balancing equations balancing, writing, and naming equations

Balancing EquationsBalancing Equations

Balancing, Writing, and Balancing, Writing, and Naming EquationsNaming Equations

2

Mass:Mass: proton = 1.00728 amuproton = 1.00728 amu

neutron = 1.0086 amuneutron = 1.0086 amu

electron = 0.0005486electron = 0.00054861212C atom = 12.00000 amuC atom = 12.00000 amu1313C atom = 13.00335 amuC atom = 13.00335 amu

amu = Atomic Mass Unitsamu = Atomic Mass Units

Atomic and Molecular MassAtomic and Molecular Mass

1 amu mass of carbon 12 atom12

3


The atomic masses as tabulated in the periodic The atomic masses as tabulated in the periodic table are the averages of the naturally table are the averages of the naturally occurring isotopes.occurring isotopes.

Mass of C = average of Mass of C = average of 1212C and C and 1313C C

= 0.9889 x 12 amu + = 0.9889 x 12 amu +

0.0111 x 13.0034 amu0.0111 x 13.0034 amu

= 12.011 amu= 12.011 amu

4


The mass of a molecule is just the sum of the masses of the atoms making up the molecule.

m(Cm(C22HH44OO22) = 2) = 2·m·mCC + 4·m + 4·mH H + 2·m+ 2·mOO

= = 22·(12.01) + 4·(1.01) + 2·(16.00)·(12.01) + 4·(1.01) + 2·(16.00)

= 60.06 amu= 60.06 amu

http://www.wfu.edu/chemistry/courses/jonesbt/111/Chapter3/MW.xls



7

Avogadro and the MoleAvogadro and the Mole

One moleOne mole of a substance is the gram of a substance is the gram mass value equal to the amu mass of the mass value equal to the amu mass of the substance.substance.

One moleOne mole of any substance contains of any substance contains 6.02 x 106.02 x 102323 units of that substance. units of that substance.

Avogadro’s Number (NAvogadro’s Number (NAA,, 6.022 x 6.022 x

10102323)) is the numerical value assigned to is the numerical value assigned to the unit, 1 mole.the unit, 1 mole.

8


9


Methionine, an amino acid used by organisms Methionine, an amino acid used by organisms to make proteins, is represented below. Write to make proteins, is represented below. Write the formula for methionine and calculate its the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)yellow = S; ivory = H)

10


The Mole:The Mole: Allows us to Allows us to

make comparisons make comparisons

between substances between substances

that have that have

different different

masses.masses.

LecturePLUS Timberlake 99 11

Writing Mole FactorsWriting Mole Factors

4 Fe + 3 O4 Fe + 3 O22 2 Fe 2 Fe22OO33

Fe and OFe and O22 4 mole4 mole Fe Fe and and 3 mole 3 mole OO22

3 mole 3 mole OO22 4 mole 4 mole FeFe

Fe and FeFe and Fe22OO33 4 mole 4 mole FeFe andand 2 mole 2 mole

FeFe22OO33

2 mole 2 mole FeFe22OO33 4 mole 4 mole FeFe

OO22 and Fe and Fe22OO3 3 3 mole 3 mole OO22 and and 2 mole 2 mole FeFe22OO33

2 mole 2 mole FeFe22OO33 3 mole 3 mole OO22


Learning Check S1Learning Check S1

3 H3 H22(g) + N(g) + N22(g) 2 NH(g) 2 NH33(g)(g)

A. A mole factor for HA. A mole factor for H2 2 and Nand N22 is is

1) 1) 3 mole N 3 mole N22 2) 2) 1 mole N 1 mole N22 3) 3) 1 mole N 1 mole N22

1 mole H1 mole H22 3 mole H 3 mole H22 2 mole H 2 mole H22

B. A mole factor for NHB. A mole factor for NH33 and H and H22 is is

1) 1) 1 mole H1 mole H2 2 2) 2) 2 mole NH 2 mole NH33 3) 3) 3 mole N 3 mole N22

2 mole NH2 mole NH33 3 mole H 3 mole H22 2 mole NH 2 mole NH33


Solution S1Solution S1

3 H3 H22(g) + N(g) + N22(g) 2 NH(g) 2 NH33(g)(g)

A. A mole factor for HA. A mole factor for H2 2 and Nand N22 is is

2) 2) 1 mole N1 mole N22

3 mole H3 mole H22

B. A mole factor for NHB. A mole factor for NH33 and H and H22 is is

2) 2) 2 mole NH2 mole NH33

3 mole H3 mole H22


Chemical CalculationsChemical Calculations

4 Fe + 4 Fe + 3 O3 O22 2 Fe2 Fe22OO33

How many How many moles of Femoles of Fe22OO33 are produced are produced

when 6.0when 6.0 moles O moles O22 react? react?

