unit 3 (chp 1,2,3): matter, measurement, & stoichiometry john d. bookstaver st. charles...
TRANSCRIPT
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Unit 3 (Chp 1,2,3):
Matter, Measurement, & Stoichiometry
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
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Chemistry:The study of matter and the changes it undergoes.
Quantitativeor
Qualitative
(has mass and takes up space)
Ni + HClnickel hydrochloric acid
NiCl2nickel(II) chloride
H2 +
hydrogen solid aqueous gas solid crystalsmetal solution
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MatterAtom:
Element:
Compound:
H H C C
C C
OO Na
simplest particle retaining properties.
same type of atom (1 or more)
different atoms bonded.H2O CO2 NaCl
H2 O2
C
molecule
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salt, baking soda, water, sugar
oxygen, iron, hydrogen, gold
Matter
MixturePure
Substance
ElementsCompoundsHeterogeneous
MixtureHomogeneou
s Mixture
separate physically
cannot separate physically
differences or unevenly
mixed
uniform or evenly mixed
separate chemically
cannot separate
Chemicalchanges
Physicalchanges
(solutions)
filtering
distillation(boiling)
(suspensions/colloids) NaCl NaHCO3
H2O C12H22O11
O2 FeH2 Au
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Density:
d =mV
Chemical Property OR Physical Property
Why?
ratio of mass to volumeor
matter to space occupied
Units: g/mLg/cm3
kg/L
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Changes of Matter
• Physical Changes: do not change the composition of a
substance.• temperature, changes of state, amount, etc.
• Chemical Changes: result in new substances.
• combustion, oxidation, decomposition, etc.
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Chemical Separation:
• Compounds can be decomposed into elements.
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Physical Separation:
Separates heterogeneous mixtures (solids from liquids).
Filtration:
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Physical Separation:
Distillation: Separates solution by boiling point differences.
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Physical Separation:Chromatography:Separates homogeneous mixture by differences in solubility (attractions).
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Power of 10 is the number of places the decimal has been moved.
Examples: 42000 = 4.2 x 104
0.0508 = 5.08 x 10–2
positive power: move decimal right to obtain the original # in standard notation.
negative power: move decimal left to obtain the original # in standard notation.
Scientific Notation
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1. Convert the numbers to scientific notation.
(i) 24500
(ii) 0.000985
(iii) 12002
2. Convert to standard notation.
(i) 4.2 x 105
(ii) 2.15 x 10-4
(iii) 3 x 10-3
2.45 x 104
9.85 x 10–4
1.2002 x 104
420,0000.000215
0.003
Scientific Notation
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Metric Prefixes
BASE UNIT: 1 m 1 L 1 g
0.01 cm
0.001 mL
0.000 001 µg0.000 000 001 nm
Prefix Symbol Multiplier Examples:
1,000,000,000 GB
1,000,000 MJ1,000 kg
(atoms)
(light wavelength)
(nuclei)
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Uncertainty in MeasurementsMeasuring devices have different uses and different degrees of precision.
(uncertainty)
% Error = |Accepted – Experimental| x100 Accepted
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Uncertainty
5.23 cm
(uncertain)
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Significant Figures• measured digits.
• last digit is estimated, but IS significant.
• do not overstate the precision
5.23 cm
5.230 cm
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Significant Zeroes
1. All nonzero digits are significant.
2. Captive Zeroes between two significant figures are significant.
3. Leading Zeroes at the beginning of a number are never significant.
4. Trailing Zeroes:
SIG, if at end AND a decimal point.
NOT, if there is no decimal point.
0.0003700400 grams
0’s
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round answers to keep the fewest decimal places
round answers to keep the fewest significant figures
Significant Figures
+ or –
x or ÷
3.48 + 2.2 =
6.40 x 2.0 =
5.68 5.7
12.8 13
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1. How many sig figs are in each number?(i) 250.0(ii) 4.7 x 10–5
(iii) 34000000(iv) 0.03400
2. Round the answer to the correct sig figs. (i) 34.5 x 23.46 (ii) 123/3 (iii) 23.888897 + 11.2 (iv) 2.50 x 2.0 – 3
Significant Figures WS 1s
4
224
80940
35.12
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WARM UP (for QUIZ!!!)
