0620 · 3.3The mole and the Avogadro constant

Chemical quantities and concentration basics

0:00A balance measures grams, but chemical reactions happen between particles. Ten grams of magnesium and ten grams of oxygen do not contain equal numbers of atoms. To predict a reaction, we need a bridge between the mass we measure and the particles taking part. That bridge is the mole, and it connects almost every calculation in this video. We will use this idea to calculate reacting masses, gas volumes and solution concentrations.

0:23It also explains how titration results reveal an unknown concentration. Finally, we will work out chemical formulae, percentage yield and percentage purity. Rather than treating these as separate tricks, we will build them from the same particle counting method. You need chemical formulae, balanced equations and relative masses as starting knowledge. A formula tells us which atoms are present and their proportions. Relative molecular mass adds the relative atomic masses of every atom in a molecule.

0:51For ionic compounds, we use relative formula mass because their formula describes a ratio, not a separate molecule. Concentration tells us how much solute is present in a particular volume of solution. The solute is the dissolved substance, and the solution includes both that substance and its solvent. Concentration can be measured in grams per cubic decimetre or moles per cubic decimetre. These describe mass or amount of substance within the same volume.

1:16A concentration of ten grams per cubic decimetre means each cubic decimetre of solution contains ten grams of dissolved substance. It does not mean ten grams was added to one cubic decimetre of water. Dissolving can change the final volume, so calculations use the volume of solution. A more concentrated sample contains more solute within the same volume. This next part is for the Extended paper.

Moles, molar mass and particles

1:39The mole is the unit of amount of substance, and its unit symbol is shown on screen. One mole contains six point zero two times ten to the twenty-three particles. This number is the Avogadro constant, and the particles can be atoms, ions or molecules. Think of a mole as a counting unit, rather like a dozen. A dozen always means twelve objects, whether they are eggs or pencils. Similarly, a mole always represents the same enormous number of specified particles.

2:06Different particles have different masses, so equal amounts in moles do not necessarily weigh the same. Always identify the particles being counted. One mole of water contains one mole of water molecules, but each molecule contains two hydrogen atoms. Therefore, that sample contains two moles of hydrogen atoms and one mole of oxygen atoms. For sodium chloride, the formula describes equal numbers of sodium ions and chloride ions. Molar mass is the mass per mole of a substance, measured in grams per mole.

2:34Its numerical value matches the relative atomic, molecular or formula mass. Water has a relative molecular mass of eighteen, so its molar mass is eighteen grams per mole. The values match, but the quantities differ because relative masses have no units. Amount of substance equals mass divided by molar mass. Dividing asks how many mole sized portions fit into the measured mass. To calculate mass instead, multiply the amount by the mass of each mole.

3:01If mass and amount are known, dividing the first by the second gives the molar mass. Suppose an exam asks you to calculate the amount in nine point zero zero grams of water. Its molar mass is eighteen point zero grams per mole. Divide nine point zero zero grams by eighteen point zero grams per mole. The result is zero point five zero zero moles, which makes sense because the sample weighs half the mass of one mole.

3:24Now calculate the mass of zero point two five zero moles of carbon dioxide. Adding twelve and twice sixteen gives a molar mass of forty-four grams per mole. Multiply zero point two five zero moles by forty-four grams per mole. The answer is eleven point zero grams, because this sample contains one in four of a mole. The same relationship can reveal an unknown relative atomic mass. Suppose zero point zero two zero zero moles of an element’s atoms have a mass of zero point five four zero grams.

3:51Dividing those values gives a molar mass of twenty-seven point zero grams per mole. Its relative atomic mass is therefore twenty-seven point zero, with no unit. For a molecular substance, suppose zero point zero five zero zero moles weigh four point four zero grams. Dividing mass by amount gives eighty-eight point zero grams per mole. The relative molecular mass is therefore eighty-eight point zero. Exactly the same calculation gives relative formula mass for an ionic substance, provided the amount refers to its formula units.

4:21To calculate a number of particles, multiply the amount by the Avogadro constant. Our zero point two five zero moles of carbon dioxide gives one point five zero five times ten to the twenty-three molecules. Rounded to three significant figures, that is one point five one times ten to the twenty-three molecules. These are molecules, not the total number of atoms. The reverse calculation divides the particle count by the Avogadro constant.

