CHEMISTRY 2202 UNIT 1 STOICHIOMETRY · STOICHIOMETRY: - the calculation of quantities USED IN or...
Transcript of CHEMISTRY 2202 UNIT 1 STOICHIOMETRY · STOICHIOMETRY: - the calculation of quantities USED IN or...
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CHEMISTRY 2202UNIT 1
STOICHIOMETRY
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Introduction
Chemistry is both a qualitative (describing with words) and a quantitative (describing with numbers) science.
Chemical equations are ‘recipes’ that show chemists what we need and what will be produced in a reaction.
eg. H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l)
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Introduction
This balanced equation helps us answer questions like:● How much sulfuric acid is needed to neutralize
30 mL of 5.00 M sodium hydroxide? ● What volume of water will be produced when
30 mL of sodium hydroxide reacts with 10 mL of
sulfuric acid?
Chemical equations help us “count” molecules or QUANTIFY chemical reactions.
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Unit Outline
STOICHIOMETRY:
- the calculation of quantities USED IN or PRODUCED by a chemical reaction.
3 Sections:
Part 1 - Mole calculations
(ch. 2 & 3 in text)
Part 2 - Using mole ratios in chemical equations (ch. 4)
Part 3 - Solution stoichiometry
(ch. 6)
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PART 1 – MOLE CALCULATIONS ● Isotopes and Atomic Mass● Average Atomic Mass ● Molar Mass ● Mole Calculations ● Molar Volume ● Percent Composition ● Empirical Formulas ● Molecular Formulas ● Formula of a Hydrate
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ISOTOPES and ATOMIC MASS Atomic number:
▫The # of protons in an atom or ion.
▫In a neutral atom: #p+ = #e-
eg.
● Atomic number of C is _____
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ISOTOPES and ATOMIC MASS •Mass number:
▫The # located in the bottom of each box on the periodic table (has decimal places). ▫Mass Number = #p+ + #no
ie. mass # - #p+ = #no
eg. ● mass number of C = _____ ● # of no in C = _____
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Your Turn!!!
Atom Atomic # Atomic Mass # #p
+#e
-#n
o
Carbon (C) 6 12
Carbon (C) 6 13
Carbon (C) 6 14
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ISOTOPES AND ATOMIC MASS . . .
•ISOTOPE
▫Atoms that have the same number of protons (ie. same element) but a different NUMBERS OF NEUTRONS are called isotopes
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ISOTOPES AND ATOMIC MASS . . . ▫Isotopes have different mass numbers.
eg. Carbon has 3 naturally occurring isotopes:
● carbon-12 98.9 % of all carbon in the world
● carbon-13 1.1 % of all carbon in the world
● carbon-14 0.0000000001 % of all carbon
▫When these isotope values are averaged, with their percent abundance in nature, we get the number that appears on the periodic table (12.01 g/mol).
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ISOTOPES AND ATOMIC MASS . . .
ISOTOPE NOTATION:
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ISOTOPES and ATOMIC MASS Example
Carbon-12 Carbon-13 Carbon-14
How many neutrons are in each isotope?
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ISOTOPES AND ATOMIC MASS . . .
Name each isotope of magnesium.
How many neutrons are in each isotope?
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YOUR TURN!!!
A. hydrogen-2
B.
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YOUR TURN!!!
Write the isotope notation for an isotope of:
A. Copper that has 29 p+ and 35 no
B. Vanadium that has 23 p+
and 28 no.
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AVERAGE ATOMIC MASS (AAM)
ANALOGY: ● The average mark you receive in this course is a
weighted average of all your scores. ● Your final grade depends on how well you do in
different pieces of work and their % of the total grading scheme.
● Your final mark is NOT a straight average. ● On the periodic table, the atomic mass numbers
are also an average, based on different types of isotopes and their relative percent abundance.
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AVERAGE ATOMIC MASS . . .
● This average is based on the concept of the ATOMIC MASS UNIT (AMU)
● The symbol for the AMU is μ. ● 1 μ = 1/12 the mass of a carbon-12 atom
(from IUPAC conference, 1961) ie. 1 carbon-12 atom has a mass of 12 μ. ● On the periodic table, carbon’s atomic mass is
an average, 12.01, expressed in g/mol.
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AVERAGE ATOMIC MASS . . . To calculate average atomic mass (AAM):
1.Change percent abundance value to decimal value (divide by 100).
2. Multiply the atomic mass by its % abundance for each isotope. 3. Add these numbers together. 4. Use significant digits.
In other words: AAM = (exact atomic mass x % abundance as a decimal) +
(exact atomic mass x % abundance as a decimal) +
(exact atomic mass x % abundance as a decimal) + .…
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AVERAGE ATOMIC MASS . . .
ISOTOPE ATOMIC MASS % ABUNDANCE
Lithium-6 6.01513 u 7.420
Lithium-7 7.01601 u 92.58
Calculate the average atomic mass of lithium from the given data:
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AVERAGE ATOMIC MASS . . .
