WEBVTT 00:00:00.500 --> 00:00:03.975 Welcome to GCSE Edexcel Science revision. 00:00:04.125 --> 00:00:07.796 Unit C C 9: Calculations involving masses. 00:00:10.467 --> 00:00:16.459 Relative formula mass, M R, is the sum of relative atomic masses of all atoms in a formula. 00:00:16.609 --> 00:00:20.211 For molecules it can also be called relative molecular mass. 00:00:20.361 --> 00:00:22.153 M R has no unit. 00:00:24.833 --> 00:00:33.594 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. 00:00:33.744 --> 00:00:37.017 Subscripts multiply the atom immediately before them. 00:00:39.667 --> 00:00:49.111 Brackets multiply a whole group: C A, open bracket O H close bracket, subscript two, has one calcium, two oxygen and two hydrogen atoms. 00:00:49.261 --> 00:00:54.934 M R equals 40 plus two times the quantity 16 plus one, which equals 74. 00:00:55.084 --> 00:00:58.027 Brackets multiply all atoms inside the group. 00:01:02.700 --> 00:01:10.496 Percentage by mass of an element equals total A R contribution from that element divided by M R times 100. 00:01:10.646 --> 00:01:18.198 In water, oxygen contributes 16 divided by 18 times 100 approximately equals 88.9 percent. 00:01:20.867 --> 00:01:25.784 Use supplied A R values rather than assuming a different precision from another table. 00:01:25.934 --> 00:01:29.421 Keep unrounded values during working and round at the end. 00:01:32.100 --> 00:01:41.229 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. 00:01:43.900 --> 00:01:48.136 A molecular formula gives actual numbers of atoms in one molecule. 00:01:48.286 --> 00:01:53.607 An empirical formula gives the simplest whole-number ratio of atoms of each element. 00:01:56.267 --> 00:02:02.719 For butane, molecular formula C 4 H 10 reduces to empirical formula C 2 H 5. 00:02:02.869 --> 00:02:07.900 Some molecular formulae, such as H 2 O, are already in the simplest ratio. 00:02:10.567 --> 00:02:19.600 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. 00:02:22.267 --> 00:02:27.919 If a ratio is 1 to 1.5, multiply all parts by 2 to obtain 2 to 3. 00:02:28.069 --> 00:02:32.144 Do not round 1.5 directly to 2 and change the composition. 00:02:34.800 --> 00:02:49.071 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. 00:02:49.221 --> 00:02:52.090 The supplied masses have rounding uncertainty. 00:02:52.240 --> 00:02:58.651 Rounded experimental data support C A C L 2; divide masses by A R before finding the ratio. 00:03:03.333 --> 00:03:08.379 Percentages can be treated as masses in a hypothetical 100 grams sample. 00:03:08.529 --> 00:03:14.713 Use the same divide-by-A R method rather than using mass percentages directly as atom ratios. 00:03:17.367 --> 00:03:25.558 Molecular formula equals empirical formula multiplied by the molecular M R divided by the empirical formula M R. 00:03:25.708 --> 00:03:34.623 For C H two with molecular M R 42, multiplier 42 divided by 14 equals three, and the formula is C three H six. 00:03:37.300 --> 00:03:41.674 Weigh a dry crucible and lid, add cleaned magnesium ribbon and reweigh. 00:03:41.824 --> 00:03:44.778 The difference gives the starting magnesium mass. 00:03:47.433 --> 00:03:54.483 Heat with the lid slightly lifted periodically to admit oxygen while limiting loss of magnesium oxide powder. 00:03:54.633 --> 00:03:59.295 Use tongs and eye protection; do not look directly at burning magnesium. 00:03:59.445 --> 00:04:04.571 School practical schematic; teacher supervision, eye protection and tongs are needed. 00:04:09.233 --> 00:04:15.098 Allow the crucible to cool before weighing, then repeat heating, cooling and weighing to constant mass. 00:04:15.248 --> 00:04:18.971 This supports completion of reaction and more reliable weighing. 00:04:21.633 --> 00:04:29.877 Final oxide mass minus starting magnesium mass gives the mass of oxygen gained, after subtracting crucible and lid masses correctly. 00:04:32.533 --> 00:04:40.565 Divide M G mass by A R(M G) equals 24 and O mass by A R(O) equals 16, then simplify the ratio. 00:04:40.715 --> 00:04:44.901 Ideal magnesium oxide has empirical formula M G O. 00:04:47.567 --> 00:04:52.397 Product loss, incomplete oxidation and side reactions can affect the result. 00:04:52.547 --> 00:04:59.053 Explain the direction of an error using how it changes the calculated oxygen or magnesium amount. 00:05:01.733 --> 00:05:08.323 Atoms are rearranged in chemical reactions, not created or destroyed, so total mass is conserved. 