EL 4

• Chris Harris: 'hello and welcome to this video on formula equations and amount of substance my name is Chris Harris and I'm from a Larry tutors comm and this video the whole point of this video is to give you an overview of this part of the elements for life topic for OCR be Salters and basically it's just designed provision to give you an idea of an overview and because it's revision and the slides that I'm using here you can have access to them as well and you can purchase them you just click on the link in the description box below and you can get ahold of them there they're really good value for money and and and can really kind of compliment your revision and help you to get that a star or a in chemistry okay so like I say this is dedicated to the Psalter specification and it meets these points taken from the syllabus to okay so just have a quick look at the atom and structure of the atom and because we're going to talk about obviously moles and and and and the like so we need to know a little bit more about the structure of the atom so just a nutshell we have a nucleus it's really small it contains protons and neutrons we have electrons that sit in orbitals and these obviously orbit the nucleus around it we need to know the charges of these things as well protons have a relative charge of plus one so that positive a massive one neutrons are neutral and they have a massive one electrons are minus one the relative charge is minus one and their relative mass is one over two thousand so we can't say it's zero because they do have a mass it's just very very light or very low and obviously in the periodic table you'll see your elements and have written are like this the bigger number is the mass number tells number of protons and neutrons in the nucleus and the bottom number is called the atomic or proton number and this tells us the number of protons in the nucleus and remember in elements the number of protons equals the number of electrons so this can also be the number of electrons in elements okay so we need to know that with our ions and isotopes so I on to have a different number electrons and protons see here oxygen has got gained two electrons this is a negative ion and it would then combine with the positive ions to form a stable compound but if you have a look at the number of protons in here number protons is eight in oxygen and as a charge of plus eight neutrons is eight but they don't have a charged electron this has 10 electrons a charge of minus two so overall it has a charge of minus a minus 10 sorry so overall it has a charge of minus T and which is this is why it's owed to minus looking at positive ions something like sodium sodium has 11 protons in the in the middle of it which is a charge of plus 11 12 these others which have a charge of zero and it has 10 electrons which means it has a charge of minus 10 so overall the overall charge is +1 so positive ions actually lose an electron to form the positive ion that they are here and this is an example which is sodium isotopes these are elements with the same number of protons but they have a different number of neutrons so you can see here that we've got a set of 3 isotopes here these are all carbon isotopes and if we have a look here with the same number of protons here across all of them and so we've got 6 protons and each cousin so you see the program at the bottom is the same but the number of neutrons is different you can see here that we've got because it's gone 12 13 14 we've got extra neutrons obviously in these isotopes up here so these ones are the isotopes of each other because they have the same number of protons but a different number of neutrons okay you've got to know your definitions as well you'll see and these might see these crop up in you exam so the relative atomic masses the AR this is the weighted mean mass of an atom of an element compared to 1/12 of the mass of an atom of carbon-12 so make sure you know that definition make sure you know the definition of relative isotopic mass as well this is just the mass not amine mass is just a mass of an atom of an isotope compared to 1/12 of the mass of an apple carbon-12 so it's always this same bit at the end here and relative molecular mass or formula mass so this is the mean mass of a molecule compared to on 12 of the mass of an atom of carbon-12 so it's a molec'
• Atomic mass

  Mass of an atom of an isotope compared to 1/12 of the mass of an apple carbon-12
• Calculating the number of moles in a solution

  Number of moles = Concentration in moles per cubic meter x Volume in decimeters cubed
• Relative molecular mass or formula mass

  Mean mass of a molecule compared to 1/12 of the mass of an atom of carbon-12
• Calculating the number of particles in a substance

  Number of particles = Avogadro's number x number of moles
• One mole of any substance will contain 6.02 x 10^23 atoms or molecules, known as Avogadro's number
• Calculating the number of moles from mass and MR or AR

  Number of moles = Mass in grams / MR or AR
• Calculating theoretical mass from a balanced equation

  Write out the balanced equation, calculate the theoretical mass based on the given mass of reactants
• Times ten to the minus three is equivalent to dividing by a thousand
• Writing ionic equations

  Splitting acids, bases, and salts into their ions. Cancel out ions that appear on both sides to simplify the equation. Ensure charges balance
• State symbols in equations: S (solids), L (liquids), G (gases), aq (aqueous - dissolved in water)
• Calculating number of moles

  Number of moles = concentration x volume. Convert volume to decimetres cubed by multiplying by 10 to the power of -3
• When calculating moles, it's useful to work out the moles of something first as it can help in various calculations
• Calculating percentage yield