Calculate the average atomic mass of Mg given that it has three isotopes: ● magnesium-24 → 23.98 μ; 78.60% abundance ● magnesium-25 → 24.99 μ; 10.11 % abundance ● magnesium-26 → 25.98 μ; 11.29 % abundance
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Naturally occurring magnesium exists as a mixture of three isotopes. Mg-24 has an atomic mass of 23.985 amu and a relative abundance of 78.70 %. Mg-25 has an atomic mass of 24.985 amu and a relative abundance of 10.13%. The average atomic mass of magnesium is 24.31 amu.
Calculate the atomic mass of the remaining isotope.
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For you to do!
Page 45-#2
Page 46-#2
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The Mole
•INTRODUCTION:
▫Chemists face the
difficult task of trying to
measure extremely small
particles, such as: ● FORMULA UNITS and IONS for ionic compounds ● MOLECULES for molecular compounds ● ATOMS for elements
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The Mole
eg.
1 teaspoon of water → 1.67 x 1023
molecules
(5.0 mL)
•A device that could count molecules at a rate of 1 million per second would take over 5 billion years to count the molecules in a teaspoon of water!!!
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The mole
Item Quantity Amount
Gloves Pair 2
Soft-drinks six-pack 6
eggs dozen 12
pens Gross (12 dozen) 144
paper ream 500
Names to represent some common quantities:
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The Mole
IUPAC to the RESCUE!!
•Chemists use the concept of the MOLE as a convenient way to deal with counting huge numbers of particles.
One mole is the number of atoms contained in exactly 12 g of carbon-12.
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The Mole
1 mol = 6.02 x 1023 particles
(atoms, ions, molecules, formula units)
•This number is referred to as Avogadro’s Number (in honour of scientist Amadeo Avogadro who discovered it), and is written:
NA = 6.02 x 1023 particles/mol
The letter n is used to represent
# of moles
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The Mole
•Just how big is this number?
•Put it this way . . .
▫If we had 6.02 x 1023
marbles ( 1 mol of marbles), it would be enough to cover all of Earth to a depth of 80 km!
▫If you won 6.02 x 1023
dollars, it would be enough to spend a billion dollars a day for over a trillion years before you had to worry about running out!
▫(In book, how big? p. 49)
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The Mole•One mole always contains 6.02 x 10
23 particles.
ie.
● 1 mol C = 6.02 x 1023 atoms
● 1 mol O2 = 6.02 x 1023 molecules of O2
● 1 mol CH4 = 6.02 x 1023 molecules of CH4
= 2.41 x 1024 atoms of H
● 1 mol NaCl = 6.02 x 1023 formula units
= 6.02 x 1023 Na+
ions
= 6.02 x 1023 Cl- ions
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Mole Calculations – Mole to Particle
● Where:
n = number of moles (mol)
N = number of particles
(formula units, ions, molecules, or atoms)
NA = Avogadro’s number
(6.02 x 1023 particles/mol)
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Mole Calculations – Mole to Particle
eg #1:
How many moles in 1.21 x 1035
molecules of CO2?
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Mole Calculations – Mole to Particle
eg #2: How many atoms in 0.35 mol of Fe?
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Mole Calculations – Mole to Particle
eg #3:
How many formula units in 5.69 mol of Na2SO4?
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Mole Calculations – Mole to Particle
eg #4:
How many Na+
ions in 5.69 mol of Na2SO4?
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HOMEWORK!!!
Pg. 52 #'s 7,8,9,10
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MOLAR MASS (M)
INTRODUCTION:
•Moles cannot be measured directly.
•Chemists measure chemical amounts using
mass for solids and volume for gases & liquids.
•To convert between moles and mass we need
the mass of one mole or the MOLAR MASS.
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Molar Mass (M) of Elements •The mass of 1 mol of a substance is referred to as its molar mass (M) and is measured in g/mol.
eg.
1 mol C → 12.01 g/mol
1 mol H → 1.01 g/mol
1 mol Pb → 207.19 g/mol
1 mol Na → 22.99 g/mol
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Molar Mass (M) of Compounds
•The molar mass of a compound is the sum of the molar masses of each element in the compound.
eg.
Calculate the molar mass of: ● H2O ● Na2SO4
● Ba(MnO4)2 .8 H2O
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For you to do!
● Pg. 57 # 17 &19 Answers on page 77
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Mole Calculations – Mole to Mass
Rearrange:
Where: n = number of moles (mol) m = mass (g) M = molar mass (g/mol)
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Mole Calculations – Mole to Mass
eg #1:
How many moles in 4.67 g of copper ?
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Mole Calculations – Mole to Mass
eg # 2:
What is the mass of 0.487 mol of H2O ?
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Mole Calculations – Mole to Mass
eg # 3:
Calculate the mass of 2.00 mol of H2(g).
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Mole Calculations – Mole to Mass
eg # 4:
How many moles are in 80.0 g of NH4Cl ?