4:46Suppose a sample contains three point zero one times ten to the twenty-three atoms. Division gives zero point five zero zero moles of atoms. Notice that particles are a count, while moles express amount of substance; they are connected but are not interchangeable units. This next part is for the Extended paper.

Gas volumes and reacting masses

5:03At room temperature and pressure, one mole of any gas occupies twenty-four cubic decimetres for these calculations. This is the molar gas volume. Equal gas volumes under the same conditions therefore contain equal amounts in moles, even when their masses differ. Gas volume divided by twenty-four cubic decimetres per mole gives the amount. Multiplying the amount by that molar volume works in the opposite direction. The condition matters because gas volume changes with temperature and pressure.

5:30Do not use this gas relationship for liquids, solids or solutions simply because their volumes are given. Calculate the amount in six hundred cubic centimetres of gas at room temperature and pressure. First divide by one thousand to obtain zero point six zero zero cubic decimetres. Then divide that volume by twenty-four cubic decimetres per mole. The result is zero point zero two five zero moles, with the conversion completed before using the formula.

5:55In the opposite direction, zero point one five zero moles of gas occupies three point six zero cubic decimetres at room temperature and pressure. We obtain this by multiplying zero point one five zero by twenty-four. Multiplying the answer by one thousand gives three thousand six hundred cubic centimetres. Keep the requested volume unit visible throughout your working. A balanced equation gives reacting ratios in particles, and therefore in moles.

6:19It does not usually give a ratio in grams because different substances have different molar masses. This explains the standard calculation route: convert the known quantity into moles, use the equation, then convert to the requested quantity. The middle step connects reactant and product. Magnesium plus oxygen gives magnesium oxide, and the balanced symbol equation is on the board. Two moles of magnesium react with one mole of oxygen molecules to produce two moles of magnesium oxide.

6:47Suppose two point four zero grams of magnesium react completely with excess oxygen. Excess means enough oxygen is available to use up all the magnesium. Divide two point four zero grams by twenty-four grams per mole to obtain zero point one zero zero moles of magnesium. The equal coefficients give zero point one zero zero moles of magnesium oxide. Multiplying by forty grams per mole produces four point zero zero grams.

7:11The product weighs more than the original metal because oxygen has joined it. Only half as many moles of oxygen molecules react, giving zero point zero five zero zero moles. Multiplying by thirty-two grams per mole gives one point six zero grams of oxygen. Adding the reacting masses gives four point zero zero grams. This confirms conservation of mass without incorrectly assuming that the equation coefficients represent grams. A limiting reactant is used up first and determines the maximum amount of product.

7:39Magnesium plus hydrochloric acid gives magnesium chloride plus hydrogen. The equation shows that one mole of magnesium requires two moles of hydrochloric acid. Suppose we have two point four zero grams of magnesium but only zero point one five zero moles of acid. The magnesium amount is zero point one zero zero moles. Complete reaction would require zero point two zero zero moles of hydrochloric acid. Only zero point one five zero moles is available, so the acid is limiting.

8:07Comparing masses alone would not identify this, because the balanced equation requires a particular mole ratio. Divide the available acid amount by two to find zero point zero seven five zero moles of magnesium reacting. The equation gives the same amount of hydrogen. Multiply zero point zero seven five zero moles by twenty-four cubic decimetres per mole. The maximum hydrogen volume is one point eight zero cubic decimetres at room temperature and pressure.

8:33Some magnesium remains because the acid runs out first. Subtracting zero point zero seven five zero from zero point one zero zero gives zero point zero two five zero moles. Multiplying by twenty-four grams per mole gives zero point six zero zero grams remaining. Always base the product calculation on the limiting reactant, not whichever starting quantity seems easier. This next part is for the Extended paper.

Solution concentration and titration

8:55Solution calculations use the same idea of dividing a total quantity by volume. Dividing the mass of solute by the solution volume gives a concentration in grams per cubic decimetre. For moles per cubic decimetre, the amount of solute in moles is used instead. Multiplying either concentration by the volume recovers the total quantity. A solution concentration does not tell us the total amount unless its volume is also known.

