Chp 1,2,3
9/8/2016
Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 3 (Chp 1,2,3):
Matter, Measurement, & Stoichiometry John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall Inc.
Chemistry: The study of matter and the changes it undergoes. Ni
+
H2
HCl
nickel hydrochloric acid solid aqueous metal solution
+
NiCl2
hydrogen nickel(II) chloride gas solid crystals
Quantitative or Qualitative
Matter Atom: simplest particle with properties of element
Element: same type of atom (1 or more) HH
H2
O O
O2
C
C
Na
C C
C
Compound: different atoms bonded molecule
H2O
CO2
NaCl
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chromatography distillation (boiling)
Matter
separate physically
Mixture differences or unevenly mixed filtering Heterogeneous Mixture
cannot separate physically
Physical changes
uniform or evenly mixed
Homogeneous Mixture
(suspensions/colloids) (solutions)
Pure Substance
separate chemically
cannot separate
Chemical changes Compounds
Elements
salt, baking soda, oxygen, iron, water, sugar hydrogen, gold NaCl NaHCO3 O2 Fe H2O C12H22O11 H2 Au
Changes of Matter • Physical Changes: do not change the composition (same substance) • temperature, changes of state, amount, etc.
• Chemical Changes: result in new substances • combustion, oxidation, decomposition, etc.
Chemical Separation: • Compounds can be decomposed into elements.
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Physical Separation: Filtration: Separates heterogeneous mixtures (solids from liquids).
Physical Separation: Distillation: Separates solution by boiling point differences.
Physical Separation: Chromatography: Separates solution by differences in solubility (attractions).
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Scientific Notation Power of 10 is the number of places the decimal has been moved.
42000 = 4.2 x 104
Examples:
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 1. Convert the numbers to scientific notation. 2.45 x 104 (i) 24500 9.85 x 10–4 (ii) 0.000985 (iii) 12002 1.2002 x 104 2. Convert to standard notation. 420,000 (i) 4.2 x 105 0.000215 (ii) 2.15 x 10-4 0.003
(iii) 3 x 10-3
Metric Prefixes Prefix
Symbol
Multiplier Examples: 1,000,000,000 GB 1,000,000 MJ 1,000 kg
BASE UNIT:
1m 1L
1g 0.01 cm 0.001 mL
(light wavelength)
0.000 001 µg 0.000 000 001 nm
(atoms) (nuclei)
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Precision in Measurements Measuring devices have different uses and different degrees of precision. (uncertainty)
% Error = |Accepted – Experimental| x100 Accepted
Significant Digits • measured digits (using marks on instrument) • last estimated digit (one digit past marks)
5.23 cm
• do not overstate the precision 5.230 cm
Significant Zeroes 0.0003700400 grams
0’s
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 there’s a decimal point. NOT, if there is no decimal point.
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Sigs Digs in Operations
+ or –
round answers to keep the fewest decimal places
3.48 + 2.2 = 5.68
x or ÷
5.7
round answers to keep the fewest significant digits
6.40 x 2.0 = 12.8
13
Sig Digs Practice
WS 1s
1. How many sig figs are in each number? 4 (i) 250.0 –5 (ii) 4.7 x 10 2 (iii) 34000000 2 4 (iv) 0.03400 2. Round the answer to the correct sig figs. (i) 34.5 x 23.46 809 (ii) 123/3 40 (iii) 23.888897 + 11.2 35.1 (iv) 2.50 x 2.0 – 3 2
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
• 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.
O
√ H2O
X H2O2
elemental formulas (composition) of pure compounds cannot vary.
H
H
H H O O
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.
__H 2 2 + __O2
2 2O __H
Balancing Equations!!!
Symbols of Elements Mass Number = p’s + n’s
12 6C
Element Symbol
Atomic Number (Z) = p’s 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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Isotopes element: same or different mass: same or different why? same # of protons (& electrons), but different # of neutrons
1 1H
2 1H
protium
3 1H
deuterium
tritium
Average Atomic Mass • average atomic mass: calculated as a weighted average of isotopes by their relative abundances. • 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%.
(6.015)(0.0750) + (7.016)(0.925) = 6.94 amu
Avg. Mass = (Mass1)(%) + (Mass2)(%) …
Mass Spectrometry atomized, ionized magnetic field
element sample
isotopes separated WS Atomic by Structure difference Cl (avg at. Mass) = in mass
~75% ~25%
(35)(~0.75) + (37)(~0.25) = ?
