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Unit 1 Chemistry succeeding in the vce, 2017 extract from the master class teaching materials Our Master Classes form a component of a highly specialised weekly program, which is designed to ensure that students reach their full potential (including the elite A and A+ scores). These classes incorporate the content and teaching philosophies of many of the top schools in Victoria, ensuring students are prepared to a standard that is seldom achieved by only attending school. These classes are guaranteed to motivate students and greatly improve VCE scores! For additional information regarding the Master Classes, please do not hesitate to contact us on (03) 9663 3311 or visit our website at www.tsfx.com.au. essential for all year 11 and 12 students! important notes Our policy at TSFX is to provide students with the most detailed and comprehensive set of notes that will maximise student performance and reduce study time. These materials, therefore, include a wide range of questions and applications, all of which cannot be addressed within the available lecture time. Where applicable, fully worked solutions to the questions in this booklet will be handed to students on the last day of each subject lecture. Although great care is taken to ensure that these materials are mistake free, an error may appear from time to time. If you believe that there is an error in these notes or solutions, please let us know asap ([email protected]). Errors, as well as additional advice, clarifications and important updates, will be posted at www.tsfx.com.au/vce-updates. The views and opinions expressed in this booklet and corresponding lecture are those of the authors/lecturers and do not necessarily reflect the official policy or position of TSFX. TSFX - voted number one for excellence and quality in VCE programs. copyright notice These materials are the copyright property of The School For Excellence and have been produced for the exclusive use of students attending this program. Reproduction of the whole or part of this document constitutes an infringement in copyright and can result in legal action. No part of this publication can be reproduced, copied, scanned, stored in a retrieval system, communicated, transmitted or disseminated, in any form or by any means, without the prior written consent of The School For Excellence (TSFX). The use of recording devices is STRICTLY PROHIBITED. Recording devices interfere with the microphones and send loud, high-pitched sounds throughout the theatre. Furthermore, recording without the lecturer’s permission is ILLEGAL. Students caught recording will be asked to leave the theatre, and will have all lecture materials confiscated. it is illegal to use any kind of recording device during this lecture QUANTITIES IN CHEMISTRY SIGNIFICANT FIGURES When performing calculations, you are required to state answers to the appropriate number of significant figures. A significant figure indicates the precision to which a value or measurement is given. A significant figure is: A non-zero digit. A zero that follows a digit. Therefore: All non-zero digits are significant. For example: 1, 2, 3, 23, 341 Zeros which follow a non-zero digit are significant. For example: 200, 0.0001500 Zeros which come before a non-zero digit are insignificant. For example: 0.0001500 Any zero between two digits is significant. For example: 20.00046 How many significant figures? Count the number of figures from the first non-zero digit. Examples: 38.4 has 3 significant figures. 20.03 has 4 significant figures. 500 has 3 significant figures. 0.007 has 1 significant figure. 0.00700 has 3 significant figures. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 1 MATHEMATICAL OPERATIONS Addition and Subtraction When adding or subtracting numbers, count the number of decimal places to determine how many significant figures should appear in the answer. The answer cannot contain more decimal places than the smallest number of decimal places in the numbers being added or subtracted. For example: The sum of the numbers 0.25, 1.3324, 23.112233 is: 23.112233 1.3324 1.25 6 decimal places 4 decimal places 2 decimal places 8 significant figures 5 significant figures 3 significant figures The sum of these numbers (25.694633) must be written to the same number of decimal places as the value with the smallest number of decimal places i.e. 1.25 Therefore, the answer is 25.69, which has 4 significant figures. Multiplication and Division When multiplying or dividing numbers, the answer must be written to the same number of significant figures as the piece of data with the smallest number of significant figures. For example: 39.100 1.0 12.00 48.0 3 decimal places 1 decimal place 2 decimal places 1 decimal place 5 significant figures 2 significant figures 4 significant figures 3 significant figures The sum of these numbers must be written correct to 1 decimal place i.e. 100.1 (which has 4 significant figures!) The product of these numbers (8445.6) must be written to 2 significant figures i.e. 8.4 10 . 3 Note: Do NOT consider counting numbers in significant figures. Significant figures are only considered in the FINAL answer. Do not round off values until the final step (i.e. round when stating the final answer). © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 2 STANDARD FORM A value written in standard form lies between 1 and 10, and is multiplied by a power of 10. For example: 0.102 is written as 1.02 10 1 102 is written as 1.02 10 2 0.000102 is written as 1.02 104 400 is written as 4.00 10 2 Standard form is used to express a large or very small number value to a small number of significant figures. For example: 371.74 becomes 3.7 10 2 , correct to 2 significant figures. Note: If you are able to eliminate zeros by writing a number in scientific notation, then these zeros 4 are not significant. For example: 0.00040 can be written as 4.0 10 , and hence has 2 significant figures. QUESTION 1 Consider the following numbers: 107.9, 16.42, 7.02, 8.30, 9.1 (a) The least number of significant figures in this data is __________________. (b) If these figures were added the answer would be: __________________. (c) The product of these numbers is: __________________. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 3 RELATIONSHIPS USED IN STOICHIOMETRY Stoichiometry involves the calculation of the amounts of chemical substances. THE MOLE The “mole” is defined as the amount of substance contains the same number of formula units as there are atoms in 12 grams of the carbon-12 isotope i.e. 6.0226 10 23 formula units. This quantity is known as Avogadro's Constant ( N A ). 1 mole of a substance contains 6.0226 10 23 formula units of that substance The symbol for the amount of substance (mole) is n and the unit of measurement is mol . RATIOS IN CHEMICAL FORMULAE In order to solve quantity based questions in chemistry, we may need to use ratios in chemical formulae. As an example, a molecule of water, H 2O , is made up of 2 atoms of hydrogen and 1 atom of oxygen. We say that these atoms have combined in a 2 to 1 ratio (2 : 1). We can use these ratios to determine the amount of any chemical species in any given scenario. For example: 100 molecules of water are made up of 200 atoms of hydrogen and 100 atoms of oxygen. 2 mole of water is made up of 4 mole of hydrogen atoms and 2 mole of oxygen atoms. Note: Ratios cannot be used directly to determine the mass of each element in a compound. For example: 1 gram of H 2 SO4 does not contain 2 grams of H , 1 gram of S atoms and 4 grams of O atoms. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 4 QUESTION 2 In one mol of aluminium hydroxide, Al(OH)3.3H2O, there are: A 3 mol of aluminium atoms. B 5 mol of hydrogen atoms. C 16 mol of atoms. D 4 mol of oxygen atoms. CALCULATIONS INVOLVING AVOGADRO’S CONSTANT Number of Particles n( particles ) N A Step 1: Calculate the amount, in mol, of the given substance. Step 2: Use mole ratios to determine the amount, in mol, of the particle being counted. For example: 1 mole of H 2 SO4 contains 6.023 10 23 H 2 SO4 molecules. 1 mole of H 2 SO4 contains 2 6.023 10 23 atoms of hydrogen. 1 mole of H 2 SO4 contains 3 6.023 10 23 ions (when in an aqueous solution). Step 3: Multiply the amount, in mol, by Avogadro's Constant. ÷ NA Amount of Substance (n) Number of Particles (N) NA © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 5 QUESTION 3 How many particles of CuSO4 are present in 2.00 mole CuSO4 ? Solution Number of Particles n( particles ) N A N A 2.00 6.023 10 23 1 1.21 10 24 particles QUESTION 4 Given 3.0 mol of ethanol, C2 H 5OH , calculate: (a) The number of mol of carbon atoms. (b) The number of mol of hydrogen atoms. (c) The number of mol of oxygen atoms. (d) The number of carbon atoms. (e) The number of hydrogen atoms. (f) The number of oxygen atoms. (g) The number of atoms. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 6 QUESTION 5 Calculate the number of chloride ions in 0.50 mol of calcium chloride ( CaCl2 ). Solution Number of Particles n( particles ) N A QUESTION 6 Calculate the amount of substance (mol) corresponding to 3.7 10 27 molecules of ethanol. Solution Number of Particles n( particles ) N A © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 7 RELATIVE MASSES Chemists have used experimental data to determine the relative masses of different isotopes. The standard of relative mass is the common isotope of carbon, 12C, which has a relative mass of 12 units. The relative isotopic mass (RIM) of an isotope is the mass of an atom of the isotope relative to the mass of an atom of 12C = 12 units. A naturally occurring sample of an element contains the same isotopes present in the same proportions, regardless of its source. The average relative mass (relative to the 12C atom) of these naturally occurring mixtures of isotopes is called the relative atomic mass (RAM) of an element, and its symbol is Ar . Since masses are relative, there are no units for RAM. The relative molecular mass (RMM, M r ) of a compound is the mass of a molecule of that substance relative to the mass of 12C, i.e. 12. The M r is calculated by adding the relative atomic masses of the elements in the molecular formula. For Example: CH 4 = 12 + (4 x 1) = 16. For non-molecular compounds, relative formula mass (RFM, M r ) is used. This mass is calculated by adding the relative atomic masses of the elements in the formula. For Example: NaCl = 23 + 35.5 = 58.5. As relative atomic masses describe the relative masses between particles, we can say that: 12C = 12 1 mol of 12C = 12 g Therefore: The mass of 1 mol of any atom = Ar in g. The mass of 1 mol of any molecule = M r in g. i.e. The mole is a measure of the amount of substance that has the same number of particles as there are atoms in 12 g of 12C. The mass of one mol of an element or compound is its molar mass. It is the relative mass of the element or compound expressed in g. The symbol for molar mass is M and the unit of measurement is g mol-1. Example: If M CO2 = 44 g mol-1 we may assume that: The molar mass of CO2 = 44 g mol-1. One mol of CO2 has a mass of 44 g. 6.02 x 1023 molecules of CO2 have a mass of 44 g. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 8 QUESTION 7 The relative formula mass of calcium hydrogen carbonate, Ca(HCO3)2 is closest to: A B C D 149.1 161.1 162.1 202.2 QUESTION 8 (a) What is the molar mass of zinc? (b) How many zinc atoms are contained in one mole of zinc? (c) Calculate the mass in grams of one zinc atom. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 9 MOLAR MASS (M) AND MOLE To link a given mass of substance ( m ), its molar mass ( M ) and the amount of substance ( n ), we apply the following rule: Amount of substance (mol) = mass of substance (g) / molar mass (g mol-1) n Where: m M n m M = The mass of substance in grams = The molar mass (g/mol) = The amount of substance in mol M Amount of Substance (mol) Mass (g) M Note: The molar mass used must correspond to the substance for which a mass has been given or is the subject of the question. QUESTION 9 Calculate the amount, in mol, of substance in 0.20 g of hydrogen fluoride ( HF ). Solution n m 0.20 0.010 mol M 19.9984 QUESTION 10 Calculate the amount, in mol, of CO2 molecules present in 8.8 g of CO2 . Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 10 QUESTION 11 Calculate the mass of 0.10 mol of CO2 molecules. Solution QUESTION 12 What amount, in mol, of oxygen atoms are there in 36 g of glucose (C 6 H 12 O6 ) ? M (C 6 H 12 O6 ) 180 g / mol . Solution n C6 H12O6 m 36 0.20 mol M 180 1 mole C6 H12O6 contains 6 mol O atoms nO 6 nC 6 H 12 O6 6 0.2 1.2 mol QUESTION 13 The amount, in mol, of chloride ions in 3.85 g of FeCl3 is closest to: A B C D 0.0079 0.024 0.036 0.072 M ( FeCl3 ) 162.206 g / mol © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 11 QUESTION 14 Find the amount of H 2 O in 20.0 g of CuSO 4 .5 H 2 O . M (CuSO4 .5 H 2 O ) 249.5 g / mol M (CuSO 4 ) 159.5 g / mol Solution QUESTION 15 A sample of Fe2O3 contains 0.60 mole of oxide ions. The total mass of the sample Fe2 O3 is closest to: A B C D 29 g 32 g 64 g 144 g M ( Fe2 O3 ) 159.694 g / mol © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 12 QUESTION 16 In 1.7 g of NH3, find: (a) The total number of atoms. (b) The number of H atoms. (c) The number of N atoms. (d) The mass of H. (e) The number of mol of N. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 13 QUESTION 17 A cup of tea contains 5.6 g of sucrose C12H22O11. Calculate the: (a) Number of sucrose molecules present. (b) Number of H atoms present. (c) Mass of C present. (d) Number of mol of O atoms present. (e) Total number of atoms present. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 14 QUESTION 18 In exactly 2.0 g of hydrogen gas, H 2 , there are approximately: A 6.023 x 1023 hydrogen molecules. B 1.2 x 1024 hydrogen molecules. C 6.023 x 1023 hydrogen atoms. D 6.023 x 1023 protons. QUESTION 19 What is the mass of 6.02 x 1023 molecules of CO2 ? Solution Number of Particles n( particles ) N A n(CO2 ) 6.02 1023 1.00 mol NA m(CO2 ) n M 1 44 44.0 g QUESTION 20 Calculate the number of ions in 5.85 g of NaCl. Solution Number of Particles n( particles ) N A n m 5.85 0.100 mol M 58.5 n(ions) 2 n(NaCl) 2 0.100 0.200 mol Number of ions 0.200 N A 1.20 1023 © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 15 QUESTION 21 Calculate the number of chloride ions in 4.75 g of AlCl3. Solution QUESTION 22 Calculate the number of O atoms in 5.7 x 1024 H 2 SO 4 molecules. Solution QUESTION 23 Calculate the mass of O in 1.73 x 1024 molecules of H2SO4. Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 16 QUESTION 24 Calculate the mass of one P4 O10 molecule. Solution QUESTION 25 An atom weighs 4.49 x 10-23 g. The element is most likely: A B C D Hydrogen Carbon Aluminium Oxygen QUESTION 26 How many molecules of water are present in 20.00 g CuSO 4 .5 H 2O ? Solution no H 2O nCuSO4. .5H 2O N A C.S. nCuSO4. .5H 2 O m M 20.00 0.080096 mol 249.7 no H 2 O 0.080096 6.02310 5 23 2 .4121 10 23 2 .412 10 23 molecules © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 17 QUESTION 27 Calculate the mass in grams of: (a) (b) 1.204 10 24 barium atoms, correct to 3 significant figures. One barium atom, correct to 3 significant figures. Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 18 DETERMINING FORMULAE OF COMPOUNDS PERCENTAGE COMPOSITION The percentage composition of an element in a compound is simply the mass of that element in 100 grams of a sample. The percentage by mass of an element in a compound is constant for all pure samples irrespective of their source. We can therefore use Relative Atomic Masses to calculate the percentage composition of an element in a compound. % of an element = mass of that element in one mole of the given compound x 100 by mass mass of one mole of the given compound or more simply: % composition = (M of element) x The number of atoms of that element in the given compound x 100 M of the given compound QUESTION 28 What is the percentage (by mass) of nitrogen in ammonium sulphate? Solution Mr (( NH 4 ) 2 SO4 ) 132 Mr ( N ) 14 %N 14 2 100% 21.2% 132 QUESTION 29 The percentage by mass of carbon in butane, C4 H10 , is approximately: A B C D 40% 41% 80% 83% © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 19 QUESTION 30 Calculate the percentage composition by mass of compounds with the formulae: (a) H 2O (b) Al2 (CO3 )3 QUESTION 31 (a) The relative atomic mass of carbon is 12.0. Explain the meaning of relative atomic mass and show clearly how it is different from relative isotopic mass. (b) Calculate the percentage composition of carbon in carbonic acid (H2CO3). Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 20 QUESTION 32 A sample of fertiliser was analysed and found to contain 80% by mass of ammonium nitrate ( NH 4 NO3 ) and 20% by mass of potassium chloride ( KCl ). The mass of nitrogen in a 1.00 kg packet of the fertiliser is: A B C D 140 g 175 g 280 g 350 g © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 21 QUESTION 33 An industrial chemist working for a mining company analysed a nickel ore sample by dissolving it in acid, then precipitating the nickel as the compound Ni(C4H7N2O2)2. A 6.372 g sample of ore yielded 2.288 g of precipitate after drying. What is the nickel content of this ore sample expressed as a percentage by mass? Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 22 EMPIRICAL FORMULAE The empirical formula of a compound is the smallest whole number ratio of the atoms of the different elements that make up the compound. Empirical formulae are determined experimentally, usually by determining the mass of each element present in a given mass of compound. Therefore, to determine the empirical formula of a compound, an experimentally determined ratio by mass must be converted to a ratio by numbers of atoms. This is done by calculating the amount, in mol, of each element. METHOD: Step 1: Set up the elements present as a ratio. Step 2: Calculate the mass or percentage composition of each element. Set up a ratio by mass. Step 3: Set up a ratio by mol (convert the answers to amounts in mole). Step 4: Simplify the ratio by dividing through by the smallest number. Step 5: If required, multiply answers by an appropriate number to obtain the simplest whole number ratio. QUESTION 34 Which of the following is not an empirical formula? A CH2O B CHO C C2H5O4 D (COOH)2 © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 23 QUESTION 35 Find the empirical formula of a compound containing: 8.80 g of Cu 2.20 g of O Solution Cu : O 8.80g : 2.20g Find n 8.80 63.5 : 2.20 16 Simplify 0.139 : 0.138 Divide by smallest ratio 0.139 0.138 : 0.138 0.138 Mass (g) Simplify 1 1 The empirical formula of the compound is CuO . © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 24 QUESTION 36 Find the empirical formula of a compound containing: 15 g of P 19.35 g of O Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 25 QUESTION 37 Zinc iodate contains 61.20% iodine and 23.13% oxygen. Calculate the empirical formula of zinc iodate. Solution Zn : I : O 15 .67 % 15.67 g : 61.20% 61.20 g : 23.13% 23.13 g Find n 15.67 65.37 : 61.20 126.90 : 23.13 15.99 Simplify 0.2397 : 0.4823 : 1.4465 Divide by smallest ratio 0.2397 0.2397 : 0.4823 0.2397 : 1.4465 0.2397 1 1 : : 6.03 6 Mass in 100 g Simplify 2.01 2 The empirical formula of the compound is ZnI 2 O6 . QUESTION 38 A compound containing only C and O contains 27.3% C. What is the compound’s empirical formula? Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 26 QUESTION 39 What is the empirical formula of a compound containing: 32.4 % Na 22.6 % S 45.0 % O Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 27 QUESTION 40 0.8 g of magnesium is heated and reacts completely with chlorine. 2.26 g of a white powder forms. What is the empirical formula of the compound? Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 28 MOLECULAR FORMULAE The molecular formula describes the actual number of atoms of each element present in a molecule of a compound. The molecular formula is always a whole number multiple of the empirical formula, and may be obtained from the empirical formula if the molar mass of a compound is known. METHOD: Step 1: Find the molar mass of the empirical formula unit. Step 2: Find the molar mass of the compound. Step 3: Divide the molar mass of the compound by the molar mass of the empirical formula unit to give the number of empirical formula units. Step 4: Multiply the empirical formula by the number of empirical formula units. Molecular Formula Empirical Formula n Where n RMM Empirical Formula ( mass ) QUESTION 41 A compound has an empirical formula of CH. If the compound’s molar mass is 78 g mol-1, what is its molecular formula? Solution QUESTION 42 Hydrazine is used as a rocket fuel. Its empirical formula is NH2 and it has a relative molecular mass of 32. Find its molecular formula. Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 29 QUESTION 43 A hydrocarbon contains 7.2 g of C and 1.5 g of H . The molar mass of the compound was found to be 58 gmol 1 . What is the molecular formula of the compound? Solution C : H 7.2 g : 1.5 g Find n 7.2 12 : 1.5 1 Simplify 0.6 : 1.5 Divide by smallest ratio 0.6 0.6 : 1.5 0.6 Mass (g) Simplify 1 2.5 2 5 The empirical formula of the compound is C2 H10 . Molecular Formula Empirical Formula n M C2 H 5 29 gmol 1 M compound 58 gmol 1 Number of empirical formula units n 58 RMM 2 Empirical Formula ( mass ) 29 Therefore, Molecular Formula C2 H5 2 C4 H10 © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 30 QUESTION 44 Ethylene glycol is a compound used as antifreeze in car motors in cold weather. Its relative molecular mass is 62 g/mol and its percentage composition is 38.7% carbon, 9.7% hydrogen and 51.6% oxygen. Find its molecular formula. Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 31 QUESTION 45 The odour of mothballs is due to naphthalene, a hydrocarbon containing 93.8 % carbon by mass. 0.200 mol of the compound has a mass of 25.6 g. Find the molecular formula of naphthalene. Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 32 QUESTION 46 The chemical analysis of a sugar that is found in blood shows that it consists of 53.34% oxygen, 40.00% carbon and 6.66% hydrogen by mass. (a) Use this information to determine the empirical formula of this sugar. (b) Given that 0.034 mol of this sugar has a mass of 6.12 g, determine the molecular formula of the sugar. Solution © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 33 QUESTION 47 A compound is known to contain only sodium, sulphur and oxygen. A 100 g sample of the compound contains 17.04 g of sodium and 47.1 g of sulphur. Calculate the empirical formula of the compound. If the compound has a relative molecular mass of 270, find its molecular formula. Solution 1. Determine the mass of each element Mass of sodium = 17.04 g in 100 g of the compound Mass of sulphur = 47.41 g in 100 g of the compound Mass of oxygen = 100 g – 17.04 g – 47.41 g = 35.55 g 2. Calculate the amount, in mole, of each element: Na 3. : mass molar mass 17.04 : 23.0 moles 0.741 : S 47.41 32.1 1.48 : O : 35.55 16.0 : 2.22 Divide through by the smaller number to find the simplest whole number ratio. Na : 0.741 : 0.741 moles 1 ratio S 1.48 0.741 : 2 : : : O 2.22 0.741 3 The empirical formula of the compound is NaS2O4. 4. Note the empirical formula, and add the relative atomic masses. RAM NaS O 23.0 2 32.1 3 16.0 135.2 amu 2 5. 3 Divide the relative molecular mass by the relative mass of the empirical formula. Groups of NaS2O3 in the compound RMM compound RMM NaS 2O3 270 2 groups of NaS2O3 135.2 There are 2 groups of NaS2O3 in each molecule, so the molecular formula of the compound is Na2S4O6. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 34 QUESTION 48 1.00 g of vitamin C was burnt forming 1.50 g of CO2 and 0.408 g of H2O. Vitamin C contains C, O and H. (a) Calculate the mass of carbon in the sample. (b) Calculate the mass of hydrogen in the sample. (c) Use your answers from (a) and (b) to determine the mass of oxygen in vitamin C. (d) Determine the empirical formula of vitamin C. © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 35 SUCCEEDING IN THE VCE 2017 UNIT 1 CHEMISTRY STUDENT SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/VCE-UPDATES QUESTION 1 (a) (b) (c) 2 1.48.7 9.4 105 QUESTION 2 Answer is C QUESTION 4 (a) 6.0 mol (b) 18 mol (c) 3.0 mol (d) 3.6 1024 atoms (e) 1.1 1025 atoms (f) 1.8 1024 atoms (g) 1.6 1025 atoms QUESTION 5 6.02 x 1023 QUESTION 6 6.2 103 mol QUESTION 7 Answer is C QUESTION 8 (a) 65.37 g mol-1 (b) 6.02 1023 (c) 1.086 10-22g QUESTION 10 0.20 mol QUESTION 11 4.4 g QUESTION 13 Answer is C QUESTION 14 0.401 mol © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 1 QUESTION 15 Answer is B QUESTION 16 (a) 2.4 1023 (b) 1.8 1023 (c) 6.0 1022 (d) 0.30 g (e) 0.10 mol QUESTION 17 (a) 9.9 1021 (b) 2.2 1023 (c) 2.4 g (d) 0.18 mol (e) 4.4 1023 QUESTION 18 Answer is A QUESTION 19 44.0 g QUESTION 20 1.20 1023 QUESTION 21 6.43 1022 QUESTION 22 2.3 1025 QUESTION 23 184 g QUESTION 24 4.72 10-22 g QUESTION 25 Answer is C QUESTION 27 (a) (b) 275 g 2.28 10-22 g QUESTION 29 Answer is D QUESTION 30 (a) %H = 11.1% %O = 88.9% (b) %Al = 23.1% %C = 15.4% %O = 61.5% © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 2 QUESTION 31 (a) The RIM is the mass of an isotope relative to 12C which is 12 exactly (1), this is the reference point (1). Naturally occurring C is a mixture of isotopes with a different RIM and abundances (1). The RAM of C is a weighted average of these RIM (1). (b) % C = 19.4% QUESTION 32 Answer is C QUESTION 33 % Mass(Ni) = 7.302% QUESTION 34 Answer is D QUESTION 36 P2 O5 QUESTION 38 CO2 QUESTION 39 Na2 SO4 QUESTION 40 Mg4Cl5 QUESTION 41 C6H6 QUESTION 42 N2H4 QUESTION 44 C2 H6 O2 QUESTION 45 C10 H8 QUESTION 46 (a) CH2O (b) C6H12O6 QUESTION 48 (a) 184 g (b) 184 g (c) 184 g (d) C3 H4 O4 © The School For Excellence 2017 Succeeding in the VCE – Unit 1 Chemistry Page 3