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For you to do!!
Pg. 59 #20 & Pg. 60 #25
Answers on page 77
Note● Molar amount = number of moles of whatever
● 1 kg = 1000 g ● 1 g = 1000 mg
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Mole Calculation – Mass to Particles
• We do not have a direct quantity that relates mass (m) and number of particles (N), but they are related through number of moles (n).
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Mole Calculation – Mass to Particles
eg.
How many molecules are in 26.9 g of H2O(l)?
m = 26.9 g
Mwater= 18.02 g/mol
NA = 6.02 x 1023 molecules/mol
Find N
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Mole Calculation – Mass to Particles
eg.
A sample of Sn contains 4.69 x 1028 atoms. Calculate
its mass.
N = 4.69 x 1028 atoms
NA = 6.022 x 1023 molecules/mol
MSn = 118.69 g/mol
Find m
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Mole Calculation – Mass to Particles
eg.
How many molecules are in 4.78 g of glucose?
m = 4.78 g
Mglucose= 180.18 g/mol
NA = 6.022 x 1023 molecules/mol
Find N
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Mole Calculation – Mass to Particles
Particles(N)
Moles(n)
Mass(m)
X NA
÷ NA
x M
÷ M
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Mole Calculation – Mass to Particles
Practice: p. 63 #’s 28 & 33
p. 64 #’s 34 & 37
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MOLAR VOLUME (gases)
Pressure (P)
•The force exerted per unit of surface area.
•SI unit is the Pascal (Pa).
(Used to be the atmosphere - atm)
Temperature (T)
•measures the kinetic energy of the particles.
•SI unit is Celsius degrees (oC).
•Other unit is Kelvin (K).
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MOLAR VOLUME (gases)
•The volume of a gas is dependent on P and T.
A ‘COOL’ balloon
Pressure & Volume
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MOLAR VOLUME (gases)
Standard Conditions
(needed to measure gas volumes)
STP – Standard Temperature and Pressure
T = 0oC (273 K)
P = 101.3 kPa (1 atm)
SATP – Standard Ambient Temperature and Pressure
T = 25oC
P = 100 kPa
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MOLAR VOLUME (gases)
•Experimental evidence shows that the volume of one mole of ANY GAS at STP is 22.4 L/mol.
ie. MV = 22.4 L/mol OR VSTP = 22.4 L/mol
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Mole Calculations – Molar Volume
Where:
n = number of moles, in mol
v = volume, in L
VSTP
= molar volume of a gas at STP
22.4 L/mol
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Mole Calculations – Molar Volume
eg 1:
What is the volume, in L, of 0.600 mol of SO2 gas at STP?
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Mole Calculations – Molar Volume
eg 2:
What is the number of moles in 33.6 L of He gas at STP?
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Mole Calculations – Molar Volume
eg 3:
What is the mass of 44.8 L of methane at STP?
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Mole Calculations – Molar Volume
eg 4:
A chemist isolates 2.99 g of a gas. The sample occupies 800.0 mL at STP. Is the gas most likely to be O2(g), Kr(g), Ne(g), or F2(g)?
(Hint: Determine the molar mass of the gas.)
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HOMEWORK!!!
p. 73 #’s 38 - 43
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SUMMARY
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PERCENT COMPOSITION
The composition of many substances is expressed in terms
of percentages.
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The London 2012 Olympic medals weigh between 375 - 400g.
Gold medal - 92.5% silver, 1.34% gold, with the remainder copper.
Silver medal - 92.5% silver, with the remainder copper.
Bronze medal - 97.0% copper, 2.5% zinc and 0.5% tin.
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3.5 % of sea water is dissolved salt
EXCEPT
the Dead Sea which is 33% salt.
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PERCENT COMPOSITION
Two types of calculations:
1.Percent Composition given mass
% element = mass of element x 100
mass of compound
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Percent Composition given mass (g)
eg. 1
8.20 g of Mg combines with 5.40 g of O to form MgO. Determine the percent composition of this compound?
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Percent Composition given mass (g)
eg. 2
29.0 g of Ag combines with 4.30 of S to
form Ag2S. Calculate the mass percent for S
in this compound?
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Your Turn!
p. 82 #’s 2 & 4
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PERCENT COMPOSITION •Percent composition is the percent by mass of each element in a compound
•Should add up to 100%.
•AKA: Mass Percent
eg.
C8H
8O
3 – the vanillin molecule
C-->63.1 %
H-->5.3 %
O-->31.6 %
TOTAL: 100 %
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PERCENT COMPOSITION
2.Percent Composition given the formula
% element = M of element x 100 M = molar mass
M of compound
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Percent Composition given the Formula
eg. 3
Calculate the percent composition of C2H6.
79.85 %
20.2 %
Based on your answer, calculate the mass
of C in an 82 g sample of C2H6.
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Percent Composition given the Formula
eg. 4
Calculate the mass percent of NaHSO4.
