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Chemical calculations grade 11 (Stoichiometry)
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Chemical calculations grade 11 (Stoichiometry)

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PowerPoint to teach Chemical Calculations (Stoichiometry) including, mass mass, mass volume, volume of reacting gases, limiting reagent, percentage composition, empiracle formula

PowerPoint to teach Chemical Calculations (Stoichiometry) including, mass mass, mass volume, volume of reacting gases, limiting reagent, percentage composition, empiracle formula

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  • 1. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Chemical Calculations (Stoichiometry) K Warne
  • 2. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Atomic Weights You must be able to… · Describe the mole as the SI unit for amount of substance · Relate amount of substance to relative atomic mass · Describe relationship between the mole and Avogadro’s number · Conceptualise the magnitude of Avogadro’s number · Describe the relationship between molar mass and relative molecular mass · Calculate the molar mass of a substance given its formula
  • 3. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net REACTIONS Hydrogen and oxygen react to produce water. (Synthesis reaction) H2 + O2  H2O Ar: H = 1, O = 16 Mr: H2 =2, O2 = 32 H2O = 16+2=18 Balanced Eqn: 2H2 + O2  2H2O LHS: RHS: 2(2) + 32 = 36 2(2+16) = 36 LHS = RHS MASS is CONSERVED!! This is true for ALL REACTIONS. This is the reaction used to power the space shuttle.
  • 4. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net ASKEDGIVEN Mole Calculations MOLES MOLES MASS MASS VOLUME VOLUME MOLAR RATIO Number Of particles Number Of particles
  • 5. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net The Mole The mole is defined as, “the amount of ………….. with the same number of ……………………… particles as ….. grams of carbon 12”. (n used as symbol for moles) 602 300 000 000 000 000 000 000 Six hundred and two thousand, three hundred, billion billion ! 6.023x1023 particles 12.00 g C Symbol (….) Number of particles = no of moles x no. particles in a mole Particles = ……………..
  • 6. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Moles & Particles ........................................................................... ........................................................................... ........................................................................... ........................................... ........................................................................... ........................................................................... ........................................................................... ........................................... ........................................................................... ........................................................................... ........................................................................... ........................................... particles 1 mol 1 mol 1 mol moles (n) X L 6.023 x 1023  18.069 x 1023 6.023 x 1023 6.023 x 1023 6.023 x 1023 ?3
  • 7. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net The Mole and Mass The mole is defined in such a way that the MASS NUMBER (A) of an element is equal to the relative atomic mass mass of one mole of the substance. (in grams) - THE MOLAR MASS • Eg Na = 23g/mol, water(H2O)=18g/mol Z A XAtomic Number (smaller) Mass Number (bigger) protons + neutrons Periodic Table Symbol Relative atomic mass or mass(g) of one mole
  • 8. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Isotopes • Isotopes - Atoms of the same element which have different numbers of neutrons. Eg: 6 13C & 6 12C Relative atomic mass is (actually) the average mass (of all the isotopes in a random sample) of the atoms of an element relative to 1/12 of the mass of a carbon-twelve atom. 6 13C • 6 protons • 6 electrons • 13-6 = 7 neutrons 6 12C • 6 protons • 6 electrons • 12-6 = 6 neutrons
  • 9. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net The Mole - equations Sodium reacts with water to form hydrogen and sodium hydroxide according to the equation. Na + H2O  H2 + NaOH If 46g of sodium are reacted with excess water what mass of hydrogen would be formed? 1. Balance the reaction 2Na + 2H2O  H2 + 2NaOH 2 Work out moles of reactant (given). n(Na)=m/Mr=46/23=2mol 3 Go through the equation to find out the number of moles reacting and being formed - the molar ratio: Na : H2 2:1 => 1 mole H2 formed 4 Work out quantity asked for. m(H2) = nxMr = 1 x 2 = 2 g
  • 10. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Mole examples - B & J p119 21 & p120 22 1. Na + Cl2  NaCl Calculate the mass of salt formed if 2.3g of sodium is reacted with XS chlorine. 2. Zn + HCl  ZnCl2 + H2 What mass of HCl is needed to produce 100g of hydrogen? 3. KClO3  KCl + O2 What mass of oxygen is produced from 1kg of potassium chlorate? 4. Fe2O3 + H2  Fe + H2O What mass of iron is produced if 3g of rust (Fe2O3) is reacted with XS(100g )of hydrogen?
  • 11. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Mole examples - B & J p119 21 & p120 22 1. 2Na + Cl2  2NaCl Calculate the mass of salt formed if 2.3g of sodium is reacted with XS chlorine. 1. n(Na) = m/Ar = 2.3/23 = 0.1mol 2. Molar Ratio Na:NaCl 2:2 ie 1:1 => n(NaCl) = 0.1mol 3. m(NaCl) = nxMr = 0.1x(23+35.5) = 5.85g
  • 12. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Limiting reagent If 20 g of Na reacts with 20 g of water – what mass of hydrogen will be formed?
