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Stoichiometry in Reactions

- Chemical Reactions
- Inorganic Chemistry
- Catalysts
- Chlorine Reactions
- Group 2
- Group 2 Compounds
- Halogens
- Ion Colours
- Period 3 Elements
- Period 3 Oxides
- Periodic Table
- Periodic Trends
- Properties of Halogens
- Properties of Transition Metals
- Reactions of Halides
- Reactions of Halogens
- Shapes of Complex Ions
- Test Tube Reactions
- Titrations
- Transition Metal Ions in Aqueous Solution
- Transition Metals
- Variable Oxidation State of Transition Elements
- Ionic and Molecular Compounds
- Bond Length
- Bonding and Elemental Properties
- Intramolecular Force and Potential Energy
- Lewis Dot Diagrams
- Limitations of Lewis Dot Structure
- Polar and Non-Polar Covalent Bonds
- Resonance Chemistry
- Sigma and Pi Bonds
- The Octet Rule
- Types of Chemical Bonds
- VSEPR
- Organic Chemistry
- Acylation
- Alcohol Elimination Reaction
- Alcohols
- Aldehydes and Ketones
- Alkanes
- Alkenes
- Amide
- Amines
- Amines Basicity
- Amino Acids
- Anti-Cancer Drugs
- Aromatic Chemistry
- Benzene Structure
- Biodegradability
- Carbon -13 NMR
- Carbonyl Group
- Carboxylic Acids
- Chlorination
- Chromatography
- Column Chromatography
- Combustion
- Condensation Polymers
- Cracking (Chemistry)
- Elimination Reactions
- Esters
- Fractional Distillation
- Gas Chromatography
- Halogenoalkanes
- Hydrogen -1 NMR
- IUPAC Nomenclature
- Infrared Spectroscopy
- Isomerism
- NMR Spectroscopy
- Nucleophilic Substitution Reactions
- Optical Isomerism
- Organic Analysis
- Organic Compounds
- Organic Synthesis
- Oxidation of Alcohols
- Ozone Depletion
- Paper Chromatography
- Polymerisation Reactions
- Preparation of Amines
- Production of Ethanol
- Properties of Polymers
- Reaction Mechanism
- Reactions of Aldehydes and Ketones
- Reactions of Alkenes
- Reactions of Benzene
- Reactions of Carboxylic Acids
- Reactions of Esters
- Structure of Organic Molecules
- Synthetic Routes
- Thin Layer Chromatography Practical
- Thin-Layer Chromatography
- Understanding NMR
- Uses of Amines
- Physical Chemistry
- Acid Dissociation Constant
- Acid-Base Indicators
- Acids and Bases
- Amount of Substance
- Application of Le Chatelier's Principle
- Arrhenius Equation
- Atom Economy
- Atomic Structure
- Avogadro Constant
- Beer-Lambert Law
- Bond Enthalpy
- Bonding
- Born Haber Cycles
- Born-Haber Cycles Calculations
- Brønsted-Lowry Acids and Bases
- Buffer Capacity
- Buffer Solutions
- Buffers
- Buffers Preparation
- Calculating Equilibrium Constant
- Calorimetry
- Carbon Structures
- Chemical Equilibrium
- Chemical Kinetics
- Chemical Thermodynamics
- Collision Theory
- Constant Pressure Calorimetry
- Covalent Bond
- Determining Rate Constant
- Deviation From Ideal Gas Law
- Dilution
- Dipole Chemistry
- Distillation
- Dynamic Equilibrium
- Electric Fields Chemistry
- Electrochemical Cell
- Electrochemical Series
- Electrochemistry
- Electrode Potential
- Electrolysis
- Electromagnetic Spectrum
- Electron Configuration
- Electron Shells
- Electronegativity
- Elemental Composition of Pure Substances
- Empirical and Molecular Formula
- Energetics
- Enthalpy Changes
- Entropy
- Equilibrium Concentrations
- Equilibrium Constant Kp
- Equilibrium Constants
- Examples of Covalent Bonding
- Factors Affecting Reaction Rates
- Free Energy
- Fundamental Particles
- Ground State
- Half Equations
- Heating Curve for Water
- Hess' Law
- Hybrid Orbitals
- Ideal Gas Law
- Ideal and Real Gases
- Intermolecular 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
- Magnitude of Equilibrium Constant
- Mass Spectrometry
- Mass Spectrometry of Elements
- Maxwell-Boltzmann Distribution
- Metallic Bonding
- Metallic Solids
- Molar Mass Calculations
- Molarity
- Molecular Orbital Theory
- Oxidation Number
- Partial Pressure
- Particulate Model
- Percentage Yield
- Photoelectric Effect
- Physical Properties
- Polarity
- Polyprotic Acid Titration
- Properties of Equilibrium Constant
- Properties of Solids
- Properties of Water
- Rate Equations
- Reaction Quotient
- Reaction Quotient and Le Chatelier's Principle
- Real Gas
- Redox
- Relative Atomic Mass
- Representations of Equilibrium
- Reversible Reaction
- SI units chemistry
- Shapes of Molecules
- Solids Liquids and Gases
- Solubility Product
- Solubility Product Calculations
- Solutions and Mixtures
- States of Matter
- Stoichiometry in Reactions
- Strength of Intermolecular Forces
- Thermodynamically Favored
- Trends in Ionisation Energy
- VSEPR Theory
- Water in Chemical Reactions
- Weak Acids and Bases
- pH
- pH Curves and Titrations
- pH Scale
- pH and Solubility
- pH and pKa
- pH and pOH