M = 120.07 g/mol
19.15 % 0.841 % 26.71 % 53.30 %
Based on your answer, calculate the mass of H in a 20.2 g sample of NaHSO4.
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For you to do!
Pg. 85 #'s 5 & 8
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EMPIRICAL and MOLECULAR FORMULAS
● An empirical formula gives the simplest ratio of elements in a compound.
● A molecular formula shows the actual number of atoms in a molecule of a compound.
Ex: glucose: molecular formula-->
empirical formula-->● Ionic compounds are always wrote as empirical
formulas – simplest ratio● A molecular formula is only used for molecular
compounds
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EMPIRICAL and MOLECULAR FORMULAS Compound Molecular Formula Empirical Formula
Butane
C4H
10 C
2H
5
Glucose
C6H
12O
6 C
6H
12O
6
Water
H2O H
2O
Benzene
C6H
6
CH
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EMPIRICAL FORMULAS
The empirical formula of a compound may be determined from the % composition of a compound.
Method:
1.Change % to mass out of 100 g.
2.Convert mass to moles.
3.Divide by the smallest number of moles to get the ratio for the empirical formula.
4.Multiply to get a whole number ratio.
(if needed)
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EMPIRICAL FORMULAS
Page 90 in book. Helpful for EF calculations.
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A FUN WAY TO REMEMBER EF calculations . . .
Percent to mass
Mass to Mole
Divide by small
Multiply till whole
By Joel Thompson
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EMPIRICAL FORMULAS
eg. 1
A compound was analyzed and found to contain 87.4% N and 12.6 % H by mass. Determine the empirical formula of the compound.
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EMPIRICAL FORMULAS
eg. 2
What is the empirical formula of a compound that is 25.9 % N and 74.1 % O ?
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EMPIRICAL FORMULAS
eg. 3
A compound contains 89.91% C and 10.08 % H by mass. Determine the empirical formula of the compound.
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MOLECULAR FORMULA CALCULATIONS
•We can determine the molecular formula of a compound using the molar mass and the empirical formula of a compound.
METHOD:
1.Determine empirical formula (may be given).
2.Divide the molar mass by the empirical formula mass.
3.Round off the result to the nearest whole number.
4.Multiply empirical formula by this whole number to get the molecular formula.
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MOLECULAR FORMULA CALCULATIONS
eg. 1
Determine the molecular formula of a compound with a molar mass of 60.00 g/mol and an empirical formula of CH4N.
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MOLECULAR FORMULA CALCULATIONS
eg. 2
The empirical formula of hydrazine is NH2. The molar mass of hydrazine is 32.06 g/mol. What is the molecular formula for hydrazine?
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For You to Try
p. 97 #’s 17 – 20
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Unit 1 Part 2
CALCULATIONS AND CHEMICAL EQUATIONS
I know the answer!
TEE HEE!
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UNIT 1 – PART 2 (CH 4)
Topics: ● Stoichiometry
● definition – what it is ● types of stoichiometry ● mole ratios ● calculations using n & m
● other calculations using v & VSTP
● Theoretical yield, actual yield and percent yield ● Limiting and Excess Reagent ● Core Lab 2 – % Yield in a Chemical Reaction
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STOICHIOMETRY
● Stoichiometry comes from the Greek words stoikheion, meaning “element” and metron, meaning “to measure”.
● Stoichiometry is the determination of quantities needed for, or quantities produced by, chemical reactions.
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STOICHIOMETRY
3 types of stoichiometry
1. Gravimetric stoichiometry ● stoichiometry using mass (g) of solids(s).
2. Solution stoichiometry ● stoichiometry using concentration & volume of
solutions.
3. Gas stoichiometry ● stoichiometry using volume (L) of gases (g) at
STP.
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STOICHIOMETRY
Q: How do we calculate quantities used
or quantities produced in chemical reactions?
A: Mole Ratios
• The coefficients from a balanced chemical equation provide the mole ratios of reactants
and products in a chemical reaction.
• Coefficients from a balanced chemical equation, work like a sandwich recipe . . .
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MOLE RATIOS
Clubhouse sandwich recipe
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MOLE RATIOS
Slices of Toast
Slices of Turkey
Strips of Bacon
# of Sandwiches
12
27
66
100
Fill in the missing quantities
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MOLE RATIOS
● Mole ratios from balanced chemical equations work in the same way
eg: equation for an air bag exploding
sodium azide sodium + nitrogen gas 2 NaN3(s) 2 Na(s) + 3 N2(g)
2 UNITS or
2 MOLESNOT 2 g
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MOLE RATIOS
2 NaN3(s) 2 Na(s) + 3 N2(g)
6
10
150
10
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MOLE RATIOS
● A mole ratio is a mathematical expression that shows the relative amounts of two species involved in a chemical change.