6.06.0 mole O mole O22 x x 2 mole Fe 2 mole Fe22OO33 = 4.0 = 4.0 mole mole

FeFe22OO33

3 mole O3 mole O22


Learning Check S2Learning Check S2

4 Fe + 3 O4 Fe + 3 O22 2 Fe 2 Fe22OO33

How many moles of Fe are needed to react How many moles of Fe are needed to react

with 12.0 mole of Owith 12.0 mole of O22??

1) 3.00 mole Fe 1) 3.00 mole Fe

2) 9.00 mole Fe2) 9.00 mole Fe

3) 16.0 mole Fe3) 16.0 mole Fe



44 Fe + Fe + 33 O O22 2 Fe 2 Fe22OO33

12.0 mole O12.0 mole O22 x x 44 mole Fe mole Fe = 16.0 = 16.0 mole Femole Fe

33 mole O mole O22


Learning Check S 3Learning Check S 3

4 Fe + 3 O4 Fe + 3 O22 2 Fe 2 Fe22OO33

How many grams of OHow many grams of O22 are needed to produce 0.400 are needed to produce 0.400

mole of Femole of Fe22OO33??

1) 38.4 g O1) 38.4 g O22

2) 19.2 g O2) 19.2 g O22

3) 1.90 g O3) 1.90 g O22


Solution S 3Solution S 3

2) 19.2 g O2) 19.2 g O22

0.400 mole Fe0.400 mole Fe22OO33 x x 3 mole O3 mole O22 x x 32.0 g O32.0 g O22

2 mole Fe2 mole Fe22OO33 1 mole 1 mole OO22

= = 19.2 g O19.2 g O22


Mass of A ReactionMass of A Reaction

The reaction between HThe reaction between H22 and O and O22

produces 13.1 g of water. How many produces 13.1 g of water. How many grams of Ograms of O22 reacted? reacted?

Write the equationWrite the equation

HH22 (g) + O (g) + O22 (g) (g) HH22O (g)O (g)

Balance the equationBalance the equation

2 H2 H22 (g) + O (g) + O22 (g) (g) 2 H2 H22O (g)O (g)


Organize dataOrganize data mole bridgemole bridge

2 H2 H22 (g) + O (g) + O22 (g) (g) 2 H2 H22O (g)O (g)

? g? g 13.1 g 13.1 g

Plan Plan g Hg H22O mole HO mole H22O mole OO mole O22 O O22

SetupSetup

13.1 g H13.1 g H22O x O x 1 mole H1 mole H22O O x x 1 mole O1 mole O2 2 x x 32.0 32.0

g Og O22 18.0 g H 18.0 g H22OO 2 mole H2 mole H22O O

1 mole O1 mole O22

= = 11.6 g O11.6 g O22


Learning Check S 4Learning Check S 4

How many grams of OHow many grams of O22 are need to are need to

react 50.0 grams of Na in the react 50.0 grams of Na in the

reactionreaction

4 Na + O4 Na + O22 2 Na 2 Na22OO

Complete the set up:Complete the set up:

50.0 g Na x 50.0 g Na x 1 mole Na1 mole Na x ________ x _______ x ________ x _______

23.0 g Na23.0 g Na


Solution S 4Solution S 4

4 Na + O4 Na + O22 2 Na 2 Na22OO

50.0 g Na x 50.0 g Na x 1 mole Na1 mole Na x x 1 mole O1 mole O22 x x 32.0 g 32.0 g

23.0 g Na 4 mole Na 1 mole 23.0 g Na 4 mole Na 1 mole

OO22

= g O= g O22


Learning Check S5 Learning Check S5

Acetylene gas CAcetylene gas C22HH22 burns in the burns in the

oxyactylene torch for welding. How oxyactylene torch for welding. How many grams of Cmany grams of C22HH22 are burned if the are burned if the

reaction produces 75.0 g of COreaction produces 75.0 g of CO22? ?

