analytical chemistry chem 3811 chapters 1 and 3 (review)
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ANALYTICAL CHEMISTRY CHEM 3811 CHAPTERS 1 AND 3 (REVIEW). DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTERS 1 AND 3 MEASUREMENTS, SIGNIFICANT FIGURES ERRORS, STOICHIOMETRY, CONCENTRATIONS. ANALYTICAL CHEMISTRY. - PowerPoint PPT PresentationTRANSCRIPT
ANALYTICAL CHEMISTRY CHEM 3811
CHAPTERS 1 AND 3 (REVIEW)
DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences
Clayton state university
CHAPTERS 1 AND 3
MEASUREMENTS, SIGNIFICANT FIGURESERRORS, STOICHIOMETRY, CONCENTRATIONS
ANALYTICAL CHEMISTRY
- Deals with the separation, identification, quantification, and statistical treatment of the components of matter
Two Areas of Analytical Chemistry
Qualitative Analysis- Deals with the identification of materials in a given sample
(establishes the presence of a given substance)
ANALYTICAL CHEMISTRY
- Deals with the separation, identification, quantification, and statistical treatment of the components of matter
Quantitative Analysis- Deals with the quantity (amount) of material
(establishes the amount of a substance in a sample)
- Some analytical methods offer both types of information (GC/MS)
ANALYTICAL CHEMISTRY
Analytical Methods
- Gravimetry (based on weight)
- Titrimetry (based on volume)
- Electrochemical (measurement of potential, current, charge, etc)
- Spectral (the use of electromagnetic radiation)
- Chromatography (separation of materials)
- Chemometrics (statistical treatment of data)
ANALYTICAL CHEMISTRY
General Steps in Chemical Analysis
- Formulating the question (to be answered through chemical measurements)
- Selecting techniques(find appropriate analytical procedures)
- Sampling(select representative material to be analyzed)
- Sample preparation(convert representative material into a suitable form for analysis)
ANALYTICAL CHEMISTRY
General Steps in Chemical Analysis
- Analysis(measure the concentration of analyte in several identical portions)
(multiple samples: identically prepared from another source)(replicate samples: splits of sample from the same source)
- Reporting and interpretation(provide a complete report of results)
- Conclusion(draw conclusions that are consistent with data from results)
Measurement
- Is the determination of the dimensions, capacity, quantity, or extent of something
- Is a quantitative observation and consists of two parts: a number and a scale (called a unit)
Examplesmass, volume, temperature, pressure, length, height, time
MEASUREMENT
MEASUREMENT SYSTEMS
Two measurement systems:
English System of Units (commercial measurements): pound, quart, inch, foot, gallon
Metric System of Units (scientific measurements)SI units (Systeme International d’Unites)
liter, meter, gramMore convenient to use
FUNDAMENTAL SI UNITS
Physical Quantity
MassLengthTimeTemperatureAmount of substanceElectric currentLuminous intensity
Name of Unit
KilogramMeterSecondKelvinMoleAmpereCandela
Abbreviation
kgms (sec)KmolAcd
DERIVED SI UNITS
Physical Quantity
Force PressureEnergyPowerFrequency
Name of Unit
NewtonPascalJouleWattHertz
Abbreviation
N (m-kg/s2)Pa (N/m2; kg/(m-s2)J (N-m; m2-kg/s2)W (J/s; m2-kg/s3)Hz (1/s)
METRIC UNITS
Prefix
GigaMegaKiloDeciCentiMilli
MicroNanoPico
Femto
Abbreviation
GMkdcmµnpf
Notation
109
106
103
10-1
10-2
10-3
10-6
10-9
10-12
10-15
UNIT CONVERSIONS
24 hours = 1 day or
Length/Distance
2.54 cm = 1.00 in.12 in. = 1 ft1 yd = 3 ft1 m = 39.4 in.1 km = 0.621 mile1 km = 1000 m
Time
1 min = 60 sec1 hour = 60 min24 hours = 1 day7 days = 1 week
Volume
1 gal = 4 qt1 qt = 0.946 L1 L = .0265 gal1 mL = 0.034 fl. oz.
