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2013. Chemistry NCEA L2 2.1 Quantitative Analysis. Version 1.1. The mole and molar mass. The number of particles in a substance is measured in moles 1 mole of particles = 6.02214 x 10 23 particles for any substance! called Avogadro’s number.

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  1. 2013 Chemistry NCEA L2 2.1 Quantitative Analysis GZ Science resources 2013 Version 1.1

  2. The mole and molar mass The number of particles in a substance is measured in moles 1 mole of particles = 6.02214 x 1023 particles for any substance! called Avogadro’s number Molar mass is calculated by adding up the atomic mass (or mass number ) or each individual atom. E.g. H2O Molar mass M = 1 + 1 + 16 = 18 GZ Science resources 2013

  3. Converting number of particles to amount, n If the actual number of particles is known (atoms, ions or molecules) then the amount, n, is easily calculated since n= For example 12 x 1024 sodium ions is equivalent to mol = 20 mol of sodium ions. How many moles of water molecules are there in 3 x 1022 molecules of H2O? GZ Science resources 2013

  4. Relative atomic and molecular mass Ar and M Sometimes information is provided in terms of the relative atomic mass, Ar, of an element rather than its molar mass. This has exactly the same numerical value but has no units e.g. Ar of oxygen is 16 and the M of oxygen is 16 g mol-1. Note: The relative atomic mass is, in fact, the mass of an atom relative to the mass of an atom of the isotope carbon -12 which has a mass of 12. This means an atom of oxygen-16 is times heavier than an atom of carbon-12. The relative molecular mass, Mr, is the mass of a molecule relative to carbon-12. e.g. Mr of CuSO4 is 159.5, the same numerical value as the molar mass but having no units, (M = 159.5 g mol-1). GZ Science resources 2013

  5. Calculating Molar Mass of a compound GZ Science resources 2013

  6. One step calculations EXAMPLECalculate the amount (in moles) of methane, CH4, in 12.5 g of the gas? M(CH4) = 16.0 g mol1 n = m / M n = c . v a) Select one of the above equations according to what information is in the question and rearrange if necessary. (i.e. mass and molar mass) b) Write down the equation n(CH4) = m/M c) Write in the values (with units) underneath and calculate n(CH4) =12.5g/16.0g mol-1 n(CH4) = 0.781mol GZ Science resources 2013

  7. One step calculations n = moles (mol) m = mass (grams) M = molar mass (gmol-1) • Mass of x moles? Given m or M • n = m /Mm = n . M • m = 0.241mol x 142.1gmol-1 • m = 34.2g n = m / M n = moles (mol) c = concentration (molL-1) v= volume (L) 2. How many moles? Given v or c n = c . v n = 0.100molL-1 x 0.010L n = 1 x 10-3 moles n = c . v GZ Science resources 2013

  8. Concentration • Concentration is a quantitative measure of how much dissolved substance there is in a given volume of solvent. The concentration can be expressed or measured in 3 different ways: • grams per litre, g L-1 • If the mass is in grams and the volume in litres then • concentration = 2) % w/v - or percentage weight per volume, is the mass of solute in 100mL of solvent. 3) moles per litre, mol L-1 This is the most common concentration unit used in chemistry. It is calculated using concentration = = GZ Science resources 2013

  9. One step calculations n = m/M mol = mass molar mass The amount n is related to mass m using relationship The mass (m) of a substance Symbol = m unit = grams -1 -1 -1 GZ Science resources 2013

  10. Two step calculations EXAMPLEIf 2.3 g of sucrose (C12H22O11) is dissolved in 12mL of distilled water, what will the concentration be? M(C12H22O11) = 342 g mol-1 n = m / M n = c . v a) Select one of the above equations depending on what 2 values you have (i.e you have mass – 2.3g, and M – 342gmol-1 but only volume – 12ml) b) Write down the equation chosen and calculate n(C12H22O11) = m/M c) Use the calculated moles to insert in the other equation and calculate c(C12H22O11)= n / v information given in question (convert to litres)

