WEBVTT 00:00:00.500 --> 00:00:03.975 Welcome to GCSE Edexcel Science revision. 00:00:04.125 --> 00:00:07.755 Unit S C 9: Calculations involving masses. 00:00:10.433 --> 00:00:16.426 Relative formula mass, M R, is the sum of relative atomic masses of all atoms in a formula. 00:00:16.576 --> 00:00:20.177 For molecules it can also be called relative molecular mass. 00:00:20.327 --> 00:00:22.119 M R has no unit. 00:00:24.800 --> 00:00:33.561 For water, H 2 O: M R equals 2 times 1 plus 16 equals 18, using A R(H) equals 1 and A R(O) equals 16. 00:00:33.711 --> 00:00:36.983 Subscripts multiply the atom immediately before them. 00:00:39.633 --> 00:00:49.078 Brackets multiply a whole group: C A, open bracket O H close bracket, subscript two, has one calcium, two oxygen and two hydrogen atoms. 00:00:49.228 --> 00:00:54.901 M R equals 40 plus two times the quantity 16 plus one, which equals 74. 00:00:55.051 --> 00:00:57.993 Brackets multiply all atoms inside the group. 00:01:02.667 --> 00:01:10.463 Percentage by mass of an element equals total A R contribution from that element divided by M R times 100. 00:01:10.613 --> 00:01:18.165 In water, oxygen contributes 16 divided by 18 times 100 approximately equals 88.9 percent. 00:01:20.833 --> 00:01:25.751 Use supplied A R values rather than assuming a different precision from another table. 00:01:25.901 --> 00:01:29.388 Keep unrounded values during working and round at the end. 00:01:32.067 --> 00:01:41.195 Do not confuse formula mass with an atom count: C O 2 contains three atoms, but M R equals 12 plus 2 times 16 equals 44. 00:01:43.867 --> 00:01:48.103 A molecular formula gives actual numbers of atoms in one molecule. 00:01:48.253 --> 00:01:53.574 An empirical formula gives the simplest whole-number ratio of atoms of each element. 00:01:56.233 --> 00:02:02.685 For butane, molecular formula C 4 H 10 reduces to empirical formula C 2 H 5. 00:02:02.835 --> 00:02:07.867 Some molecular formulae, such as H 2 O, are already in the simplest ratio. 00:02:10.533 --> 00:02:19.567 To obtain an empirical formula from masses, divide each element's mass by its A R, then divide all results by the smallest to find the mole ratio. 00:02:22.233 --> 00:02:27.886 If a ratio is 1 to 1.5, multiply all parts by 2 to obtain 2 to 3. 00:02:28.036 --> 00:02:32.111 Do not round 1.5 directly to 2 and change the composition. 00:02:34.767 --> 00:02:49.038 For 10.0 grams C A and 17.8 grams C L with A R 40 and 35.5, amounts are 0.25 and about 0.501 moles, approximately 1 to 2, giving C A C L 2. 00:02:49.188 --> 00:02:52.057 The supplied masses have rounding uncertainty. 00:02:52.207 --> 00:02:58.617 Rounded experimental data support C A C L 2; divide masses by A R before finding the ratio. 00:03:03.300 --> 00:03:08.345 Percentages can be treated as masses in a hypothetical 100 grams sample. 00:03:08.495 --> 00:03:14.680 Use the same divide-by-A R method rather than using mass percentages directly as atom ratios. 00:03:17.333 --> 00:03:25.525 Molecular formula equals empirical formula multiplied by the molecular M R divided by the empirical formula M R. 00:03:25.675 --> 00:03:34.590 For C H two with molecular M R 42, multiplier 42 divided by 14 equals three, and the formula is C three H six. 00:03:37.267 --> 00:03:41.641 Weigh a dry crucible and lid, add cleaned magnesium ribbon and reweigh. 00:03:41.791 --> 00:03:44.745 The difference gives the starting magnesium mass. 00:03:47.400 --> 00:03:54.449 Heat with the lid slightly lifted periodically to admit oxygen while limiting loss of magnesium oxide powder. 00:03:54.599 --> 00:03:59.262 Use tongs and eye protection; do not look directly at burning magnesium. 00:03:59.412 --> 00:04:04.537 School practical schematic; teacher supervision, eye protection and tongs are needed. 00:04:09.200 --> 00:04:15.065 Allow the crucible to cool before weighing, then repeat heating, cooling and weighing to constant mass. 00:04:15.215 --> 00:04:18.937 This supports completion of reaction and more reliable weighing. 00:04:21.600 --> 00:04:29.844 Final oxide mass minus starting magnesium mass gives the mass of oxygen gained, after subtracting crucible and lid masses correctly. 00:04:32.500 --> 00:04:40.532 Divide M G mass by A R(M G) equals 24 and O mass by A R(O) equals 16, then simplify the ratio. 00:04:40.682 --> 00:04:44.868 Ideal magnesium oxide has empirical formula M G O. 00:04:47.533 --> 00:04:52.364 Product loss, incomplete oxidation and side reactions can affect the result. 00:04:52.514 --> 00:04:59.020 Explain the direction of an error using how it changes the calculated oxygen or magnesium amount. 00:05:01.700 --> 00:05:08.290 Atoms are rearranged in chemical reactions, not created or destroyed, so total mass is conserved. 