Chapter 4Chapter 4

Chemical Equations and Chemical Equations and StoichiometryStoichiometry

Chapter goalsChapter goals

• Balance equations for simple chemical Balance equations for simple chemical reactions.reactions.

• Perform stoichiometry calculations Perform stoichiometry calculations using balanced chemical equations.using balanced chemical equations.

• Understand the meaning of a limiting Understand the meaning of a limiting reagent.reagent.

• Calculate the theoretical and percent Calculate the theoretical and percent yields of a chemical reaction.yields of a chemical reaction.

• Use stoichiometry to analyze a mixture Use stoichiometry to analyze a mixture of compounds or to determine the of compounds or to determine the formula of a compound. formula of a compound.

Chemical EquationsChemical Equations

• shorthand notation for chemical shorthand notation for chemical changechange

• based on the based on the law of conservation of law of conservation of mattermatter (A. Lavoisier, 18 (A. Lavoisier, 18thth century): valid century): valid for ordinary reactions. For extraordinary for ordinary reactions. For extraordinary reactions, such as nuclear reactions, the reactions, such as nuclear reactions, the more general more general law of conservation of law of conservation of energyenergy holds holds

SymbolismSymbolism• starting materials (reactants) on leftstarting materials (reactants) on left• products on rightproducts on right• reactants separated from products by reactants separated from products by

or or • reactants separated from each other with reactants separated from each other with

++, same for products, same for products heatheat• hh light light• balanced by both mass and charge by balanced by both mass and charge by

using using coefficientscoefficients• use correct molecular formulasuse correct molecular formulas

Symbolism (contd…)Symbolism (contd…)

• trailing subscriptstrailing subscripts (g) gas(g) gas ((ll) liquid) liquid (s) solid(s) solid (aq) aqueous solution(aq) aqueous solution• catalysts, special solvents, special catalysts, special solvents, special

conditions written above and/or conditions written above and/or below arrow, below arrow, heat, heat, hh light light

Balancing Chemical EquationsBalancing Chemical Equations• done by inspectiondone by inspection

• usually best to begin with largest usually best to begin with largest moleculemolecule

• for reactions of for reactions of organic compoundsorganic compounds, , such as such as combustion reactionscombustion reactions, balance , balance C C atoms firstatoms first. Secondly balance H atoms. . Secondly balance H atoms. Then, balance O atoms Then, balance O atoms

• convert any fractional coefficient to convert any fractional coefficient to whole number by multiplying all equation whole number by multiplying all equation times the denominator of fractiontimes the denominator of fraction

Example, BalanceExample, BalanceAl(s) + OAl(s) + O22(g) Al2O3(s)

Balancing Al

2 Al(s) + O2 Al(s) + O22(g)(g) Al2O3(s)

Balancing O

2 Al(s) + 1.5 O2 Al(s) + 1.5 O22(g)(g) Al2O3(s)

fractional coefficients not allowedthen, multiply 2 (2 1.5 = 3)

4 Al(s) + 3 O4 Al(s) + 3 O22(g)(g) 2Al2O3(s)

Example, BalanceExample, Balance

NHNH33(g)(g) + O2(g) NO2(g) + H2O(l)

Balancing H

22 NH NH33(g)(g) + O2(g) NO2(g) + 3 H2O(l)

Balancing N

22 NH NH33(g)(g) + O2(g) 2 NO2(g) + 3 H2O(l)

Balancing O

2 NH2 NH33(g)(g) + 3.5 O2(g) 2 NO2(g) + 3 H2O(l)

4 NH4 NH33(g)(g) + 7 O2(g) 4 NO2(g) + 6 H2O(l)

Types of ReactionsTypes of Reactions

• combinationcombination A + B A + B C C

• decompositiondecomposition A A B + C B + C

• combustioncombustion

A + OA + O22 products products

Combustion ReactionsCombustion Reactions• reaction with molecular oxygenreaction with molecular oxygen

• of hydrocarbons of hydrocarbons producesproduces carbon dioxide and watercarbon dioxide and water

e.g.. CHe.g.. CH44(g) + O(g) + O22(g) (g) CO CO22(g) + H(g) + H22O(O(ll))

(methane)(methane)

CHCH44(g) + 2 O(g) + 2 O22(g) (g) CO CO22(g) + 2H(g) + 2H22O(O(ll))

Also: 2 HAlso: 2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O (O (ll))

Combustion ReactionsCombustion Reactions CC55HH1212(g) + O(g) + O22(g) (g) CO CO22(g) + H(g) + H22O(O(ll))

Balance C atomsBalance C atoms CC55HH1212(g) + O(g) + O22(g) (g) 55COCO22(g) + H(g) + H22O(O(ll))

Balance H atomsBalance H atoms CC55HH1212(g) + O(g) + O22(g) (g) 5CO 5CO22(g) + (g) + 66HH22O(O(ll))