  11. Two step calculations EXAMPLE500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1 NaHCO3 solution. What is the concentration of the final solution? n = c . v c = n / v a) Calculate n of each solution (n = c.v) and add together to get n (total) b) Add volumes together to get v (total) c) Calculate concentration of final solution concentration (final) = n (total) / v (total) GZ Science resources 2013

  12. Two step calculations 3. What is the concentration? ( M and m given) c = n / v but n = m / M a) n = 2.12g /142.1gmol-1 n = 0.015mol b) c = 0.015mol / 0.250L c = 0.059molL-1 GZ Science resources 2013

  13. Two step calculations • 4. How much mass of x to make up solution? Given v + c • n = c . V conc = 0.200molL-1 vol = 250ml • x = CuSO4 M(CuSO4) = 249.5gmol-1 • a) n = 0.200molL-1 x 0.250L • n = 5.00 x 10-3 mol • m = M . n • m = 249.5gmol-1 x 5.00 x 10-3 mol • m = 1.248g n = c . v GZ Science resources 2013

  14. Percentage composition EXAMPLEAllicin is the compound responsible for the characteristic smell of garlic. Its formula is C6H10S2O. What is the % of sulfur (S) in this compound? m = n x M • Place each element from the formula into a column and write the above equation under each and calculate • C H S • m(C) = n x M m(H) = n x M m(S) = n x M • b) Calculate the mass of the compound by adding all the masses together • c) Calculate % of element in the compound • i.e. % S = mass (S) ÷ Mass (compound) x 100 • d) All of the percentages should add up to 100% C6 H10 S20 C=12 H=1 S= 32.1 n (moles) is found from compound M(molar mass) is found in periodic table

  15. Percentage composition 5a. What is the Percentage Composition of x in a compound? Na2B4O7 x = B Na B O m = n . M m = n . M m = n . M m = 2 x 23.0 m = 4 x 10.8 m = 7 x 16.0 m(Na) = 46g m(B) = 43.2g m(O) = 112.0g m(Na) + m(B) + m(O) = 201.2g %B = 43.2g / 201.2g x 100 %B = 21.5 Shortcut Note – you can calculate the Molar mass of the element in question (B) and divide it by the Molar mass of the compound but you lose the opportunity to gain partial marks if you make a mistake. GZ Science resources 2013

  16. Percentage composition The mass of the different elements within a compound can be expressed as a percentage of the total mass – this can be used to identify a compound GZ Science resources 2013

  17. Mass and Percentage composition EXAMPLEA precipitate of 1.34 g of Fe2O3 was obtained by treating a vitamin supplement dissolved in water with NaOH and then heating. What is the mass of iron in the tablet? m = n x M • Place each element from the formula into a column and write the above equation under each and calculate mass. • Fe O • m(Fe) = n x M m(O) = n x M • b) Calculate the mass of the compound by adding all the masses from the elements together i.e. m (Fe2O3) = m(Fe) + m(O) • c) Calculate % of element in the compound • i.e. % Fe = mass (Fe) ÷ Mass (compound) x 100 • d) Find the % Fe in the mass given (i.e 1.34g) • (mass given ÷100) x % Make sure your calculated mass is less than the given mass (unless 100% then it will be the same. Use units, especially in your final answer.