● A mole ratio ● comes from a balanced chemical equation ● show the relative amounts of reactants &
products in moles
looks like ---> coefficient required
coefficient given
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MOLE RATIOS
● Mole ratios from balanced chemical equations represent relative number of particles (atoms, molecules, ions, formula units or moles).
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
15 mol
18 molecules
35 molecules
7.38 mol
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MOLE RATIOS
NOTE:
Mole ratios DO NOT REPRESENT
MASS
P4 + 5 O2 → 2 P2O5
1 g of phosphorus reacts with 5 g of oxygen
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MOLE TO MOLE STOICHIOMETRY
STEPS:
1.Write a balanced chemical equation with subscripts
(Identify given and required values).
2.Use MOLE RATIOS from balanced chemical equation .
nrequired = ngiven x coefficient required
coefficient given
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MOLE TO MOLE STOICHIOMETRY
eg.1
How many moles of ammonia gas are produced when 0.60 mol of nitrogen gas reacts with hydrogen gas?
(1.2mol)
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MOLE TO MOLE STOICHIOMETRY
eg.2
How many moles of CO2 are produced when 31.5 mol of C3H8 is burned? (94.5 mol)
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MOLE TO MOLE STOICHIOMETRY
eg.3
Calculate the number of moles of magnesium needed to produce 0.586 mol of Mg3N2. (1.76 mol)
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MOLE TO MOLE STOICHIOMETRY
eg.4
How many moles of sulfur are produced when 4.25 mol of SO3 decomposes? (0.531 mol)
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FOR YOU TO TRY
● Pg. 114-117 #'s 2,5,9
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MORE STOICHIOMETRY CALULATIONS
● Other types of Stoichiometry: ● Mole to Mass (or Mass to Mole) ● Mass to Mass ● Volume to Volume (for gases at STP) ● SolutionStoichiometry (Part 3 of Unit 1...)
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MOLE TO MASS STOICHIOMETRY
● Mole to Mass Calculations are the same as Mole to Mole but also require conversion between mass and moles:
● Use n=m or m= n x M
M
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MOLE TO MASS STOICHIOMETRY
STEPS:
1.Balanced chemical - identify given and required
2.Convert mass to moles
3.Mole Ratio
nrequired = ngiven x coefficient required
coefficient given
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MOLE TO MASS STOICHIOMETRY
eg. 1:
Calculate the number of moles of oxygen that will react with 6.49 g of aluminum. (0.180 mol)
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MOLE TO MASS STOICHIOMETRY
eg. 2:
How many moles of water are produced when 20.6 g of CH4 burns? (2.57 mol)
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MOLE TO MASS STOICHIOMETRY
eg. 3: Calculate the mass of aluminum oxide that can be produced from the reaction between 0.1804 mol of oxygen gas and excess aluminum. (18.39g)
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MOLE TO MASS STOICHIOMETRY
eg. 4: What mass of Ca3P2 is produced when
4.38 mol of Ca reacts with phosphorus? (266g)
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For you to Try
Pg. 149 #'s 6 & 7
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Mass to Mass Stoichiometry
STEPS:
1.Balanced chemical (given!! required!! subscripts!!)
2.Convert mass to moles.
3.Mole Ratio
nrequired = ngiven x coefficient required
coefficient given
4.Convert moles to mass
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Mass to Mass Stoichiometry
NOTE: The coefficients from balanced equations
represent MOLE RATIOS.
MASSES are NOT to be used
in the MOLE RATIO step.
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Mass to Mass Stoichiometry
eg1:
How many grams of NH3 gas are produced when 5.40 g of hydrogen gas reacts with nitrogen gas? (30.4g)
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Mass to Mass Stoichiometry
eg 2:
What mass of nitrogen gas is needed to react completely with hydrogen gas and produce 30.6 g of ammonia gas?
(24.72g)
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Mass to Mass Stoichiometry
eg 3:
What mass of CaCl2(aq) is produced when Ca(NO3)2(aq)
reacts with 4.39 g of NaCl(s)? (4.17g)
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Mass to Mass Stoichiometry
eg 4:
Given that 45 g of sulfur (S8) gas reacts with aluminum solid, find the mass of product. (7.0x101g)
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For you to try
Page 121 #'s 11-14
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LAW OF CONSERVATION OF MASS
● Mass is conserved in chemical reaction.
OR
total mass of = total mass of
the reactants the products. ● We can verify this law by using the mole ratios
and molar masses of reactants and products from balanced chemical equations.
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Law of Conversation of Mass
Quantity 2 Mg(s) + O2(g) 2 MgO(s)
n (mol) 2 1 2
m (g) 2(24.31) 32.00 2(40.31)
The total mass of the reactants is 2(24.31) + 32.00 = 80.62gThe total mass of the products is 2(24.31+16.00) = 2(40.31) = 80.62g
The mass of the products and reactants are the same. So mass is conserved!!
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Law of Conservation of Mass
Ex:
The decomposition of 500.00 g of Na3N produces
323.20 g of N2. How much Na is produced in this
decomposition?
