C4 - Quantitative Chemistry

Chemistry Study Notes

Relative Atomic Mass

  • Definition: The relative atomic mass is the weighted average mass of an atom of an element compared to 1/12 the mass of an atom of carbon-12.

  • Relative Formula Mass:

    • Calculated by adding together the relative atomic masses of all the atoms in the molecular formula.

  • Example Calculation (MgCl₂):

    • Relative atomic mass of Mg (Ar) = 24

    • Relative atomic mass of Cl (Ar) = 35.5

    • Formula:

    • extRelativeFormulaMassofMgCl2=24+(2imes35.5)ext{Relative Formula Mass of MgCl}_2 = 24 + (2 imes 35.5)

    • Calculation: 24+(2imes35.5)=24+71=9524 + (2 imes 35.5) = 24 + 71 = 95

    • Therefore, extMr(MgCl2)=95ext{Mr(MgCl}_2) = 95

Calculate the Mass of an Element in a Compound

  • Percentage Mass Formula:

    • extPercentagemassofanelementinacompound=<spanstyle="color:yellow;">ext{Percentage mass of an element in a compound} = <span style="color: yellow;">\frac{\text{Ar of element} \times \text{number of atoms of that element}}{\text{Molar mass of the compound}} \times 100 </span></p></li></ul></li><li><p><strong>ExampleCalculation:</strong></p><ul><li><p>Amixturecontains20</span> </p></li></ul></li><li><p><strong>Example Calculation:</strong></p><ul><li><p>A mixture contains 20% iron ions by mass. Find the mass of iron chloride (FeCl₂) needed to provide the iron:</p></li></ul><ol><li><p>Ar of Fe = 56</p></li><li><p>Find mass of iron in a 50g mixture:</p><ul><li><p>Calculation: 50g \times 0.20 = 10g ext{ of iron} </p></li></ul></li><li><p>Calculatethemassofironinironchloride:</p><ul><li><p></p></li></ul></li><li><p>Calculate the mass of iron in iron chloride:</p><ul><li><p> ext{Molar mass of FeCl}_2 = 56 + (2 imes 35.5) = 56 + 71 = 127 </p></li><li><p>PercentageofironinFeCl2:</p></li><li><p></p></li><li><p>Percentage of iron in FeCl₂:</p></li><li><p> \frac{56}{127} \times 100 = 44.1 \% </p></li><li><p>Amountofironchlorideneededfor10giron:</p></li><li><p></p></li><li><p>Amount of iron chloride needed for 10g iron:</p></li><li><p> \text{Mass of FeCl}_2 = \frac{10g}{0.441} = 22.6g </p></li></ul></li><li><p>Thus,youwouldneedapproximately<strong>23gofironchloride</strong>.</p></li></ol></li></ul><h4>theMole</h4><ul><li><p></p></li><li><p><strong>Avogadro′sConstant:</strong></p><ul><li><p>Definedas</p></li></ul></li><li><p>Thus, you would need approximately <strong>23g of iron chloride</strong>.</p></li></ol></li></ul><h4> the Mole</h4><ul><li><p></p></li><li><p><strong>Avogadro's Constant:</strong></p><ul><li><p>Defined as 6.02 \times 10^{23} particlespermole.</p></li></ul></li><li><p><strong>DefinitionofaMole:</strong></p><ul><li><p>1moleofanysubstancecontains<spanstyle="color:blue;">Avogadro′sarrayofparticles,whichcanbeatoms,ions,ormolecules.</span>The<spanstyle="color:yellow;">unitusedtomeasurethesizeofparticles</span>inchemistry.</p></li></ul></li><li><p><strong>MassEquivalence:</strong></p><ul><li><p>The<strong><em><u>massofonemoleofatomsormoleculesisnumericallyequaltoitsrelativeatomicorformulamassingrams.</u></em></strong></p></li></ul></li><li><p><strong>Example:</strong></p><ul><li><p>Carbon(C)hasanatomicmass(Ar)of12,thus:</p></li><li><p>1moleofcarbon=12g.</p></li></ul></li><li><p><strong>FormulaforMolesCalculation:</strong></p><ul><li><p><spanstyle="color:yellow;"><strong>particles per mole.</p></li></ul></li><li><p><strong>Definition of a Mole:</strong></p><ul><li><p>1 mole of any substance contains <span style="color: blue;">Avogadro's array of particles, which can be atoms, ions, or molecules.</span> The <span style="color: yellow;">unit used to measure the size of particles</span> in chemistry.</p></li></ul></li><li><p><strong>Mass Equivalence:</strong></p><ul><li><p>The <strong><em><u>mass of one mole of atoms or molecules is numerically equal to its relative atomic or formula mass in grams.</u></em></strong></p></li></ul></li><li><p><strong>Example:</strong></p><ul><li><p>Carbon (C) has an atomic mass (Ar) of 12, thus:</p></li><li><p>1 mole of carbon = 12g.