  18. Mass and Percentage composition 5b. Find Mass of element in compound given mass of compound % of O in BaSO4 a) Find Percentage Composition of each atom in a given compound Ba S O m(Ba) = n . Mm(S) = n.Mm(O) = n. M m(Ba) = 1mol x 137.3gmol-1 m(S) = 1mol x 32.0gmol-1 m(O) = 4mol x 16.0 m(Ba) = 137.3g m(S) = 32.0g m(O) = 64g b) Calculate m of compound - m(BaSO4) = m(Ba) + m(S) + m(O) = 233.5g m(BaSO4) = 233.5g c) Calculate Percentage % O = 64g/233.5g x 100 %O = 27.4% d) Calculate percentage of O in given mass of compound mass = 0.192g m(O) = 27.4% of 0.192g m(O) = 0.0526g GZ Science resources 2013

  19. Empirical formula EXAMPLECaffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20% hydrogen, 28.9% nitrogen and 16.5% oxygen. Calculate the empirical formula of caffeine. n= m/ M • All % should add to 100%, change each into grams so they all total 100g (i.e. 28.9% = 28.9g) • b) Calculate the number of moles of each element (place each element in a column and write formula underneath, then calculate n) • c) Divide each n value by the smallest value – You may have to multiply all of the numbers up until you get the lowest common whole number of each • i.e n(C) = 4.13mol n(H) = 5.20mol n(N) = 2.06mol n(O) = 1.03mol -- • 1.03mol 1.03mol1.03mol1.03mol • d) Write the empirical formula underneath with the element followed by the number of moles i.e C4H5N2O

  20. Empirical formula • 6. Empirical Formula( Find ratio of each atom in a compound) • Convert % of each atom to Mass (grams) - total g = 100g • C = 40% H = 6.67% O = 55.33% • b) Calculate n of each • C H O • n(C) = m/ M n(H) = m /M n(O) = m/ M • n(C) = 40g/12.0gmol-1 n(H) = 6.67g/1.0gmol-1n(O) 55.33g/16.0gmol-1 • n(C) = 3.33mol n(H) = 6.67mol n(O) = 3.33mol • Divide each n by the smallest n value to obtain ratio • 3.33/3.33 : 6.67/3.33 : 3.33/3.33 • 1 : 2 : 1 • d) Convert to formula C H2O GZ Science resources 2013

  21. Empirical formula An empirical can be calculated from the percentage composition. It is the simplest whole number ratio of the atoms in a compound. -1 -1 GZ Science resources 2013

  22. Molecular formula EXAMPLEThe porphryn molecule forms an important part of the haemoglobin structure. This porphryn molecule is 78.5% carbon, 3.2% hydrogen and 18.3% nitrogen. (i) Calculate the empirical formula of porphryn. (ii) If the molar mass of porphryn is 306 g mol1, calculate the molecular formula. • Calculate the empirical formula so you have a whole number formula - i.e porphryn = C10H5N2 molar mass of carbon • Calculate the molar mass of the empirical formula i.e C = (10 x 12) + … • c) Divide the molar mass of the molecular formula (given in question) by the molar mass of the empirical formula to calculate the multiplier i.e (molecular)306gmol-1 / (empirical) 153gmol-1 = 2 • d) If the multiplier value is 1 then the empirical formula is the same as the molecular formula. Otherwise multiply each element by the value • C10H5N2 x 2 = C20H10N4 GZ Science resources 2013

  23. Molecular formula • 7. Molecular Formula( Find moles of each atom in a compound) • Given the Molecular M = 180.0gmol-1 • Find M of the empirical formula (add up all M x each atom) • Calculate multiplier = Molecular M = 180.0gmol-1 • Empirical M 30gmol-1 • = 6 • c) Multiply empirical formula by multiplier • CH2O x 6 = C6H12O6 GZ Science resources 2013

  24. Molecular formula The Molecular Formula can be determined from the empirical formula and molar mass It gives the actual number of atoms present GZ Science resources 2013

  25. Gravimetric Analysis This is an experimental procedure involving the measurement of masses of substances to determine the composition of unknown substances. • The technique is: • Highly accurate (since masses can be measured accurately) • Useful for substances involved in chemical reactions that go to completion. Calculations should be carried out to three significant figures. If the masses of the elements combined to make a compound are known accurately, than the empirical formula can be calculated. Masses of compounds can be determined from precipitation. GZ Science resources 2013