Mass
1 Ib = 454 g1 Ib = 16 oz1 kg = 2.20 Ib1 oz = 28.3 g
day1
hours24
hours24
day1»
Convert 34.5 mg to g
How many gallons of juice are there in 20 liters of the juice?
gallon0.50.53liter1
gallon0.0265xliters20
g10x3.54org0.0354mg1000
g1xmg34.5 2
UNIT CONVERSIONS
Convert 4.0 gallons to quarts
quarts16quarts15.9559liter0.946
quart1x
gallons0.265
liter1xgallons4.0
SIGNIFICANT FIGURES
Exact Numbers - Values with no uncertainties
- There are no uncertainties when counting objects or people(24 students, 4 chairs, 10 pencils)
- There are no uncertainties in simple fractions(1/4, 1/7, 4/7, 4/5)
Inexact Numbers - Associated with uncertainties
- Measurement has uncertainties (errors) associated with it- It is impossible to make exact measurements
SIGNIFICANT FIGURES
Measurements contain 2 types of information- Magnitude of the measurement- Uncertainty of the measurement
- Only one uncertain or estimated digit should be reported
Significant Figures digits known with certainty + one uncertain digit
RULES FOR SIGNIFICANT FIGURES
1. Nonzero integers are always significant
2. Leading zeros are not significant 0.0045 (2 sig. figs.) 0.00007895 (4 sig. figs.) The zeros simply indicate the position of the decimal point
3. Captive zeros (between nonzero digits) are always significant 1.0025 (5 sig figs.) 12000587 (8 sig figs)
RULES FOR SIGNIFICANT FIGURES
4. Trailing zeros (at the right end of a number) are significant only if the number contains a decimal point 2.3400 (5 sig figs) 23400 (3 sig figs)
5. Exact numbers (not obtained from measurements) are assumed to have infinite number of significant figures
RULES FOR SIGNIFICANT FIGURES
Rounding off Numbers
1. In a series of calculations, carry the extra digits through to the final result before rounding off to the required significant figures
2. If the first digit to be removed is less than 5, the preceding digit remains the same (2.53 rounds to 2.5 and 1.24 rounds to 1.2)
RULES FOR SIGNIFICANT FIGURES
Rounding off Numbers
3. If the first digit to be removed is greater than 5, the preceding digit increases by 1 (2.56 rounds to 2.6 and 1.27 rounds to 1.3)
4. If the digit to be removed is exactly 5- The preceding number is increased by 1 if that results in an even number (2.55 rounds to 2.6 and 1.35000 rounds to 1.4)- The preceding number remains the same if that results in an odd number(2.45 rounds to 2.4 and 1.25000 rounds to 1.2)
RULES FOR SIGNIFICANT FIGURES
Multiplication and Division
- The result contains the same number of significant figures as the measurement with the least number of significant figures
2.0456 x 4.02 = 8.223312 = 8.22
3.20014 ÷ 1.2 = 2.6667833 = 2.7
- The certainty of the calculated quantity is limited by the least certain measurement, which determines the final number of
significant figures
RULES FOR SIGNIFICANT FIGURES
Addition and Subtraction
- The result contains the same number of decimal places as the measurement with the least number of decimal places
- The certainty of the calculated quantity is limited by the least certain measurement, which determines the final number of
significant figures
5.479
0.234
3.2
2.045
4.028
3.52
7.548
= 5.5= 4.03
SCIENTIFIC NOTATION
- Used to express too large or too small numbers (with many zeros)in compact form
- The product of a decimal number between 1 and 10 (the coefficient)and 10 raised to a power (exponential term)
24,000,000,000,000 = 2.4 x 1013
coefficient
Exponential term
Exponent (power)0.000000458 = 4.58 x 10-7
SCIENTIFIC NOTATION
- Provides a convenient way of writing the required number of significant figures
6300000 in 4 significant figures = 6.300 x 106
2400 in 3 significant figures = 2.40 x 103
0.0003 in 2 significant figures = 3.0 x 10-4
SCIENTIFIC NOTATION
- Add exponents when multiplying exponential terms
(5.4 x 104) x (1.23 x 102) = (5.4 x 1.23) x 10 4+2
= 6.6 x 106
- Subtract exponents when dividing exponential terms
(5.4 x 104)/(1.23 x 102) = (5.4/1.23) x 10 4-2
= 4.4 x 102
DENSITY