Balance O atomsBalance O atoms, 5x2 + 6 = 16 O, 5x2 + 6 = 16 O CC55HH1212(g) + (g) + 88OO22(g) (g) 5CO 5CO22(g) + 6H(g) + 6H22O(O(ll))

Combustion ReactionsCombustion Reactions CC66HH1414(g) + O(g) + O22(g) (g) CO CO22(g) + H(g) + H22O(O(ll)) Balance C atomsBalance C atoms CC66HH1414(g) + O(g) + O22(g) (g) 66COCO22(g) + H(g) + H22OO(l(l)) Balance H atomsBalance H atoms CC66HH1414(g) + O(g) + O22(g) (g) 6CO 6CO22(g) + (g) + 77HH22O(O(ll)) Balance O atomsBalance O atoms, 6x2 + 7 = 19 O, 6x2 + 7 = 19 O CC66HH1414(g) + (g) + 9.59.5OO22(g) (g) 6CO 6CO22(g) + 7H(g) + 7H22O(O(ll))

Multiply both sides of equation by 2Multiply both sides of equation by 2 2C2C66HH1414(g) + 19O(g) + 19O22(g) (g) 12CO 12CO22(g) + 14H(g) + 14H22O(O(ll))

Combustion ReactionsCombustion Reactions NHNH33(g) + O(g) + O22(g) (g) NO(g) + H NO(g) + H22O(g)O(g) N atoms are balancedN atoms are balanced Balance H atomsBalance H atoms 22NHNH33(g) + O(g) + O22(g) (g) NO(g) + NO(g) + 33HH22O(g)O(g)

N atoms are not balanced nowN atoms are not balanced now 2NH2NH33(g) + O(g) + O22(g) (g) 22NO(g) + 3HNO(g) + 3H22O(g)O(g)

Balance O atomsBalance O atoms, 2 + 3 = 5 O, 2 + 3 = 5 O 2NH2NH33(g) + (g) + 2.52.5OO22(g) (g) 2NO(g) + 3H 2NO(g) + 3H22O(g)O(g) Multiply both sides of equation by 2Multiply both sides of equation by 2 4NH4NH33(g) + (g) + 55OO22(g) (g) 4NO(g) + 6H 4NO(g) + 6H22O(g)O(g)

StoichiometryStoichiometry

• calculations involving chemical calculations involving chemical equationsequations

StepsSteps• write balanced chemical equationwrite balanced chemical equation• convert given quantities to molesconvert given quantities to moles• determine determine limiting reagentlimiting reagent if necessary if necessary• calculate number of moles of desired calculate number of moles of desired

substancesubstance• convert to desired unitsconvert to desired units

StoichiometryStoichiometryCalculations are based on relationships similar to Calculations are based on relationships similar to those in the following examplethose in the following example

CHCH44(g)(methane) + 2 O(g)(methane) + 2 O22(g) (g) CO CO22(g) + 2H(g) + 2H22O(O(ll))

1 molecule of CH1 molecule of CH44 reactsreacts with 2 molecules of O with 2 molecules of O22 to to

produceproduce 1 molecule of CO 1 molecule of CO22 + 2 molecules of H + 2 molecules of H22OO

1 mole of CH1 mole of CH44 reactsreacts with 2 moles of O with 2 moles of O22 to to

produceproduce 1 mole of CO 1 mole of CO22 + 2 moles of H + 2 moles of H22OO

16.04 g of CH16.04 g of CH44 reactsreacts with 2x32.00 of O with 2x32.00 of O22 to to

produceproduce 44.01 g of CO 44.01 g of CO22 + 2x18.02 g of H + 2x18.02 g of H22OO

Example: Iron reacts with steam to form Example: Iron reacts with steam to form FeFe33OO44 and H and H22. Calculate mol H. Calculate mol H22 produced produced

by reaction of 10.0 g iron with excess by reaction of 10.0 g iron with excess steam.steam.

Fe(s) + HFe(s) + H22O(g) O(g) Fe Fe33OO44(s) + H(s) + H22(g)(g)

3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)

First, we convert g of Fe into moles Fe

1 mol Fe10.0 g Fe x —————— = 0.179 mol Fe 55.85 g Fe

Second, we use Fe moles to calculate H2’s.Water is not used because it is excess

4 mol H2

0.179 mol Fe x ————— = 0.239 mol 3 mol Fe

Example: Tin(IV) oxide reacts with carbon at Example: Tin(IV) oxide reacts with carbon at high temperature to form elemental tin and high temperature to form elemental tin and carbon monoxide. How many kg of carbon carbon monoxide. How many kg of carbon are needed to convert 50.0 kg of tin(IV) are needed to convert 50.0 kg of tin(IV) oxide to elemental tin?oxide to elemental tin?

• SnSn4+4+

• OO2–2–

• SnOSnO22

• SnOSnO22 + C + C Sn + CO Sn + CO

To balance the equation we need 2 O on To balance the equation we need 2 O on the right and so on.the right and so on.

