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MassesofAtomsandMoles

Atomsaretoosmalltoweigh,sotherefore'relativeatomicmassesRelativeAtomicMass(RAM-torememberit),isanaveragevaluethat

dependsontheisotopestheelementcontains-whenroundedtoawhole

number.RelativeFormulaMass(RFM)isfoundbyaddinguptherelativeatomic

massesoftheatomicsinitsformula-

MrofCaCl(2),solution:ArofCa=40,ArofCl=35.5,so40+35.5

=111RelativeFormulaMassiscalledONEMOLE.

Allowsustocalculateandweighoutingramsmassesofsubstanceswith

thesamenumberofparticles.

MassofNaOH?.solution:ArofNa=30,ArofO=16,ArofH=1,

so23g+16g+g+1g=40g

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Mass Numbers

Number of neutrons = mass number - atomic number Protons/Neutals EQUAL MASSES Electrons have VERY SMALL MASSES Mass number = TOTAL of protons+neutrons

Atoms of the same element have the same ATOMIC NUMBER Number of protons and electrons must ALWAYS BE THE SAME Atoms of the same element with different numbers of neutrons are

called ISOTOPES

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EmpiricalFormula

EMPIRICALFORUMULAisthesimplestratiooftheatomsorionsin

acompound.WeworkthisoutbydividingtheMASSofeachelementin100gof

thecompoundbyitsMASSNUMBERtogivetheratioofatoms,then

convertthistoaWHOLENUMBERRATIO

Example-100gofhydrocarboncontains80gofCand20gofHNumberofmolesofcarbon-80/12=6.67Numberofmolesinhydrogen=20/1=20Ratioofatoms=6.67C:20H

Simplestratiois1C:3HSoEMPIRICALFORMULAisCH(3)

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Percentages

Percentage of any of the elements in a COMPOUND RELATIVE ATOM MASS / RELATIVE FORMULA MASS x 100 =

PERCENTAGE

Example- Mr of CO(2) = 12+(16x2)=44

Therefore percentage of carbon = (12/44) x 100 = 27.3%

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MakingAsMuchAsWeWant

MakingAsMuchAsWeWant

PERCENTAGEYIELD=(amountofproductcollected/maximumamount

ofproductpossible)x100PERCENTAGEATOMECONOMY=(relativeformulamassofuseful

product/relativeformulamassofallproducts)x100Toavoidwastebothpercentageyieldandatomeconomyshouldbeas

HIGHaspossible

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Equations and Calculations

Chemical equations show the REACTANTS and PRODUCTS of a

reaction.

E.g. 2Mg + O(2) > 2MgO Shows TWO magnesium react with ONE molecule of oxygen to

form TWO magnesium ions and TWO oxide ions In RELATIVE MASS this becomes

(2xAr of Mg)+(2xAr of O) gives (2xMr of MgO) or

(2x24+2x16=2x40) In MOLES this tells us TWO moles of Mg react with ONE mole of O(2),

to produce TWO moles of MgO

This means 48g of Mg react with 32g of O(2) to give 80g of MgO If the known mass of Mg is 5g, we can work out the MASS

(5/25) x 40 = 8.33g of MgO

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MakingAmmonia-TheHaberProcess

UsedtomakeFERTILISERSandotherCHEMICALS

TheHaberprocessalsousesanIRONCATALYST

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Reversible Reactions

So we then reach an equilibrium