- The amount of mass in a unit volume of a substance
Ratio of mass to volume =Density =
UnitsSolids: grams per cubic centimeter (g/cm3)
Liquids: grams per milliliter (g/mL)Gases: grams per liter (g/L)
- Density usually changes with change in temperature
- Density of 2.3 g/mL implies 2.3 grams per 1 mL
volume
mass
For a given liquid
- Objects with density less than that of the liquid will float
- Objects with density greater than that of the liquid will sink
- Objects with density equal to that of the liquid will remainstationary (neither float nor sink)
DENSITY
The amount of mass in a unit volume of a substance
TEMPERATURE
- The degree of hotness or coldness of a body or environment
- 3 common temperature scales
Metric system Celsius and Kelvin
English system Fahrenheit
TEMPERATURE
Celsius Scale (oC) - Reference points are the boiling and freezing
points of water (0oC and 100oC) - 100 degree interval
Kelvin Scale (K) - Is the SI unit of temperature (no degree sign)
- The lowest attainable temperature on the Kelvin scale is 0 (-273 oC) referred to as the absolute zero
Fahrenheit Scale (oF)- Water freezes at 32oF and boils at 212oF
- 180 degree interval
or
10o, 40o, 60o are considered as 2 significant figures
100o is considered as 3 significant figures
32F9
5C oo 32C
5
9F oo
273KCo 273CK o or
TEMPERATURE
LOGARITHMS
n = 10a implies log n = a
- The logarithm (base 10) of n is equal to a(written as log on calculators)
log 1000 = 3 since 1000 = 103
log 0.01 = -2 since 0.01 = 10-2
log 436 = 2.6392 is the characteristic0.639 is the mantissa
- The number of digits in the mantissa should be equal tothe number of significant figures in the original number (436)
log 4368 = 3.6403 log 0.4368 = -0.3597
LOGARITHMS
ANTILOGARITHMS
n = 10a implies antilog a = n
- n is the antilogarithm of a(written as antilog or 10x or INV log on calculators)
antilog 3 = 1000 since 103 = 1000antilog -2 = 0.01 since 10-2 = 0.01
ANTILOGARITHMS
antilog 2.639 = 436
- The number of significant figures in the answer should be equalto the number of digits in the mantissa
antilog 6.65 = 4466835.922 = 4.5 x 106
antilog -3.230 = 0.0005888436 = 5.89 x 10-4
ERRORS
- Two classes of experimental errors: systematic and random
Systematic Error- Also called determinate error
- Repeatable in a series of measurements- Can be detected and corrected
Examplesuncalibrated buret, pipet, analytical balance, pH meter
power fluctuations, temperature variations
ERRORS
- Two classes of experimental errors: systematic and random
Randon Error- Also called indeterminate error
- Always present and cannot be corrected
ExamplesTaking readings from an instrument, reading between
markings (interpolation), electrical noise in instruments
Precision - Provides information on how closely individual
measurements agree with one another(measure of reproducibility of a result)
Accuracy - Refers to how closely individual measurements
agree with the true value (correct value)(systematic errors reduce the accuracy of a measurement)
- Precise measurements may NOT be accurate
- Our goal is to be accurate and precise
ERRORS
Absolute Uncertainty- The margin of uncertainty associated with a measurement
- If estimated uncertainty in a buret reading is ± 0.05 mL then absolute uncertainty = ± 0.05 mL
- If estimated uncertainty in an analytical balance is ± 0.0001 g then absolute uncertainty = ± 0.0001g
ERRORS
Relative Uncertainty- Compares absolute uncertainty with its associated measurement
- Dimensionless
tMeasuremenofMagnitude
yUncertaintAbsoluteyUncertaintRelative
Percent Relative Uncertainty = Relative Uncertainty x 100
ERRORS
001.0mL 41.45
mL 0.05yUncertaintRelative
Percent Relative Uncertainty = 0.001 x 100 = 0.1 %
For a buret reading of 41.45 ± 0.05 mL
Absolute uncertainty = ± 0.05 mL
ERRORS
- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set of wholenumbers that balance the equation
C2H5OH(l) + O2(g) 2CO2(g) + H2O(g)
2 C atoms 2 C atoms
Place the coefficient 2 in front of CO2 to balance C atoms
BALANCING CHEMICAL EQUATIONS