2 C2 C22HH22 + 5 O + 5 O22 4 CO 4 CO22 + 2 + 2

HH22OO



2 C2 C22HH22 + 5 O + 5 O22 4 CO 4 CO22 + 2 H + 2 H22OO

75.0 g CO75.0 g CO22 x x 1 mole CO1 mole CO22 x x 2 mole C2 mole C22HH22 x x 26.0 g 26.0 g

CC22HH22 44.0 g CO 44.0 g CO2 2 4 mole CO4 mole CO2 2 1 mole 1 mole

CC22HH22 = = 22.2 g C22.2 g C22HH2 2

25

Balancing Chemical EquationsBalancing Chemical Equations A A balancedbalanced chemical equation represents chemical equation represents

the conversion of the reactants to products the conversion of the reactants to products such that the number of atoms of each such that the number of atoms of each element is conserved.element is conserved.

reactants reactants products products

limestone limestone quicklime + gas quicklime + gas

Calcium carbonate Calcium carbonate calcium oxide + carbon dioxide calcium oxide + carbon dioxide

CaCOCaCO33(s)(s) CaO CaO(s)(s) + CO + CO22(g)(g)

26

Balancing Chemical EquationsBalancing Chemical Equations

CaCOCaCO33(s)(s) CaO CaO(s)(s) + CO + CO22(g)(g)

The letters in parentheses following each The letters in parentheses following each

substance are called substance are called State SymbolsState Symbols

(g) (g) → gas→ gas (l) → liquid(l) → liquid (s) → solid(s) → solid (aq) → (aq) →

aqueousaqueous

27


A balanced equation MUST have the same number A balanced equation MUST have the same number

of atoms of each element on both sides of the of atoms of each element on both sides of the

equation.equation.

HH22 + O + O22 → H→ H22OO Not BalancedNot Balanced

HH22 + + ½½OO22 → H→ H22OO BalancedBalanced

2H2H22 + O + O22 → 2H→ 2H22OO BalancedBalanced

28


The numbers multiplying chemical formulasThe numbers multiplying chemical formulas

in a chemical equation are called:in a chemical equation are called:

Stoichiometric Coefficients (S.C.)Stoichiometric Coefficients (S.C.)

2H2H22 + O + O22 → 2H→ 2H22OO BalancedBalanced

Here 2, 1, and 2 are stoichiometric coefficients.Here 2, 1, and 2 are stoichiometric coefficients.

29


Hints for Balancing Chemical Equations:Hints for Balancing Chemical Equations:

1)1) Save single element molecules for last.Save single element molecules for last.

2)2) Try not to change the S.C. of a molecule Try not to change the S.C. of a molecule

containing an element that is already balanced.containing an element that is already balanced.

3)3) If possible, begin with the most complex If possible, begin with the most complex

molecule that has no elements balanced.molecule that has no elements balanced.

30


Example 1:Example 1: CHCH44 + O + O22 → CO→ CO22 + H + H22OO

Balance OBalance O22 last last

C is already balancedC is already balanced

Start by changing S.C. of HStart by changing S.C. of H22O to balance HO to balance H

CHCH44 + O + O22 → CO→ CO22 + 2H + 2H22OO

31


Example 1:Example 1: CH CH44 + O + O22 → CO→ CO22 + 2H + 2H22OO

Now C and H are balancedNow C and H are balanced

Balance O by changing the S.C. of OBalance O by changing the S.C. of O22

CHCH44 + 2O + 2O22 → CO→ CO22 + 2H + 2H22OO

BALANCED!BALANCED!

32


Example 2:Example 2: B B22HH66 + O + O22 → → BB22OO33 + H + H22OO

Balance O lastBalance O last

B is already balancedB is already balanced

Start by changing S.C. of HStart by changing S.C. of H22O:O:

BB22HH66 + O + O22 → → BB22OO33 + 3H + 3H22OO

33


Example 2:Example 2: B B22HH66 + O + O22 → → BB22OO33 + 3H + 3H22OO

B and H are balancedB and H are balanced

Balance O by changing S.C. of OBalance O by changing S.C. of O22

BB22HH66 + 3O + 3O22 → → BB22OO33 + 3H + 3H22OO

BALANCED!BALANCED!

