Calculations Specification points Year 0 Moles I Quantitative Chemistry Relative formula mass The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of the atoms in the numbers shown in the formula. In a balanced chemical equation, the sum of the relative formula masses of the reactants is equal to the products. Chemical amounts are measured in moles. The symbol for the unit mole is mol. One mole of a substance in grams is numerically equal to its relative formula mass. The number of atoms, molecules, ions or electrons in a mole of this given substance is the Avogadro constant. The value of the Avogadro constant is 6.0 x 0 3 per mole. Students should be able to use the relative formula mass of a substance to calculate the number of moles in a given mass of that substance and vice versa. Amounts of substances in equations The masses of reactants and products can be calculated from balanced symbol equations. Chemical equations can be interpreted in terms of moles. Students should be able to: calculate the masses of substances shown in a balanced symbol equation calculate the masses of reactants and products from the balanced symbol equation and the mass of a given reactant or product Students to balance an equation given the masses of reactants and products. Limiting reactants In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used. The reactant that is completely used up is called the limiting reactant because it limits the amount of products. Students should be able to explain the effect of a limiting quantity of a reactant on the amount of products it is possible to obtain in terms of amounts in moles or masses in grams. Percentage yield Even though no atoms are gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of a product because: the reaction may not go to completion because it is reversible some of the product may be lost when it is separated from the reaction mixture some of the reactants may react in ways different to the expected reaction. The amount of a product obtained is known as the yield. When compared with the maximum theoretical amount as a percentage, it is called the percentage yield. % Yield = Mass of product actually made Maximum theoretical mass of product 00 Students should be able to: calculate the theoretical amount of a product from a given amount of reactant and the balanced equation for the reaction calculate the percentage yield of a product from the actual yield of a reaction.
Atom economy The atom economy (atom utilisation) is a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and for economic reasons to use reactions with high atom economy. The percentage atom economy = Relative formula mass of desired product Sum of relative formula masses of all reactants from equation 00 Students should be able to: calculate the atom economy of a reaction from the balanced equation explain why a particular reaction is chosen to produce a product given appropriate data such as atom economy, yield, rate, equilibrium position and usefulness of by-products. Independent Study suggestions. Look at the specification points above use the textbook pages first edition (74-84) second edition (76-80 and 84-90) to make a few notes/spider diagram/revision cards. Watch the Fuse School short 3-4 minute explanation videos on any areas you need extra help with: Balancing equations: https://www.fuseschool.org/topics/64/contents/038 Balancing equations II: https://www.fuseschool.org/topics/64/contents/039 State symbols in equations: https://www.fuseschool.org/topics/64/contents/380 Atom economy: https://www.fuseschool.org/topics/64/contents/93 Introduction to the mole: https://www.fuseschool.org/topics/64/contents/305 Avogadros number: https://www.fuseschool.org/topics/64/contents/85 Moles in equations: https://www.fuseschool.org/topics/64/contents/35 Using moles part I: https://www.fuseschool.org/topics/64/contents/034 Using moles part : https://www.fuseschool.org/topics/64/contents/035 Using moles part 3: https://www.fuseschool.org/topics/64/contents/036 Empirical and molecular formula: https://www.fuseschool.org/topics/64/contents/406 Law on conservation of mass: https://www.fuseschool.org/topics/64/contents/08 Calculating masses in reactions: https://www.fuseschool.org/topics/64/contents/385 Theoretical yield and losses: https://www.fuseschool.org/topics/64/contents/304 % yield: https://www.fuseschool.org/topics/64/contents/386 Measuring loss of mass in a reaction: https://www.fuseschool.org/topics/64/contents/99 3. Have a go at the Balancing equation quiz https://www.fuseschool.org/topics/64/contents/6 4. Have a go at the past paper question below which you can mark using the markscheme
Q.Magnesium reacts with steam to produce hydrogen gas and magnesium oxide. A teacher demonstrated the reaction to a class. The figure below shows the apparatus the teacher used. (a) (i) The hydrogen produced was collected. Describe how to test the gas to show that it is hydrogen. Test... Result... (ii) Explain why the magnesium has to be heated to start the reaction. (b) The equation for the reaction is: Mg(s) + H O(g) MgO(s) + H (g) (i) The teacher used.00 g of magnesium. Use the equation to calculate the maximum mass of magnesium oxide produced. Give your answer to three significant figures. Relative atomic masses (A r ): O = 6; Mg = 4 Maximum mass =... g (3)
(ii) The teacher s demonstration produced.50 g of magnesium oxide. Use your answer from part (b)(i) to calculate the percentage yield. If you could not answer part (b)(i), use.8 g as the maximum mass of magnesium oxide. This is not the answer to part (b)(i). Percentage yield =... % (iii) Give one reason why the percentage yield is less than 00%. () (Total 0 marks) Q. Aspirin tablets have important medical uses. A student carried out an experiment to make aspirin. The method is given below.. Weigh.00 g of salicylic acid.. Add 4 cm 3 of ethanoic anhydride (an excess). 3. Add 5 drops of concentrated sulfuric acid. 4. Warm the mixture for 5 minutes. 5. Add ice cold water to remove the excess ethanoic anhydride. 6. Cool the mixture until a precipitate of aspirin is formed. 7. Collect the precipitate and wash it with cold water. 8. The precipitate of aspirin is dried and weighed.