• Review WS 1s #1, 3, 10
• Complete WS 1a #1, 2, 8, 9, 10
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Law of Definite Proportions
elemental formulas (composition) of pure compounds cannot vary.
• 2 H’s & 1 O is ALWAYS water.• Water is ALWAYS 2 H’s & 1 O.
• 2 H’s & 2 O’s is NOT water.
√ H2O
X H2O2
HO
H
H H
O
O
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Law of Conservation of Mass
The total mass of substances present at the end of a chemical process is the same as the mass of substances present before the process took place.
__H2 + __O2 __H2O
22
Balancing Equations!!!
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Symbols of Elements
126C
Mass Number= p’s + n’s
Atomic Number (Z)= p’s
Element Symbol
All atoms of the same element have the same number of protons (same Z), but…
can have different mass numbers. HOW?
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11H
21H
31H
protium deuterium tritium
element:mass:
why?
same or differentsame or differentsame # of protons (& electrons), but different # of neutrons
Isotopes
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Average Atomic Mass• average atomic mass: calculated as a
weighted average of isotopes by their relative abundances.
(6.015)(0.0750) + (7.016)(0.925) = 6.94 amu
Avg. Mass = (Mass1)(%) + (Mass2)(%) …
• lithium-6 (6.015 amu), which has a relative abundance of 7.50%, and
• lithium-7 (7.016 amu), which has a relative abundance of 92.5%.
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WS Atomic Structure
element sample
atomized,ionized
magnetic field
~75%
~25%
isotopes separated
by difference in mass
(35)(~0.75) + (37)(~0.25) = ?Cl (avg at. Mass) =
Mass Spectrometry
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Molecular (Covalent) Compounds
Covalent compounds contain nonmetals that “share” electrons to form molecules.
(molecular compounds)
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Diatomic Molecules
These seven elements occur naturally as molecules containing two atoms.
“H-air-ogens”
7
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Binary Molecular Compounds
• list less electronegative atom first. (left to right on PT)
• use prefix for the number of atoms of each element.
• change ending to –ide.
CO2: carbon dioxide
CCl4: carbon tetrachloride
N2O5: ________________
CuSO4∙5H2Ocopper(II) sulfate pentahydrate
dinitrogen pentoxide
(ionic & covalent)
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IonsCationsmetals lose e’spositive (+)(metal) ion
Anionsnonmetals
gain e’snegative (–)
(nonmetal)ide
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Ionic BondsAttraction between +/– ions formed by metals & nonmetals transferring e–’s.
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Formulas of Ionic Compounds
• Compounds are electrically neutral, so the formulas can be determined by:
– Crisscross the charges as subscripts (then erase)– If needed, reduce to lowest whole number ratio.
Pb4+ O2– Pb2O4 PbO2
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Naming Ionic Compounds
1) Cation: Write metal name (ammonium NH4+)
For transition metals with multiple charges, write charge as Roman numeral in parentheses.
Iron(II) chloride, FeCl2Iron(III) chloride, FeCl3
2) Anion: Write nonmetal name with –ide
OR the polyatomic anion name. (–ate, –ite)
Iron(II) sulfide, FeS
Magnesium sulfate, MgSO4
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Common Polyatomic
Ions
Name Symbol Charge*ammonium NH4
+ 1+*acetate (ethanoate)
C2H3O2–
(CH3COO–)1–
*hydroxide OH– 1–*perchlorate ClO4
– 1–*chlorate ClO3
– 1– chlorite ClO2
– 1– hypochlorite ClO– 1– bromate BrO3
– 1– iodate IO3
– 1–*nitrate NO3
– 1– nitrite NO2
– 1– cyanide CN– 1–*permanganate MnO4
– 1–*bicarbonate(hydrogen carbonate)
HCO3– 1–
*carbonate CO32– 2–
*sulfate SO42– 2–
sulfite SO32– 2–
*chromate CrO42– 2–
dichromate Cr2O72– 2–
*phosphate PO43– 3–
* these 12 will be on Quiz 1
- all 20 Polyatomic Ions will be on Quiz 2
WS 2d
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perchlorate ClO4–
chlorate ClO3–
chlorite ClO2–
hypochlorite ClO–
C N O FSi P S Cl
As Se BrTe I
nitrate NO3–
nitrite NO2–
In Out Ion Name
–
–
“Oxyanion” Names (elbO’s)
4321
43
___-ate___-ite
sulfate SO42–
sulfite SO32–
phosphate PO43–
per-___-ate
hypo-___-ite
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nitrate NO3–
nitrite NO2–
sulfate SO42–
sulfite SO32–
perchlorate ClO4–
chlorate ClO3–
chlorite ClO2–
hypochlorite ClO–
Ion Acidadd H+
Name Acids from these oxyanions:
In Out Ion Name Acid Name
4 – per-___-ate3 4 ___-ate2 3 ___-ite1 – hypo-___-ite
per-___-ic acid ___-ic acid ___-ous acid hypo-___-ous acid
Naming Acids
WS 2e
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Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
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Anatomy of a Chemical Equation
Reactants appear on the left side of the equation.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
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Anatomy of a Chemical Equation
Products appear on the right side of the equation.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
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Anatomy of a Chemical Equation
States (s, l, g, aq) written in parentheses next to each compound
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
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Anatomy of a Chemical Equation
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
Subscripts show how many atoms of each element
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Anatomy of a Chemical Equation
Coefficients show the amount of each particle and are inserted to balance the equation.