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Mole to Volume Stoichiometry
If gas volume at STP is GIVEN or REQUIRED you must use VSTP in your calculations.
eg 1:
What volume of oxygen at STP is needed to react with 2.34 g of NO2 ?
2 N2O5(s) 4 NO2(g) + O2(g)
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Mole to Volume Stoichiometry
eg 2:
Calculate the mass of N2H4(g) needed to produce 157 L of nitrogen gas at STP in the reaction below. 2 N2H4(l) + N2O4(l) 3 N2(g) + 4 H2O(l)
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For you to try
p. 123 # 16
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LIMITING AND EXCESS REAGENTS
Sandwich Time Again!
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LIMITING AND EXCESS REAGENTS
Here’s what we find in our kitchen:
◦How many sandwiches can we make??
6 Slices of Toast
12 Pieces of Turkey
20 slices of bacon
What limits the # of sandwiches??
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LIMITING AND EXCESS REAGENTS
Chemical reactions with 2 or more reactants, stop when 1 reactant is completely used up.
Limiting reagent ● AKA limiting reactant ● The reactant that is completely used in a
chemical reaction, thereby limiting the amount of product that can form.
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LIMITING AND EXCESS REAGENTS
● Excess reagent ● AKA excess reactant ● the reactant that is left over after the reaction
is complete. ● there is more than is needed for complete
reaction. ● We can predict which reactant is the LR by
using mole ratios from balanced chemical equations.
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DETERMING THE LIMITING REAGENT
How to Solve:
1.Write balanced chemical equation.
2.Calculate n(mols) for EACH reactant.
3.Use the mole ratio for each reactant. ● The LR is the reactant that produces less moles
of product
4.Calculate the amount of product using the LR.
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DETERMINE THE LIMITING REAGENT
eg 1:
Determine the Limiting Reagent and Excess Reagent if 14.8 g of C3H8(g) reacts with 2.14 g of O2(g).
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NOTE:
When you are given the quantity (mass, volume or moles) of 2 or more reactants in a stoichiometry question, YOU MUST ALWAYS determine the LR and use it to determine the amount of product.
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Limiting Reagent
eg 2.
What mass of NaCl is produced when 6.70 mol of Na reacts with 3.20 mol of Cl2?
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Limiting Reagent
eg 3.
What mass of CaCO3 will be produced when 20.0 g of Ca3(PO4)2 reacts with 15.0 g of Na2CO3 ?
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Limiting Reagent
eg 4.
What volume of H2 at STP will be produced when 10.0 g of Zn reacts with 20.0 g of HCl(aq)?
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For you to try
p. 134
#’s 27ab & 30a
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Percent Yield Calculations
● The term YIELD refers to the mass of product formed in a chemical reaction.
eg. Burning 99.5 g of propane in sufficient oxygen produces 298 g of carbon dioxide gas.
THEORETICAL YIELD ● what you are supposed to get ● The amount of product predicted or calculated.
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Percent Yield Calculations
● A question that asks you to find the theoretical yield is asking you to calculate the amount expected.
Actual yield ● what you actually get in an experiment ● The amount actually produced in a chemical
reaction in the LAB
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Percent Yield Calculations
For most reactions, your actual yield is less than your theoretical yield, for many reasons, including:
Reaction is slow and was incomplete. Impurities are present. Reaction goes to equilibrium (never really
finishes). Sample is wet when weighed. Experimental error.
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Percent Yield Calculations
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Percent Yield Calculations
eg 1.
Calculate the theoretical yield of copper if 1.87 g of Al reacts with excess aqueous copper (II) sulfate?
2 Al + 3 CuSO4 → Al2(SO4)3 + 3 Cu
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Percent Yield Calculations
eg 1.(cont’d)
If the reaction produces 3.74 g of copper, what is the percent yield of copper?
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Percent Yield Calculations
eg 2.
20.0 g of bromic acid, HBrO3 (aq) , is reacted with excess HBr.
HBrO3(aq) + 5 HBr(aq) 3 H2O(l) + 3 Br2(aq)
(a) What is the theoretical yield of Br2 for this reaction? (74.4 g)
(b) If 47.3 g of Br2 are produced, what is the percentage yield of Br2? (63.6%)
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Percent Yield Calculations
eg 3.
Barium sulfate forms a precipitate in this reaction:
Ba(NO3)2 (aq) + Na2SO4 (aq) BaSO4(s) + 2 NaNO3(aq)
When 35.0 g of Ba(NO3)2 reacts with excess Na2SO4,
29.8 g of BaSO4 are recovered by the chemist.
(a) Calculate the theoretical yield of BaSO4. (31.3 g)
(b) Calculate the percentage yield of BaSO4. (95.2 %)
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Percent Yield Calculations
eg 4.
Yeasts can act on a sugar, such as glucose, C6H12O6, to produce ethyl alcohol, C2H5OH, and carbon dioxide.