(a) The equation for this reaction is shown below. C 7 H 6 O 3 + C 4 H 6 O 3 C 9 H 8 O 4 + CH 3 COOH salicylic acid aspirin Calculate the maximum mass of aspirin that could be made from.00 g of salicylic acid. The relative formula mass (M r ) of salicylic acid, C 7 H 6 O 3, is 38 The relative formula mass (M r ) of aspirin, C 9 H 8 O 4, is 80............ Maximum mass of aspirin =... g (b) The student made.0 g of aspirin from.00 g of salicylic acid. Calculate the percentage yield of aspirin for this experiment. (If you did not answer part (a), assume that the maximum mass of aspirin that can be made from.00 g of salicylic acid is.50 g. This is not the correct answer to part (a).)............ Percentage yield of aspirin =... % (c) Suggest one possible reason why this method does not give the maximum amount of aspirin....... () (d) Concentrated sulfuric acid is a catalyst in this reaction. Suggest how the use of a catalyst might reduce costs in the industrial production of aspirin....... () (Total 6 marks)
Q3. Some students investigated magnesium oxide. (a) Magnesium oxide has the formula MgO. (i) Calculate the relative formula mass (M r ) of magnesium oxide. Relative atomic masses: O = 6; Mg = 4. Relative formula mass =... (ii) Calculate the percentage by mass of magnesium in magnesium oxide. Percentage by mass of magnesium in magnesium oxide =...% (iii) Calculate the mass of magnesium needed to make 5 g of magnesium oxide. Mass of magnesium =... g () (b) The students calculated that if they used 0. g of magnesium they should make 0.0 g of magnesium oxide. They did this experiment to find out if this was correct.
The students weighed 0. g of magnesium ribbon into a crucible. They heated the magnesium ribbon. They lifted the lid of the crucible slightly from time to time to allow air into the crucible. The students tried to avoid lifting the lid too much in case some of the magnesium oxide escaped. When all of the magnesium appeared to have reacted, the students weighed the magnesium oxide produced. The results of the experiment are shown below. Mass of magnesium used in grams Mass of magnesium oxide produced in grams 0. 0.8 (i) The mass of magnesium oxide produced was lower than the students had calculated. They thought that this was caused by experimental error. Suggest two experimental errors that the students had made. (ii) The students only did the experiment once. Give two reasons why they should have repeated the experiment. (Total 9 marks)
Q4. (a) A chemist was asked to identify a nitrogen compound. The chemist carried out an experiment to find the relative formula mass (M r ) of the compound. The M r of the compound was 44. Relative atomic masses: N = 4, O = 6 Draw a ring around the formula of the compound. NO NO N O 4 N O () (b) Potassium nitrate is another nitrogen compound. It is used in fertilisers. It has the formula KNO 3. The M r of potassium nitrate is 0. Calculate the percentage of nitrogen by mass in potassium nitrate. Relative atomic mass: N = 4....... Percentage of nitrogen =... % (Total 3 marks) Q5. Toothpastes often contain fluoride ions to help protect teeth from attack by bacteria. Some toothpastes contain tin(ii) fluoride. This compound has the formula SnF. (a) Calculate the relative formula mass (M r ) of SnF. Relative atomic masses: F = 9; Sn = 9............ Relative formula mass (M r ) =...