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
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Reaction Types
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Combination
2 Mg(s) + O2(g) 2 MgO(s)
Demo: MgO 2 → 1A + B → AB
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2 NaN3(s) 2 Na(s) + 3 N2(g)
Decomposition1 → 2AB → A + B
(50 milliseconds!)
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Replacement Reactions(or “Displacement”)
Single ReplacementAB + C → A + CB
Pb(NO3)2(aq) + KI(aq) PbI2(s) + KNO3(aq)
Double ReplacementAB + CD → AD + CB
video clip
Demo: PbI2
AgNO3(aq) + Cu(s) Ag(s) + CuNO3(aq)
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CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
•Often involve hydrocarbons reacting with oxygen in the air
WS 4a
CxHy + _O2 _CO2 + _H2O
Combustion
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
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Formula Weights
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Formula Weight (FW)
• Sum of the atomic weights for the atoms in a chemical formula
• Formula Weight of calcium chloride, CaCl2, is…
Ca: 1(40.08 amu)
+ Cl: 2(35.45 amu)
110.98 amu
• Sum of the atomic weights for the atoms in a molecule or compound
• Molecular Weight of ethane, C2H6, is…
Molecular Weight (MW)
C: 2(12.01 amu)
+ H: 6(1.008 amu)
30.07 amu
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Percent Composition
One can find the percent by mass of a compound of each element in the compound by using this equation.
% element =(# of atoms)(AW)
(FW)x 100
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Percent Composition
So the percentage of carbon in ethane (C2H6) is…
%C =(2)(12.01)
(30.07)
24.0230.07
= x 100
= 79.88% C
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Moles
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Avogadro Constant• One mole of particles
contains the Avogadro constant of those particles 6.022 x 1023
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Mole Relationships• One mole of atoms, ions, or molecules contains the
Avogadro constant of those particles 6.022 x 1023
In 1 mol Na2CO3 , how many…
• Na atoms?• C atoms?• O atoms?• How many donuts in 1 mol of donuts?• How many boogers in 1 mol of boogers?
Which has more atoms, 1 mol CH3 or 1 mol NH3 ?How about CH3CH2OH or H2SO4 ?
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Molar Mass• the mass of 1 mol of a substance (g/mol)
–molar mass (in g/mol) of an element is the atomic mass (in amu) on the periodic table
– formula weight (amu) of a compound
same number as the
molar mass (g/mol) of 1 mole of particles of that compound
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Using MolesMoles are the bridge from
the particle (micro) scale to
the real-world (macro) scale.
Mass(grams)
Particles(atoms)
(molecules)(units)
Moles(groups of 6.022x1023 particles)
molar mass
# g1 mol
1 mol# g
Avogadro constant
6.022x1023
1 mol
1 mol6.022x1023
macro-bridge micro-
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Using Moles1.What is the mass of 1 mole of copper(II)
bromide, CuBr2?
2.How many moles are there in 112 g of copper(II) bromide, CuBr2?
3.How many particles present in each of the questions #1 & #2 above?