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Molecular (Covalent) Compounds Covalent compounds contain nonmetals that “share” electrons to form molecules. (molecular compounds)
Diatomic Molecules “H-air-ogens”
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These seven elements occur naturally as molecules containing two atoms.
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 pentoxide N2O5: dinitrogen ________________ CuSO4·5H2O (ionic & covalent)
copper(II) sulfate pentahydrate
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Cations Anions Ions nonmetals metals lose e’s gain e’s positive (+) negative (–) (metal) ion (nonmetal)ide
Ionic Bonds Attraction between +/– ions formed by metals & nonmetals transferring e–’s.
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, FeCl2 Iron(III) chloride, FeCl3 2) Anion: Write nonmetal name with –ide OR the polyatomic anion name. (–ate, –ite) Iron(II) sulfide, FeS Magnesium sulfate, MgSO4
Common Polyatomic Ions * these 12 will be on Quiz 1 - all 20 Polyatomic Ions will be on Quiz 2 WS 2d
Name Symbol Charge *ammonium NH4+ 1+ *acetate C2H3O2– 1– – (ethanoate) (CH3COO ) *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 1– HCO3– (hydrogen carbonate) *carbonate CO32– 2– *sulfate SO42– 2– 2– sulfite SO3 2– 2– *chromate CrO4 2– dichromate Cr2O72– 2– *phosphate PO43– 3–
“Oxyanion” Names (elbO’s) C Si
N O P S As Se Te
perchlorate chlorate chlorite hypochlorite
ClO4– ClO3– ClO2– ClO–
nitrate nitrite
NO3– NO2–
In Out 4 –
sulfate sulfite
SO42– SO32–
phosphate
PO43–
3 2 1
4 3 –
F Cl Br I
Ion Name
per-___-ate ___-ate ___-ite hypo-___-ite
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Naming Acids Ion
add H+
Acid
In Out Ion Name Acid Name 4 – per-___-ate per-___-ic acid 3 4 ___-ate ___-ic acid 2 3 ___-ite ___-ous acid 1 – hypo-___-ite hypo-___-ous acid perchlorate Name Acids chlorate from these chlorite oxyanions: hypochlorite WS 2e
ClO4– ClO3– ClO2– ClO–
nitrate NO3– nitrite NO2– sulfate SO42– sulfite SO32–
Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Reactants appear on the left side of the equation.
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Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Products appear on the right side of the equation.
Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
States (s, l, g, aq) written in parentheses next to each compound
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 CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Coefficients show the amount of each particle and are inserted to balance the equation.
Reaction Types
Combination A + B → AB
Demo: MgO
2 Mg(s)
+
2→1
O2(g) → 2 MgO(s)
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Decomposition
1→2
AB → A + B
(50 milliseconds!)
2 NaN3(s) 2 Na(s) + 3 N2(g)
Replacement Reactions (or “Displacement”)
Single Replacement AB + C → A + CB (aq)
+
(s) →
(s)
video clip
+
(aq)
Double Replacement AB + CD → AD + CB
Pb(NO3)2(aq) + KI(aq) → PbI2(s) + KNO3(aq) Demo:
Combustion CxHy + _O2 _CO2 + _H2O • Often involve hydrocarbons reacting with oxygen in the air WS 4a
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
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Formula Weights
Formula Weight (FW)
Molecular Weight (MW)
• Sum of the atomic weights for the atoms in a chemical formula • Formula Weight of calcium chloride, CaCl2, is…
• Sum of the atomic weights for the atoms in a molecule or compound • Molecular Weight of ethane, C2H6, is…
Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu
C: 2(12.01 amu) + H: 6(1.008 amu) 30.07 amu
Percent Composition % element =
(# of atoms)(AW) x 100 (FW)
One can find the percent by mass of a compound of each element in the compound by using this equation.
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Percent Composition So the percentage of carbon in ethane (C2H6) is… %C = =
(2)(12.01) (30.07) 24.02 x 100 30.07
= 79.88% C
Moles
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 ?