C2H5OH(l) + O2(g) 2CO2(g) + 3H2O(g)
(5+1)=6 H atoms 3(1x2)=6 H atoms
Place 3 in front of H2O to balance H atoms
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set of wholenumbers that balance the equation
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)
1+(3x2)=7 O atoms (2x2)+3=7 O atoms
Place 3 in front of O2 to balance O atoms
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set of wholenumbers that balance the equation
C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(g)
2 C atoms(5+1)=6 H atoms
1+(3x2)=7 O atoms
2 C atoms(3x2)=6 H atoms
(2x2)+3=7 O atoms
Check to make sure equation is balancedWhen the coefficient is 1, it is not written
BALANCING CHEMICAL EQUATIONS
- Whole numbers are placed on the left side of the formula (called coefficients) to balance the equation (subscripts remain unchanged)
- The coefficients in a chemical equation are the smallest set of wholenumbers that balance the equation
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g)
BALANCING CHEMICAL EQUATIONS
- States of reactants and products- Physical states of reactants and products are represented by
(g): gas(l): liquid(s): solid
(aq): aqueous or water solution
Balance the following chemical equations
Fe(s) + O2(g) → Fe2O3(s)
C12H22O11(s) + O2(g) → CO2(g) + H2O(g)
(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + H2O(g)
BALANCING CHEMICAL EQUATIONS
MOLAR MASS
- Add atomic masses to get the formula mass (in amu) = molar mass (in g/mol)
- That is the mass, in g, of 1 mole of the substance
1 mole = 6.02214179 x 1023 entities (atoms or molecules) Usually rounded to 6.022 x 1023 (Avogadro’s number)
This implies that 6.022 x 1023 amu = 1.00 gAtomic mass (amu) = mass of 1 atom
molar mass (g) = mass of 6.022 x 1023 atoms
MOLAR MASS
Calculate the mass of 2.4 moles of NaNO3
Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00)= 85.00 g /mol NaNO3
= 204 g NaNO3
= 2.0 x 102 g NaNO3
3
333 NaNOmole1
NaNOg85.00xNaNOmole2.4NaNOg
CHEMICAL FORMULA
Consider Na2S2O3:
- Two atoms of sodium, two atoms of sulfur, and three atoms ofoxygen are present in one molecule of Na2S2O3
- Two moles of sodium, two moles of sulfur, and three moles ofoxygen are are present in one mole of Na2S2O3
CHEMICAL FORMULA
How many moles of sodium atoms, sulfur atoms, and oxygenatoms are present in 1.8 moles of a sample of Na2S2O3?
I mole of Na2S2O3 contains 2 moles of Na, 2 moles of S, and 3 moles of O
atomsNamoles3.6OSNamole1
atomsNamoles2xOSNamoles1.8atomsNamoles
322322
atomsSmoles3.6OSNamole1
atomsSmoles2xOSNamoles1.8atomsSmoles
322322
atomsOmoles5.4OSNamole1
atomsOmoles3xOSNamoles1.8atomsOmoles
322322
CHEMICAL CALCULATIONS
Calculate the number of molecules present in 0.075 g of urea,(NH2)2CO
Given mass of urea: - convert to moles of urea using molar mass
- convert to molecules of urea using Avogadro’s number
= 7.5 x 1020 molecules (NH2)2CO
CO)(NHmole1
CO)NH(molecules10x6.022x
CO)(NHg60.07
CO)(NHmole1xCO)(NHg0.075
22
2223
22
2222
CHEMICAL CALCULATIONS
How many grams of carbon are present in a 0.125 g of vitamin C,C6H8O6
Given mass of vitamin C: - convert to moles of vitamin C using molar mass
- convert to moles of C (1 mole C6H8O6 contains 6 moles C)- convert moles carbon to g carbon using atomic mass
= 0.0511 g carbon
Cmol1
Cg12.01x
OHCmol1
Cmol6x
OHCg176.14
OHCmol1xOHCg0.125
686686
686686
CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)
Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
- 1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O
- 1 mole of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O
- make sure the equation is balanced- calculate moles of propane from given mass and molar mass- determine moles of oxygen from mole ratio (stoichiometry)
- calculate mass of oxygen
= 349 g O2
CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)
2
2
83
2
83
8383 Omol1
Og32.00x
HCmol1
Omol5x
HCg44.11
HCmol1xHCg96.1
Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
What mass of oxygen will react with 96.1 g of propane?
