Chemical EquationsChemical Equations

Balancing equations and Balancing equations and applications applications

ObjectivesObjectives

Write and balance chemical equationsWrite and balance chemical equations Perform calculations using moles and molar Perform calculations using moles and molar

massesmasses Determine percent composition from formulaDetermine percent composition from formula Determine empirical formula from percent Determine empirical formula from percent

compositioncomposition Determine molecular formulaDetermine molecular formula

The chemical equationThe chemical equation

aaA + A + bbB = B = ccC + C + ddDD

The Law of Conservation of Matter states that matter The Law of Conservation of Matter states that matter is neither created nor destroyedis neither created nor destroyed

All the atoms on the left must be the same as those All the atoms on the left must be the same as those on the righton the right

Reactantside

Productside

coefficient ELEMENT orCOMPOUND

Chemical book-keepingChemical book-keeping

The key to writing correct equations is to The key to writing correct equations is to ask the question, “Have I gained or lost any ask the question, “Have I gained or lost any atoms?”atoms?”

Another thing is to put down the correct Another thing is to put down the correct formula for each reactant or productformula for each reactant or product

Formulas cannot be changed in order to Formulas cannot be changed in order to balance the equationbalance the equation

In the reaction of hydrogen with oxygen to produce In the reaction of hydrogen with oxygen to produce

water, the reactants are the elements Hwater, the reactants are the elements H22 and O and O2, 2,

and the product is Hand the product is H22OO

Count the atoms: 4 H and 2 O 4 H and 2 OCount the atoms: 4 H and 2 O 4 H and 2 O

The big number multiplies every

atom after it

The subscript only multiplies the atom before

it

Balance the equationsBalance the equations

CHCH4 4 ++ OO22 = CO = CO22 + H + H22OO CHCH44 + + 22OO22 = CO = CO22 + + 22HH22OO

CC33HH88 + O + O22 = CO = CO2 2 + H+ H22OO CC33HH88 + + 55OO22 = = 33COCO22 + + 44HH22OO

NN22 + H + H22 = NH = NH33

NN22 + + 33HH22 = = 22NHNH33

Do balancing equation exercisesDo balancing equation exercises

Molecules or molesMolecules or moles

The numbers (coefficients) in chemical The numbers (coefficients) in chemical equation can refer to moleculesequation can refer to molecules

But for practical applications, we need a But for practical applications, we need a more useful number: we cannot count more useful number: we cannot count moleculesmolecules

The MoleThe Mole The mole is a unit of quantity used in The mole is a unit of quantity used in

chemistry to measure the number of atoms chemistry to measure the number of atoms or moleculesor molecules

DEFINITION:DEFINITION: The number of atoms in exactly 12 g of The number of atoms in exactly 12 g of 1212CC

A mole of anything always has the same A mole of anything always has the same number of particles: atoms, molecules or number of particles: atoms, molecules or potatoes – 6.02 x 10potatoes – 6.02 x 1023 23 – Avogadro’s number– Avogadro’s number

Mole conversionsMole conversions

Atomic and molecular weightsAtomic and molecular weights

Two scales:Two scales: Atomic mass unit scaleAtomic mass unit scale

The mass of an individual atom or molecule in The mass of an individual atom or molecule in atomic mass units (amu)atomic mass units (amu)

Molar mass scaleMolar mass scale The mass of a mole of atoms or molecules in The mass of a mole of atoms or molecules in

gramsgrams

Confusing?...Confusing?...

The Good NewsThe Good News

The weight of an atom in amu has the The weight of an atom in amu has the same same numerical value numerical value as its molar mass in gramsas its molar mass in grams

The atomic mass of carbon is 12 amuThe atomic mass of carbon is 12 amu The molar mass of carbon is 12 gThe molar mass of carbon is 12 g

The formula mass of HThe formula mass of H22O is 18 amuO is 18 amu

The molar mass of HThe molar mass of H22O is 18 gO is 18 g

ExamplesExamples

How many atoms are in 6.94 g of lithium if How many atoms are in 6.94 g of lithium if the atomic mass of Li is 6.94 amu?the atomic mass of Li is 6.94 amu?