(63.55) + 2(79.90) = 223.35 g
112 g CuBr2 x1 mol CuBr2
223.35 g CuBr2
= 0.501 mol CuBr2
0.501 mol x6.022 x 1023 particles
1 mol= 3.02 x 1023
particles
= 6.022 x 1023 particles
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•Balanced chemical equations show the amount of:
Most important are the ratios of reactants and products in moles, or…
mol-to-mol ratios
Stoichiometry:
calculations of quantities in chemical rxns
–how much reactant is consumed or–how much product is formed
atoms, molecules, moles, and mass
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g A
g B mol B
g A1 mol A
g B1 mol B
molar mass A
molar mass B
mol A
Rxn: A(aq) + 2 B(aq) C(aq) + 2 D(aq)
Stoichiometric Calculations
1 mol Ag A
OR mol-to-mol ratio
2 mol B1 mol A
1 mol A2 mol B
OR
Coefficients of balanced equation
???
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Stoichiometric problems have 1-3 Steps: (usually)
1) Convert grams to moles (if necessary)
using the molar mass (from PT)
2) Convert moles (given) to moles (wanted)using the mol ratio (from coefficients)
3) Convert moles to grams (if necessary)
using the molar mass (from PT)
grams A x 1 mol A .
grams A=_ mol B
mol Ax x grams B
1 mol B
1) molar mass 2) mole ratio 3) molar mass
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Example : g of A g of BSolid magnesium is added to an aqueoussolution of hydrochloric acid. What mass of H2 gas will be produced from completely reacting 18.0 g of HCl with magnesium metal?
Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)
= ____ g H2
18.0 g HCl xg HCl mol HCl
mol HCl x mol H2
mol H2
g H2
36.46x 2.016
12
1 1
0.498 g H2
molar mass A
molar mass B
g of A
Stoichiometric CalculationsHW p. 114 #58
mole ratio B/A
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Finding Empirical Formulas
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Types of Formulas
• Empirical formulas:
the lowest ratio of atoms of each element in a compound.
• Molecular formulas:
the total number of atoms of each element in a compound.
CH3
C2H6
C2H4O
C6H12O3
molecular mass = emp. form. empirical mass multiple
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Percent to Mass
Mass to Mole
Divide by Small
Times ‘till Whole
Steps (rhyme)
÷ moles by smallest to get mole ratio of atoms
MM from PT
assume 100 g
x (if necessary) to get whole numbers of atoms
Calculating Empirical Formulas
75 % C 75 g C 6.2 mol C
25 % H 25 g H 24.8 mol H
1 C
4 H
CH4
from Mass % Composition
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Butane is 17.34% H and 82.66% C by mass.Determine its empirical formula.
If molecular mass is 58 g∙mol–1, what is theMolecular Formula?
82.66 g C
17.34 g H
1) Percent to Mass
2) Mass to Mole
82.66 g C x = 6.883 mol C
17.34 g H x = 17.20 mol H
1 mol C12.01 g C
1 mol H 1.008 g H
6.883 mol
6.883 mol
3) Divide by Small
4) Times ’till Whole
= 1 1 C
= 2.499 2.5 H
x 2= 2 C
C2H5 x 2= 5 H
C4H10
5829.06
= 2 2 (C2H5) =molecular massempirical mass
HW p. 113 #43a, 48
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Percent to Mass
Mass to Mole
Divide by Small
Times ‘till Whole
Calculating Empirical Formulas
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• Hydrocarbons with C and H are analyzed through combustion with O2 in a chamber.
g C is from the g CO2 produced
g H is from the g H2O produced
g X is found by subtracting (g C + g H) from g sample
Combustion Analysis
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When 4-ketopentenoic acid is analyzed by combustion, a 0.3000 g sample produces 0.579 g of CO2 and
0.142 g of H2O.
The acid contains only C, H, and O.
What is the empirical formula of the acid?
Combustion Analysis
Example 1
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0.579 g CO2 x1 mol CO2
44.01 g CO2
1 mol H2O18.02 g H2O
1 mol C1 mol CO2
x12.01 g C1 mol C
x
2 mol H1 mol H2O
x 1.008 g H1 mol H
x0.142 g H2O x
= 0.158 g C
= 0.0159 g H
0.3000 g sample – (0.158 g C) – (0.0159 g H) =
= 0.126 g O
? g C
? g H
? g O
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0.00788 mol
0.00788 mol
0.00788 mol0.158 g C x
1 mol C12.01 g C
1 mol H1.008 g H
0.0132 mol C=
0.0158 mol H=0.0159 g H x
=
=
1.67 C
2 H
0.126 g O x1 mol O
16.00 g O0.00788 mol O= = 1 O
x 3 = 5 C
x 3 = 6 H
x 3 = 3 OC5H6O3
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A sample of a chlorohydrocarbon with a mass of 4.599 g, containing C, H and Cl, was combusted in excess oxygen to yield 6.274 g of CO2 and 3.212 g of H2O.