</p></li></ul></li><li><p><strong>Formula for Moles Calculation:</strong></p><ul><li><p><span style="color: yellow;"><strong> \text{Number of moles} = \frac{\text{mass in grams}}{\text{Molar mass (Mr)}} </strong></span></p></li></ul></li></ul><h4>ConservationofMass</h4><ul><li><p><strong>LawofConservationofMass:</strong></p><ul><li><p><strong><u>Massisconservedduringchemicalreactions</u></strong>;therefore,the<strong><u>massofthe</u></strong><strong><u>reactantsequalsthemassoftheproducts</u></strong>.</p></li></ul></li><li><p><strong>ExampleReaction:</strong></p><ul><li><p></strong></span></p></li></ul></li></ul><h4>Conservation of Mass</h4><ul><li><p><strong>Law of Conservation of Mass:</strong></p><ul><li><p><strong><u>Mass is conserved during chemical reactions</u></strong>; therefore, the <strong><u>mass of the</u></strong> <strong><u>reactants equals the mass of the products</u></strong>.</p></li></ul></li><li><p><strong>Example Reaction:</strong></p><ul><li><p> 2\text{Li} + \text{F}_2 \rightarrow 2\text{LiF} </p></li><li><p>CalculationforLeftSide:</p></li><li><p></p></li><li><p>Calculation for Left Side:</p></li><li><p> \text{Li:} 2 \times 7 = 14 </p></li><li><p></p></li><li><p> \text{F:} 2 \times 19 = 38 </p></li><li><p>Total=52g.</p></li><li><p>RightSide:</p></li><li><p></p></li><li><p>Total = 52g.</p></li><li><p>Right Side:</p></li><li><p> 2 \times (7 + 19) = 2 \times 26 = 52g </p></li></ul></li><li><p><strong>ImplicationsofMassChanges:</strong></p><ul><li><p><spanstyle="color:red;">Massincrease</span>indicates<spanstyle="color:red;">gasfromairisabsorbed</span>;gasreactsandremainsinthereactionvessel.</p></li><li><p><spanstyle="color:red;">Massdecrease</span>indicatesa<spanstyle="color:red;">gasproductescapesfromthevessel</span>andcan′tbeaccountedfor.</p></li></ul></li></ul><h4>TheMoleandChemicalEquations</h4><ul><li><p><strong>BigNumbersinChemicalEquations:</strong></p><ul><li><p>Coefficientsinfrontofchemicalformulas<spanstyle="color:blue;">indicatethenumberofmoles</span>ofeachsubstanceinthereaction.</p></li></ul></li><li><p><strong>StepstoBalanceEquationsUsingReactingMasses:</strong></p><ol><li><p>Dividethemassofeachsubstancebyitsrelativeformulamassto<markdata−color="yellow"style="background−color:yellow;color:inherit;">findthenumber</mark><markdata−color="yellow"style="background−color:yellow;color:inherit;">ofmoles.</mark></p></li><li><p><markdata−color="yellow"style="background−color:yellow;color:inherit;">Dividethenumberofmolesofeachsubstancebythesmallestnumberofmolesinthereaction.</mark></p></li><li><p>Convertto<markdata−color="yellow"style="background−color:yellow;color:inherit;">wholenumbers</mark>ifnecessarybymultiplying.</p></li><li><p><markdata−color="yellow"style="background−color:yellow;color:inherit;">Writethebalancedequation</mark>.</p></li></ol></li></ul><h4>LimitingReactants</h4><ul><li><p><strong>Definition:</strong>Limitingreactantisthereactantthatiscompletely<spanstyle="color:yellow;"><strong>consumedfirst</strong></span><strong>ina</strong><strong>reaction,</strong><spanstyle="color:yellow;"><strong>haltingthereaction</strong></span>.</p></li><li><p><strong>ExampleProcess:</strong></p><ul><li><p>Whenmagnesiumcarbonatereactswithhydrochloricacid,thereactionfizzlesoutwhenonereactantisusedup.</p></li></ul></li><li><p><strong>ProductFormation:</strong></p><ul><li><p>The<spanstyle="color:purple;">amountofproductformedisdirectlyproportionaltotheamountoflimiting</span><spanstyle="color:purple;">reactantpresent</span>.</p></li></ul></li><li><p><strong>CalculatingMassofProduct:</strong></p><ol><li><p><markdata−color="purple"style="background−color:purple;color:inherit;">Writethebalancedequation</mark>.</p></li><li><p>Determine<markdata−color="purple"style="background−color:purple;color:inherit;">relativemolarmassofcorrectproductsandreactants</mark>.</p></li><li><p><markdata−color="purple"style="background−color:purple;color:inherit;">Usethemoleratio</mark>fromthebalancedequationtofindhowmanymolesareneeded.</p></li><li><p><markdata−color="purple"style="background−color:purple;color:inherit;">Calculatethemassoftheproductusingmoles</mark>.</p></li><li><p><strong>YieldofReaction:</strong>This<spanstyle="color:red;">referstothemassofactualproductobtainedcomparedtotheoreticalcalculations.