6.02 x 106.02 x 102323

What is the molar mass of HWhat is the molar mass of H22O if the atomic O if the atomic

mass of H = 1 amu and O = 16 amumass of H = 1 amu and O = 16 amu 18 g18 g

How many H How many H atomsatoms in 1 mol of CH in 1 mol of CH44?? 1 mol = 6.02 x 101 mol = 6.02 x 102323 particles particles

How many moles of OHow many moles of O22 in 64 g of oxygen? in 64 g of oxygen?

Atomic mass of O = 16 AMUAtomic mass of O = 16 AMU

Calculate the formula mass of NaCalculate the formula mass of Na22SOSO44 in AMU. in AMU.

Use atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMUUse atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMU

Significance of formula unitSignificance of formula unit

Ionic compounds do not contain molecules. Ionic compounds do not contain molecules. Simplest formula is the formula unitSimplest formula is the formula unit

Covalent compounds, the molecular formula is Covalent compounds, the molecular formula is the formula unitthe formula unit

Percent composition and empirical Percent composition and empirical formulaformula

Chemical analysis gives the mass % of Chemical analysis gives the mass % of each element in the compoundeach element in the compound

Molar masses give the number of molesMolar masses give the number of moles Obtain mole ratiosObtain mole ratios Determine empirical formulaDetermine empirical formula

Determining percent compositionDetermining percent composition

Percent composition is obtained from the actual Percent composition is obtained from the actual masses.masses.

Example: Example: Sample contained 0.4205 g of C and 0.0795 g of H. Sample contained 0.4205 g of C and 0.0795 g of H. Total mass = 0.5000 g (0.4205 + 0.0795)Total mass = 0.5000 g (0.4205 + 0.0795)Therefore: in 100 g there are:Therefore: in 100 g there are: (84.10 %) (84.10 %)

(15.90 %)(15.90 %)

Percent composition: 84.10 % C, 15.90 % HPercent composition: 84.10 % C, 15.90 % H

100 g0.4205 = 84.10 g C

.5000 gx

100 g0.0795 = 15.90 g H

.5000 gx

Percent composition from formulaPercent composition from formula

What is percent composition of CWhat is percent composition of C55HH1010OO22?? 1 mol C1 mol C55HH1010OO22 contains 5 mol C, 10 mol H and 2 mol O contains 5 mol C, 10 mol H and 2 mol O

atomsatoms

Mass of each elementMass of each element

Total mass = 102.13 gTotal mass = 102.13 g

12.01 g C5 mol C = 60.05 g C

1 mol C1.008 g H

10 mol H = 10.08 g H1 mol H

16.00 g O2 mol O = 32.00 g O

1 mol O

Convert masses into percentsConvert masses into percents

Percent composition:Percent composition:58.80 % C + 9.870 % H + 31.33 % O = 100.00%58.80 % C + 9.870 % H + 31.33 % O = 100.00%

5 10 2

60.05 g C% C = x100 = 58.80 % C

102.13 g C H O

5 10 2

10.08 g H% H = x100 = 9.870 % H

102.13 g C H O

5 10 2

32.00 g O% O = x100 = 31.33 % O

102.13 g C H O

Empirical formula from percent Empirical formula from percent composition: 84.1 % C, 15.9 % Hcomposition: 84.1 % C, 15.9 % H

1.1. Convert percents into molesConvert percents into moles84.10 g of C 84.10 g of C ≡ 7.00 mol C ≡ 7.00 mol C

15.9 g of H ≡ 15.8 mol H15.9 g of H ≡ 15.8 mol H

2.2. Determine mole ratioDetermine mole ratioMole ratio H:C = Mole ratio H:C =

Simplest formula (decimal form): CSimplest formula (decimal form): C11HH2.262.26

Make Make smallestsmallest integers by multiplying integers by multiplying

CC44HH99

May require rounding. Errors in real data cause problemsMay require rounding. Errors in real data cause problems Do percent composition and empirical formula exercisesDo percent composition and empirical formula exercises