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
Using Moles Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. macro- molar Mass
mass
bridge Moles
microAvogadro constant
Particles
1 mol #g
(groups of (atoms) 6.022x1023 (molecules) 23 particles) 6.022x10 (units) 1 mol
#g 1 mol
1 mol 6.022x1023
(grams)
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Using Moles 1.What is the mass of 1 mole of copper(II) bromide, CuBr2?(63.55) + 2(79.90) = 223.35 g = 6.022 x 1023 particles 2.How many moles are there in 112 g of copper(II) bromide, CuBr2? 1 mol CuBr2 = 0.501 mol 112 g CuBr2 x 223.35 g CuBr2 CuBr2 3.How many particles present in each of the questions #1 & #2 above? 6.022 x 1023 particles = 3.02 x 1023 0.501 mol x 1 mol particles
Stoichiometry: calculations of quantities in chemical rxns –how much reactant is consumed or –how much product is formed
•Balanced chemical equations show the amount of: atoms, molecules, moles, and mass Most important are the ratios of reactants and products in moles, or…
mol-to-mol ratios
Stoichiometric Calculations Rxn: gA
A(aq) + 2 B(aq) C(aq) + 2 D(aq) molar mass A g A 1 mol A
? 1 mol A OR ? gA ? g B 1 mol B molar gB mass B
mol A
Coefficients of balanced equation OR
mol-to-mol 2 mol B 1 mol A ratio
1 mol A 2 mol B
mol B
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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) 1 mol A x _ mol B x grams A mol A
grams A x
.
grams B = 1 mol B
1) molar mass 2) mole ratio 3) molar mass
Stoichiometric Calculations Example : g of A g of B
HW p. 114 #58
Solid magnesium is added to an aqueous solution 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) mole ratio B/A 18.0 g HCl x g of A
molar mass B
2.016 g H2 1 mol HCl x 1 mol H2 x 36.46 g HCl 2 mol HCl 1 mol H2 molar mass A
= ____ g Hg2 H2 0.498
Finding Empirical Formulas
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Types of Formulas • Empirical formulas: the lowest ratio of atoms of CH 3 each element in a compound.
C2H4O
• Molecular formulas: C2H6 C6H12O3 the total number of atoms of each element in a compound. molecular mass = emp. form. empirical mass multiple
Calculating Empirical Formulas Steps (rhyme)
from Mass % Composition
Percent to Mass assume 100 g Mass to Mole
MM from PT
÷
moles by smallest to Divide by Small get mole ratio of atoms CH4 x (if necessary) to get
Times ‘til Whole whole numbers of atoms 75 % C
75 g C
6.2 mol C
1C
25 % H
25 g H
24.8 mol H
4H
1) Percent to Mass
3) Divide by Small
2) Mass to Mole
4) Times ’til Whole
Butane is 17.34% H and 82.66% C by mass. Determine its empirical formula. 82.66 g C x 1 mol C = 6.883 mol C = 1 ≈ 1 C x 2 12.01 g C 6.883 mol =2C 17.34 g H x 1 mol H = 17.20 mol H 1.008 g H 6.883 mol
= 2.499 ≈ 2.5 H C2H5 x 2 =5H If molecular mass is 58 g·mol–1, what is the Molecular Formula? HW p. 113 #43a, 48
molecular mass empirical mass
58 =2 29.06
2 (C2H5) = C4H10
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Calculating Empirical Formulas Percent to Mass Mass to Mole Divide by Small Times ‘til Whole
Combustion Analysis
• Hydrocarbons with C and H are analyzed through combustion with O2 in a chamber. Step 1 is “combustion g H is from the g H2O produced to mass”
g C is from the g CO2 produced
g X is found by subtracting (g C + g H) from g sample
Combustion Analysis Example 1 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?
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0.579 g CO2 x ?gC
1 mol CO2 1 mol C 12.01 g C x x 44.01 g CO2 1 mol CO2 1 mol C = 0.158 g C
Step 1: “combustion to mass”
1 mol H2O x 2 mol H x1.008 g H 0.142 g H2O x 18.02 g H2O 1 mol H2O 1 mol H ?gH = 0.0159 g H 0.3000 g sample – (0.158 g C) – (0.0159 g H) = ?gO
= 0.126 g O
0.158 g C x
1 mol C = 0.0132 mol C = 1.67 C 12.01 g C 0.00788 mol x3=5C
0.0159 g H x
0.126 g O x
1 mol H = 0.0158 mol H = 2 H 1.008 g H 0.00788 mol x3=6H
1 mol O = 0.00788 mol O = 1 O 16.00 g O 0.00788 mol x3=3O
C5H6O3
Combustion Analysis Example 2 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?
