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Quantitative C hemistry

Quantitative C hemistry. Calculations in Chemistry: part 2. Compound F ormation. A particular compound always contains the same elements . Independent of how you make the compound the elements are always present in the same proportion by mass .

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Quantitative C hemistry

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  1. Quantitative Chemistry Calculations in Chemistry: part 2

  2. CompoundFormation • A particularcompoundalwayscontainsthe same elements. • Independent ofhowyoumakethecompoundtheelementsarealwayspresent in the same proportionbymass. • These proportionscannotbechanged. • Magnesium oxidealwayscontains 60% Mg and 40% O bymass.

  3. Reactingamountsofsubstance • Relative formulamassescan also beusedtocalculatetheamountsofcompoundsreactedtogetherorproduced in reactions. • If 0.24g of Mg reactwith 0.16g of O toproduce 0.40g ofMgO.......... HowmuchMgO will beproducedbyburning 12g of Mg? • 0.24g Mg produces 0.40g ofMgO • so 1g of Mg produces 0.40/0.24 gofMgO = 1.67 gofMgO • so 12g of Mg produces 12 x 1.67 gMgO = 20g ofMgO

  4. Important • Calculationsofquantitiesliketheseare a veryimportantpartofchemistry. • Thereis a great deal ofinformationstored in theformulaeandtheequation.

  5. Activity • Question 3 on page 181 • Extension: Read throughpg 178 – 179 (compoundformationandchamicalformulea. • Answerquestion 4 on page 181

  6. The Mole • Whencarrying out an experiment a chemistcan not weigh out singleatomsormolecules. • A countingunit was found. • The standardunitof a substanceisthe relative atomicmass in grams. • This unitiscalled 1 mole (1 mol) • Carbon Ar = 12 • 1 moleofcarbonis 12g In a similarwaybanksweighcoinsbecausetheyknowhowmuchonecoinweighs

  7. Avogadro‘s Constant (L) • 1 moleofanysubstancecontainsthe same numberofatoms. • 6.02 x 1023 atoms per mole 6.02 x 1023 cokecansstackedtogetherwouldcoverthesurfaceoftheEarth to a depthof 200 miles.

  8. Calcualtionsinvolvingthe Mole • Howtocalculatethemolar mass. • Write theformulaforethanol C2H5OH • CalcualteMr = (2x12) + (5 x 1) + 16 + 1 = 46 • The molar massofethanolis 46g/mol

  9. Calculations • Foranygivenmassof a substanceyoucancalculatethemolespresent. Mass NumberMr OfMoles

  10. Example • Howmanymolesarethere in 60g ofSodium Hydroxide? • MrofNaOH = 23 + 16 + 1 = 40 • Molar massofNaOH = 40g /mol • Numberofmoles = mass / molar mass = 60 / 40 = 1.5 mols

  11. Activity • Worksheet 6.2

  12. Working out theEmpiricalFormula • Wecanwork out thechemicalformulafrom experimental data. • Magnesium reactswithoxygento form magnesiumoxide. In an experimentifweknowthat 0.24g of Mg reactswith 0.16g of O2 wecandeterminetheformula.

  13. Activity: Calculatetheempiricalformulaforsiliconoxide Completequestions on page 185

  14. A Chemical Footbridge: calculatingreactingamounts • Whatmassofaluminiumoxideisproducedwhen 9.2g ofaluminuimmetalreactwithoxygen? • Balancedequation: • 4Al + 3O2 2Al2O3 Ratio 4:2 • Convert 9.2 g of Al intomoles. • Numberofmoles = mass/ molar mass = 9.2 / 27 = 0.34 mol • Usetheratiofromtheequationtowork out howmanymolesof Al2O3areproduced. • 4 molof Al produce 2 molof Al2O3 • So 0.34 molof Al produce 0.17 molof Al2O3

  15. A Chemical Footbridge: calculatingreactingamounts…..continued • Work out themassof Al2O3 • Mr = 102 • Mass = molar mass x numberofmoles = 102 x 0.17 = 17.3 g

  16. Concentrationsof Solutions • Twotypesofconcentration • Massconcentration - g/dm³ • Molar concentration - mol / dm³ • 1dm³ = 1000 cm³ = 10 cm x 10 cm x 10cm = 1 Litre • Concetration = massofsolute volumeofsolution

  17. Concentration • A 1 mol /dm³ solutionofsodiumchloridecontains 58.5g ofNaCl (1 mol) dissolved in waterandmadeupto a final volumeof 1 dm³ or 1 litre.

  18. Concentrationcalculations • Howmanymolesofsugerarethere in 500 cm³of a 3.0 mol/dm³sugarsolution? • Numberofmoles = concentration x volume 1000 = 3.0/ 1000 x 500 = 1.5 mol

  19. Example • Calculatetheconcentrationof a solutionofsodiumhydrioxide , NaOH, thatcontains 10g ofNaOH in a final volumeof 250 cm³ (= 0.25 dm³) • Calculatenumberofmoles • Mr = 23 + 16 + 1 = 40 • Numberofmoles = mass/ Mr = 10/40 = 0.25 mol • Find theconcentration • Numberofmoles = concentration x volume • Concentration = numberofmoles / volume = 0.25 / 0.25 = 1 mol /dm³

  20. Activity • Read pages 192 and 193

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