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
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