  26. Water of Crystalisation • The crystals of some salts contain molecules of water. • These molecules fit into the crystal lattice. • The ratio of water to salt is always a whole number • The water present in hydrated salts is known as water of crystallisation • When a hydrated salt is heated, the water is driven off and the salt is said to become anhydrous GZ Science resources 2013

  27. Water of Crystalisation EXAMPLEHydrated magnesium sulfate is heated in a crucible. The following data is collected. mass of crucible and lid = 26.49 g mass of hydrated magnesium sulfate = 2.12 g mass crucible, lid and magnesium sulfate after first heating = 27.55 g mass of crucible, lid and magnesium sulfate after second heating = 27.52 g Use these results to determine the formula of hydrated magnesium sulfate. M(MgSO4) = 120 g mol1 M(H2O) = 18.0 g mol1 • Write down two columns : Anhydrous salt and Water • Calculate the mass of each from the data given. • > Anhydrous salt = (mass of crucible, lid and MgSO4 after second heating) – (mass of crucible and lid) • > Water = (mass of hydrated salt) – (mass of salt calculated) • c) Calculate n of each (n=m/M) • d) Divide each by smallest value and write down formula - MgSO4 .(number of moles) H2O

  28. Water of Crystalisation Anhydrous salt Water mass = m(heated) – m(crucible) mass = m(hydrated) – m(heated) m(CaCO3) = 1.56g m(H2O) = 1.52g n= m/M n = m/M n=1.56g/90.08gmol-1 n= 1.52g/18.0gmol-1 n=0.017mol n = 0.080mol n(CaCO3) : n(H2O) 0.017mol : 0.080mol 0.017mol : 0.017mol 1 : 5 CaCO3.5H2O Data from experiment or given in question Divide each by the smallest value –this will be n (anhydrous salt) The dot (.) means + water GZ Science resources 2013

  29. Equations and mole ratios n = m/M EXAMPLEWine makers convert sugars, such as glucose, to ethanol and carbon dioxide. C6H12O6 2C2H5OH + 2CO2 Starting with 540g of glucose what is the maximum amount of ethanol, in moles and also in grams that could be produced. M (C2H5OH) = 46 g mol-1M (C6H12O6) = 180g mol-1 C6H12O6 2C2H5OH + 2CO2 M = 180gmol-1 M = 46gmol-1 - m = 540g m = ? Write down the equation with the known information underneath n = m/M n = 540g / 180gmol-1 n = 3.0mol m = n x M m = 6.0 mol x 46gmol-1 m = 276g 1 3 1. Calculate moles of known. 2. convert mole ratios. 3. calculate mass of unknown. 2 1 mole of C6H12O6 forms 2 moles of C2H5OH So n (C2H5OH) = n(C6H12O6) x U/K = 3.0mol x 2/1 = 6.0mol GZ Science resources 2013

  30. Equations and mole ratios n = m/M • 9. Calculate mass of x given equation and mass of a compound y x = NaCl y = Cl2 m(Cl2) = 12.8g • Cl2 + 2NaI 2NaCl + I2 • Calculate moles of compound y • n = m/M • n(Cl2) = 12.8g/71.0gmol-1 • n(Cl2)=0.180moles • b) Convert moles y to x • 1mol Cl2 forms2mol NaClso n(NaCl) = 2 x n(Cl2) = 0.361moles • c) Calculate mass (or concentration) of NaCl • m = n . M c = n/v • m(NaCl) = 0.361mol x 58.5gmol-1 c(NaCl) = 0.361mol x 0.100L • m(NaCl)= 21.12g c(NaCl) = 3.61molL-1