34


Example 3:Example 3: MnOMnO22 + KOH + O + KOH + O22 → → KK22MnOMnO44 + H + H22OO

Balance O lastBalance O last

Mn is already balancedMn is already balanced

Change S.C. of KOH to balance KChange S.C. of KOH to balance K

MnOMnO22 + 2KOH + O + 2KOH + O22 → → KK22MnOMnO44 + H + H22OO

35


Example 3:Example 3: MnOMnO22 + 2KOH + O + 2KOH + O22 → → KK22MnOMnO44 + H + H22OO

Mn, K, and H are balanced (H was balanced by chance)Mn, K, and H are balanced (H was balanced by chance)

Balance OBalance O

MnOMnO22 + 2KOH + + 2KOH + ½½OO22 → → KK22MnOMnO44 + H + H22OO

oror

2MnO2MnO22 + 4KOH + O + 4KOH + O22 → 2→ 2KK22MnOMnO44 + 2H + 2H22OO

36


Example 4:Example 4: NaNONaNO22 + H + H22SOSO44→→

NO + HNONO + HNO33 + H + H22O + NaO + Na22SOSO44

Hard one (no single element molecules)Hard one (no single element molecules)

S is balancedS is balanced

Start with NaNOStart with NaNO22 to balance Na to balance Na

2NaNO2NaNO22 + H + H22SOSO44→ → NO + HNONO + HNO33 + H + H22O + NaO + Na22SOSO44

37


Example 4:Example 4: 2NaNO2NaNO22 + H + H22SOSO44→→

NO + HNONO + HNO33 + H + H22O + NaO + Na22SOSO44

S, Na, and N are balancedS, Na, and N are balanced

Cannot balance H without changing S.C. for HCannot balance H without changing S.C. for H22SOSO44!!

Boo!Boo! Option 1: trial and errorOption 1: trial and error

Option 2: Go on to next problem!Option 2: Go on to next problem!

38


Balance the following equations:Balance the following equations:

CC66HH1212OO66 →→ C C22HH66O + COO + CO22

Fe + OFe + O22 →→ Fe Fe22OO33

NHNH33 + Cl + Cl22 →→ N N22HH44 + NH + NH44ClCl

KClOKClO33 + C + C1212HH2222OO1111 →→ KCl + CO KCl + CO22 + H + H22OO

39



CC66HH1212OO66 →→ 2C 2C22HH66O + 2COO + 2CO22

Fe + OFe + O22 →→ Fe Fe22OO33



40




4Fe + 3O4Fe + 3O22 →→ 2Fe 2Fe22OO33 (balance O first) (balance O first)



41




4Fe + 3O4Fe + 3O22 →→ 2Fe 2Fe22OO33 (balance O first) (balance O first)


N:H is 1:3 on left, must get 1:3 on right!N:H is 1:3 on left, must get 1:3 on right!

42



N:H is 1:3 on left, must get 1:3 on right!N:H is 1:3 on left, must get 1:3 on right!

4NH4NH33 + Cl + Cl22 →→ N N22HH44 + 2NH + 2NH44ClCl

43

Balancing Chemical EquationsBalancing Chemical Equations Balance the following equations:Balance the following equations:


4Fe + 3O4Fe + 3O22 →→ 2Fe 2Fe22OO33

4NH4NH33 + Cl + Cl22 →→ N N22HH44 + 2NH + 2NH44ClCl

KClOKClO33 + C + C1212HH2222OO1111 →→ KCl + CO KCl + CO22 + H + H22O (tough!)O (tough!)

44




balance Cbalance C

KClOKClO33 + C + C1212HH2222OO1111 →→ KCl + 12CO KCl + 12CO22 + H + H22OO

balance Hbalance H

KClOKClO33 + C + C1212HH2222OO1111 →→ KCl + 12CO KCl + 12CO22 + 11H + 11H22OO

balance Obalance O

8KClO8KClO33 + C + C1212HH2222OO1111 →→ KCl + 12CO KCl + 12CO22 + 11H + 11H22OO

45



8KClO8KClO33 + C + C1212HH2222OO1111 →→ KCl + 12CO KCl + 12CO22 + 11H + 11H22OO

balance K (and hope Cl is balanced)balance K (and hope Cl is balanced)

8KClO8KClO33 + C + C1212HH2222OO1111 →→ 8KCl + 12CO 8KCl + 12CO22 + +

11H11H22OO

Balanced!Balanced!

46


Write a balanced equation for the reaction of Write a balanced equation for the reaction of element A (red spheres) with element B element A (red spheres) with element B (green spheres) as represented below:(green spheres) as represented below:

47

StoichiometryStoichiometry Stoichiometry: Stoichiometry: Relates the moles of Relates the moles of

products and reactants to each other and products and reactants to each other and to measurable quantities.to measurable quantities.