SnOSnO22 + 2 C + 2 C Sn + 2 CO Sn + 2 CO

1x 103 g 1 mol SnO2

50.0 kg SnO2 x ———— x —————— 1 kg 150.7 g SnO2

2 moles C 12.01 g C 1 kg Cx —————— x ———— x ———— 1 mol SnO2 1 mol C 1x 103 g C

7.97 kg C (we may use kmol too)

(332 moles SnO2 664 moles C)

The air in a closed container consists of The air in a closed container consists of

1.40 mol O1.40 mol O22, 7.00 mol N, 7.00 mol N22, with traces of other , with traces of other

gases. A small lamp fueled by methanol, gases. A small lamp fueled by methanol,

CHCH33OH, is lighted in the container. How OH, is lighted in the container. How

many mL of methanol (d=0.791 g/mL) will be many mL of methanol (d=0.791 g/mL) will be

consumed when the lamp goes out?consumed when the lamp goes out?

• CHCH33OH(l) + OOH(l) + O22(g)(g)

• CHCH33OH(l) + OOH(l) + O22(g)(g) CO CO22(g) + H(g) + H22O(l)O(l)

• 2 CH2 CH33OH(l) + 3 OOH(l) + 3 O22(g)(g) 2 CO 2 CO22(g) + 4 H(g) + 4 H22O(l)O(l)

I recommend you to practice this balancingI recommend you to practice this balancing

2 CH2 CH33OH(l) + 3 OOH(l) + 3 O22(g)(g) 2 CO 2 CO22(g) + 4 H(g) + 4 H22O(l)O(l)

We can determine the amount of CH3OH out of O2’s

2 mol CH3OH1.40 mol O2 x —————— = 0.933 mol CH3OH

3 mol O2

32.04 g CH3OH 1 mL CH3OH 0.933 mol x —————— x —————— 1 mol CH3OH 0.791 g CH3OH

= 37.8 mL CH3OH

• the reactant that is present in quantity the reactant that is present in quantity smaller to completely react other reactant; smaller to completely react other reactant; is is consumed completelyconsumed completely during the reaction; during the reaction;

determines amount of product yieldeddetermines amount of product yieldedIt can be also seen as the reagent that It can be also seen as the reagent that theoretically produces the smallest amount of theoretically produces the smallest amount of product(s).product(s).

• excess reagent (reactant)excess reagent (reactant): the reactant : the reactant present in quantity greater than needed for present in quantity greater than needed for the reaction; the reaction;

part of it remainspart of it remains after the reaction is after the reaction is completedcompleted

Limiting Reagent (reactant)Limiting Reagent (reactant)

• CuOCuO

• NHNH33

• CuO(s) + NHCuO(s) + NH33(g) (g) Cu(s) + H Cu(s) + H22O(l) + NO(l) + N22(g)(g)

• 3CuO + 2NH3CuO + 2NH33(g) (g) 3Cu(s) + 3H 3Cu(s) + 3H22O(l) + NO(l) + N22(g)(g)

Example: Copper(II) oxide reacts with Example: Copper(II) oxide reacts with ammonia to form copper, water, and ammonia to form copper, water, and nitrogen. If 236.1 g copper(II) oxide are nitrogen. If 236.1 g copper(II) oxide are treated with 64.38 g ammonia, how much treated with 64.38 g ammonia, how much copper (copper (gramsgrams) is produced? How many g ) is produced? How many g of excess reagent remain?of excess reagent remain?

3CuO + 2NH3CuO + 2NH33(g) (g) 3Cu(s) + 3H 3Cu(s) + 3H22O(l) + NO(l) + N22(g)(g)

1 mol CuO236.1 g CuO x —————— = 2.968 mol CuO 79.55 g CuO

1 mol NH3 64.38 g NH3 x —————— = 3.780 mol NH3 17.03 g NH3

Now, let’s see which one of the two, CuO or NH3, theoretically produces the smallest amount of a product, Cu in this case.

3 mol Cu2.968 mol CuO ————— = 2.968 mol Cu 3 mol CuO And, now, ammonia, NH3, …

3CuO + 2NH3CuO + 2NH33(g) (g) 3Cu(s) + 3H 3Cu(s) + 3H22O(l) + NO(l) + N22(g)(g)

3 mol Cu 3.780 mol NH3————— = 5.670 mol Cu 2 mol NH3

Then, the amount of Cu we can obtain with CuO, 2.968 mol, is less than what we can get with NH3, i.e., CuO is the LR.

Conclusion: we must use the amount of CuOto calculate the amount of Cu yielded and for the amount of NH3 that has reacted.

2.968 mol CuO 3.780 mol NH3 ───────── = 0.989 < ───────── = 1.89 3 (coeff.) 2 (coeff.)

