- Chemical Analysis
- Chemical Reactions
- Chemistry Branches
- Inorganic Chemistry
- Ionic and Molecular Compounds
- Kinetics
- Making Measurements
- Nuclear Chemistry
- Organic Chemistry
- Physical Chemistry
- Absolute Entropy And Entropy Change
- Acid Dissociation Constant
- Acid-Base Indicators
- Acid-Base Reactions and Buffers
- Acids and Bases
- Alkali Metals
- Allotropes of Carbon
- Amorphous Polymer
- Amount of Substance
- Application of Le Chatelier's Principle
- Arrhenius Equation
- Arrhenius Theory
- Atom Economy
- Atomic Structure
- Autoionization of Water
- Avogadro Constant
- Avogadro's Number and the Mole
- Beer-Lambert Law
- Bond Enthalpy
- Bonding
- Born Haber Cycles
- Born-Haber Cycles Calculations
- Boyle's Law
- Brønsted-Lowry Acids and Bases
- Buffer Capacity
- Buffer Solutions
- Buffers
- Buffers Preparation
- Calculating Enthalpy Change
- Calculating Equilibrium Constant
- Calorimetry
- Carbon Structures
- Cell Potential
- Cell Potential and Free Energy
- Chalcogens
- Chemical Calculations
- Chemical Equations
- Chemical Equilibrium
- Chemical Thermodynamics
- Closed Systems
- Colligative Properties
- Collision Theory
- Common-Ion Effect
- Composite Materials
- Composition of Mixture
- Constant Pressure Calorimetry
- Constant Volume Calorimetry
- Coordination Compounds
- Coupling Reactions
- Covalent Bond
- Covalent Network Solid
- Crystalline Polymer
- De Broglie Wavelength
- Determining Rate Constant
- Deviation From Ideal Gas Law
- Diagonal Relationship
- Diamond
- Dilution
- Dipole Chemistry
- Dipole Moment
- Dissociation Constant
- Distillation
- Dynamic Equilibrium
- Electric Fields Chemistry
- Electrochemical Cell
- Electrochemical Series
- Electrochemistry
- Electrode Potential
- Electrolysis
- Electrolytes
- Electromagnetic Spectrum
- Electron Affinity
- Electron Configuration
- Electron Shells
- Electronegativity
- Electronic Transitions
- Elemental Analysis
- Elemental Composition of Pure Substances
- Empirical and Molecular Formula
- Endothermic and Exothermic Processes
- Energetics
- Energy Diagrams
- Enthalpy Changes
- Enthalpy For Phase Changes
- Enthalpy of Formation
- Enthalpy of Reaction
- Enthalpy of Solution and Hydration
- Entropy
- Entropy Change
- Equilibrium Concentrations
- Equilibrium Constant Kp
- Equilibrium Constants
- Examples of Covalent Bonding
- Factors Affecting Reaction Rates
- Finding Ka
- Free Energy
- Free Energy Of Dissolution
- Free Energy and Equilibrium
- Free Energy of Formation
- Fullerenes
- Fundamental Particles
- Galvanic and Electrolytic Cells
- Gas Constant
- Gas Solubility
- Gay Lussacs Law
- Giant Covalent Structures
- Graham's Law
- Graphite
- Ground State
- Group 3A
- Group 4A
- Group 5A
- Half Equations
- Heating Curve for Water
- Heisenberg Uncertainty Principle
- Henderson-Hasselbalch Equation
- Hess' Law
- Hybrid Orbitals
- Hydrogen Bonds
- Ideal Gas Law
- Ideal and Real Gases
- Intermolecular Forces
- Introduction to Acids and Bases
- Ion And Atom Photoelectron Spectroscopy
- Ion dipole Forces
- Ionic Bonding
- Ionic Product of Water
- Ionic Solids
- Ionisation Energy
- Ions: Anions and Cations
- Isotopes
- Kinetic Molecular Theory
- Lattice Structures
- Law of Definite Proportions
- Le Chatelier's Principle
- Lewis Acid and Bases
- London Dispersion Forces
- Magnitude Of Equilibrium Constant
- Mass Spectrometry
- Mass Spectrometry of Elements
- Maxwell-Boltzmann Distribution
- Measuring EMF
- Mechanisms of Chemical Bonding
- Melting and Boiling Point
- Metallic Bonding
- Metallic Solids
- Metals Non-Metals and Metalloids
- Mixtures and Solutions
- Molar Mass Calculations
- Molarity
- Molecular Orbital Theory
- Molecular Solid
- Molecular Structures of Acids and Bases
- Moles and Molar Mass
- Nanoparticles
- Neutralisation Reaction
- Oxidation Number
- Partial Pressure
- Particulate Model
- Partition Coefficient
- Percentage Yield
- Periodic Table Organization
- Phase Changes
- Phase Diagram of Water
- Photoelectric Effect
- Photoelectron Spectroscopy
- Physical Properties
- Polarity
- Polyatomic Ions
- Polyprotic Acid Titration
- Prediction of Element Properties Based on Periodic Trends
- Pressure and Density
- Properties Of Equilibrium Constant
- Properties of Buffers
- Properties of Solids
- Properties of Water
- Quantitative Electrolysis
- Quantum Energy
- Quantum Numbers
- RICE Tables
- Rate Equations
- Rate of Reaction and Temperature
- Reacting Masses
- Reaction Quotient
- Reaction Quotient And Le Chateliers Principle
- Real Gas
- Redox
- Relative Atomic Mass
- Representations of Equilibrium
- Reversible Reaction
- SI units chemistry
- Saturated Unsaturated and Supersaturated
- Shapes of Molecules
- Shielding Effect
- Simple Molecules
- Solids Liquids and Gases
- Solubility
- Solubility Curve
- Solubility Equilibria
- Solubility Product
- Solubility Product Calculations
- Solutes Solvents and Solutions
- Solution Representations
- Solutions and Mixtures
- Specific Heat
- Spectroscopy
- Standard Potential
- States of Matter
- Stoichiometry In Reactions
- Strength of Intermolecular Forces
- The Laws of Thermodynamics
- The Molar Volume of a Gas
- Thermodynamically Favored
- Trends in Ionic Charge
- Trends in Ionisation Energy
- Types of Mixtures
- VSEPR Theory
- Valence Electrons
- Van der Waals Forces
- Vapor Pressure
- Water in Chemical Reactions
- Wave Mechanical Model
- Weak Acid and Base Equilibria
- Weak Acids and Bases
- Writing Chemical Formulae
- pH
- pH Change
- pH Curves and Titrations
- pH Scale
- pH and Solubility
- pH and pKa
- pH and pOH
- The Earths Atmosphere