  31. Equations and mole ratios n = m/M • 9. Calculate Mass of x given Formula and equation and mass of y m(CH4) = 6800g m(CH3OH) = ? • 3CH4 + CO2 +2H2O 4CH3OH • Calculate using mass ratios – write under formulas • 3CH4 + CO2 + 2H2O 4CH3OH • Mass= 6800g = x • M = 48gmol-1 =128.0gmol-1 • m(CH3OH) = m(CH4) x M(CH3OH) • M(CH4) • m(CH3OH) = 6800g x 128.0gmol-1 • 48.0gmol-1 • m(CH3OH) = 18133.3g or 18.133kg GZ Science resources 2013

  32. Equations and mole ratios n = m/M EXAMPLE20.0 mL of an aqueous solution of 0.105 mol L1 sodium carbonate is titrated with a 0.250 mol L1 hydrochloric acid solution. What volume of hydrochloric acid is required to reach the equivalence point of the titration. 2 HCl(aq) + Na2CO3(aq)→ 2 NaCl(aq) + CO2(g) + H2O(l) Na2CO3 + 2HCl → 2NaCl + CO2 + H2O c = 0.105molL-1 c = 0.250molL-1 v =0.020L v = ? Write down the equation with the known information underneath n = c x v n = 0.105 x 0.020 n = 2.10 x 10-2mol v = n / c v = 4.20 x 10-2 mol / 0.250 v = 0.0168L or 16.8mL 1 1. Calculate moles of known. 2. convert mole ratios. 3. calculate volume of unknown. 3 2 1 mole of Na2CO3 forms 2 moles of HCl So n (HCl) = 2 x n(Na2CO3) = 2 x (2.10 x 10-2)mol = 4.20 x 10-2mol GZ Science resources 2013

  33. Equations and mole ratios n = m/M • 13. Calculate concentration of x given conc. of y and formula • Na2CO3 + 2HCl 2NaCl + H2O + CO2 • 10.0ml of 0.106molL-1 sodium carbonate reacted with 21.56ml HCl • Calculate mols of y (Na2CO3) • n(Na2CO3) = c x v • = 0.106molL-1 x 0.0100L • = 0.00106mol • b) Convert mols y to x • 1mol Na2CO3 = 2molHCl • n(HCl) = 2 x n(Na2CO3) • =0.00212mol • c) Calculate concentration • c = n/v • c(HCl) = 0.00212mol / 0.02156L = 0.0983molL-1

  34. Empirical formula and mole ratios n = m/M EXAMPLEA chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of gas. Determine the empirical formula of the compound. ?S + ?F  SF? m = 4.69g m = m(SF?) – m(S) = 11.12g m = 15.81g M = 32.1gmol-1 M = 19.0gmol-1 1 Write down the equation with the known information underneath n(S) = m/M n(F) = m/M n(S) = 4.69/32.1 n(F) = 11.12/19.0 n(S) = 0.146mol n(F) = 0.585mol 2 Calculate number of moles of each element making up compound n(S) = 0.146 n(F) = 0.585mol 0.146 0.146mol = 1 : 4 Divide each number of moles by the smallest value to get the smallest common whole number 3 Use those numbers to write formula of compound S F4 4 GZ Science resources 2013

  35. Empirical formula and mole ratios n = m/M 10. Calculate compound Formula given mass and equation m(product) = 43.5g m(Fe) =15.0g Fe + Cl2 Fe Clx a) Calculate moles of reactant (for which given mass) n(Fe) = m/M n(Fe) = 15.0g/55.9gmol-1 n(Fe)=0.268moles b) Convert moles reactant to product 1mol Fe forms 1mol FeClx c) Calculate mass of each possible molecular formula of FeClx m = n .M m = n . M m(FeCl2) =0.268mol x 126.9gmol-1 m(FeCl3)= 0.268mol x 162.3gmol-1 M(FeCl2) = 34.1g m(FeCl3) = 43.5g d) Match m(product) with calculated mass = FeCl3