3CuO + 2NH3CuO + 2NH33(g) (g) 3Cu(s) + 3H 3Cu(s) + 3H22O(l) + NO(l) + N22(g)(g)

How many moles of Cu? 2.968 mol Cu

63.55 g Cu2.968 mol Cu x ————— = 188.6 g Cu 1 mol Cu

How many g of NH3 in excess?

started with 3.780 mol NH3. The reacted from CuO, 2 mol NH3 2.968 mol CuO x ————— = 1.979 mol NH3 used up 3 mol CuO

Then, by subtraction 3.780 – 1.979 = 1.801 mol NH3

17.03 g left over 1.801 mol NH3 x ———— = 30.67 g NH3 1 mol

4 Li(s)4 Li(s) + + O O22(g)(g) 2 Li2 Li22O(s)O(s)

startstart ( (givengiven) 4.0 mol) 4.0 mol 1.0 mol 1.0 mol 0 mol 0 molreacted -4.0 mol -1.0 mol +2.0 molreacted -4.0 mol -1.0 mol +2.0 molFinish 0 molFinish 0 mol 0 mol 0 mol 2.0 mol 2.0 mol NoLRNoLR

Which one is the limit reagent, Li or OWhich one is the limit reagent, Li or O22?? 2 mol Li2 mol Li22OO4.0 mol Li 4.0 mol Li ───────= ───────= 2 mol Li2 mol Li22O O theoret. producedtheoret. produced 4 mol Li 4 mol Li

2 mol Li2 mol Li22OO1.0 mol O1.0 mol O22───────=───────=2 mol Li2 mol Li22O theoret. producedO theoret. produced 1 mol O1 mol O22

Then, there is no L. R. Both react completelyThen, there is no L. R. Both react completely

4 Li(s) +4 Li(s) + O O22(g)(g) 2 Li2 Li22O(s)O(s)

Start (Start (givengiven)4.0 mol 0.5 mol 0 mol)4.0 mol 0.5 mol 0 molreacted -2.0 mol -0.5 mol +1.0 molreacted -2.0 mol -0.5 mol +1.0 mol

Finish 2.0 molFinish 2.0 mol 0 mol 0 mol 1.0 mol 1.0 mol OO22 LR LR

Which one is the limit reagent, Li or OWhich one is the limit reagent, Li or O22?? 2 mol Li2 mol Li22OO4.0 mol Li 4.0 mol Li ───────= ───────= 2 mol Li2 mol Li22O O theoret. producedtheoret. produced 4 mol Li 4 mol Li

2 mol Li2 mol Li22OO0.5 mol O0.5 mol O22───────= ───────= 1 mol Li1 mol Li22O theort. producedO theort. produced 1 mol O1 mol O22

Then, Then, OO22 is the L. R. 1.0 mole of Li is the L. R. 1.0 mole of Li22O formed; 2 mole of O formed; 2 mole of Li excessLi excess

4 Li(s)4 Li(s) + + O O22(g)(g) 2 Li2 Li22O(s)O(s)

startstart 8.0 mol 8.0 mol 3.0 mol 3.0 mol 0 mol 0 mol

reacted -8.0 mol -2.0 mol +4.0 molreacted -8.0 mol -2.0 mol +4.0 mol

Finish 0 molFinish 0 mol 1.0 mol1.0 mol 4.0 mol 4.0 mol Li Li LRLR

Homework…Homework…

Example: Dihydrogen sulfide and sulfur Example: Dihydrogen sulfide and sulfur dioxide react to form sulfur & water. How dioxide react to form sulfur & water. How much sulfur is formed when 5.00 g much sulfur is formed when 5.00 g dihydrogen sulfide are mixed with 5.00 g of dihydrogen sulfide are mixed with 5.00 g of sulfur dioxide? How many g of excess sulfur dioxide? How many g of excess reagent remain after reaction?reagent remain after reaction?Equation

HH22S(g) + SOS(g) + SO22(g) S S88(s) + H(s) + H22O(l)O(l)

balancing (try it)

16 H16 H22S(g) + 8 SOS(g) + 8 SO22(g) (g) 3 S 3 S88(s) + 16H(s) + 16H22O(l)O(l)

Firstly: calculate moles (of molecules) of HFirstly: calculate moles (of molecules) of H22S S

and SOand SO22. Then, determine Limiting Ractant. Then, determine Limiting Ractant 1 mol H2S5.00 g H2S x ————— = 0.147 mol H2S 34.1 g H2S

1 mol SO2 5.00 g SO2 x ————— = 0.0780 mol SO2 64.1 g SO2

How much S8?

16 H16 H22S(g) + 8 SOS(g) + 8 SO22(g) (g) 3 S 3 S88(s) + 16 H(s) + 16 H22O(l) O(l)

3 mol S8 0.147 mol H2S x ————— = 0.0276 mol S8 16 mol H2S (smaller) 3 mol S8 0.0780 mol SO2 x ————— = 0.0292 mol S8 8 mol SO2 HH22S is the L.R.S is the L.R.