84.10 g C

12.00 g/mol15.9 g H

1.008 g/mol15.8 mol H

2.26 :17.00 mol C

What is percent C content of CWhat is percent C content of C22HH66?? Molar mass C = 12 g/mol; molar mass H = 1 g/molMolar mass C = 12 g/mol; molar mass H = 1 g/mol

Empirical formula with more than two Empirical formula with more than two elementselements

Percent composition of vitamin C is:Percent composition of vitamin C is: 40.9 % C, 4.58 % H, 54.5 % O40.9 % C, 4.58 % H, 54.5 % O

1.1. Convert into molesConvert into moles

2.2. Determine mole ratiosDetermine mole ratios

3.3. Find lowest whole numbersFind lowest whole numbers

Inaccuracy can lead to ambiguous or Inaccuracy can lead to ambiguous or incorrect formulasincorrect formulas

What if H:C is 2.20 rather than 2.26? An What if H:C is 2.20 rather than 2.26? An error of only 3 %error of only 3 %

Formula becomes CFormula becomes C55HH1111 rather than C rather than C44HH99

What if H:C is 2.30 rather than 2.26? An What if H:C is 2.30 rather than 2.26? An error of only 2 %error of only 2 %

Formula becomes CFormula becomes C33HH77

Sometimes chemical intuition is required: we Sometimes chemical intuition is required: we know there is FeO, Feknow there is FeO, Fe33OO44 and Fe and Fe22OO33; so a ; so a formula FeOformula FeO33 would indicate an error would indicate an error

Rounding or not: the role of chemical Rounding or not: the role of chemical intuitionintuition

Formulae are always written with integersFormulae are always written with integers Experimental ratios are always fractionsExperimental ratios are always fractions Two choices:Two choices:

Round to nearest whole numberRound to nearest whole number Multiply top and bottom to find ratio of whole Multiply top and bottom to find ratio of whole

numbers with same valuenumbers with same value

Choice depends on the type of substanceChoice depends on the type of substance

Hydrocarbons: A case for not Hydrocarbons: A case for not roundingrounding

There are millions of different hydrocarbonsThere are millions of different hydrocarbons What if H:C is 2.20 rather than 2.26? An error of What if H:C is 2.20 rather than 2.26? An error of

only 3 %only 3 % Formula becomes CFormula becomes C55HH1111 rather than C rather than C44HH99

What if H:C is 2.30 rather than 2.26? An error of What if H:C is 2.30 rather than 2.26? An error of only 2 %only 2 %

Formula becomes CFormula becomes C33HH77

All formulae are reasonableAll formulae are reasonable So how do I know what the composition is?So how do I know what the composition is? Additional knowledge about the substance is Additional knowledge about the substance is

helpful: melting point, boiling point, molar masshelpful: melting point, boiling point, molar mass

Inorganic compounds: Rounding Inorganic compounds: Rounding makes sensemakes sense

Inorganic compounds tend to have few Inorganic compounds tend to have few compositionscompositions

Iron forms three oxides: FeO, FeIron forms three oxides: FeO, Fe33OO44 and and

FeFe22OO33

Experimental formula FeOExperimental formula FeO1.751.75 would indicate would indicate

FeFe22OO33 not FeO not FeO22

Practice empirical formula problemPractice empirical formula problem A compound contains 62.1 % C, 5.21 % H, 12.1 % N and A compound contains 62.1 % C, 5.21 % H, 12.1 % N and

20.7 % O. What is the empirical formula?20.7 % O. What is the empirical formula?

Empirical and molecular formulaEmpirical and molecular formula

Percent composition gives the Percent composition gives the empiricalempirical (simplest) formula. It says nothing about the (simplest) formula. It says nothing about the molecularmolecular formula. formula.