- make sure the equation is balanced- calculate moles of propane from given mass and molar
mass- determine moles of CO2 from mole ratio (stoichiometry)
- calculate mass of CO2
= 288 g CO2
CHEMICAL EQUATIONS(STOICHIOMETRIC CALCULATIONS)
2
2
83
2
83
8383 COmol1
COg44.01x
HCmol1
COmol3x
HCg44.11
HCmol1xHCg96.1
Given: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
What mass of CO2 will be produced from 96.1 g of propane?
CONCENTRATION OF SOLUTIONS
- The amount of solute dissolved in a given quantity of solvent or solution
Molarity (M)- The number of moles of solute per liter of solution
Lsolutionofvolume
solutemolesMolarity
- A solution of 1.00 M (read as 1.00 molar) contains 1.00 mole of solute per liter of solution
CONCENTRATION OF SOLUTIONS
Calculate the molarity of a solution made by dissolving 2.56 g ofNaCl in enough water to make 2.00 L of solution
- Calculate moles of NaCl using grams and molar mass- Convert volume of solution to liters
- Calculate molarity using moles and liters
NaClmol0.0438NaClg58.44
NaClmol1xNaClg2.56
mol/L)(orM0.0219solutionL2.00
NaClmol0.0438Molarity
CONCENTRATION OF SOLUTIONS
After dissolving 1.56 g of NaOH in a certain volume of water, the resulting solution had a concentration of 1.60 M. Calculate the
volume of the resulting NaOH solution
- Convert grams NaOH to moles using molar mass- Calculate volume (L) using moles and molarity
NaOHmol0.0380NaOHg41.00
NaOHmol1xNaOHg1.56
solutionL0.0237NaOHmol1.60
solutionLxNaOHmol0.0380solution Volume
CONCENTRATION OF IONS
Consider
1.00 M NaCl: 1.00 M Na+ and 1.00 M Cl-
1.00 M ZnCl2: 1.00 M Zn2+ and 2.00 M Cl-
1.00 M Na2SO4: 2.00 Na+ and 1.00 M SO42-
CONCENTRATION OF IONS
Calculate the number of moles of Na+ and SO42- ions in 1.50 L
of 0.0150 M Na2SO4 solution
0.0150 M Na2SO4 solution contains:2 x 0.0150 M Na+ ions and 0.0150 M SO4
2- ions
Moles Na+ = 2 x 0.0150 M x 1.50 L = 0.0450 mol Na+
Moles SO42- = 0.0150 M x 1.50 L = 0.0225 mol SO4
2-
PERCENT COMPOSITION
100%xsolutionofmasstotal
soluteofmass(wt%)PercentMass
- May also be represented by %(m/m)
mass of solution = mass of solute + mass of solvent
PERCENT COMPOSITION
A sugar solution is made by dissolving 5.8 g of sugar in82.5 g of water. Calculate the percent by mass concentration
of sugar.