00:05:08.440 --> 00:05:12.561 A balanced equation has the same number of each atom on both sides. 00:05:15.233 --> 00:05:23.328 In a closed system, measured total mass before and after a reaction is unchanged, including any precipitate or gas kept inside. 00:05:26.000 --> 00:05:29.717 In an open flask, mass can appear to fall when a gas escapes. 00:05:29.867 --> 00:05:37.290 Burning magnesium can gain mass because oxygen from the air joins the product; neither observation breaks conservation. 00:05:39.967 --> 00:05:51.560 For 2 M G plus O 2 produces 2 M G O, formula-mass totals give 48 grams magnesium plus 32 grams oxygen produces 80 grams magnesium oxide. 00:05:51.710 --> 00:05:56.307 Coefficients give particle or mole ratios, not equal gram ratios. 00:05:56.457 --> 00:06:00.725 Coefficients are mole ratios; masses also depend on M R. 00:06:05.400 --> 00:06:14.924 Scale a reacting-mass ratio proportionally: 12 grams M G gives 20 grams M G O if magnesium reacts completely and oxygen is in excess. 00:06:17.600 --> 00:06:21.445 The limiting reactant is used up first and limits product formed. 00:06:21.595 --> 00:06:26.737 An excess reactant has some left over; use the limiting amount in the balanced equation. 00:06:29.400 --> 00:06:36.556 To work out the reacting ratio (stoichiometry) from masses, convert each mass to moles and simplify the mole ratio. 00:06:36.706 --> 00:06:42.081 The masses alone usually do not give the numbers in front of the formulae in the balanced equation. 00:06:44.733 --> 00:06:53.841 Concentration in grams per cubic decimetre equals mass of dissolved solute in grams divided by volume of solution in cubic decimetres. 00:06:53.991 --> 00:06:57.621 Use solution volume, not just the starting solvent volume. 00:07:00.300 --> 00:07:13.825 1 cubic decimetre equals 1000 cubic centimetres, so divide cubic centimetres by 1000 before substituting. 250 cubic centimetres is 0.250 cubic decimetres. 00:07:16.500 --> 00:07:29.001 For 5.0 grams solute in 250 cubic centimetres solution: concentration equals 5.0 divided by 0.250 equals 20 grams per cubic decimetre. 00:07:29.151 --> 00:07:31.626 Convert volume units before dividing. 00:07:36.300 --> 00:07:44.554 Rearrange as mass equals concentration times volume and volume equals mass divided by concentration, with consistent units. 00:07:47.233 --> 00:07:52.801 Dilution adds solvent while solute amount remains unchanged, lowering concentration. 00:07:52.951 --> 00:07:57.495 If solution volume doubles with no solute lost, concentration halves. 00:08:00.167 --> 00:08:03.821 A mole is a way of counting a very large number of particles. 00:08:03.971 --> 00:08:10.966 One mole contains about six point zero two times ten to the power twenty-three particles: the Avogadro constant. 00:08:11.116 --> 00:08:16.120 State which particles you mean, such as atoms, molecules, ions or formula units. 00:08:18.800 --> 00:08:27.203 Molar mass has units grams per mole and is numerically equal to the appropriate relative particle mass for GCSE calculations. 00:08:27.353 --> 00:08:32.109 One mole of H 2 O has mass 18 grams, not one molecule's mass. 00:08:34.767 --> 00:08:40.035 Amount in moles, n equals mass in grams divided by molar mass in grams per mole. 00:08:40.185 --> 00:08:45.540 For 9 grams water, n equals 9 divided by 18 equals 0.5 moles. 00:08:48.200 --> 00:08:50.749 Mass equals amount times molar mass. 00:08:50.899 --> 00:08:57.853 For 0.25 moles C O 2, mass equals 0.25 times 44 equals 11 grams. 00:09:00.533 --> 00:09:06.580 Number of particles equals amount times six point zero two times ten to the power twenty-three. 00:09:06.730 --> 00:09:13.044 Amount equals particle count divided by six point zero two times ten to the power twenty-three. 00:09:13.194 --> 00:09:20.435 For zero point five moles of molecules there are three point zero one times ten to the power twenty-three molecules. 00:09:20.585 --> 00:09:23.838 Reverse each step to work backwards from a particle count. 00:09:28.500 --> 00:09:34.258 In 2 H 2 plus O 2 produces 2 H 2 O, the mole ratio is 2 to 1 to 2. 00:09:34.408 --> 00:09:40.197 Four moles H 2 need two moles O 2 and produce four moles water if reaction is complete. 00:09:42.867 --> 00:09:51.078 For a limiting-reactant calculation, convert both masses to moles and compare with the equation ratio before calculating product. 00:09:51.228 --> 00:09:56.454 Excess material cannot create extra product after the limiting reactant is used. 00:09:59.133 --> 00:10:02.035 That completes Calculations involving masses. 00:10:02.185 --> 00:10:05.530 Revisit the notes and test yourself on the revision website.