48

StoichiometryStoichiometry

Aqueous solutions of NaOCl (household bleach) are Aqueous solutions of NaOCl (household bleach) are

prepared by the reaction of NaOH with Clprepared by the reaction of NaOH with Cl22::

2 NaOH(2 NaOH(aqaq) + Cl) + Cl22((gg) ) NaOCl( NaOCl(aqaq) + NaCl() + NaCl(aqaq) + H) + H22O(O(ll))

How many grams of NaOH are needed to react with 25.0 g How many grams of NaOH are needed to react with 25.0 g

of Clof Cl22??

49


2 NaOH + Cl2 NaOH + Cl22 → NaOCl + NaCl + H→ NaOCl + NaCl + H22OO

25.0 g Cl25.0 g Cl22 reacts with ? g NaOH reacts with ? g NaOH

22

22 353.0

90.70

10.25 Clmoles

Clg

ClmoleClg

NaOHgNaOHmole

NaOHg

Clmole

NaOHmoles

Clg

ClmoleClg 2.28

1

0.40

1

2

90.70

10.25

22

22

50


Calculate the molar mass of the following:Calculate the molar mass of the following:

FeFe22OO33 (Rust) (Rust)

CC66HH88OO77 (Citric acid) (Citric acid)

CC1616HH1818NN22OO44 (Penicillin G) (Penicillin G)

Balance the following, and determine how many Balance the following, and determine how many

moles of CO will react with 0.500 moles of Femoles of CO will react with 0.500 moles of Fe22OO33..

Fe Fe22OO33(s)(s) + CO + CO(g)(g) Fe Fe(s)(s) + CO + CO22(g)(g)

51


FeFe22OO33 + CO + CO →→ Fe + CO Fe + CO22

Balance (not a simple one)Balance (not a simple one)

Save Fe for lastSave Fe for last

C is balanced, but can’t balance OC is balanced, but can’t balance O

In the products the ratio C:O is 1:2 and can’t changeIn the products the ratio C:O is 1:2 and can’t change

Make the ratio C:O in reactants 1:2Make the ratio C:O in reactants 1:2

FeFe22OO33 + 3CO + 3CO →→ 2Fe + 3CO 2Fe + 3CO22

52


FeFe22OO33 + 3CO + 3CO →→ 2Fe + 3CO 2Fe + 3CO22

COmolesOFemole

COmoleOFemoles 50.1

1

3500.0

3232

53


54


Aspirin is prepared by reaction of salicylic acid Aspirin is prepared by reaction of salicylic acid

(C(C77HH66OO33) with acetic anhydride (C) with acetic anhydride (C44HH66OO33) to form ) to form

aspirin (Caspirin (C99HH88OO44) and acetic acid (CH) and acetic acid (CH33COCO22H). Use H). Use

this information to determine the mass of this information to determine the mass of

acetic anhydride required to react with 4.50 g acetic anhydride required to react with 4.50 g

of salicylic acid. How many grams of aspirin will of salicylic acid. How many grams of aspirin will

result? How many grams of acetic acid will be result? How many grams of acetic acid will be

produced as a by-product? produced as a by-product?

55


Salicylic acid + Acetic anhydride Salicylic acid + Acetic anhydride →→

Aspirin + acetic acidAspirin + acetic acid

CC77HH66OO3 3 + C+ C44HH66OO3 3 → → CC99HH88OO44 + CH + CH33COCO22HH

CC77HH66OO3 3 + C+ C44HH66OO3 3 → → CC99HH88OO44 + C + C22HH44OO22

Balanced!Balanced!

Equal # moles for allEqual # moles for all

56


4.50 g Salicylic acid (C4.50 g Salicylic acid (C77HH66OO33) = ? moles) = ? moles

MW CMW C77HH66OO33 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00

= 138.12 g/mole= 138.12 g/mole

..0326.0..12.138

..1..50.4 ASmoles

ASg

ASmoleASg

57


Since all compounds have the same S.C., there Since all compounds have the same S.C., there

must be 0.0326 moles of all 4 of them involved in must be 0.0326 moles of all 4 of them involved in

the reaction.the reaction.

g Aspirin (Cg Aspirin (C99HH88OO44) = 0.0326 moles x MW Aspirin) = 0.0326 moles x MW Aspirin

= .0326 x [9x12.01 + 8x1.008 + 4x16.00]= .0326 x [9x12.01 + 8x1.008 + 4x16.00]

=.0326 mole x 180.15 g/mole=.0326 mole x 180.15 g/mole

5.87 g Aspirin5.87 g Aspirin

58


Yields of Chemical Reactions:Yields of Chemical Reactions: If the actual If the actual

amount of product formed in a reaction is less amount of product formed in a reaction is less

than the theoretical amount, we can calculate than the theoretical amount, we can calculate

aa percentage yieldpercentage yield..