% 6.6100%xg) 82.5 g (5.8
g5.8(wt%)PercentMass
PERCENT COMPOSITION
100%xsolutionofvolumetotal
soluteofvolume(vol%)PercentVolume
- May also be represented by %(v/v) - Due to the way molecules are packed and differences in distances between molecules (bond lengths), the volume
of the resulting solution is almost always less than the sum of the volume of solute and the volume of solvent
PERCENT COMPOSITION
Calculate the volume percent of solute if 345 mL of ethyl alcohol is dissolved in enough water to produce 1257 mL
of solution
% 27.4 100%x mL 1257
mL345(vol%)PercentVolume
PARTS PER MILLION (PPM)
610xsampleofmass
substanceofmassppm
Percent can be defined as parts per hundred
1 ppm ≈ 1 µg/mL or 1 mg/L
PARTS PER MILLION (PPM)
If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express
the concentration of pesticide in ppm
ppm = µg/mL0.250 L = 250 mL
Density = 1.00 g/mL Implies mass solution = 250 g
ppm0.0548mL250
μg13.7ppm
PARTS PER BILLION (PPB)
1 ppb ≈ 1 ng/mL or 1 µg/L
910xsampleofmass
substanceofmassppb
PARTS PER MILLION (PPB)
If 0.250 L of aqueous solution with a density of 1.00 g/mL contains 13.7 μg of pesticide, express
the concentration of pesticide in ppb
ppm = µg/LVolume of solution = 0.250 L
Density = 1.00 g/mL Implies mass solution = 250 g
ppb5.48L 0.250
μg13.7ppb
DILUTION
- Consider a stock solution of concentration M1 and volume V1
- If water is added to dilute to a new concentration M2 and volume V2
- moles before dilution = moles after dilution
- Implies that M1V1 = M2V2
DILUTION
Calculate the volume of 3.50 M HCl needed to prepare 500.0 mL of 0.100 M HCl
(3.50 M)(V1) = (0.100 M)(500.0 mL)
V1 = 14.3 mL
Mole Fraction (χ)
- Fraction of moles of a component of solution
CONCENTRATION OF SOLUTIONS
components all of moles total
component of moles
The sum of mole fractions of all components = 1
Given that the total moles of an aqueous solution of NaCl andother solutes is 1.75 mol. Calculate the mole fraction of NaCl
if the solution contains 4.56 g NaCl.
NaClmol0.0780NaClg58.44
NaClmole1xNaClg4.56NaClMoles
CONCENTRATION OF SOLUTIONS
0446.0 totalmol 1.75
NaCl mol 0.0780fraction mole
MOLALITY (m)
Moles of solute per kg of solvent
Unit: m or molal
solvent kg
solute molesm
MOLALITY (m)
What is the molality of a solution that contains 2.50 g NaCl in 100.0 g water?
- Calculate moles NaCl- Convert g water to kg water
- Divide to get molality
NaClmol0.0428NaClg58.44
NaClmole1xNaClg2.50NaClmoles
NaCl428.0solventkg0.1000
NaClmol0.0428molality m
CONVERTING CONCENTRATION UNITS
Calculate the molality of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water
Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water
solutiong 100
ethanolg6.75ethanolpercentMass
ethanol57.1kg 0)(93.25/100
mol /46.08)(6.75
solvent kg
ethanol molmolality m
CONVERTING CONCENTRATION UNITS
Calculate the mole fraction of a 6.75 %(m/m) solution of ethanol (C2H5OH) in water
Mass water = 100 g solution – 6.75 g ethanol = 93.25 g water
mol0.146g 46.08
mol1xethanolg6.75ethanolmoles
mol5.17g 18.02
mol1xwaterg93.25watermoles
0.0275mol5.17)(0.146
mol0.146
moles total
componentmolfractionmol
CONVERTING CONCENTRATION UNITS
Practice Question
Given that the mole fraction of ammonia (NH3) in water is 0.088Calculate the molality of the ammonia solution
CONVERTING CONCENTRATION UNITS
- Molarity is temperature dependent (changes with change in temperature)
- Volume increases with increase in temperature hence molarity decreases
On the other hand- Molality
- Mass percent- Mole fraction
are temperature independent
CHEMICAL EQUILIBRIUM
- Occurs when there is product build-up during a chemical reaction- The product molecules interact with one another to
re-produce reactants
Chemical Equilibrium - When the rate of product formation (forward reaction)
is equal to the rate of reactant formation (reverse reaction)
A + B C + D
forward reaction
reverse reaction
- Reactant and product concentrations are usually not equal
- Such reactions are known as reversible reactions
- Forward reaction rate decreases with time as reactants are used up