In this article, you will learn how stoichiometry is used in reactions, more importantly how avogadro's constant is exploited for different means. Stoichiometry is crucial to all parts of chemistry, as without specific calculations, different experiments would not be possible. Stoichiometry underpins the whole of chemistry.

- In this article, you will learn what is stoichiometry and its definition
- Avogadro and his constant (the mole concept)
- Calculations involving the mole concept using gases, volumes, and masses
- Example calculations using mole formulae

What is Stoichiometry?

**Stoichiometry**: the relationship between the reacting quantities of compounds.

Here we will explore stoichiometry in the context of **the mole** concept and Avogadro's constant, the **ideal gas equation**, and **mass spectra**. This will allow us to make** balanced equations**, and determine **empirical and molecular formulae**.

Where do we start thinking about "how much" of a substance is in a particular weighted amount? This was a question puzzling scientists for centuries, one being Amedeo Avogadro.

Avogadro began to look at the relationship between the atomic number of atoms and the amount of atoms present. **Atomic number** defines the amount of protons present in the nucleus of the atom in question.

You can find out the **atomic number** of any compound by looking at its positioning in the periodic table. Elements in the periodic table are arranged by increasing atomic number.

Suppose we consider carbon with atomic number of 12. How many atoms are there in 12 grams of carbon-12? Avogadro found the number to be 6.02214179 x 10^{23}.

**Avogadro's constant** (or **Avogadro's number**) is 6.02214179 x 10^{23}

This constant is often simplified to 6.022 x 10^{23} for making calculations simpler. The constant is **unitless** as it refers to a specific amount, so you can say 6.02214179 x 10^{23} atoms if needed.

Please note when using the rounded form of Avogadro's number (6.022 x 10^{23}) you need to be mindful how this will affect your future calculation and manipulation of errors.

**The mole** is a concept extrapolated from Avogadro's constant. The number Avogadro discovered (6.02214179 x 10^{23}) is often referred to as a mole.

1 mole = 6.02214179 x 10^{23} atoms

This is a simple analogy you use in daily life, such as a "dozen of ..." or a "pair of ...". Here you can use it exchangeably in chemical contexts. Such as "one mole of carbon-12" will refer to 6.02214179 x 10^{23} of carbon-12 atoms.

Here we will explore how to use the defined concept of the mole for calculations and conversions involving mass and concentrations.

How do we know how many moles is in a particular weighted amount? We need to know precisely this to measure out the precise mass, and thus the exact amount of moles we need for balanced reactions to take place accordingly.

The amount of moles is determined by the mass divided by the relative atomic mass of the element.

\[ \mbox{moles (n)} = \frac{ \mbox{mass in grams}}{ \mbox{atomic relative mass in grams per mole}} \]

or simply \( n= \frac{m}{A_r} \)

Can you see how the highlighted units in the equation above cancel out to give a number of moles?

In fact, the **relative atomic mass** of elements is related to the 1/12 of the grams of one mole of carbon-12 modules. Hence the units of relative atomic mass are grams per mole.

In fact, you can use the formula above to d**etermine the amount of moles of any substance** or compound, by replacing the relative atomic mass to the molar mass of the compound.

\[ \mbox{moles (n)} = \frac{ \mbox{mass in grams}}{ \mbox{molar mass in grams per mole}} \]

or simply \( n= \frac{m}{M_r} \)

With the formula above, you can find the amount of moles of any compound to use in reactions and experiments.

Alternatively, you can also manipulate the formula to **determine the mass of a compound** for a specific reaction. This is especially useful if you have a specific number of moles you need for an experiment and you need to know how much to weight out a specific element or substance.

\[ \mbox{mass in grams} = \mbox{moles (n)} \times \mbox{molar mass in grams per mole} \]

\[ \mbox{molar mass in grams per mole} = \mbox{mass in grams)} \times \mbox{moles (n)} \]

You can use the concept of a mole to **calculate the concentrations of solutions**. Instead of using the formulae above with the molar mass or relative atomic mass, we can use a different formula that is based on volumes and concentrations.

