

Stoichiometry uses balanced equations to calculate quantities of reactants and products in chemical reactions.

The Stoichiometry Roadmap

Mass A ÷ Molar Mass A

Moles A × Mole Ratio

Moles B × Molar Mass B

Mass B

Example: Combustion of Methane

Problem: How many grams of CO�?are produced when 32g of CH�?burns completely?

CH�?+ 2O�?�?CO�?+ 2H₂O

Step 1: Moles of CH�?= 32g ÷ 16 g/mol = 2 mol

Step 2: Mole ratio: 1 mol CH�?: 1 mol CO�?br>So 2 mol CH�?�?2 mol CO�?/span>

Step 3: Mass of CO�?= 2 mol × 44 g/mol = 88g

Answer: 88 grams of CO�?produced

Limiting Reagents

The limiting reagent is the reactant that runs out first, determining the maximum amount of product that can form.

Finding the Limiting Reagent

Convert all reactant masses to moles
Divide each by its coefficient in the balanced equation
The smallest result indicates the limiting reagent
Use the limiting reagent to calculate product amounts

Example: Making Water

Problem: 4g H�?reacts with 32g O�? Which is limiting?

2H�?+ O�?�?2H₂O

Moles: H�? 4g ÷ 2g/mol = 2 mol
O�? 32g ÷ 32g/mol = 1 mol

Divide by coefficients: H�? 2 mol ÷ 2 = 1
O�? 1 mol ÷ 1 = 1

Compare: They're equal! Neither is limiting (stoichiometric amounts)

If they were unequal, the smaller value indicates the limiting reagent.

Solution Concentration

Concentration describes how much solute is dissolved in a given amount of solution.

Concentration Formulas

Molarity (M)

M = n / V

moles of solute / liters of solution

Dilution

M₁V�?= M₂V�?/div> Moles before dilution = moles after

Mass Percent

% = (m_solute / m_solution) × 100

Example: Making a Solution

Problem: How many grams of NaCl are needed to make 500 mL of 0.5 M solution?

Find moles: n = M × V = 0.5 mol/L × 0.5 L = 0.25 mol

Find mass: m = n × M = 0.25 mol × 58.5 g/mol = 14.6g

Answer: Dissolve 14.6g NaCl in water to make 500 mL solution

Gas Laws

Gas laws describe the relationships between pressure, volume, temperature, and amount of gas.

Boyle's Law

P₁V�?= P₂V�?/div>

At constant T, pressure and volume are inversely proportional.

Charles's Law

V�?T�?= V�?T�?/div>

At constant P, volume and temperature are directly proportional.

Gay-Lussac's Law

P�?T�?= P�?T�?/div>

At constant V, pressure and temperature are directly proportional.

Ideal Gas Law

PV = nRT

Combines all relationships. R = 8.314 J/(mol·K) or 0.0821 L·atm/(mol·K)

Temperature in Kelvin! All gas law calculations require absolute temperature. Convert °C to K by adding 273.15.