100% yieldproduct lTheoretica

yieldproduct Actual yield%

59


Dichloromethane (CHDichloromethane (CH22ClCl22) is prepared by ) is prepared by

reaction of methane (CHreaction of methane (CH44) with chlorine (Cl) with chlorine (Cl22) )

giving hydrogen chloride as a by-product. giving hydrogen chloride as a by-product.

How many grams of dichloromethane result How many grams of dichloromethane result

from the reaction of 1.85 kg of methane if from the reaction of 1.85 kg of methane if

the yield is 43.1%?the yield is 43.1%?

60


CHCH44 + Cl + Cl22 → CH→ CH22ClCl22 + HCl + HCl

BalanceBalance

CHCH44 + 2Cl + 2Cl22 → CH→ CH22ClCl22 + 2HCl + 2HCl

1.85 kg CH1.85 kg CH44 = ? moles CH = ? moles CH44

61



1.85 kg CH1.85 kg CH44 = ? moles CH = ? moles CH44

MW CHMW CH44 = 1x12.01 + 4x1.008 = 16.04 g/mole = 1x12.01 + 4x1.008 = 16.04 g/mole

44

44 115

4.16

1100085.1 CHmoles

CHg

CHmole

kg

gCHkg

62



115 moles CH115 moles CH44

in theory we should produce:in theory we should produce:

115 moles of CH115 moles of CH22ClCl22 and 230 moles of HCl and 230 moles of HCl

And use up 230 moles of ClAnd use up 230 moles of Cl22

63



115 moles of CH115 moles of CH22ClCl22 = ? g = ? g

MW CHMW CH22ClCl22 = 12.01 + 2x1.008 + 2x35.45 = 12.01 + 2x1.008 + 2x35.45

= 84.93= 84.93

115 moles x (84.03 g/mole) = 9770 g115 moles x (84.03 g/mole) = 9770 g

64



Expect 9770 g CHExpect 9770 g CH22ClCl22

but the yield is 43.1%but the yield is 43.1%

So we produced just 0.431 x 9770 gSo we produced just 0.431 x 9770 g

4.21 kg CH4.21 kg CH22ClCl22

65



Suppose the reaction went to completionSuppose the reaction went to completion

(100% yield)(100% yield)

Is mass conserved?Is mass conserved?

66



Start with 115 moles CHStart with 115 moles CH44 and 230 moles Cl and 230 moles Cl22

total mass = 115x16.04 + 230x70.90total mass = 115x16.04 + 230x70.90

= 1850 + 16300 = 18150= 1850 + 16300 = 18150

only 3 sig. figs. → 18.2 kgonly 3 sig. figs. → 18.2 kg

67



End with 115 moles CHEnd with 115 moles CH22ClCl22 and 230 moles HCl and 230 moles HCl

total mass = 115x84.93 + 230x36.46total mass = 115x84.93 + 230x36.46

= 9770 + 8390 = 18160= 9770 + 8390 = 18160

only 3 sig. figs → 18.2 kgonly 3 sig. figs → 18.2 kg


Hydrogen and oxygen are diatomic Hydrogen and oxygen are diatomic elements.elements.

Their subscripts cannot be changed.Their subscripts cannot be changed. The subscripts on water cannot be The subscripts on water cannot be

changed.changed.

Hydrogen + oxygen Hydrogen + oxygen waterwater

HH22 + O + O22 H H22OO

Balancing EquationBalancing Equation

Count the atoms on each side.Count the atoms on each side. Reactant side: 2 atoms H and 2 atoms OReactant side: 2 atoms H and 2 atoms O Product side: 2 atoms H and 1 atom OProduct side: 2 atoms H and 1 atom O

HH2 2 + O+ O22 H H22OO

Balancing EquationsBalancing Equations HH2 2 + O+ O22 H H22OO

If the subscripts cannot be altered, If the subscripts cannot be altered, how can the atoms be made how can the atoms be made equal?equal?

Adjust the number of molecules by Adjust the number of molecules by changing the changing the coefficientscoefficients. .


Reactants: 2 atoms of H and 2 Reactants: 2 atoms of H and 2 atoms of Oatoms of O

Products: 4 atoms of H and 2 Products: 4 atoms of H and 2 atoms of Oatoms of O

H is no longer balanced!H is no longer balanced!