  Work out the theoretical mass, balance the equation, determine the molar masses of the species involved, convert masses to grams, divide the mass of the desired product by the molar mass of the reactant, and calculate the maximum amount of product that can be produced
• Determining empirical formula


  Write out the elements involved, convert percentages to masses, divide by the relative atomic mass to get the number of moles, find the ratio of moles, and write the final formula using the smallest ratio
• Empirical formula calculation for a hydrocarbon

  Write down the combustion products (carbon dioxide and water) and their masses

  2. Calculate the number of moles of each molecule by dividing the masses by their relative molecular masses

  3. Determine the number of moles of carbon atoms and hydrogen atoms based on the moles of carbon dioxide and water

  4. Find the ratio of carbon to hydrogen atoms by dividing the number of atoms by the smallest number of moles

  5. The empirical formula is derived from the ratio of carbon and hydrogen atoms, e.g., CH
• Calculation of water of crystallization

  Heat the hydrated salt to remove water of crystallization

  2. Weigh the anhydrous compound after heating

  3. Subtract the mass of the anhydrous compound from the mass of the hydrated compound to find the mass of water lost

  4. Determine the value of X (amount of water of crystallization) by comparing the mass of water lost to the total mass of the hydrated compound
• Conical flasks used for standard solutions can typically hold 250 cm³ or 250 ml
• Actual yield is the amount of product obtained in an experiment
• In a reaction involving the complete combustion of calcium, if 32.6 grams of calcium oxide was produced and the theoretical mass is 47.6 grams, the percentage yield is calculated as 32.6 / 47.6 * 100 = 68.5%
• Calculating percentage yield

  1. Actual yield divided by theoretical yield multiplied by 100

  2. Theoretical yield is the amount of product produced assuming no products are lost and all reactants react fully
• Standard solutions are used in titrations and have a known concentration
• Calculating water content in calcium sulfate

  1. Subtracting the mass of anhydrous calcium sulfate from the mass of hydrated calcium sulfate to find the mass of water present

  2. Dividing the masses by the relative molecular masses of calcium sulfate and water to determine the number of moles

  3. Dividing the number of moles by the smallest number of moles to find the ratio of water molecules to calcium sulfate molecules

  4. Determining the value of x as 5, indicating the number of water molecules in the formula caso4 · 5H2O
• Percentage yield is never 100% due to losses in the reaction process
• Making a standard solution

  Weighing out the precise amount of solid, dissolving it in water to a fixed volume in a conical flask
• Standard solutions can be made from solids or more concentrated liquids dissolved in water to a fixed volume
• Theoretical yield is the maximum amount of product that can be obtained in a reaction
• Percentage yield is calculated as actual yield divided by theoretical yield multiplied by 100
• The mass and volume of the standard solution are used to calculate the concentration
• Preparing a standard solution from a solid

  1. Weigh out the amount of solid precisely using a balance and a plastic or glass weighing boat

  2. Transfer the solid from the weighing boat into a beaker and wash any solid left behind into the beaker using deionized water

  3. Dissolve the solid fully in deionized water by stirring with a glass rod

  4. Transfer the dissolved solid into a volumetric flask using a funnel to avoid spillage

  5. Fill the volumetric flask with deionized water to the graduation line, using a pipette near the line for accuracy

  6. Mix the solution by inverting the flask to ensure even distribution
• The relative formula mass of sodium hydroxide is 40
• Calculating the mass of solid sodium hydroxide required for a specific solution

  1. Calculate the number of moles of the solution by multiplying the concentration by the volume

  2. Calculate the mass by multiplying the number of moles by the relative formula mass of sodium hydroxide
• Deionized water is used because it doesn't contain any ions that could interfere with the reaction
• Calculating the volume of a more concentrated solution to use in making a standard solution

  Volume to use = Final concentration / Initial concentration x Volume required
• Final concentration

  Concentration of the final solution needed
• Example of dilution calculation

  • Given values: 250 cm^3 solution of 0.75 mol/dm^3 HCl from 2.5 mol/dm^3 HCl
• Indicators used in titrations

  • Phenolphthalein and methyl orange, know how they change in acids and alkalis
• Readings in titrations should be taken from the bottom of the meniscus and at eye level, recorded to two decimal places, and repeated for concordant results
• Lab technicians need to know how to make up solutions and substances, often diluting concentrated solutions to required strengths
• Initial concentration


  Concentration of the solution being diluted down