</span></p></li></ol></li></ul><h4>GasesandSolutions</h4><ul><li><p><strong>MolarVolumeofaGas:</strong></p><ul><li><p>Atroomtemperature<markdata−color="red"style="background−color:red;color:inherit;">(20°C),1moleofanygasoccupies</mark><strong><markdata−color="red"style="background−color:red;color:inherit;">24dm3</mark></strong><markdata−color="red"style="background−color:red;color:inherit;">.</mark></p></li></ul></li><li><p><strong>FormulaforVolumeofGasCalculation:</strong></p><ul><li><p><markdata−color="blue"style="background−color:blue;color:inherit;"></p></li></ul></li><li><p><strong>Implications of Mass Changes:</strong></p><ul><li><p><span style="color: red;">Mass increase</span> indicates <span style="color: red;">gas from air is absorbed</span>; gas reacts and remains in the reaction vessel.</p></li><li><p><span style="color: red;">Mass decrease</span> indicates a <span style="color: red;">gas product escapes from the vessel</span> and can't be accounted for.</p></li></ul></li></ul><h4>The Mole and Chemical Equations</h4><ul><li><p><strong>Big Numbers in Chemical Equations:</strong></p><ul><li><p>Coefficients in front of chemical formulas <span style="color: blue;">indicate the number of moles</span> of each substance in the reaction.</p></li></ul></li><li><p><strong>Steps to Balance Equations Using Reacting Masses:</strong></p><ol><li><p>Divide the mass of each substance by its relative formula mass to <mark data-color="yellow" style="background-color: yellow; color: inherit;">find the number</mark> <mark data-color="yellow" style="background-color: yellow; color: inherit;">of moles.</mark></p></li><li><p><mark data-color="yellow" style="background-color: yellow; color: inherit;">Divide the number of moles of each substance by the smallest number of moles in the reaction.</mark></p></li><li><p>Convert to <mark data-color="yellow" style="background-color: yellow; color: inherit;">whole numbers</mark> if necessary by multiplying.</p></li><li><p><mark data-color="yellow" style="background-color: yellow; color: inherit;">Write the balanced equation</mark>.</p></li></ol></li></ul><h4>Limiting Reactants</h4><ul><li><p><strong>Definition:</strong> Limiting reactant is the reactant that is completely <span style="color: yellow;"><strong>consumed first</strong></span><strong> in a</strong> <strong>reaction, </strong><span style="color: yellow;"><strong>halting the reaction</strong></span>.</p></li><li><p><strong>Example Process:</strong></p><ul><li><p>When magnesium carbonate reacts with hydrochloric acid, the reaction fizzles out when one reactant is used up.</p></li></ul></li><li><p><strong>Product Formation:</strong></p><ul><li><p>The <span style="color: purple;">amount of product formed is directly proportional to the amount of limiting</span> <span style="color: purple;">reactant present</span>.</p></li></ul></li><li><p><strong>Calculating Mass of Product:</strong></p><ol><li><p><mark data-color="purple" style="background-color: purple; color: inherit;">Write the balanced equation</mark>.</p></li><li><p>Determine <mark data-color="purple" style="background-color: purple; color: inherit;">relative molar mass of correct products and reactants</mark>.</p></li><li><p><mark data-color="purple" style="background-color: purple; color: inherit;">Use the mole ratio</mark> from the balanced equation to find how many moles are needed.</p></li><li><p><mark data-color="purple" style="background-color: purple; color: inherit;">Calculate the mass of the product using moles</mark>.</p></li><li><p><strong>Yield of Reaction:</strong> This <span style="color: red;">refers to the mass of actual product obtained compared to theoretical calculations.</span></p></li></ol></li></ul><h4>Gases and Solutions</h4><ul><li><p><strong>Molar Volume of a Gas:</strong></p><ul><li><p>At room temperature <mark data-color="red" style="background-color: red; color: inherit;">(20°C), 1 mole of any gas occupies </mark><strong><mark data-color="red" style="background-color: red; color: inherit;">24 dm³</mark></strong><mark data-color="red" style="background-color: red; color: inherit;">.