  36. Dilution calculations EXAMPLE20.0 mL of a 0.00500 mol L–1 FeCl3.6H2O.solution was pipetted into a volumetric flask and diluted to 100 mL. What is the concentration of the diluted solution? concentration (new) = concentration (original) x volume (original) volume (new) c(new) = 0.00500molL-1 x (0.020L / 0.100L) c(new) = 0.00500molL-1 x 0.2 c(new) = 0.00100molL-1 Make sure you convert all of your volumes into litres first Because this is a dilution the new concentration must be less than the original GZ Science resources 2013

  37. Dilution calculations 11. Calculate new concentration given old conc. and volumes e.g. 40ml of 2.0molL-1 diluted by adding 60ml water c(new) = c(old) x old volume new volume c(new) = 2.0molL-1 x 40ml (40ml + 60ml) C(new) =0.80molL-1 Also can calculate new concentration through evaporation GZ Science resources 2013

  38. Mixed solution calculations EXAMPLE500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1 NaHCO3 solution. What is the concentration of the final solution? n(solution 1)= c1 x v1+n(solution 2)= c2 x v2 = total moles Final concentration = total mols / total volume (v1 + v2) n(solution 1 ) = 0.253molL-1 x 0.50L = c(new) = 0.00500molL-1 x 0.2 c(new) = 0.00100molL-1 Make sure you convert all of your volumes into litres first GZ Science resources 2013

  39. Volumetric Analysis GZ Science resources 2013 Volumetric analysis involves a titration. >In a titration, a solution, of unknown concentration, is typically titrated (added) from a burette into a flask. >The flask typically contains a known amount of solution. A pipette is used to measure a precise set volume called an aliquot. >In an acid-base titration the end point of the titration is reached when there is a neutralisation. Indicator (usually phenolphthalein) is added into the flask and a colour change (lasting more than 10 seconds) will indicate a neutralisation. >Using c=n/v and a balanced equation, the concentration of the solutions can be calculated.

  40. Titration equipment GZ Science resources 2013 Pipette >calibrated to provide an exact volume. >clean and rinsed with solution to be pipetted. >reading from the bottom of the meniscus.

  41. Titration equipment GZ Science resources 2013 Burette >burette must be cleaned and rinsed with small amount of solution that will go into it. >The volume of the solution in the burette is read from the bottom of the meniscus. Volume is recorded before and after the titration. >the volume of solution delivered by the burette is called a titre.

  42. Titration equipment GZ Science resources 2013 Conical Flask >rinsed with distilled water before use. >aliquot of solution (unknown or known concentration) placed into the flask from the pipette. >a few drops of indicator are placed into the flask >the flask is swirled during the titration.

  43. Reading the burette Reading is to be taken from the bottom of the meniscus curve. a) Read the ml 29ml b) Read the 1/10 of ml 29.9ml c) Divide the 1/10ml up into 0.00, 0.02, 0.04…0.08 and read to closest 29.98ml GZ Science resources 2013

  44. Titration equipment setup Make sure you label your beaker with the solutions they contain Rinse all glassware appropriately GZ Science resources 2013

  45. Indicator selection Before starting the titration a suitable pH indicator must be chosen. The endpoint of the reaction, when all the products have reacted, will have a pH dependent on the relative strengths of the acids and bases. The pH of the endpoint can be roughly determined using the following rules: A strong acid reacts with a strong base to form a neutral (pH=7) solution. A strong acid reacts with a weak base to form an acidic (pH<7) solution. A weak acid reacts with a strong base to form a basic (pH>7) solution. When a weak acid reacts with a weak base, the endpoint solution will be basic if the base is stronger and acidic if the acid is stronger. If both are of equal strength, then the endpoint pH will be neutral. A suitable indicator should be chosen, that will experience a change in colour close to the end point of the reaction. Indicator Acid Base pH range Methyl Orange Red Yellow 3.1 - 4.4 Phenolphthalein Colourless Pink 8.3 - 10.0 GZ Science resources 2013