9:20Two samples can have identical concentrations but different amounts of solute. A larger sample contains more because there is more solution at that concentration. This is why a concentration must be multiplied by volume before using a reacting mole ratio. Suppose five point zero zero grams of sodium chloride is present in two hundred and fifty cubic centimetres of solution. Convert the volume to zero point two five zero cubic decimetres.

9:44Dividing the mass by that volume gives twenty point zero grams per cubic decimetre. The sample is one in four of a cubic decimetre, so a full cubic decimetre would hold four times that mass. For amount concentration, suppose zero point zero five zero zero moles is dissolved in the same solution volume. Two hundred and fifty cubic centimetres equals zero point two five zero cubic decimetres. Dividing the amount by that volume gives zero point two zero zero moles per cubic decimetre.

10:11Both concentration calculations describe how much solute is present per unit volume. To convert between the concentration units, use molar mass. Sodium hydroxide has a molar mass of forty point zero grams per mole. Multiplying zero point two zero zero moles per cubic decimetre by this value gives eight point zero zero grams per cubic decimetre. Dividing eight point zero zero by forty point zero reverses the conversion. Calculate the amount in one hundred and fifty cubic centimetres of this sodium hydroxide solution.

10:39Convert the volume to zero point one five zero cubic decimetres. Multiply that by zero point two zero zero moles per cubic decimetre. The solution contains zero point zero three zero zero moles of solute, ready to use in a reaction calculation. To find the volume containing zero point zero four zero zero moles, divide by the concentration. Zero point zero four zero zero divided by zero point two zero zero gives zero point two zero zero cubic decimetres.

11:06Multiplication by one thousand converts this to two hundred cubic centimetres. A lower concentration would require a larger volume to supply the same amount. A titration measures the volume required for substances in solution to react in their balanced equation ratio. One known concentration then allows an unknown amount or concentration to be calculated. Experimental readings can appear in practical papers as well as calculation questions. The detailed method belongs in the video on acid base titrations.

11:33Sulfuric acid of concentration zero point one zero zero moles per cubic decimetre is run from a burette into sodium hydroxide solution. The flask holds twenty-five point zero cubic centimetres of the alkali. The burette reads one point two zero cubic centimetres at the start and twenty-one point two zero at the end-point. The titre is the delivered volume, so subtracting the initial reading from the final one gives twenty point zero zero cubic centimetres.

11:58Sulfuric acid plus sodium hydroxide gives sodium sulfate plus water. Convert the acid volume to zero point zero two zero zero zero cubic decimetres. Multiplying by its concentration gives two point zero zero times ten to the minus three moles of acid. One mole of this acid reacts with two moles of sodium hydroxide. The alkali amount is therefore four point zero zero times ten to the minus three moles.

12:22Convert the sodium hydroxide volume to zero point zero two five zero cubic decimetres. Divide four point zero zero times ten to the minus three moles by this volume. Its concentration is zero point one six zero moles per cubic decimetre. We used the equation ratio between the amount calculations, rather than assuming equal volumes or equal concentrations must react. The same titration data can instead predict a required solution volume.

12:45Suppose a different sodium hydroxide solution, of concentration zero point two zero zero moles per cubic decimetre, must neutralise the same acid. Dividing the required four point zero zero times ten to the minus three moles by this gives zero point zero two zero zero cubic decimetres. That equals twenty point zero cubic centimetres. This next part is for the Extended paper.

Empirical and molecular formulae

13:06An empirical formula shows the simplest whole number ratio of atoms of each element in a compound. A molecular formula shows the actual number of atoms of each element in a molecule. Because formulae describe atom ratios, experimental masses must first be converted into amounts in moles. A compound contains two point four zero grams of carbon, zero point six zero zero grams of hydrogen and three point two zero grams of oxygen.

13:32Dividing carbon’s mass by twelve gives zero point two zero zero moles of atoms. Hydrogen’s mass divided by one gives zero point six zero zero moles. For oxygen, dividing by sixteen gives zero point two zero zero moles. Divide every amount by the smallest value, zero point two zero zero moles. This gives one for carbon, three for hydrogen and one for oxygen. The empirical formula is CH₃O. Comparing the original masses directly would be wrong because a gram of hydrogen contains many more atoms than a gram of oxygen.

