Module 2: Foundations in Chemistry - Amount of substance

Sorry to anyone trying to use this that it's a little cluttered, I really only made it for personal use. If it helps you regardless though then that's great, good luck in your exams!

• Amount of a substance
• The Mole
• The amount of any substance containing as many particles as there are carbon atoms in exactly 12g of the carbon-12 isotope.
• Can refer to anything, not just atoms. 1 mole of H - 1 mole of hydrogen ATOMS. 1 mole of H2 - 1 mole of hydrogen MOLECULES.
• Number of particles in a mole of any substance.
• Number of particles = Avagadro's constant X moles of particles
• Molar mass - mass per mole of a substance.
• Units - g mol^-1
• Mass (g) = Moles (mol) X Molar Mass (g mol^-1)
• Determination of formulae
• Empirical formula - SIMPLEST whole number ratio of atoms of each element present in a compound.
• Molecular formula - ACTUAL number of atoms of each element present in a compound.
• Water of crystallisation
• Hydrated - A crystalline compund which contains water molecules in its structure.
• Anhydrous - A compund when all the water of crystallisation has been removed.
• Water of crystallisation - Water molecules that form an essential part of the crystalline structure of a compund.
• Calculating the formula of a hydrated salt -https://www.youtube.com/watch?v=t3A69QIlpwM
• Calculations of reacting masses
• Solids
• Mass (g) = Moles (mol) X Molar Mass (g mol^-1)
• Gases
• At room temperature and pressure (1 mole of gas occupies approximately 24dm^3)
• dm^3 - Volume = Moles X 24
• cm^3 - Volume = Moles X 24000
• NOT at room temperature and pressure
• pV = nRT
• p = Pressure (Pa)
• V = Volume (m^3)
• n = Moles (mol)
• R = 8.314 J mol^-1 K^-1
• T = Temperature (K)
• Moles in solution
• Moles (mol) = Concentration (mol dm^3) X Volume (dm^3)
• Moles (mol) = Concentration (mol dm^3) X (Volume (cm^3) ÷ 1000)
• Percentage yield, atom economy and the limiting reagent
• Percentage yield
• Maximum amount of products - theoretical yield. This yield is 100%.
• 100% yields are rare.
• Reaction may be at equilibrium and not go to completion.
• Side reactions may lead to by-products.
• Reactants may not be pure.
• Some reactants or some products may be left im the apparatus used in the experiment.
• Percentage yield = (actual amount of product in mol ÷ theoretical amount of product in mol) X 100
• Atom economy
• Considers desired products as well as any by-products.
• Determines efficiency of a reaction in terms of the atoms involved.
• Addition reaction - Has 100% atom economy as there are no by-products, only useful products.
• Considers desired products as well as any by-products.
• Atom economy = (molecular mass of the desired product ÷ sum of molecular masses of all products) X 100
• Limiting Reagent
• The reactant that is completely used up first.
• Caluclations are made using the limiting reagent.
• You can find out which reactant is the limiting reagent by working out the moles of each reactanr and comparing them, the limiting reagent is the one that would run out first if equal amounts of reactant were added.
• 2H2 + O2 -> 2H20
• If equal amounts were added this equation would be H2 + O2 -> H20 and the Hydrogen would be used up first.
• Standard Solutions
• Has a known concentration
• To find the volume needed: Work out the amount in moles of the solution required then calculate the mass of solute needed for the chosen volume of water,
• Amount of a substance
• The Mole
• The amount of any substance containing as many particles as there are carbon atoms in exactly 12g of the carbon-12 isotope.
• Can refer to anything, not just atoms. 1 mole of H - 1 mole of hydrogen ATOMS. 1 mole of H2 - 1 mole of hydrogen MOLECULES.
• Number of particles in a mole of any substance.
• Number of particles = Avagadro's constant X moles of particles
• Molar mass - mass per mole of a substance.
• Units - g mol^-1
• Determination of formulae
• Empirical formula - SIMPLEST whole number ratio of atoms of each element present in a compound.
• Molecular formula - ACTUAL number of atoms of each element present in a compound.
• Water of crystallisation
• Hydrated - A crystalline compund which contains water molecules in its structure.
• Anhydrous - A compund when all the water of crystallisation has been removed.
• Water of crystallisation - Water molecules that form an essential part of the crystalline structure of a compund.
• Calculating the formula of a hydrated salt -https://www.youtube.com/watch?v=t3A69QIlpwM
• Calculations of reacting masses
• Solids
• Gases
• At room temperature and pressure (1 mole of gas occupies approximately 24dm^3)
• dm^3 - Volume = Moles X 24
• cm^3 - Volume = Moles X 24000
• NOT at room temperature and pressure
• pV = nRT
• p = Pressure (Pa)
• V = Volume (m^3)
• n = Moles (mol)
• R = 8.314 J mol^-1 K^-1
• T = Temperature (K)
• Moles in solution
• Moles (mol) = Concentration (mol dm^3) X Volume (dm^3)
• Moles (mol) = Concentration (mol dm^3) X (Volume (cm^3) ÷ 1000)
• Percentage yield, atom economy and the limiting reagent
• Percentage yield
• Maximum amount of products - theoretical yield. This yield is 100%.
• 100% yields are rare.
• Reaction may be at equilibrium and not go to completion.
• Side reactions may lead to by-products.
• Reactants may not be pure.
• Some reactants or some products may be left im the apparatus used in the experiment.
• Percentage yield = (actual amount of product in mol ÷ theoretical amount of product in mol) X 100
• Atom economy
• Determines efficiency of a reaction in terms of the atoms involved.
• Addition reaction - Has 100% atom economy as there are no by-products, only useful products.
• Atom economy = (molecular mass of the desired product ÷ sum of molecular masses of all products) X 100
• Limiting Reagent
• The reactant that is completely used up first.
• Caluclations are made using the limiting reagent.
• You can find out which reactant is the limiting reagent by working out the moles of each reactanr and comparing them, the limiting reagent is the one that would run out first if equal amounts of reactant were added.
• 2H2 + O2 -> 2H20
• If equal amounts were added this equation would be H2 + O2 -> H20 and the Hydrogen would be used up first.
• Acid-base titrations
• Volumetric analysis - You measure the volume of one solution that reacts with a measured volume of a second solution.
• Acid-base titrations are a type of volumetric analysis, you react a solution of an acid with a solution of a base in the presence of an indicator.
• Indicator must be a different colour in the acid solution and the base solution.
• When the indicator changes colour we assume that all of the moles of acid have reacted with all of the moles of base.
• Untitled

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