Theoretically, 0.147 mol of HTheoretically, 0.147 mol of H22S produces 0.0276 S produces 0.0276

mol Smol S88, and , and 0.0780 mol SO2 produces 0.0292 produces 0.0292

mol Smol S88. Then, H. Then, H22S is the S is the Limiting Reactant, and Limiting Reactant, and

SOSO22 is the excess reagent. is the excess reagent.To determine the amount of sulfur we use H2S:

16 H16 H22S(g) + 8 SOS(g) + 8 SO22(g) (g) 3 S 3 S88(s) + 16 H(s) + 16 H22O(l)O(l)

3 mol S8 already done

0.147 mol H2S x ————— = 0.0276 mol S8 16 mol H2S MW (S8) = 8 x 32.066 = 256.6 g/mol

256.5 g S8 0.0276 mol S8 x ————— = 7.08 g S8 1 mol S8

How many g of excess reagent remain after How many g of excess reagent remain after reaction? That is SOreaction? That is SO22

We need to determine the amount of SO2 that reacts with 0.147 mol of H2S (the LR):

16 H16 H22S(g) + 8 SOS(g) + 8 SO22(g) (g) 3 S 3 S88(s) + 16 H(s) + 16 H22O(l)O(l)

8 mol SO2 0.147 mol H2S x ————— = 0.0735 mol SO2 16 mol H2S

0.0780 mol SO2 initial – 0.0735 mol reacted = 0.0045 mol remaining after reaction MW (SO2) = 32.066 + 2 x 16.00 = 64.07 g/mol

64.07 g SO2 0.0045 mol SO2 x —————— = 0.29 g SO2 (excess) 1 mol SO2

Percent YieldPercent Yield• Theoretical Yield: the maximum amount of Theoretical Yield: the maximum amount of

product that product that can becan be obtained from a chemical obtained from a chemical reaction. It is the one we calculate from the reaction. It is the one we calculate from the chemical equations chemical equations

• Actual Yield: the amount of product that Actual Yield: the amount of product that isis experimentally obtained from a reactionexperimentally obtained from a reaction—it —it is is less than the theoretical yield (due to …)less than the theoretical yield (due to …)

actual yieldactual yield% yield = % yield = ———————————————— x 100 x 100 theoretical yieldtheoretical yield

% yield % yield 100% 100%

Example: Benzene, CExample: Benzene, C66HH66, reacts with , reacts with

bromine to form bromobenzene, Cbromine to form bromobenzene, C66HH55Br, and Br, and

hydrogen bromide. Reaction of 8.00 g hydrogen bromide. Reaction of 8.00 g benzene with excess bromine yielded 12.85 benzene with excess bromine yielded 12.85 g bromobenzene. Calculate % yield of g bromobenzene. Calculate % yield of bromobenzene.bromobenzene.

• actual yield = 12.85 g Cactual yield = 12.85 g C66HH55BrBr

• CC66HH66 + Br + Br22 C C66HH55Br + HBrBr + HBr

• To calculate %yield we need to know the To calculate %yield we need to know the theoretical yield. We will use 8.0 g Ctheoretical yield. We will use 8.0 g C66HH66 ( (LRLR) )

because bromine is the excess reagentbecause bromine is the excess reagent

CC66HH66 + Br + Br22 C C66HH55Br + HBrBr + HBr

MW (MW (C6H6) = 6 x 12.011 + 6 x 1.008 = 78.11 g/mol

How many moles of CHow many moles of C66HH66??

1mol C6H6 8.00 g C6H6 x —————— = 0.102 mol C6H6 78. 11 g C6H6

How many moles of How many moles of C6H5Br are produced?

1mol C6H5Br0.102 mol C6H6 x —————— = 0.102 mol C6H5Br 1 mol C6H6

Converting moles of Converting moles of C6H5Br to grams… The theoretic.

yieldyield 157.0 g C6H5Br 0.102 mol C6H5Br x ——————— = 16.0 g C6H5Br 1 mol C6H5Br

the percent yield of Cthe percent yield of C66HH55BrBr

actual yieldactual yield% yield = % yield = ———————————————— x 100 x 100 theoretical yieldtheoretical yield

12.85 g C6H5Br % yield =% yield = ——————— x 100 = 80.3% 16.0 g C6H5Br

64.0 g of methanol, CH64.0 g of methanol, CH33OH, were expected to be OH, were expected to be

produced through the reaction produced through the reaction CO(g) + 2H CO(g) + 2H22(g) (g) CH CH33OH(OH(ll) )

One student got 56.0 g of methanol for that reaction One student got 56.0 g of methanol for that reaction in the laboratory. What was the %yield of methanol?in the laboratory. What was the %yield of methanol?

64.0 g of methanol is the theoretical yield 64.0 g of methanol is the theoretical yield (expected) (expected)

56.0 g is the actual yield (in the laboratory)56.0 g is the actual yield (in the laboratory)

56.0 g % Yield = ———— 100 = 87.5 % 64.0 g

Analysis of mineral sampleAnalysis of mineral sample

The mineral cerussite is mostly PbCOThe mineral cerussite is mostly PbCO33, but , but other substances are present. To analyze for other substances are present. To analyze for the PbCOthe PbCO33 content, a sample of mineral is content, a sample of mineral is first treated with nitric acid to dissolve PbCOfirst treated with nitric acid to dissolve PbCO33

PbCOPbCO33(s) + 2 HNO(s) + 2 HNO33(aq) (aq) Pb(NO Pb(NO33))22(aq) + H(aq) + H22O(l) + COO(l) + CO22(g)(g)

On adding sulfuric acid, lead(II) sulfate precipitates.