14:02If the question gives percentages instead, imagine a one hundred gram sample. Each percentage then becomes the corresponding mass in grams, and the same method applies. A ratio containing one point five needs every value multiplied by two to produce whole numbers. Do not round a genuine fractional ratio to the nearest integer, because that changes the composition. Suppose this compound has a relative molecular mass of sixty-two. The empirical formula mass is twelve plus three plus sixteen, giving thirty-one.

14:30Divide sixty-two by thirty-one to obtain a multiplier of two. Doubling every atom count gives the molecular formula C₂H₆O₂, while preserving the original elemental ratio. This next part is for the Extended paper.

Percentage yield, purity and recap

14:43Percentage composition, yield and purity all compare a part with a relevant total. Divide the part by that total, then multiply by one hundred. Their equations look similar, but selecting the correct quantities matters more than remembering their arrangement. Percentage composition by mass describes how much of a compound’s mass comes from a particular element. Water has a relative molecular mass of eighteen, including sixteen contributed by oxygen. Dividing sixteen by eighteen and multiplying by one hundred gives eighty-eight point nine per cent oxygen.

15:13The two hydrogen atoms do not outweigh oxygen simply because they are more numerous. Theoretical yield is the maximum product predicted from the balanced equation and the limiting reactant. Actual yield is the quantity obtained experimentally. Our magnesium oxide calculation predicted four point zero zero grams, but suppose only three point two zero grams is collected. Divide three point two zero by four point zero zero, then multiply by one hundred.

15:37The percentage yield is eighty point zero per cent. A yield below one hundred per cent can result from incomplete reaction or product lost during collection. An apparent value above one hundred per cent suggests extra material, such as retained water or impurities. It does not mean more pure product formed than the reactants could supply. Compare actual and theoretical quantities using matching units. Percentage purity compares the mass of the desired substance with the total sample mass.

16:04Suppose five point zero zero grams of impure calcium carbonate reacts with excess hydrochloric acid. Calcium carbonate plus hydrochloric acid gives calcium chloride, water and carbon dioxide. The gas volume is zero point nine six zero cubic decimetres at room temperature and pressure, and the impurities do not react. Divide the gas volume by twenty-four cubic decimetres per mole to obtain zero point zero four zero zero moles of carbon dioxide.

16:29The equation gives the same amount of calcium carbonate. Forty plus twelve plus three times sixteen gives a molar mass of one hundred grams per mole. Multiplication gives four point zero zero grams of pure calcium carbonate within the sample. Divide four point zero zero grams by the original five point zero zero grams, then multiply by one hundred. The percentage purity is eighty point zero per cent. Here, the reaction measures how much useful substance was present before the experiment.

16:56Yield instead compares the collected product with the maximum predicted amount, so the denominators are different. Common examination errors include using equation coefficients as mass ratios and forgetting to convert cubic centimetres into cubic decimetres. Another is finding an empirical formula directly from masses without calculating moles. Write the quantity and unit at each stage to prevent these mistakes. Keep extra calculator digits during working, then round the final answer appropriately.

17:23An exam may ask you to state a definition, explain the limiting reactant, then calculate a product quantity. One mole contains six point zero two times ten to the twenty-three particles. Two point four zero grams of magnesium is zero point one zero zero moles, which needs zero point two zero zero moles of hydrochloric acid. Only zero point one five zero moles of acid is available, so hydrochloric acid is the limiting reactant.

17:46For the calculation, the acid supplies zero point zero seven five zero moles of hydrogen using the equation ratio. Multiplying by twenty-four cubic decimetres per mole gives one point eight zero cubic decimetres at room temperature and pressure. Show the mole ratio explicitly rather than presenting an unexplained answer. Each step should make clear which substance the quantity describes. The mole connects measurable quantities with numbers of particles. Molar mass links grams to moles, while the Avogadro constant links that amount to a particle count.

18:17Gas volume and solution concentration offer two more routes into the same central quantity. A balanced equation then connects the substances taking part through their mole ratios. Formula calculations use those amounts to uncover atom ratios, while percentage calculations compare a part with the correct total. Before calculating, identify the substance, the quantity and its unit. Then choose the relationship that connects what you know with what you need. This approach turns stoichiometry into a connected explanation of reactions, rather than a collection of memorised procedures.