C6H12O6 (s) 2 C2H5OH(l) + 2CO2(g)
If 223 g of ethyl alcohol are recovered after 1.63 kg of glucose react, what is the percentage yield of the reaction? (834 g, 26.7 %)
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Solutions Chemistry
Chapter 7
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Solutions
● Terms ● Molar Concentration (mol/L)● Dilutions● % Concentration (pp. 255 – 263)● Solution Process● Solution Preparation● Solution Stoichiometry● Dissociation
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Terms
solutionsolventsoluteconcentrateddiluteaqueousmiscibleimmisciblealloysolubility
molar solubilitysaturatedunsaturatedsupersaturateddissociationelectrolytenon-electrolytesolubleinsoluble
limiting reagentexcess reagentactual yieldtheoretical yielddecantingpipettingfiltrateprecipitatedynamic equilibrium
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Terms
solution - is a homogeneous mixture solute - the substance that dissolves OR the
substance in lesser quantitysolvent - the substance which dissolves the
solute OR the substance in greater quantityconcentrated - a large amount of solute relative
to the amount of solvent dilute - a small amount of solute relative to the
amount of solvent
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Terms
saturated – contains the maximum amount of dissolved solute at a given temperature and pressure
unsaturated – contains less than the maximum amount of dissolved solute at a given temperature and pressure
supersaturated – contains more than the maximum amount of dissolved solute for a given temperature and pressure
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Terms
miscible – liquids that dissolve in each other immiscible – liquids that do not dissolve in each
otheraqueous - the solvent is water alloy - a solid solution of two or more metals
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Terms
Solubility - the maximum amount of solute that can be dissolved under specific temperature and pressure conditions
eg. the solubility of HCl at 25 °C is 12.4 mol/L
eg. 100.0 mL of water at 25°C dissolves 36.2 g of sodium chloride
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Terms
soluble – solubility is greater than 1 g per 100 mL of solvent.
insoluble - solubility is less than 0.1 g per 100 mL of solvent.
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9 types of solution
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Rate of dissolving
● for most solids, the rate of dissolving is greater at higher temperatures
● stirring a mixture or by shaking the container increases the rate of dissolving.
● decreasing the size of the particles increases the rate of dissolving.
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Solubility
● small molecules are often more soluble than larger molecules.
● solubility of solids increases with temperature.● the solubility of most liquids is not affected by
temperature.● the solubility of gases decreases as
temperature increases● an increase in pressure increases the solubility
of a gas in a liquid.
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“Like dissolves like”
● ionic solutes and polar covalent solutes both dissolve in polar solvents
● non-polar solutes dissolve in non-polar solvents.
● compounds with very strong ionic bonds, such as AgCl, tend to be less soluble in water than compounds with weak ionic bonds, such as NaCl.
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Surprise Review
● Find the molar mass of Ca(OH)2
● How many moles in 45.67 g of Ca(OH)2?
● Find the mass of 0.987 mol of Ca(OH)2.
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Molar Concentration
The terms concentrated and dilute are qualitative descriptions of solubility.
A quantitative measure of solubility uses numbers to describe how much solute is dissolved or the concentration of a solution.
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Molar Concentration
The MOLAR CONCENTRATION of a solution is the number of moles of solute (n) per litre of solution (v).
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Molar Concentration
FORMULA:Molar Concentration = number of moles volume in litres
C = n V
mol/L OR M (molar)
C
n V orVxCn Rearranged Formulae
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Molar Concentration
Calculate the molar concentration of:
1). 4.65 mol of NaOH is dissolved to prepare 2.83 L of solution. (1.64 mol/L)
2). 15.50 g of NaOH is dissolved to prepare 475 mL of solution.
( 0.3875 mol → 0.816 mol/L)
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Molar Concentration
3). the volume of 6.00 mol/L HCl(aq) that can be made using 0.500 mol of HCl. (0.083L)
4). the volume of 1.60 mol/L HCl(aq) that can be made using 20.0 g of HCl. (0.343L)
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Dilution (p. 272)When a solution is diluted:- The concentration decreases- The volume increases- The number of moles remains the
same
ni = nf Number of moles after dilution
Number of moles before dilution
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Dilution (p. 272) ni = nf
Ci Vi = Cf Vf
eg. Calculate the molar concentration of a vinegar solution prepared by diluting 10.0 mL of a 17.4 mol/L solution to a final volume of 3.50 L. (0.0497 mol/L)
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p. 273 #’s 25 – 27
p. 276 #’s 1, 2, 4, & 5
I will actually be checking these questions. So please do them!!
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June 2010 # 12.
A lab technician prepares a dilute solution of hydrochloric acid. If 50.0 mL of 2.50 mol/L hydrochloric acid is added to 450.0 mL of water, what is the new concentration?