Molecular formula describes number of Molecular formula describes number of atoms in the moleculeatoms in the molecule May be much larger than the empirical formula May be much larger than the empirical formula

in the case of molecular covalent compoundsin the case of molecular covalent compounds For ionic compounds empirical formula = For ionic compounds empirical formula =

“molecular” formula“molecular” formula

Elements and compounds can have molecular Elements and compounds can have molecular formula different from simplest formulaformula different from simplest formula

SubstanceSubstance Empirical Empirical formulaformula

Molecular Molecular formulaformula

SubstanceSubstance Empirical Empirical formulaformula

Molecular Molecular formulaformula

SulphurSulphur SS SS88PhosphorousPhosphorous PP PP44

BenzeneBenzene CHCH CC66HH66AcetyleneAcetylene CHCH CC22HH22

EthyleneEthylene CHCH22 CC22HH44CyclohexaneCyclohexane CHCH22 CC66HH1212

Determination of molecular formulaDetermination of molecular formula

Require:Require:1.1. Empirical formula from percent composition Empirical formula from percent composition

analysisanalysis2.2. Molar mass from some other sourceMolar mass from some other source Number of empirical formula units in molecule:Number of empirical formula units in molecule:

There are There are nn (A (AaaBBbbCCcc) in molecule:) in molecule: Molecular formula is AMolecular formula is AnanaBBnbnbCCncnc

Molar mass

Empirical formula massn

Molecular formula of vitamin CMolecular formula of vitamin C

Empirical formula of vitamin C is CEmpirical formula of vitamin C is C33HH44OO33

Molar mass vitamin C is 176.12 g/molMolar mass vitamin C is 176.12 g/mol Mass of empirical formula = 88.06 g/molMass of empirical formula = 88.06 g/mol

(3 x 12.01 + 4 x 1.008 + 3 x 16.00)(3 x 12.01 + 4 x 1.008 + 3 x 16.00)

Number of formula units per molecule =Number of formula units per molecule =

Molecular formula = 2(CMolecular formula = 2(C33HH44OO33) = C) = C66HH88OO66

Molar mass vitamin C 176.122

Empirical formula mass vitamin C 88.06n

Ionic solids do not have molecular Ionic solids do not have molecular formulasformulas

Infinite lattices like ionic solids and Infinite lattices like ionic solids and covalently bonded lattices are not molecular.covalently bonded lattices are not molecular.

The formula used for an ionic compound is The formula used for an ionic compound is the same as the empirical formula – with the same as the empirical formula – with one or two exceptionsone or two exceptions HgHg22ClCl22 rather than HgCl rather than HgCl

Structural formula provides more Structural formula provides more informationinformation

The molecular formula indicates the number of The molecular formula indicates the number of atoms in the moleculeatoms in the molecule

The structural formula indicates how those atoms The structural formula indicates how those atoms are arrangedare arranged

CC22HH66O is the molecular formula for ethanol O is the molecular formula for ethanol andand

dimethyl etherdimethyl ether Structural formula for ethanol is CHStructural formula for ethanol is CH33CHCH22OHOH

Structural formula for ether is CHStructural formula for ether is CH33OCHOCH33

In a recipe we would need to use the structural In a recipe we would need to use the structural formula to identify the correct reagentformula to identify the correct reagent

In some cases the mole contents of the In some cases the mole contents of the compound are obtained indirectlycompound are obtained indirectly

Analysis of the Analysis of the hydrocarbon is performed hydrocarbon is performed by combustion. by combustion.

Mole ratios of the elements Mole ratios of the elements are derived from the mole are derived from the mole ratios of the combustion ratios of the combustion products COproducts CO22 + H + H22O O

(1 mol H(1 mol H22O O ≡ 2 mol H)≡ 2 mol H)

(1 mol CO(1 mol CO22 ≡ 1 mol C) ≡ 1 mol C)

The molecular or empirical formula can The molecular or empirical formula can be used to determine the percent be used to determine the percent

compositioncomposition Formula of aspirin is CFormula of aspirin is C99HH88OO44

Molar mass is 180 gMolar mass is 180 g Mass of C = 108 gMass of C = 108 g Mass of H = 8 gMass of H = 8 g Mass of O = 64 gMass of O = 64 g

% C = 108/180 x 100 %% C = 108/180 x 100 % % H = 8/180 x 100 %% H = 8/180 x 100 % % O = 64/180 x 100 %% O = 64/180 x 100 %