  46. Standard solutions A standard solution is a solution whose concentration is known accurately. Its concentration is usually given in mol L–1. When making up a standard solution it is important that the correct mass of substance is accurately measured and all of this is successfully transferred to the volumetric flask used to make up the solution. • Making a 1molL-1 of NaCl solution • Calculate the mass of 1mole of NaCl • m (NaCL) = n x M • m (NaCl) = 58.5g • b) Determine the volume of the volumetric flask being used and mass of substance added. • 1L flask = 1mol • 100mL flask = 1/10mol • c) Weigh and add substance to the volumetric flask with wash bottle • d) Continue filling flask with distilled water to line – bottom of meniscus must touch line GZ Science resources 2013

  47. Titration method • First, the burette should be rinsed with the standard solution, the pipette with the unknown solution, and the conical flask with distilled water. • 2. Secondly, a known volume of the unknown concentration solution should be taken with the pipette and placed into the conical flask, along with a small amount of the indicator chosen. The burette should be filled to the top of its scale with the known solution with a funnel. Remove the funnel after the burette is filled. Record the starting volume. • 3. The known solution should then be allowed out of the burette, into the conical flask. At this stage we want a rough estimate of the amount of this solution it took to neutralize the unknown solution. Let the solution out of the burette until the indicator changes colour and then record the value on the burette. This is the first titre and should be excluded from any calculations. • 4. Perform at least three more titrations, this time more accurately, taking into account roughly of where the end point will occur. Record of each of the readings on the burette at the end point. Endpoint is reached when the indicator just changes colour for more than 10 seconds. The titres must be concordant to 0.2ml to gain Excellence ( 0.4ml Merit and Achieved)

  48. Recording Titrations • Read your instructions carefully. Understand clearly which solution goes into the conical flask and which goes into the burette. Make sure you convert all volumes into litres i.e. 10ml aliquots from a pipette is 0.010L • Record the given concentration on your sheet • Concentration of standard sodium hydroxide solution = _____________ mol L1 • 3. Rule up a table to record your titrations. Titre = final reading – start reading GZ Science resources 2013

  49. Concordant titres • At least 3 titrations must be within 0.2ml of each other to be concordant at Excellence level • Any titres not falling within this range must be discarded and not used in the average. • If more there are more than 3 concordant titres within this range they will be included in the average calculation EXAMPLE successive titrations – 24.50 mL, 23.25 mL, 23.35 mL and 23.28 mL. 23.35mL – 23.25mL = 0.10ml This difference is less than 0.2ml allowed, and 23.28 falls within this range so all three are used. 24.50mL falls outside this range so is discarded. v = (23.35mL + 23.28mL + 23.25mL) ÷ 3 = 23.29mL = 0.0233L GZ Science resources 2013

  50. Titration calculations c = n/v EXAMPLEA standard solution of 0.180 mol L1 hydrochloric acid was titrated against 25.0 mL samples of a solution of sodium carbonate. The following volumes of hydrochloric acid solution were used in successive titrations – 24.50 mL, 23.25 mL, 23.35 mL and 23.28 mL. The equation for the reaction is Na2CO3 + 2 HCl 2 NaCl + CO2 + H2O Use this information to determine the concentration of the sodium carbonate solution. Give your answer to three significant figures. Na2CO3 + 2HCl → 2NaCl + CO2 + H2O v = 0.0250L v = 0.0233L (concordant titres) c =? c = 0.180molL-1 Write down the equation with the known information underneath n = c x v n = 0.180molL-1x 0.0233L n = 4.19 x 10-3mol c = n / v c =2.10 x 10-3 mol /0.0250L c = 8.39 x 10-2molL-1 1 1. Calculate moles of known. 2. convert mole ratios. 3. calculate concentration of unknown. 3 2 2 moles of HCl forms 1 mole of Na2CO3 So n (Na2CO3) = n(HCl) ÷ 2 = (4.19 x 10-3) ÷ 2 = 2.10 x 10-3mol GZ Science resources 2013

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