(A) 0.250 mol/L (C) 3.60 mol/L
(B) 0.278 mol/L (D) 4.00 mol/L
Answer: (B)
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Percent Concentration● Concentration may also be given as a %.● The amount of solute is a percentage of the
total volume or mass of solution. liquids in liquids - % v/v solids in liquids - % m/v solids in solids - % m/m
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Percent Concentration
100x(mL) solutionofvolume
(g) soluteofmass(m/v)Percent
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100x(g) solutionofmass
(g) soluteofmass(m/m)Percent
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100x(mL) solutionofvolume
(mL) soluteofvolume(v/v)Percent
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Concentration in ppm and ppb
Parts per million (ppm) and parts per billion (ppb) are used for extremely small concentrations
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610solution
solute
mxppmm
910solution
solute
mxppbm
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eg. 5.00 mg of NaF is dissolved in 100.0 kg of solution. Calculate the concentration in:
a) ppm
b) ppb
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ppm = 0.005 g x 106
100,000 g
= 0.05 ppm
ppb = 0.005 g x 109
100,000 g
= 50.0 ppb
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For you to do as homework!!
● p. 265 #’s 15 – 17
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Solution Preparation & Dilution● standard solution – a solution of known
concentration● volumetric flask – a flat-bottomed glass vessel that
is used to prepare a standard solution● delivery pipet – a pipet that accurately measures
one volume● graduated pipet – pipets that have a series of lines
that can be used to measure many different volumes
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To prepare a standard solution:
1. calculate the mass of solute needed
2. weigh out the desired mass
3. dissolve the solute in a beaker using less than the final volume
4. transfer the solution to a volumetric flask (rinse the beaker into the flask)
5. add water until the bottom of the meniscus is at the etched line
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To dilute a standard solution:● 1. Rinse the pipet several times with deionized
water.● 2. Rinse the pipet twice with the standard
solution.● 3. Use the pipet to transfer the required
volume.● 4. Add enough water to bring the solution to
its final volume.
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Solution Stoichiometry
1. Write a balanced equation
2. Calculate moles given
n=m/M
3. Mole ratios
4. Calculate required quantity
m = n x M
V
nCOR
C
nVOR
OR n = CV
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Solution Stoichiometry
eg. 45.0 mL of a HCl(aq) solution is used to neutralize 30.0 mL of a 2.48 mol/L NaOH solution.
Calculate the molar concentration of the HCl(aq) solution.
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Sample Problems1. What mass of copper metal is needed to
react with 250.0 mL of 0.100 mol/L silver nitrate solution? (0.794g)
2. Calculate the volume of 2.00 M HCl(aq) needed to neutralize 1.20 g of dissolved NaOH. (0.0150L)
3. What volume of 3.00 mol/L HNO3(aq) is needed to neutralize 450.0 mL of 0.100 mol/L Sr(OH)2(aq)? (30.0mL)
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Sample Problem Solutions
Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)
Step 2 n = 0.02500 mol AgNO3
Step 3 n = 0.01250 mol Cu
Step 4 m = 0.794 g Cu
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Sample Problem Solutions
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
Step 2 n = 0.0300 mol NaOH
Step 3 n = 0.0300 mol HCl
Step 4 V = 0.0150 L HCl
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Sample Problem Solutions
2 HNO3(aq) + Sr(OH)2(aq) →
2 H2O(l) + Sr(NO3)2(aq)
Step 2 n = 0.04500 mol Sr(OH)2
Step 3 n = 0.0900 mol HNO3
Step 4 V = 0.0300 mol/L HNO3
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The Solution Process (p. 299)
● Dissociation occurs when an ionic compound breaks into ions as it dissolves in water.
● A dissociation equation shows what happens to an ionic compound in water.
eg. NaCl(s) → Na+(aq) + Cl-(aq)
K2SO4(s) → 2 K+(aq) + SO4
2-(aq)
Ca(NO3)2(s) → Ca2+(aq) + 2 NO3
-(aq)
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The Solution Process (p. 299)
● Solutions of ionic compounds conduct electric current.
● A solute that conducts an electric current in an aqueous solution is called an electrolyte.
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The Solution Process (p. 299)
● Acids are also electrolytes.● Acids form ions when dissolved in water.
eg. H2SO4(aq) → 2 H+(aq) + SO4
2-(aq)
HCl(s) → H+(aq) + Cl-(aq)
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● Molecular Compounds DO NOT dissociate in water.
eg. C12H22O11(s) → C12H22O11(aq)
● Because they DO NOT conduct electric current in solution, molecular compounds are non-electrolytes.
The Solution Process (p. 299)
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The molar concentration of any dissolved ion is calculated using the ratio from the dissociation equation.
eq. What is the molar concentration of each ion in a 5.00 mol/L MgCl2(aq) solution:
5.00 mol/L 5.00 mol/L 10.00 mol/L
The Solution Process (p. 299)
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What mass of calcium chloride is required to What mass of calcium chloride is required to prepare 2.00 L of a solution in which the Clprepare 2.00 L of a solution in which the Cl--
(aq)(aq) concentration is 0.120 mol/L ? concentration is 0.120 mol/L ?