Pb(NO3)2(aq) + H2SO4(aq) PbSO PbSO44(s) + 2 HNO(s) + 2 HNO33(aq)(aq)

Solid PbSOSolid PbSO44(s) is isolated and weighed. (s) is isolated and weighed. Suppose aSuppose a0.583 g sample of mineral produced 0.628 g PbSO0.583 g sample of mineral produced 0.628 g PbSO44. . What is the mass % of PbCOWhat is the mass % of PbCO33 in the mineral sample? in the mineral sample?KEY: 1 mol PbCOKEY: 1 mol PbCO33 1 mol Pb(NO 1 mol Pb(NO33))22 1 mol PbSO 1 mol PbSO44

Calculating moles and gramsCalculating moles and grams

First we calculate moles of PbSOFirst we calculate moles of PbSO44 ( (FW = 303.3)FW = 303.3)

1 mol PbSO1 mol PbSO44

0.628 g PbSO0.628 g PbSO44x x —————— = 0.00207 mol PbSO4

303.3 g PbSO4

Now, we determine moles of PbCO3

1 mol Pb(NO1 mol Pb(NO33))22 1 mol PbCO 1 mol PbCO33

0.00207 mol PbSO0.00207 mol PbSO44xx——————— x ——————— 1 mol PbSO4 1 mol Pb(NO3)2

= 0.00207 mol PbCO3 that must be converted to g

grams and % of PbCOgrams and % of PbCO33 (FW = 267.2 g/mol) (FW = 267.2 g/mol)

g PbCOg PbCO33 out of mol PbCO out of mol PbCO33

267.2 g267.2 g PbCO PbCO33

0.00207 mol PbCO3 x x —————— = 0.553 g PbCO3

1 mol PbCO3

Now the % PbCO3 in the mineral sample (0.583 g)

0.553 g PbCO0.553 g PbCO33 Mass percent PbCOMass percent PbCO33= = ——————— x 100 = 94.9% 0.583 g sample

Combustion AnalysisCombustion Analysis

• sample is burned in oxygensample is burned in oxygen

• C C CO CO22

• H H H H22OO

Example:Example:

• compound containing only C, H, & Ocompound containing only C, H, & O

• combustion of 28.64 mg sample of combustion of 28.64 mg sample of compound compound

88.02 mg CO88.02 mg CO22

27.03 mg H27.03 mg H22OO

• MW = 286.5 MW = 286.5 (it is a given data)(it is a given data)

• determine molecular formuladetermine molecular formula

Example:Example:

• CCxxHHyyOOzz + O + O22 CO CO22 + H + H22OO

• C C CO CO22, H , H H H22O, O O, O CO CO22 & H & H22O O

• det. mass of C and of Hdet. mass of C and of H

• det. mass of Odet. mass of O

• calc. mol C, H, & Ocalc. mol C, H, & O

• det. empirical formuladet. empirical formula

• det. molecular formuladet. molecular formula

Example:Example:

1 mmol CO2 1 mmol C88.02 mg CO2 x —————— x —————— x 44.01 mg CO2 1 mmol CO2

12.01 mg C x —————— = 24.02 mg C = 2.000 mmol of C 1 mmol C

… now H

Example:Example: 1 mmol H2O 2 mmol H27.03 mg H2O x —————— x —————— x 18.02 mg H2O 1 mmol H2O 1.008 mg H x —————— = 3.02 mg H = 3.000 mmol of H 1 mmol H

… now mg of oxygen

mg sample = mg C + mg H + mg Omg sample = mg C + mg H + mg O28.64 mg = 24.02 mg C + 3.024 mg H + mg O28.64 mg = 24.02 mg C + 3.024 mg H + mg O

mg O = 28.64 – 24.02 – 3.024 = 1.60 mg Omg O = 28.64 – 24.02 – 3.024 = 1.60 mg O

Example:Example:

1 mmol C1 mmol C24.02 mg C x 24.02 mg C x ————— = 2.000 mmol of C = 2.000 mmol of C 12.01 mg12.01 mg

1 mmol H1 mmol H3.024 mg H x 3.024 mg H x ————— = 3.000 mmol of H = 3.000 mmol of H 1.008 mg H1.008 mg H

1 mmol O1 mmol O1.60 mg O x 1.60 mg O x ————— = 0.100 mmol of O = 0.100 mmol of O 16.00 mg O16.00 mg O

Now, we will divide all by the smallest…Now, we will divide all by the smallest…

Example:Example:CC2.02.0HH3.03.0OO0.10.1

0.10.1 0.1 0.10.1 0.1

CC2020HH3030O is the empirical formulaO is the empirical formula

weight of E.F. = 20weight of E.F. = 2012.01 + 3012.01 + 301.008 + 16.001.008 + 16.00 = 286.4= 286.4

MW 286.5MW 286.5n = n = —————— = ————— = 1 W of EF 286.4 W of EF 286.4

molecular formula = Cmolecular formula = C2020HH3030O (O (same as empsame as emp))

Problem: Methane, CHProblem: Methane, CH44, burns in oxygen., burns in oxygen.