- Reverse reaction rate increases with time as
products are being formed
- Concentrations are reached when both forward and reverse rates become equal
CHEMICAL EQUILIBRIUM
EQUILIBRIUM CONSTANT
- Describes the extent of reaction in a given system
- For a chemical reaction of the form
aA + bB → cC + dD
- The equilibrium constant (Keq) is given by
ba
dc
eq [B][A]
[D][C]K
- [ ] denotes concentration in moles/liter (M)
- Product concentrations in the numerator
- Reactant concentrations in the denominator
- Concentrations are raised to the powers of the respective coefficients
- Gases (in bars) and substances in solution (in mol/L) are written in Keq expressions
EQUILIBRIUM CONSTANT
- Pure solids, pure liquids (e.g. water), and solvents are not written since they are constant
- Keq changes with change in temperature
- For exothermic forward reactions (heat released) Keq decreases with increasing temperature
- For endothermic forward reactions (heat absorbed) Keq increases with increasing temperature
EQUILIBRIUM CONSTANT
Large Keq
- Greater product concentrations than reactant concentrations - Equilibrium position lies to the right
Small Keq
- Smaller product concentrations than reactant concentrations - Equilibrium position lies to the left
Intermediate Keq (near unity) - Both products and reactants are in significant amounts
- Equilibrium position lies neither to the right nor to the left
EQUILIBRIUM CONSTANT
- Longer arrows can be used to indicate the predominant species
- Longer forward reaction arrow for large Keq
- Longer reverse reaction arrow for small Keq
CO2 + H2O H2CO3
EQUILIBRIUM CONSTANT
LE CHATELIER’S PRICIPLE
- If a stress (change of conditions) is applied to a system in equilibrium the system will readjust (change the equilibrium position) in the direction that best reduces the stress imposed
on the system
- If more products form as a result of the applied stress the equilibrium is said to have shifted to the right
- If more reactants form as a result of the applied stress the equilibrium is said to have shifted to the left
Concentration Changes
For a reaction mixture at equilibrium
- Addition of reactant(s) shifts the equilibrium position to the right
- Removal of product(s) shifts the equilibrium position to the right
- Addition of product(s) shifts the equilibrium position to the left
- Removal of reactant(s) shifts the equilibrium position to the left
LE CHATELIER’S PRICIPLE
Temperature Changes
Exothermic Reactions
- Heat is a product
- Increase in temperature shifts the equilibrium position to the left
- Decrease in temperature shifts the equilibrium position to the right
LE CHATELIER’S PRICIPLE
Temperature Changes
Endothermic Reactions
- Heat is a reactant
- Increase in temperature shifts the equilibrium position to the right
- Decrease in temperature shifts the equilibrium position to the left
LE CHATELIER’S PRICIPLE
Pressure Changes
- Gases must be involved in the chemical reaction
- The total number of moles of the gaseous state must change
- Equilibrium is shifted in the direction of fewer moles
LE CHATELIER’S PRICIPLE
Pressure Changes
Higher moles of gaseous reactants than products- Increase in pressure shifts the equilibrium position to the right- Decrease in pressure shifts the equilibrium position to the left
Higher moles of gaseous products than reactants- Increase in pressure shifts the equilibrium position to the left
- Decrease in pressure shifts the equilibrium position to the right
LE CHATELIER’S PRICIPLE
Pressure Changes
No change in equilibrium position occurs if
- There is no reactant nor product in the gaseous state
- Number of moles of gaseous reactants equals number of moles of gaseous products
- Pressure is increased by adding a nonreactive (inert) gas
LE CHATELIER’S PRICIPLE
Addition of Catalysts
- Catalysts do not change equilibrium positions
- Catalysts speed up both forward and reverse reactions so have no net effect
- Catalysts allow equilibrium to be established more quickly by lowering the activation energy
LE CHATELIER’S PRICIPLE