  • 13. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Percentage Composition Analysis of a compound by mass makes it possible to work out the % mass of each element. eg Table salt: NaCl mass analysis: One mole of NaCl would have a mass of 23 + 35.5 = 58.5g • The % composition can be found using the formula: Mass element X x100 Total Mass Compound • %Na = […../ (…..) ]x100 = …………..% (by mass) • %Cl = (…../ (…….) )x100 = …………% % Mass Element X =
  • 14. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Empirical and Molecular Formula. A compound consists of carbon, hydrogen and oxygen only. The % by mass are Carbon 40.0% and 6.7% hydrogen. Calculate the empirical and molecular formula of the compound if Mr = 60g·mol-1 %(O) = 100 – (40+6.7) = 53.3 C H O In 100g: …….g ……..g ….…g n=m/Mr: …/… 6.7/…. 53.3/…… …… …… …….. ……. …… ……. Simplest: … …… …. Empirical Formulae: ……. (12+2+16 = …..) Molecular Formula: 2(CH2O) ……… (Mr = …. X 30)
  • 15. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Empirical and Molecular Formula. A compound consists of carbon, hydrogen and oxygen only. The % by mass are Carbon 40.0% and 6.7% hydrogen. Calculate the empirical and molecular formula of the compound if Mr = 60g·mol-1 %(O) = 100 – (40+6.7) = 53.3 C H O In 100g: 40.0g 6.7g 53.3g n=m/Mr: 40/12 6.7/1 53.3/16 3.33 6.7 3.33 3.33 3.33 3.33 Simplest: 1 2.01 1 Empirical Formulae: CH2O (12+2+16 = 30) Molecular Formula: 2(CH2O) C2H4O2 (Mr = 2x30)
  • 16. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Concentration - Molarity The concentration of a solution is defined as the ………………. of ……………………… per ………………. (dm3) of …………………. solute solute Final volume of …………….. 500cm3 =+ Concentration = Amount of ……… (……….) Volume of ……………… 30g of NaCl C = n v
  • 17. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net ASKEDGIVEN Mole Calculations MOLES MOLES MASS MASS VOLUME VOLUME CONCENTRATION CONCENTRATION MOLAR RATIO Number Of particles Number Of particles
  • 18. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Volume Conversions 1 dm = ….. cm 1 dm3 = ………… cm3 1 m3 = …………….. dm3 = ………………….cm3 (10….) 1cm3 1 dm3 (1 litre) 10 cm3 10 cm3 10 cm3
  • 19. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Decimal Conversions King Henry Died a miserable death called measles Kilo Hecta Decca m(unit) deci centi milli 1000 100 10 1 1/10 1/100 1/1000 1m ?km 0. 0 0 1km 1 m = 0.001 km 1 km = 1000 m
  • 20. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Conversions 1 cm = 0.1 dm 1 cm2 = (0.1)2 dm2 1 cm2 = 0.01 dm2 1 cm3 = (0.1)3 dm3 1 cm3 = 0.001 dm3 1 dm3= 1000 cm3 25 cm3 = 0.025 dm3
  • 21. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Molar Volumes One mole of an ideal (ANY) gas occupies a volume of ………….3 at ………………………… temperature and pressure. (STP) STP: T= ….ºC, ……K P =1 atmosphere (……...kPa) Fe2O3 + 3H2  2Fe + 3H2O What volume of hydrogen reacts with 50g of Fe2O3 Fe2O3 : H2 … : ….. n(H2) =…..n(Fe2O3) = …………………… v(H2) = …………………………… dm3 n(Fe2O3) = m/Mr = ………………….= ………………mol moles = volume/molar volume ==> n = …./…..
  • 22. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Concentration Calculations Hydrochloric acid reacts with calcium carbonate according to the following equation: HCl + CaCO3  CaCl2 + H2O + CO2 What mass of calcium carbonate would be needed to react completely with 25 cm3 of a 0.1 mol.dm-3 hydrochloric acid solution?
  • 23. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Volume - Volume Calculations 1. Balance the equation 2. Calculate the moles of the substance given. 3. Work through the molar ratio to find out the moles of the substance asked. 4. Calculate the quantity asked for. (Volume V = n x Mv) Mv = 22.4dm3 At STP EG: H2 + N2 --> NH3 If 3.00 dm3 of nitrogen are reacted to produce ammonia, what volume of hydrogen will be required? (At STP) H2 + N2 --> NH3
  • 24. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Mass Volume Calculations 1. KClO3  KCl + O2 What volume of oxygen is produced by the decomposition of 1kg of potassium chlorate? 1. Balance the equation - 2KClO3  2KCl + 3O2 (1) 2. Calculate the moles of the substance given. n(KClO3) = m/Mr = 1000/(39+35.5+3(16)) = 8.16mol (1) 3. Work through the molar ratio to find out the moles of the substance asked. KClO3 : O2 2 : 3 n(O2) = 3/2n(KClO3) = 3/2(8.16) = 12.24 mol (1) 4. Calculate the quantity asked for. (Volume V = n x Mv) Mv = 22.4dm3 At STP v(O2) = n(O2)Mv = 12.24(22.4) = 275 dm3 (2)
  • 25. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Standard Solution A standard solution is one for which the concentration is precisely known. Since c = n(solute)/v(solvent) = m/Mr V • The number of moles of solute (Mass) • The volume of solution. These values must be accurately determined. 2.45g Mass is determined accurately using an electronic balance. • Possible accuracies of 0.1 - 0.0001g KMnO4 Volume is measured using a volumetric flask. • 250 cm3 • 100 cm3, 200 cm3,
  • 26. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Volumetric Flask Making a standard solution. 1. Rinse a clean & dry 100 cm3 beaker with a little distilled water. 2. Transfer the correctly weighed amount of salt to the beaker. Ensure NO SALT IS LOST. 3. Add 50 - 80 cm3 water the salt and stir gently with a glass rod until all salt is dissolved. DO NOT REMOVE THE ROD FROM THE SOLUTION NOR ALLOW ANY DROPS OF SOLUTION TO ESCAPE. 4. Add ALL the solution to volumetric flask via funnel. Ensure glass rod and beaker are thoroughly rinsed. (Include rinsings.) 5. Add enough solvent to bring the level up to the mark.
  • 27. SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY For FULL presentation click HERE >> www.warnescience.net Hi - This is a SAMPLE presentation only. My FULL presentations, which contain a lot more more slides and other resources, are freely available on my resource sharing website: www.warnescience.net (click on link or logo) Have a look and enjoy! WarneScience