</mark></p></li></ul></li><li><p><strong>Formula for Volume of Gas Calculation:</strong></p><ul><li><p><mark data-color="blue" style="background-color: blue; color: inherit;"> \text{Volume of gas (dm}^3) = \frac{\text{mass of gas (g)}}{\text{Mr of gas}} \times 24 </mark></p></li></ul></li><li><p><strong>SolutionTerminology:</strong></p><ul><li><p><strong>Solute:</strong><spanstyle="color:green;"><strong>Substancethatdissolves</strong></span><strong>(solid)inthesolvent</strong>.</p></li><li><p><strong>Solvent:</strong><spanstyle="color:green;"><strong>Liquidthatdissolves</strong></span><strong>thesolid</strong>.</p></li><li><p><strong>Concentration:</strong>Measureofthe<spanstyle="color:green;">amountofsoluteinagivenvolumeofsolvent</span>.</p></li><li><p>Higherconcentrationindicatesmoresoluteinthesamevolume.</p></li></ul></li><li><p><strong>ConcentrationCalculations:</strong></p><ul><li><p>Formolarity:<spanstyle="color:green;"><strong></mark></p></li></ul></li><li><p><strong>Solution Terminology:</strong></p><ul><li><p><strong>Solute:</strong> <span style="color: green;"><strong>Substance that dissolves</strong></span><strong> (solid) in the solvent</strong>.</p></li><li><p><strong>Solvent:</strong> <span style="color: green;"><strong>Liquid that dissolves</strong></span><strong> the solid</strong>.</p></li><li><p><strong>Concentration:</strong> Measure of the <span style="color: green;">amount of solute in a given volume of solvent</span>.</p></li><li><p>Higher concentration indicates more solute in the same volume.</p></li></ul></li><li><p><strong>Concentration Calculations:</strong></p><ul><li><p>For molarity: <span style="color: green;"><strong> \text{Concentration (mol/dm}^3) = \frac{\text{number of moles of solute}}{\text{volume of solvent in dm}^3} </strong></span></p></li></ul></li></ul><h4>AtomEconomy</h4><ul><li><p><strong>Definition:</strong>Atomeconomymeasures<spanstyle="color:green;"><strong><em><u>theproportionofreactantsthatbecomeusefulproducts</u></em></strong></span>,indicating<strong><u>howefficientlyresources</u></strong>areusedinchemicalprocesses.</p></li><li><p><strong>FormulaforAtomEconomy:</strong></p><ul><li><p><spanstyle="color:yellow;"></strong></span></p></li></ul></li></ul><h4>Atom Economy</h4><ul><li><p><strong>Definition:</strong> Atom economy measures <span style="color: green;"><strong><em><u>the proportion of reactants that become useful products</u></em></strong></span>, indicating <strong><u>how efficiently resources</u></strong> are used in chemical processes.</p></li><li><p><strong>Formula for Atom Economy:</strong></p><ul><li><p><span style="color: yellow;"> \text{Atom Economy} = \frac{\text{Relative Formula Mass of Desired Products}}{\text{Relative Formula Mass of Total Reactants}} \times 100 </span></p></li></ul></li><li><p><strong>ImplicationsofHighAtomEconomy:</strong></p><ul><li><p>Ahigheratomeconomy<markdata−color="red"style="background−color:red;color:inherit;">contributestosustainability,reducingwasteandbalancingcostefficiency.</mark></p></li><li><p>Lowatomeconomyreactions<markdata−color="red"style="background−color:red;color:inherit;">leadtoresourcedepletionandpollutionaswastematerialsaccumulate.</mark></p></li></ul></li></ul><h4></span></p></li></ul></li><li><p><strong>Implications of High Atom Economy:</strong></p><ul><li><p>A higher atom economy <mark data-color="red" style="background-color: red; color: inherit;">contributes to sustainability, reducing waste and balancing cost efficiency.</mark></p></li><li><p>Low atom economy reactions <mark data-color="red" style="background-color: red; color: inherit;">lead to resource depletion and pollution as waste materials accumulate.</mark></p></li></ul></li></ul><h4>% Yield</h4><ul><li><p><strong>Definition:</strong> The % yield is <span style="color: yellow;">the ratio of the actual yield to the theoretical maximum yield possible in a chemical reaction.</span></p></li><li><p><strong>Formula for % Yield Calculation:</strong></p><ul><li><p> ext{% Yield} = \frac{\text{mass of products actually made}}{\text{maximum theoretical mass of product}} \times 100 $$

  • Observations:

    • % yield can only range between 0% and 100%. A 100% yield indicates that there were no losses or inefficiencies in the process, while a lower percentage indicates losses due to side reactions, incomplete reactions, or inefficiencies during product separation.