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6.274 g CO2 x ?gC
1 mol CO2 1 mol C 12.01 g C x x 44.01 g CO2 1 mol CO2 1 mol C = 1.712 g C
Step 1: “combustion to mass”
1 mol H2O x 2 mol H x1.008 g H 3.212 g H2O x 18.02 g H2O 1 mol H2O 1 mol H ?gH = 0.3593 g H 4.599 g sample – (1.712 g C) – (0.3593 g H) = ? g Cl = 2.528 g Cl
1.712 g C x
1 mol C = 0.1425 mol C = 2 C 12.01 g C 0.07131 mol
0.3593 g H x
1 mol H = 0.3564 mol H = 5 H 1.008 g H 0.07131 mol
1 mol Cl 0.07131 mol Cl 2.528 g Cl x = 1 Cl = 35.45 g Cl 0.07131 mol
C2H5Cl If the compound has a MW of 193 g·mol–1, what MW 193 =3 is the molecular formula? EW 64.51 HW p. 114 #52b
C6H15Cl3
How Many Cookies Can I Make?
• Which ingredient will run out first? • If out of sugar, you should stop making cookies. • Sugar is the limiting ingredient, because it will limit the amount of cookies you can make.
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Before
H2
After O2
Which is limiting?
2 H2 + O2 Initial: 10 ? mol Change: –10 End: 0 mol limiting
Before
?7 mol –5 2 mol
?0 mol +10 10 mol
excess
H2
After O2
2 H2 + O2 Initial: 10 ? mol Change: –10 End: 0 mol
2 H2O
?7 mol –5 2 mol
2 H2O ?0 mol +10 10 mol
Does limiting mean smallest amount of reactant? No!
O2 is in smallest amount, but… H2 is in smallest “stoichiometric” amount
Limiting Reactant • convert
reactant A to reactant B to compare • If available < needed (limiting) • If available > needed (excess) Solid aluminum metal is reacted with aqueous copper(II) chloride in solution
2 Al + 3 CuCl2 2 AlCl3 + 3 Cu 54.0 g Al 4.50 mol CuCl2 (Which is limiting?) 54.0 g Al x 1 mol Al x 3 mol CuCl2 = 3.00 mol CuCl 2 26.98 g Al 2 mol Al (4.50 mol CuCl2) available > needed (3.00 mol CuCl2) CuCl2 is excess Al is limiting
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Theoretical Yield theoretical yield: the maximum amount of product that can be formed – calculated by stoichiometry – limited by LR (use LR only to calculate) limiting 54.0 g Al x 1 mol Al x 3 mol Cu x 63.55 g Cu = 191 g Cu 26.98 g Al 2 mol Al 1 mol Cu
produced
• different from actual yield (or experimental), amount recovered in the experiment HW p. 115 #72
Percent Yield A comparison of the amount actually obtained to the amount it was possible to make
%Yield =
Actual x 100 Theoretical
(calculate using the LR only)
NOT % Error: % Error = |Accepted – Experimental| x100 Accepted
Percent Yield 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?
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Percent Yield 4 Al + 3 O2
HW p. 116 #79, 77
2 Al2O3
1mol Al 2 mol Al2O3 101.96 g Al2O3 23.4 g Al x x x 26.98 g Al 4 mol Al 1 mol Al2O3 = 44.2 g Al2O3 39.3 g of aluminum oxide are formed. What is the percentage yield?
%Yield =
39.3 g 44.2 g
x 100
88.9 %
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Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Unit 3 (Chp 1,2,3):
Matter, Measurement, & Stoichiometry John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall Inc.
Chemistry: The study of matter and the changes it undergoes. Ni
+
H2
HCl
nickel hydrochloric acid solid aqueous metal solution
+
NiCl2
hydrogen nickel(II) chloride gas solid crystals
Quantitative or Qualitative
Matter Atom: simplest particle with properties of element
Element: same type of atom (1 or more) HH
H2
O O
O2
C
C
Na
C C
C
Compound: different atoms bonded molecule
H2O
CO2
NaCl
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chromatography distillation (boiling)
Matter
separate physically
Mixture differences or unevenly mixed filtering Heterogeneous Mixture
cannot separate physically
Physical changes
uniform or evenly mixed
Homogeneous Mixture
(suspensions/colloids) (solutions)
Pure Substance
separate chemically
cannot separate
Chemical changes Compounds
Elements
salt, baking soda, oxygen, iron, water, sugar hydrogen, gold NaCl NaHCO3 O2 Fe H2O C12H22O11 H2 Au
Changes of Matter • Physical Changes: do not change the composition (same substance) • temperature, changes of state, amount, etc.
