chemical equations balancing equations and applications
TRANSCRIPT
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 %