HH22 + O + O22 22HH22OO


Reactant side: 4 atoms of H and 2 atoms of Reactant side: 4 atoms of H and 2 atoms of OO

Product side: 4 atoms of H and 2 atoms of OProduct side: 4 atoms of H and 2 atoms of O It’s Balanced!It’s Balanced!

22HH22 + O + O22 22HH22OO


Count atoms.Count atoms. Reactants: 2 atoms N and 2 atoms HReactants: 2 atoms N and 2 atoms H Products: 1 atom N and 3 atoms of NHProducts: 1 atom N and 3 atoms of NH33

NN22 + H + H22 NH NH33

Nitrogen + hydrogen ammoniaNitrogen + hydrogen ammonia

Balancing EquationsBalancing Equations Nothing is balanced.Nothing is balanced. Balance the nitrogen first by placing Balance the nitrogen first by placing

a coefficient of 2 in front of the NHa coefficient of 2 in front of the NH33..

NN22 + H + H22 22NHNH33

Balancing EquationsBalancing Equations Hydrogen is not balanced.Hydrogen is not balanced. Place a 3 in front of HPlace a 3 in front of H2.2.

Reactant side: 2 atoms N, 6 atoms HReactant side: 2 atoms N, 6 atoms H Product side: 2 atoms N, 6 atoms HProduct side: 2 atoms N, 6 atoms H

NN22 + + 33HH22 22NHNH33


Count atoms.Count atoms. Reactants: Ca – 3 atoms, P – 2 Reactants: Ca – 3 atoms, P – 2

atoms, O – 8 atoms; H – atoms, S – atoms, O – 8 atoms; H – atoms, S – 1 atom, O – 4 atoms1 atom, O – 4 atoms

CaCa33(PO(PO44))22 + H + H22SOSO44 CaSOCaSO44 + H + H33POPO44

Balancing EquationsBalancing Equations Side note on CaSide note on Ca33(PO(PO44))22

The subscript after the The subscript after the phosphate indicates two phosphate indicates two phosphate groups.phosphate groups.

This means two POThis means two PO443-3- groups with groups with

two P and eight O atoms.two P and eight O atoms.


CaCa33(PO(PO44))22 + H + H22SOSO44 CaSOCaSO4 4 + H + H33POPO44

Count atoms in the product.Count atoms in the product. Ca atoms – 1, S atom – 1, O atoms – Ca atoms – 1, S atom – 1, O atoms –

4; H atoms – 3, P atom – 1, O atoms - 4; H atoms – 3, P atom – 1, O atoms - 44

Balancing EquationsBalancing Equations In this equation, the ion groups do In this equation, the ion groups do

not break up.not break up. Instead of counting individual atoms, Instead of counting individual atoms,

ion groups may be counted.ion groups may be counted.




Reactants: CaReactants: Ca2+2+ – 3, PO – 3, PO443- 3- - 2, H- 2, H++ – 2, – 2,

SOSO442+ 2+ - 1- 1

Products: CaProducts: Ca2+2+ - 1, SO - 1, SO442- 2- - 1, H- 1, H++ - 3, PO - 3, PO44

3-3- - 1- 1

Balancing EquationsBalancing Equations Balance the metal first by placing a Balance the metal first by placing a

coefficient of 3 in front of CaSOcoefficient of 3 in front of CaSO44.. Products: Ca – 3 atoms, SOProducts: Ca – 3 atoms, SO44

2-2- - 3 - 3 groupsgroups

CaCa33(PO(PO44))22 + H + H22SOSO44 33CaSOCaSO4 4 + H + H33POPO44

Balancing EquationsBalancing Equations Three sulfate groups are needed on the reactant Three sulfate groups are needed on the reactant

side so place a coefficient of 3 in front of Hside so place a coefficient of 3 in front of H22SOSO44.. 3H3H22SOSO44 gives 6 H gives 6 H++ and 3 SO and 3 SO44

2-2-.. Neither phosphate nor calcium is balanced.Neither phosphate nor calcium is balanced.

CaCa33(PO(PO44))22 + + 33HH22SOSO44 33CaSOCaSO4 4 + H + H33POPO44

Balancing EquationsBalancing Equations A coefficient of 2 placed in front A coefficient of 2 placed in front

of Hof H33POPO44 which balances both which balances both hydrogen and phosphate.hydrogen and phosphate.