(a) What are the products of the reaction? CO(a) What are the products of the reaction? CO22 + H + H22OO(b) Write the balance equation for the reaction.(b) Write the balance equation for the reaction. CHCH44(g) + 2 O(g) + 2 O22(g) (g) CO CO22(g) + 2 H(g) + 2 H22O(g)O(g) (c) What mass (g) of O(c) What mass (g) of O2 2 is required for complete is required for complete

combustion of 25.5 g of methane?combustion of 25.5 g of methane?Firstly, we need to calculate moles of methaneFirstly, we need to calculate moles of methane

1 mol CH1 mol CH44

25.5 g CH25.5 g CH44 x ─────── = 1.59 mol CH x ─────── = 1.59 mol CH44 Then, useThen, use 16.04 g CH16.04 g CH44 coeffientscoeffients and molar mass Oand molar mass O22

2 mol O2 mol O22 32.00 g O 32.00 g O221.59 mol CH1.59 mol CH4 4 ————— ————— = 102 g O2

1 mol CH1 mol CH44 1 mol O 1 mol O22

Problem: Hexane, CProblem: Hexane, C66HH1414, burns in oxygen to give , burns in oxygen to give

COCO22 and H and H22O.O. (a) Write a balance equation for the reaction.(a) Write a balance equation for the reaction.

2 C2 C66HH1414((ll) + 19 O) + 19 O22(g) (g) 12 CO 12 CO22(g) + 14 H(g) + 14 H22O(O(ll))

(b) If 215 g C(b) If 215 g C66HH1414 are mixed with 215 g of O are mixed with 215 g of O22, what , what

masses of COmasses of CO22 and H and H22O are produced?O are produced?

Firstly, we need to calculate moles of CFirstly, we need to calculate moles of C66HH1414 and O and O22

1 mol C1 mol C66HH1414 2.502.50215 g C215 g C66HH1414x ─────── = 2.50 mol Cx ─────── = 2.50 mol C66HH1414 ───= 1.25───= 1.25 86.17 g C86.17 g C66HH1414 22 1 mol O1 mol O22 6.726.72215 g O215 g O22 ————— = 6.72 mol O2 ─── = 0.354─── = 0.354 32.00 g O32.00 g O22 19 19 Now, the Limiting Reagent (Now, the Limiting Reagent (It seems to be OIt seems to be O22))

2 C2 C66HH1414((ll) + 19 O) + 19 O22(g) (g) 12 CO 12 CO22(g) + 14 H(g) + 14 H22O(O(ll))

12 mol CO12 mol CO22 2.50 mol C2.50 mol C66HH1414 x ──────── = x ──────── = 15.0 mol CO15.0 mol CO22 2 mol C2 mol C66HH1414

12 mol CO12 mol CO22 6.72 mol O6.72 mol O22 x ─────── = x ─────── = 4.24 mol CO4.24 mol CO22 OO22 is the L.R. is the L.R. 19 mol O19 mol O66

44.01 g CO44.01 g CO22 4.24 mol CO4.24 mol CO2 2 —————— = 187 g CO2

1 mol CO1 mol CO22

14 mol H14 mol H22O 18.02 g HO 18.02 g H22O O 6.72 mol O6.72 mol O22————— —————— = 89.2 g H2O 19 mol O19 mol O22 1 mol H 1 mol H22OO

Prob…Prob…2 C2 C66HH1414((ll) + 19 O) + 19 O22(g) (g) 12 CO 12 CO22(g) + 14 H(g) + 14 H22O(O(ll))

What mass of excess reagent remains after reaction? What mass of excess reagent remains after reaction? CC66HH1414

The 6.72 moles of The 6.72 moles of OO22 react with react with CC66HH1414

2 mol C2 mol C66HH1414 6.72 mol O6.72 mol O22—————— = 0.707 mol CC66HH1414

19 mol O19 mol O22

initialinitial amount of C amount of C66HH1414 was 2.50 mol was 2.50 mol − 0.707 mol − 0.707 mol reactedreacted = =

1.79 mol of C1.79 mol of C66HH1414 remainingremaining

86.17 g C86.17 g C66HH1414 1.79 mol C1.79 mol C66HH14 14 ——————— = 154 g CC66HH1414 remaining remaining 1 mol C1 mol C66HH1414