Lerne mit deinen Freunden und bleibe auf dem richtigen Kurs mit deinen persönlichen Lernstatistiken

Jetzt kostenlos anmeldenNie wieder prokastinieren mit unseren Lernerinnerungen.

Jetzt kostenlos anmeldenAtoms are teeny-tiny little things. They are so small that millions of hydrogen atoms can fit on the tip of a needle!

Scientists can't spend all day staring at a microscope counting atoms, so we count them in groups based on **Avogadro's number**. We also use the unit of **moles.**

In this article, we will look at the many ways that **Avogadro's number and the mole** are important.

- This article is about
**Avogadro's number and the mole.** - We will look at an Avogadro's number example to calculate the number of atoms
- We will cover the importance of the mole,
**stoichiometry** - Lastly, we will learn to convert from grams to mols and from mols to atoms.

Let's start by looking at the definition of Avogadro's number and the mole.

**Avogadro's number** was calculated by Jean Perrin in 1909, and was named after Amedeo Avogadro, who originally proposed the concept. It is equal to 6.022×10^{23}, and is used in chemistry to count tiny objects like atoms or protons.

Avogadro's number is unitless. We can technically use it to count anything, but it is such a massive number that it is really only practical for small things like atoms or molecules. It is similar to how "a dozen" can refer to 12 things, but we typically use it just to count eggs. When we count atoms/molecules, we use the unit **mole** (not the animal, though they are cute!).

A **mole (mol) **is the standard unit used to count large amounts of atoms, molecules, or particles. 1 mol is equal to Avogadro's number.

The mole is the standard way to count things in chemistry. It is used in our chemical formulas and equations, and it is used to convert from AMU (atomic mass unit) to the much more practical unit of grams.