CaCa33(PO(PO44))22 + + 33HH22SOSO44 33CaSOCaSO4 4 + + 22HH33POPO44


The sulfate group breaks up. Each atom The sulfate group breaks up. Each atom must be counted individually. Ugh!must be counted individually. Ugh!

Reactants: Cu – 1, H – 2, S – 1, O – 4Reactants: Cu – 1, H – 2, S – 1, O – 4 Products: Cu – 1, S – 1, O - 4, H – 2, O – Products: Cu – 1, S – 1, O - 4, H – 2, O –

1, S – 1, O - 21, S – 1, O - 2

Cu + HCu + H22SOSO44

CuSOCuSO4 4 + H + H22O + SOO + SO22

Balancing EquationsBalancing Equations Sulfur is not balanced.Sulfur is not balanced. Place a two in front of sulfuric acid.Place a two in front of sulfuric acid. Count atoms: 2 HCount atoms: 2 H22SOSO44 H – 4, S – 2, O - H – 4, S – 2, O -

88

Cu + Cu + 22HH22SOSO44

CuSOCuSO44 + H + H22O + SOO + SO22

Balancing EquationsBalancing Equations Hydrogen needs to be balanced Hydrogen needs to be balanced

so place a 2 in front of the Hso place a 2 in front of the H22O. O. Count the number of atoms.Count the number of atoms.


CuSOCuSO44 + + 22HH22O + SOO + SO22

Balancing EquationsBalancing Equations Reactants: Cu – 1, H – 4, S – 2, O – 8Reactants: Cu – 1, H – 4, S – 2, O – 8 Products: Cu – 1, S – 1, O – 4, H – 4, Products: Cu – 1, S – 1, O – 4, H – 4,

O – 2, S – 1, O – 2 = Cu – 1, S – 2, H O – 2, S – 1, O – 2 = Cu – 1, S – 2, H – 4, O – 8– 4, O – 8

It’s balanced!It’s balanced!



Balancing EquationsBalancing Equations Balancing hints:Balancing hints:

Balance the metals first.Balance the metals first. Balance the ion groups next.Balance the ion groups next. Balance the other atoms.Balance the other atoms. Save the non ion group oxygen Save the non ion group oxygen

and hydrogen until the end.and hydrogen until the end.

Balancing EquationsBalancing Equations This method of This method of

balancing balancing equations is equations is the inspection the inspection method.method.

The method is The method is trial and error.trial and error.

Practice.Practice.

Writing and NamingWriting and Naming

Write the corresponding formula Write the corresponding formula equation and then balance the equation and then balance the equation.equation.

Nickel + hydrochloric acid Nickel + hydrochloric acid

Nickel(II) chloride + hydrogenNickel(II) chloride + hydrogen

Writing and NamingWriting and Naming Write each formula independently.Write each formula independently. Ignore the rest of the equation.Ignore the rest of the equation. Balance the equation after writing Balance the equation after writing

the formulasthe formulas..

Ni + HCl NiClNi + HCl NiCl22 + H + H22

Ni + Ni + 22HCl NiClHCl NiCl22 + H + H22

Writing and NamingWriting and Naming Remember the diatomic elements: Remember the diatomic elements:

HH22, N, N22, O, O22, F, F22, Cl, Cl22, Br, Br22, and I, and I2.2.


Balance the formula equation.Balance the formula equation. Write the word equation.Write the word equation.

Cu + HCu + H22SOSO44

CuSOCuSO44 + H + H22O + SOO + SO22

Writing and NamingWriting and Naming Cu + Cu + 22HH22SOSO44


Write the names:Write the names: Cu by itself is just Cu by itself is just coppercopper. Copper(I) . Copper(I)

or copper(II) would be incorrect.or copper(II) would be incorrect. HH22SOSO44 should be named as an acid. should be named as an acid. Sulfuric acidSulfuric acid

Writing and NamingWriting and Naming CuSOCuSO44 has a SO has a SO44

2-2- group so Cu must group so Cu must be 2+. Some metals must have be 2+. Some metals must have Roman Numerals. Roman Numerals. Copper(II) Copper(II) sulfatesulfate

HH22O is known as O is known as waterwater.. SOSO22 is a nonmetal compound. Its is a nonmetal compound. Its

name is either name is either sulfur dioxidesulfur dioxide or or sulfur(IV) oxidesulfur(IV) oxide..


Copper + sulfuric acid Copper + sulfuric acid Copper(II) sulfate + water + Copper(II) sulfate + water +

sulfur dioxidesulfur dioxide



balancing equations balancing, writing, and naming equations

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