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OCR A Jun 2011 Paper 1 Q2
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![2 Magnesium is the eighth most abundant element in the Earths crust and many rocks are a source of magnesium compounds. Magnesium carbonate, MgCO3, is present in dolomite, a rock found in the Dolomite mountains in Italy. A student collected two equal-sized samples of dolomite. These samples were put into two labelled test-tubes, A and B. Tube A was heated until there was no further change in mass and was then allowed to cool. Tube B was left unheated. (a) Write the equation for the action of heat on the magnesium carbonate present in tube A.[1] (b) The student wanted to make magnesium chloride crystals. The student added an excess of warm dilute hydrochloric acid to tube A and to tube B. (i) Write the equation for the reaction of magnesium carbonate in tube B with dilute hydrochloric acid. Include state symbols.[2] (ii) State one similarity and one difference the student would see between the reactions in the two tubes. similaritydifference[2] (iii) From the solution in each tube, the student obtained crystals with the formula MgCl26H2O. Calculate the relative formula mass of MgCl26H2O. Give your answer to one decimal place. relative formula mass =[1]OCR 2011<br />
(iv) Draw a dot-and-cross diagram to show the bonding in MgCl2. Show outer electrons only. (c) A compound containing magnesium, silicon and oxygen is also present in rock types in Italy. A sample of this compound weighing 5.27 g was found to have the following composition by mass: Mg, 1.82 g; Si, 1.05 g; O, 2.40 g. [2] Calculate the empirical formula of the compound. Show your working. empirical formula =[2]OCR 2011 Turn over<br />
(d) Pharmacists sell tablets containing magnesium hydroxide, Mg(OH)2, to combat indigestion. A student carried out an investigation to find the percentage by mass of Mg(OH)2 in an indigestion tablet. The student reacted the tablet with dilute hydrochloric acid. Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l) The student found that 32.00 cm3 of 0.500 mol dm3 HCl was needed to react with the Mg(OH)2 in a 500 mg tablet. [1 g = 1000 mg]. (i) Calculate the amount, in mol, of HCl used. (ii) Determine the amount, in mol, of Mg(OH)2 present in the tablet. amount =mol [1] (iii) Determine the percentage by mass of Mg(OH)2 present in the tablet. amount =mol [1] answer =% [3] [Total: 15]OCR 2011<br />
BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGEOCR 2011 Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_1_QP_JUN11_question_2-0.jpg)
![F321 Question (i) MgCO3(s) + 2HCl(aq) (cid:198)MgCl2(aq) + H2O(l) + Answer MgCO3 (cid:198)MgO + CO2(cid:57) (a) (b) (ii) CO2(g) Correct balanced equation (cid:57) Correct states for correct species (cid:57) Similarity: (Both) dissolve OR disappear. (cid:57) Difference: One effervesces OR fizzes OR bubbles OR gas produced (cid:57) (iii) (iv) 203.3 2+ Mg Cl Cl magnesium (ion) with 8 (or no) outermost electrons AND 2 x chloride (ions) with dot-and- cross outermost octet (cid:57) correct charges (cid:57) Mark Scheme June 2011 Mark Guidance 1 1 1 IGNORE state symbols ALLOW states mark if MgCl used in place of MgCl2 ALLOW (both) go clear ALLOW CO2 produced DO NOT ALLOW incorrect gases DO NOT ALLOW responses which suggest A will effervesce e.g. as B will fizz more DO NOT ALLOW 203 or 203.0 IGNORE units For 1st mark, if 8 electrons shown around cation then extra electron around anion must match symbol chosen for electrons in cation Shell circles not required IGNORE inner shell electrons ALLOW correct diagram of a [Cl] ion with 2 x OR 2 in front OR x 2 after the diagram. ALLOW correct diagram of [Cl] ion with subscript 2. i.e. [Cl]2. DO NOT ALLOW [Cl i.e. for first mark charges do not need to be seen 2 2] [Cl] 1 1 1 1<br />
F321 Question (c) Answer 1.82 1.05 2.40 24.3 28.1 16.0 To give 0.0749 0.0374 0.150 Ratio of moles (cid:57) Answer = Mg2SiO4 (cid:57) (d) (i) 32.00 x 0.500 = 1.60 x 102 (mol) 1000 OR 0.0160 (mol)(cid:57) 1.60 x 10-2 = 8.00 x 103 (mol) (ii) OR 0.00800 (mol)(cid:57) (iii) Molar mass Mg(OH)2 = 58.3 (cid:57) mass Mg(OH)2 = 58.38.00 x 103 = 0.466(4) g (cid:57) % Mg(OH)2 = 0.4664100 = 93.3% (cid:57) 0.500 Total 15 Mark Scheme Mark Guidance June 2011 1 1 1 1 1 ALLOW 24 for Mg (giving 0.0758) and 28 for Si (giving 0.0375) ALLOW any correct ratios of moles as calculator value OR correct rounding to 2 sig figs or more ALLOW method from masses being converted to percentages ALLOW correct answer from a ratio of moles where it is clear that the candidate has divided by the atomic numbers. ALLOW ECF for formula from incorrect ratio of moles due to over-rounding calculator error or upside down mole calculation ALLOW 0.016 (mol) IGNORE trailing zeroes ALLOW ECF for answer d(i) 2 ALLOW 0.008 or 8103 (mol) Ignore trailing zeroes ALLOW 0.0080 or 8.0103 DO NOT ALLOW 58 OR 58.0 ALLOW answer to d(ii)58.3 ALLOW 0.47 ALLOW ECF for d(ii)incorrect molar mass as calculator value OR correct rounding to 2 sig figs or more ALLOW 93% OR 93.2% OR 93.28% DO NOT ALLOW d(ii)/0.5 x 100 ALLOW (answer to second marking point/0.500)100 as calculator value OR correct rounding to 2 sig figs or more ALLOW moles method for 3 marks Molar mass = 58.3 0.500/58.3 = =0.00857(6) 0.00800/0857(6)100 = 93.3% ALLOW correct answer without working for 3 marks<br />](https://learnah.org/wp-content/uploads/2018/08/OCR_1_QP_JUN11_answer_2-0.jpg)