(b) Calculate the percentage by mass of fluorine in SnF............. Percentage by mass of fluorine =... % (c) A tube of toothpaste contains. g of SnF. Calculate the mass of fluorine in this tube of toothpaste............. Mass of fluorine =... g () (d) The diagram represents the electron arrangement of a fluorine atom. Explain how a fluorine atom can change into a fluoride ion, F............. (Total 7 marks)
(i) lit splint or ignite the gas (squeaky) pop / explosion (ii) because it provides energy (for the reaction) to break bonds (in the reactants) or so the particles collide successfully ignore reference to frequency or rate of collisions because it provides the activation energy gains marks (b) (i).67(g) allow.66-.68 correct answer (to 3 significant figures) with or without working gains 3 marks if answer incorrect allow up to marks for the following steps: 4 40.00 40 / 4 or moles magnesium = / 4 or 0.04(7) multiply by 40 allow ecf from incorrect ratio or incorrect number of moles 3 (ii) if correct answer from part (b)(i) used allow ecf from part (b)(i) 89.8 or 90 if.8 g used 8.4 or 8 correct answer with or without working gains marks if answer incorrect, allow the following for mark:.50 /.67 (or their answer from part (b)(i)) if.8 g used:.50 /.8
(iii) any one from: ignore measurement errors not all the magnesium reacted allow the reaction may be reversible some of the magnesium oxide / product may have been left in the tube or may have been lost ignore magnesium lost different / unexpected reaction magnesium not pure [0] M. (a).6 / range.5 to.7 correct answer with or without or with wrong working gains marks (accept answers between.5 and.7) if answer incorrect moles of salicylic acid = /38 = 0.045 moles ie /38 or 0.045 gains mark or (80/38) gains mark or g 80/38 = (.304 g) gains mark (not.304g alone) (b) 4. range 40.7 to 4.3 accept correct answer with or without or with wrong working for marks ecf ie (. / their answer from (a)) 00 correctly calculated gains marks if answer incorrect percentage yield =. /.6 00 gains mark if they do not have an answer to part (a) or they choose not to use their answer then: yield = (. /.5) 00 () = 44 accept 44 for marks with no working (c) any one from: errors in weighing some (of the aspirin) lost do not allow lost as a gas
not all of the reactant may have been converted to product eg reaction didn t go to completion allow loss of some reactants the reaction is reversible accept other products / chemicals side reactions ignore waste products reactants impure not heated for long enough not hot enough for reaction to take place (d) any one from: use lower temperature use less fuel / energy ignore references to use of catalyst produce product faster or speed up reaction more product produced in a given time (owtte) increased productivity lowers activation energy [6] M3. (a) (i) 40 correct answer with or without working or incorrect working if the answer is incorrect then evidence of 4 + 6 gains mark ignore units (ii) 60 correct answer with or without working or incorrect working if the answer is incorrect then evidence of 4/40 or 4/(i) gains mark ecf allowed from part(i) ie 4/(i) 00 ignore units
(iii) 5 ecf allowed from parts(i) and (ii) 4/(i) 5 or (ii)/00 5 ignore units (b) (i) any two from: ignore gas is lost error in weighing magnesium / magnesium oxide allow some magnesium oxide left in crucible loss of magnesium oxide / magnesium allow they lifted the lid too much allow loss of reactants / products not all of the magnesium has reacted allow not heated enough allow not enough oxygen / air (ii) any two from: ignore fair test check that the result is not anomalous to calculate a mean / average allow improve the accuracy of the mean / average improve the reliability allow make it reliable reduce the effect of errors [9] M4. (a) N O (b) 3.8 to 4 gains full marks without working if answer incorrect 3 gains mark or
4/0 00 gains mark [3] M5. (a) 57 correct answer with or without working ( 9 + 9) for mark only allow (9 + 9 =) 38 for mark only ignore units (b) 4. accept answers in the range 4 to 4.038... ignore incorrect rounding after correct answer 5 only without working gains mark or 38/57 00 gains mark or (9/57 00 =) to. gains mark allow error carried forward from part(a) 38/(a) 00 gains marks if calculated correctly (9/38 00 =) 3.8 gains mark (c) 0.9 accept answers in the range 0.8 to 0.3 allow error carried forward from part (b) (b)/00. correctly calculated ignore units (d) an electron allow electrons allow electron shared / lost for mark apply list principle for additional particles is gained owtte must be linked to electron accept can hold / take in if in correct context eg it can hold another electron (in its outer shell) = marks it can take an electron (from another atom) = marks ignore reference to fluoride ions
incorrect number of electrons gained does not gain the second mark [7]