00:05:08.473 --> 00:05:12.594 A balanced equation has the same number of each atom on both sides. 00:05:15.267 --> 00:05:23.361 In a closed system, measured total mass before and after a reaction is unchanged, including any precipitate or gas kept inside. 00:05:26.033 --> 00:05:29.750 In an open flask, mass can appear to fall when a gas escapes. 00:05:29.900 --> 00:05:37.323 Burning magnesium can gain mass because oxygen from the air joins the product; neither observation breaks conservation. 00:05:40.000 --> 00:05:51.593 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. 00:05:51.743 --> 00:05:56.341 Coefficients give particle or mole ratios, not equal gram ratios. 00:05:56.491 --> 00:06:00.758 Coefficients are mole ratios; masses also depend on M R. 00:06:05.433 --> 00:06:14.957 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. 00:06:17.633 --> 00:06:21.478 The limiting reactant is used up first and limits product formed. 00:06:21.628 --> 00:06:26.770 An excess reactant has some left over; use the limiting amount in the balanced equation. 00:06:29.433 --> 00:06:36.589 To work out the reacting ratio (stoichiometry) from masses, convert each mass to moles and simplify the mole ratio. 00:06:36.739 --> 00:06:42.115 The masses alone usually do not give the numbers in front of the formulae in the balanced equation. 00:06:44.767 --> 00:06:53.874 Concentration in grams per cubic decimetre equals mass of dissolved solute in grams divided by volume of solution in cubic decimetres. 00:06:54.024 --> 00:06:57.654 Use solution volume, not just the starting solvent volume. 00:07:00.333 --> 00:07:13.859 1 cubic decimetre equals 1000 cubic centimetres, so divide cubic centimetres by 1000 before substituting. 250 cubic centimetres is 0.250 cubic decimetres. 00:07:16.533 --> 00:07:29.035 For 5.0 grams solute in 250 cubic centimetres solution: concentration equals 5.0 divided by 0.250 equals 20 grams per cubic decimetre. 00:07:29.185 --> 00:07:31.659 Convert volume units before dividing. 00:07:36.333 --> 00:07:44.587 Rearrange as mass equals concentration times volume and volume equals mass divided by concentration, with consistent units. 00:07:47.267 --> 00:07:52.834 Dilution adds solvent while solute amount remains unchanged, lowering concentration. 00:07:52.984 --> 00:07:57.529 If solution volume doubles with no solute lost, concentration halves. 00:08:00.200 --> 00:08:03.854 A mole is a way of counting a very large number of particles. 00:08:04.004 --> 00:08:10.999 One mole contains about six point zero two times ten to the power twenty-three particles: the Avogadro constant. 00:08:11.149 --> 00:08:16.153 State which particles you mean, such as atoms, molecules, ions or formula units. 00:08:18.833 --> 00:08:27.237 Molar mass has units grams per mole and is numerically equal to the appropriate relative particle mass for GCSE calculations. 00:08:27.387 --> 00:08:32.142 One mole of H 2 O has mass 18 grams, not one molecule's mass. 00:08:34.800 --> 00:08:40.069 Amount in moles, n equals mass in grams divided by molar mass in grams per mole. 00:08:40.219 --> 00:08:45.573 For 9 grams water, n equals 9 divided by 18 equals 0.5 moles. 00:08:48.233 --> 00:08:50.782 Mass equals amount times molar mass. 00:08:50.932 --> 00:08:57.887 For 0.25 moles C O 2, mass equals 0.25 times 44 equals 11 grams. 00:09:00.567 --> 00:09:06.613 Number of particles equals amount times six point zero two times ten to the power twenty-three. 00:09:06.763 --> 00:09:13.077 Amount equals particle count divided by six point zero two times ten to the power twenty-three. 00:09:13.227 --> 00:09:20.469 For zero point five moles of molecules there are three point zero one times ten to the power twenty-three molecules. 00:09:20.619 --> 00:09:23.871 Reverse each step to work backwards from a particle count. 00:09:28.533 --> 00:09:34.291 In 2 H 2 plus O 2 produces 2 H 2 O, the mole ratio is 2 to 1 to 2. 00:09:34.441 --> 00:09:40.231 Four moles H 2 need two moles O 2 and produce four moles water if reaction is complete. 00:09:42.900 --> 00:09:51.111 For a limiting-reactant calculation, convert both masses to moles and compare with the equation ratio before calculating product. 00:09:51.261 --> 00:09:56.487 Excess material cannot create extra product after the limiting reactant is used. 00:09:59.167 --> 00:10:02.068 That completes Calculations involving masses. 00:10:02.218 --> 00:10:05.563 Revisit the notes and test yourself on the revision website.