Composition and formulaeComposition and formulae

Of moles and menOf moles and men

Learning objectivesLearning objectives

Count atoms in formulaCount atoms in formula Define the moleDefine the mole Determine numbers of atoms or molecules in Determine numbers of atoms or molecules in

molar quantitiesmolar quantities Determine molar mass from chemical formulaDetermine molar mass from chemical formula Determine moles from mass of substanceDetermine moles from mass of substance Perform calculations of:Perform calculations of:

Percent compositionPercent composition Empirical formulaEmpirical formula Molecular formulaMolecular formula

Molecules or molesMolecules or moles

The numbers (coefficients) in chemical The numbers (coefficients) in chemical equations can refer to moleculesequations 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

Atomic and molecular massAtomic and molecular mass

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

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

Molar mass scaleMolar mass scale Mass of mole of atoms or molecules in gramsMass of mole of atoms or molecules in grams

Confusing?...Confusing?...

The Good NewsThe Good News

Molar mass in grams of element has Molar mass in grams of element has same same numerical value numerical value as mass of atom in amuas mass of atom in amu

Atomic mass of carbon = 12 amuAtomic mass of carbon = 12 amu Molar mass of carbon = 12 gMolar mass of carbon = 12 g

Formula mass of HFormula mass of H22O molecule = 18 amuO molecule = 18 amu

Molar mass of HMolar mass of H22O = 18 gO = 18 g

ExamplesExamples

How many moles of Li are in 6.94 g if atomic How many moles of Li are in 6.94 g if atomic mass of Li is 6.94 amu?mass of Li is 6.94 amu?

1.00 mol1.00 mol

What is the molar mass of NHWhat is the molar mass of NH33 if atomic if atomic

mass of H = 1 amu and N = 14 amumass of H = 1 amu and N = 14 amu 17 g/mol17 g/mol

Particle – mole conversionsParticle – mole conversions

23 1No particles = Moles (mol) x 6.02 x 10 ( )

mol

23

No particles Moles (mol) =

16.02 x 10 ( )

mol

Gram – mole conversionsGram – mole conversionsMass (g)

Moles(mol) = Molar mass (g/mol)

Mass(g) = Moles(mol) x Molar mass (g/mol)

Particle – gram conversionsParticle – gram conversions23Mass (g)

Particles = x 6.02 x 10 (/mol)Molar mass (g/mol)

23

ParticlesMass (g) = x Molar mass (g/mol)

6.02 x 10 (/mol)

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

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

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

Practice molecular formula problemPractice molecular formula problem Ibuprofen contains 75.69 % C, 8.80 % H and 15.51 % O. Ibuprofen contains 75.69 % C, 8.80 % H and 15.51 % O.

What is the molecular formula if molar mass is 206 g/mol?What is the molecular formula if molar mass is 206 g/mol?

MolarityMolarity

Concentration is usually expressed in terms Concentration is usually expressed in terms of molarity:of molarity:

Moles of solute/liters of Moles of solute/liters of solutionsolution (M) (M)

Moles of solute = molarity x volume of solutionMoles of solute = molarity x volume of solution

ExampleExample

What is molarity of 50 ml solution containing What is molarity of 50 ml solution containing 2.355 g H2.355 g H22SOSO44?? Molar mass HMolar mass H22SOSO44 = 98.1 g/mol = 98.1 g/mol

Moles HMoles H22SOSO44 = .0240 mol = .0240 mol

Volume of solution = 50/1000 = .050 LVolume of solution = 50/1000 = .050 L Concentration = moles/volumeConcentration = moles/volume

= .0240/.050 = 0.480 M= .0240/.050 = 0.480 M

DilutionDilution

More dilute solutions are prepared from More dilute solutions are prepared from concentrated ones by addition of solventconcentrated ones by addition of solvent

MM11VV11 = M = M22VV22

Molarity of new solution MMolarity of new solution M22 = M = M11VV11/V/V22

To dilute by factor of ten, increase volume by factor of To dilute by factor of ten, increase volume by factor of tenten