Calculate the empirical formula of the compound.
If the compound has a MW of 193 g∙mol–1, what is the molecular formula?
Example 2Combustion Analysis
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6.274 g CO2 x1 mol CO2
44.01 g CO2
1 mol H2O18.02 g H2O
1 mol C1 mol CO2
x12.01 g C1 mol C
x
2 mol H1 mol H2O
x 1.008 g H1 mol H
x3.212 g H2O x
= 1.712 g C
= 0.3593 g H
4.599 g sample – (1.712 g C) – (0.3593 g H) =
= 2.528 g Cl
? g C
? g H
? g Cl
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0.07131 mol
0.07131 mol
0.07131 mol1.712 g C x
1 mol C12.01 g C
1 mol H1.008 g H
0.1425 mol C=
0.3564 mol H=0.3593 g H x
=
=
2 C
5 H
2.528 g Cl x1 mol Cl
35.45 g Cl0.07131 mol Cl= = 1 Cl
C2H5ClIf the compound has a MW of 193 g∙mol–1, what is the molecular formula?
MWEW
19364.51
= 3
C6H15Cl3HW p. 114 #52b
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• Which ingredient will run out first?• If out of sugar, you should stop making cookies.
How Many Cookies Can I Make?
• Sugar is the limiting ingredient, because it will limit the amount of cookies you can make.
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Before After
limiting
2 H2 + O2 2 H2O
Initial: ? mol ? mol ? molChange: End:
10 7 0
0 mol 2 mol 10 mol–10 –5 +10
H2
O2
Which is limiting?
excess
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Before After
2 H2 + O2 2 H2O
Initial: ? mol ? mol ? molChange: End:
10 7 0
0 mol 2 mol 10 mol–10 –5 +10
H2
O2
O2 is in smallest amount, but…H2 is in smallest “stoichiometric” amount
Does limiting mean smallest amount of reactant? No!
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Solid aluminum metal is reacted with aqueous copper(II) chloride in solution.
• If 0.030 g Al are reacted with 0.0060 mol CuCI2, which is the limiting reactant?
• How much product will be produced?
Limiting Reactant
demo
Al(s) + CuCl2(aq) Cu(s) + AlCl3(aq)2 3 3 2
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2 Al(s) + 3 CuCl2(aq) 3 Cu(s) + 2 AlCl3(aq)
Limiting Reactant
0.030 g Al
0.0060 mol CuCl2
1 mol Al
26.98 g Al
3 mol Cu
3 mol CuCl2
x
x
x3 mol Cu
2 mol Al= 0.0017 mol Cu
= 0.0060 mol Cu
Al is limiting
HW p. 115 #72
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Theoretical Yieldtheoretical yield: the maximum amount of product that can be formed
–calculated by stoichiometry
(using LR only)
• This is different from the actual yield, the amount one actually produces and measures (or experimental)
0.030 g Al1 mol Al
26.98 g Alx x
3 mol Cu
2 mol Al= 0.0017 mol Cu
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Percent Yield
A comparison of the amount actually obtained to the amount it was possible to make
%Yield = x 100Actual
Theoretical
(calculate using the LR only)
% Error = |Accepted – Experimental| x100 Accepted
NOT % Error:
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Aluminum will react with oxygen gas
according to the equation below
4 Al + 3 O2 2 Al2O3
• In one such reaction, 23.4 g of Al are allowed to burn in excess oxygen.
39.3 g of aluminum oxide are formed. What is the percentage yield?
Percent Yield
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101.96 g Al2O3
1 mol Al2O3
4 Al + 3 O2 2 Al2O3
Percent Yield HW p. 116 #79
39.3 g of aluminum oxide are formed.
What is the percentage yield?
23.4 g Al1mol Al
26.98 g Alx
2 mol Al2O3
4 mol Alx
= 44.2 g Al2O3
x
%Yield = x 10039.3 g44.2 g
88.9 %