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Understanding Compound Formation and Quantitative Calculations in Chemistry

This segment covers essential concepts in chemistry regarding the consistent composition of compounds, the calculations of reacting amounts, and the application of the mole concept. Key topics include the relationship between mass proportions in compounds, how to calculate the yield of products from given reactant masses, and methods for determining molar mass and concentrations of solutions. Through practical examples and activities, learners will gain insight into how to perform calculations involving compound formation and the molar quantities of substances.

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Understanding Compound Formation and Quantitative Calculations in Chemistry

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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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