Problem Problem A mixture of CuSOA mixture of CuSO4 4 and CuSOand CuSO44•5H•5H22O has a mass of 1.245 g. O has a mass of 1.245 g. After heating to drive off all the water, the mass is only 0.832 After heating to drive off all the water, the mass is only 0.832 g. What is the mass percent of g. What is the mass percent of CuSOCuSO44•5H•5H22O in the mixture?O in the mixture?FW CuSOFW CuSO44•5H•5H22O = 159.61 + 5O = 159.61 + 518.02 = 249.69 g/mol CuSOCuSO44 H H22OO

initial mass initial mass − final mass = 1.245 g − 0.832 g = 0.413 g H− final mass = 1.245 g − 0.832 g = 0.413 g H22OO

The amount of CuSOThe amount of CuSO44•5H•5H22O in the mixture can be calculated:O in the mixture can be calculated:

249.69 g CuSO249.69 g CuSO44•5H•5H22OO0.413 g H0.413 g H22O O —————————— = 1.14 g CuSOCuSO44•5H•5H22OO 5518.02 g H2O

1.14 g CuSOCuSO44•5H•5H22OO% CuSO% CuSO44•5H•5H22O = O = ————————— 100 = 91.6% 1.245 g mixture 1.245 g mixture

ProblemProblemNickel forms a compound with CO, Ni(CO)Nickel forms a compound with CO, Ni(CO)xx. To determine its . To determine its formula, you carefully heat a 0.0973-g sample in air to convert formula, you carefully heat a 0.0973-g sample in air to convert the Ni in 0.0426 g NiO and the CO in 0.100 g of COthe Ni in 0.0426 g NiO and the CO in 0.100 g of CO22. What is . What is the empirical formula of Ni(CO)the empirical formula of Ni(CO)xx??

From moles of NiO and COFrom moles of NiO and CO22 we can calculate moles of Ni and CO: we can calculate moles of Ni and CO:

molar mass of NiO = 74.71 g/mol of COmolar mass of NiO = 74.71 g/mol of CO22 = 44.01 g/mol = 44.01 g/mol 1 mol NiO 1 mol Ni1 mol NiO 1 mol Ni0.0426 g NiOx ────── x ────── = 0.000570 mol Ni0.0426 g NiOx ────── x ────── = 0.000570 mol Ni 74.71 g NiO 1 mol NiO74.71 g NiO 1 mol NiO

1 mol CO1 mol CO22 1 mol CO 1 mol CO0.100 g CO0.100 g CO22 x ─────── x ─────── = 0.00227 mol CO x ─────── x ─────── = 0.00227 mol CO 44.01 g CO44.01 g CO22 1 mol CO 1 mol CO22

0.00227 mol CO0.00227 mol CO────────── ────────── = 4 Emp. Form. = Ni(CO)= 4 Emp. Form. = Ni(CO)44

0.000570 mol Ni0.000570 mol Ni

Aluminum chloride is made by treating scrap aluminum with chlorine. 2 Al(s) + 3 Cl2(g) 2 AlCl3(s)

If you begin with 2.70 g of Al and 4.05 g of Cl2,

a) Which reactant is limiting? 1 mol Al2.70 g Al=0.100 mol Al 4.05 g Cl2 = 0.0571 mol 27.0 g Al

2 mol AlCl3

0.0571 mol Cl2 = 0.0381 mol AlCl3

3 mol Cl2

Cl2 is the L.R. 2 mol AlCl3

0.100 mol Al = 0.100 mol AlCl3 2 mol Al

Contd… 2 Al(s) + 3 Cl2(g) 2 AlCl3(s)

b) What mass of AlCl3 can be produced?

We must use Cl2, the LR, to calculate that

0.0381 mol AlCl3 is produced from the 0.0571 molCl2

133.35 g AlCl3

0.0381 mol AlCl3 = 5.08 g AlCl3

1 mol AlCl3

Contd… 2 Al(s) + 3 Cl2(g) 2 AlCl3(s)

b) What mass of the excess reactant remains when the reaction is finished? That is Al.

The remaining Al will be the initial minus the reacted.

For calculating the reacted we must use Cl2, the LR.

2 mol Al0.0571 mol Cl2 = 0.0381 mol Al (reacted) 3 mol Cl2

0.100 mol (initial) − 0.0381 mol (reacted) = 0.0619 mol

remaining

27.0 g Al0.0619 mol Al = 1.67 g Al 1 mol Al

Styrene consists of only C and H. If 0.438 g styrene is burned in excess oxygen and produces 1.481 g CO2 and 0.303 g H2O, what is the empirical formula of styrene?

(styrene) CxHy(l) + O2(g) CO2(g) + H2O(l)mol of C in CO2 = mol of C in styrenemol of H in H2O = mol of H in styrene

1 mol CO2 1 mol C1.481 g CO2 = 0.03365 mol C 44.01 g CO2 1 mol CO2

1 mol H2O 2 mol H0.303 g H2O = 0.0336 mol H 18.02 g H2O 1 mol H2O

Cntd…

To determine the empirical formula we must

divide the two number of moles by the smallest

C0.03365 H0.0336

0.0336 0.0336

The empirical formula is CH