When we write chemical formulas and equations, the units we are using are mols. Let's look at a reaction.

$$2H_{2\,(g)} + O_{2\,(g)} \rightarrow 2H_2O$$

The coefficient (number in front) for each molecule is equal to the number of mols. If there is no coefficient written, it is 1. So in this case, we have 2 mols of H₂ and H_{2}O and 1 mol of O_{2}. The subscript (little number) tells us the number of moles per element. So for O_{2}, there are 2 mols of atomic oxygen, O, per 1 mol of molecular oxygen, O_{2}. Now that we know the basics, let's work on a problem.

$$Ba(OH)_{2\,(aq)}+2HCl_{aq} \rightarrow BaCl_{2\,(aq)}+2H_2O_{(l)}$$

**1) How many O atoms are in Ba(OH) _{2}? **

When we have a parenthesis, we multiply the coefficient by the whole thing, so we have 2 mols of O atoms. Next, we know that 1 mol = 6.022 x 10 ^{23}, so we have 12.044 x 10^{23} atoms of O.

2) How many H atoms are in 2H_{2}O ?

2) When counting total mols, we always need to multiply the subscript by the coefficient. There are 2 mols of H atoms *per molecule*, but since we have 2 mols of H_{2}O and not 1, we actually have 4 mols of H. That means in total we have 2.4088 x 10^{24 }(24.088 x 10^{23}) atoms of H.

Chemical reactions must follow the **law of conservation of mass**. This law says that mass cannot be created or destroyed in a reaction (i.e. our products need to have the same mass as our reactants.) When writing chemical reactions, we have to abide by this law. To achieve this, we balance the number of moles for each element using **stoichiometry (s**toichiometry is just a fancy word for the ratio between products and reactants). Here's an example:

**Balance the reaction:**

$$C_3H_{8\,(g)} + O_{2\,(g)} \rightarrow CO_{2\,(g)} + H_2O_{(l)}$$

Let's start by counting the mols on each side.

On the left side (reactants) we have: 3 mols of C, 8 mols of H, and 2 mols of O

And on the right side (products): 1 mol of C, 3 mols of O (2 from CO_{2} and 1 from H_{2}O), and 2 mols of H

So off the bat, we see that nothing is balanced. We can start by multiplying CO_{2} by 3 to balance our C

$$C_3H_{8\,(g)} + O_{2\,(g)} \rightarrow 3CO_{2\,(g)} + H_2O_{(l)}$$

Now we can balance H by multiplying H_{2}O by 4

$$C_3H_{8\,(g)} + O_{2\,(g)} \rightarrow 3CO_{2\,(g)} + 4H_2O_{(l)}$$

Last is O. Now we have 10 O on our right (6 from CO_{2} and 4 from H_{2}O), so we multiply O_{2} by 5 to get our final, balanced reaction

$$C_3H_{8\,(g)} + 5O_{2\,(g)} \rightarrow 3CO_{2\,(g)} + 4H_2O_{(l)}$$

When we do experiments, we are typically measuring in grams, not mols. So now we need to look at how we convert between these units.

The way we convert from mols to grams (or vice-versa) is by using an element's **atomic mass**.

The **atomic mass **of an element is the average mass of all isotopes of an element measured in amu. 1 amu = 1 g/mol.

When we are referring to the mass of a whole compound, it is called the **molecular weight/molar mass**. The molecular weight is the sum of the atomic masses of each element in a compound. Being able to convert from moles to grams is important when experimenting for many reasons.

One reason is that we can calculate the yield of a reaction** **(mass of products), based on the mass of reactants. It is also important since we measure things in grams, but all of our chemical equations/formulas are in mols.

**Given the compounds below, what is their molar mass?**

a. SF_{6}

b. Ba(OH)_{2}

a.$$S:32.06 \frac{g}{mol}\,\,\,F:19.00 \frac{g}{mol}$$$$32.06 \frac{g}{mol}+(19.00 \frac{g}{mol}*6)=146.06 \frac{g}{mol}$$

b. $$Ba:137.33 \frac{g}{mol}\,\,\,O:16.00 \frac{g}{mol}\,\,\,H:1.01 \frac{g}{mol}$$

$$137.33 \frac{g}{mol}+(16.00 \frac{g}{mol}*2)+(1.01 \frac{g}{mol}*2)=171.35 \frac{g}{mol}$$

Now that we know how to calculate molecular weight, let's move on to converting from gram to mols and vice versa