To start off, how do we define **concentration**? It is based on the amount of substance in a particular volume of liquid. For a solution, it would be a precise amount of **solute** in an exact volume of **solvent**. To put moles into this concept, we can write out the equation:

\[ \mbox{ concentration in mol dm} ^{-3} = \frac{ \mbox{moles (n)}}{ \mbox{volume in dm}^3} \]

or simply \( C= \frac{n}{V} \)

The unit dm^{3} refers to 1 liter. Therefore 1 dm^{3} = 1 L = 1000 mL

Alternatively, you can use the above formula to determine the amount of moles in a particular solution from a known concentration.

\[ \mbox{moles (n)} = \mbox{concentration in mol dm} ^{-3} \times \mbox{volume in dm}^3 \]** **

This concept is well described with the infographic on the side. You can use it to know the formula of the value you are looking to calculate, be it moles or concentration or volume.

So how does the mole concept fit into calculations that are not based on mass or concentrations, but rather gases? Here we will explore the **ideal gas law** and how it applies to **the mole concept**.

Think about gases, what does the volume they occupy depend on? That's the **pressure and temperature**. An ideal gas' volume will proportionally and linearly increase with increasing temperature or decreasing pressure.

Considering the behaviours of ideal gases based on temperature and pressure, a formula can be made (which we call the ideal gas law or **general gas law**):

\[ pV=nRT \]

In the formula above p = pressure in pascals (**Pa**), V = volume in **m**^{3}, n = number of **moles**, R = is the gas constant (**8.31 JK ^{-1} mol**

The gas constant (**R = ****8.31 JK ^{-1} mol**

You can use the above equation to calculate some physical constant regarding a specific gaseous atom or compound, be it the volute from the number of moles known under standard conditions, or the other way round.

Here we will explore **examples of stoichiometry in reactions**. Stoichiometry can be used in different types of reactions and below are some common problems and stoichiometry calculations in **chemical reactions** you will come across.

We will explore how to make **calculations relating to converting mass** and weighted amounts into moles, and vice versa.

**How many moles are in 20 grams of NaOH? **

The molar mass of NaOH is 39.997 g/mol, which you can round to 40 g/mol.

Now divide the weighted amount by the molar mass:

\( \frac{20g}{40gmol^{-1}} = 0.5mol \)

Thus there are 0.5 moles in 20 grams of NaOH.

Now consider if you need to weigh out a specific amount for a reaction. Take the example below as a reference.

**You need 0.75 moles of magnesium (Mg) for a reaction. How many grams should you weight out? **

The relative atomic mass of Mg is 24.305 g/mol, which you can round off to 24.

You can use the first infographic to determine which formula you need to use.

Now multiply the number of moles by the relative atomic mass.

\( 0.75mol \times 24gmol^{-1} = 18g \)

Thus, you need to weigh out 18 grams of Mg for the reaction.

Great! Now you can use the same principle to calculate concentrations and volumes for solutions.

Here we will explore how to **calculate concentrations** from a given amount of moles, or vice versa.

**What is the concentration of a solution consisting of 0.5 moles of NaCl in 0.25 liters? **

Use the second infographic to determine which formula you can use.

Divide the amount of moles by the volume.

\( \frac{0.5mol}{0.25L} = 2molL^{-1} \)

Thus the concentration is 2 mol dm^{-3}

Explore how you can use the ideal gas law to **calculate the volume of any gas** if the conditions are provided or mentioned.

**If a gas takes up 10 cubic meters of volume, how many moles does it have? Assume conditions of 101 kPa and 300K.**

Use the equation \( pV=nRT \) , to rearrange it into \( n= \frac{pV}{RT} \)

Plug in the numbers: \( n= \frac{101 \times 10}{8.314 \times 300} \)

n = 0.405 moles

Thus you can manipulate the basic general gas formula to get any desired unknown, be it the volume, moles, or even the conditions the gas in under.

- Stoichiometry is the measurement of substances and compounds based on the mole concept and physical properties.
- important formulae to remember
- \( \mbox{moles (n)} = \frac{ \mbox{mass in grams}}{ \mbox{atomic relative mass in grams per mole}} \)
- \( \mbox{ concentration in mol dm} ^{-3} = \frac{ \mbox{moles (n)}}{ \mbox{volume in dm}^3} \)
- \( pV=nRT \)

You can manipulate the formulae to know how to calculate any unknown for chemical reactions involving stoichiometry.

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Calculate stoichiometry in reactions using formulas based on the mole concept.

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