**How many mols of Fe _{3}(PO_{4})_{2} is 143.2 g?**

$$Fe: 55.85 \frac{g}{mol}\,\,\,P: 30.97\frac{g}{mol} \,\,\, O: 16.00 \frac{g}{mol}$$

$$(55.85\frac{g}{mol}*3)+(30.97\frac{g}{mol}*2)+(16.00\frac{g}{mol}*8)=357.49\frac{g}{mol}$$

$$\frac{143.2\,g}{357.49\frac{g}{mol}}=0.401\,mol$$

It's helpful to consider these conversions to be a ladder or staircase. You can either climb up or down, but you need to take each step to reach your destination.

By following the "steps" we can convert between units. StudySmarter Original.

For some elements/compounds, the number of atoms is equal to the number of molecules. For example, in NaCl, both Na and Cl have no subscripts (so a subscript of 1), so these values are equal.

Mols and molar mass can also be used to calculate **percent composition.**

**Percent composition** is the percentage of a compound that an element makes up by mass. The formula is:

$$\% \text{mass}=\frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}}*100\%$$

Here's an example:

What is the percent composition of H and O in H_{2}O?

The first thing we need to do is figure out the molar mass.

$$H: 1.01\frac{g}{mol}\,\,\,O: 16.00\frac{g}{mol}$$

$$(1.01\frac{g}{mol}*2)+16.00\frac{g}{mol}=18.02\frac{g}{mol}$$

Now that we have the total mass, we can determine each element's percent mass.

$$H: \frac{2.02\frac{g}{mol}}{18.02\frac{g}{mol}}*100\%=11.2\%\,\text{(because we have 2 H, we need to double the atomic mass)}$$

$$O: \frac{16.00\frac{g}{mol}}{18.02\frac{g}{mol}}*100\%=88.8\%$$

We can also use percent composition to calculate the mass of an element. So for our example above, H_{2}O has an 11.2% composition of H:

$$\frac{11.2}{100}*18.02\frac{g}{mol}=2.02\frac{g}{mol}$$

When calculating percent composition, you should make sure your percentages add to 100. For the previous example, If we forgot to multiply the mass of H by 2, the total percentage would not have equaled 100.

**Avogadro's number**is equal to 6.022 × 10^{23}and is used to count things like molecules or atoms.- 1 mole(mol) is equal to Avogadro's number and is the standard unit used to count molecules or atoms.
- The mol is important for writing and balancing chemical equations.
- We can use Avogadro's number and the mol to calculate the number of atoms or molecules when given grams.
- The formula for percent composition is: \(\% \text{mass}=\frac{\text{mass of element in 1 mol}}{\text{molar mass of compound}}*100\%\) It is used to determine what percentage an element makes up of a compound based on mass.

A mol is a unit used to count atoms, molecules, or particles. 1 mol is equal to Avogadro's number.

^{23}. It is a unitless number which is equal to one mol.

^{23}. It is useful for grouping/counting atoms and molecules.

What is the value for Avogadro's number?

6.022x10^{23}

How many moles of oxygen are in Au(NO_{3})_{3}?

9

What the molecular weight of a compound?

The total mass of the compound

What is the molecular weight of C_{6}H_{5}Br?

157.01 g/mol

How many molecules are in 3 mols of XeF_{6}? How many atoms of F?

a) 1.807 x 10^{24} molecules

b) 1.084 x 10^{25} atoms

How many grams are in 2 mols of ZnCl_{2}?

272.60 g

Already have an account? Log in

Open in App
More about Avogadro's Number and the Mole

The first learning app that truly has everything you need to ace your exams in one place

- Flashcards & Quizzes
- AI Study Assistant
- Study Planner
- Mock-Exams
- Smart Note-Taking

Sign up to highlight and take notes. It’s 100% free.

Save explanations to your personalised space and access them anytime, anywhere!

Sign up with Email Sign up with AppleBy signing up, you agree to the Terms and Conditions and the Privacy Policy of StudySmarter.

Already have an account? Log in

Already have an account? Log in

The first learning app that truly has everything you need to ace your exams in one place

- Flashcards & Quizzes
- AI Study Assistant
- Study Planner
- Mock-Exams
- Smart Note-Taking

Sign up with Email

Already have an account? Log in