• Chemical Changes: result in new substances • combustion, oxidation, decomposition, etc.
Chemical Separation: • Compounds can be decomposed into elements.
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Physical Separation: Filtration: Separates heterogeneous mixtures (solids from liquids).
Physical Separation: Distillation: Separates solution by boiling point differences.
Physical Separation: Chromatography: Separates solution by differences in solubility (attractions).
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Scientific Notation Power of 10 is the number of places the decimal has been moved.
42000 = 4.2 x 104
Examples:
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 1. Convert the numbers to scientific notation. 2.45 x 104 (i) 24500 9.85 x 10–4 (ii) 0.000985 (iii) 12002 1.2002 x 104 2. Convert to standard notation. 420,000 (i) 4.2 x 105 0.000215 (ii) 2.15 x 10-4 0.003
(iii) 3 x 10-3
Metric Prefixes Prefix
Symbol
Multiplier Examples: 1,000,000,000 GB 1,000,000 MJ 1,000 kg
BASE UNIT:
1m 1L
1g 0.01 cm 0.001 mL
(light wavelength)
0.000 001 µg 0.000 000 001 nm
(atoms) (nuclei)
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Precision in Measurements Measuring devices have different uses and different degrees of precision. (uncertainty)
% Error = |Accepted – Experimental| x100 Accepted
Significant Digits • measured digits (using marks on instrument) • last estimated digit (one digit past marks)
5.23 cm
• do not overstate the precision 5.230 cm
Significant Zeroes 0.0003700400 grams
0’s
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 there’s a decimal point. NOT, if there is no decimal point.
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Sigs Digs in Operations
+ or –
round answers to keep the fewest decimal places
3.48 + 2.2 = 5.68
x or ÷
5.7
round answers to keep the fewest significant digits
6.40 x 2.0 = 12.8
13
Sig Digs Practice
WS 1s
1. How many sig figs are in each number? 4 (i) 250.0 –5 (ii) 4.7 x 10 2 (iii) 34000000 2 4 (iv) 0.03400 2. Round the answer to the correct sig figs. (i) 34.5 x 23.46 809 (ii) 123/3 40 (iii) 23.888897 + 11.2 35.1 (iv) 2.50 x 2.0 – 3 2
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
• 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.
O
√ H2O
X H2O2
elemental formulas (composition) of pure compounds cannot vary.
H
H
H H O O
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.
__H 2 2 + __O2
2 2O __H
Balancing Equations!!!
Symbols of Elements Mass Number = p’s + n’s
12 6C
Element Symbol
Atomic Number (Z) = p’s 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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Isotopes element: same or different mass: same or different why? same # of protons (& electrons), but different # of neutrons
1 1H
2 1H
protium
3 1H
deuterium
tritium
Average Atomic Mass • average atomic mass: calculated as a weighted average of isotopes by their relative abundances. • 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%.
(6.015)(0.0750) + (7.016)(0.925) = 6.94 amu
Avg. Mass = (Mass1)(%) + (Mass2)(%) …
Mass Spectrometry atomized, ionized magnetic field
element sample
isotopes separated WS Atomic by Structure difference Cl (avg at. Mass) = in mass
~75% ~25%
(35)(~0.75) + (37)(~0.25) = ?
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Molecular (Covalent) Compounds Covalent compounds contain nonmetals that “share” electrons to form molecules. (molecular compounds)
Diatomic Molecules “H-air-ogens”
7
These seven elements occur naturally as molecules containing two atoms.
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 pentoxide N2O5: dinitrogen ________________ CuSO4·5H2O (ionic & covalent)
copper(II) sulfate pentahydrate
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Cations Anions Ions nonmetals metals lose e’s gain e’s positive (+) negative (–) (metal) ion (nonmetal)ide
Ionic Bonds Attraction between +/– ions formed by metals & nonmetals transferring e–’s.
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, FeCl2 Iron(III) chloride, FeCl3 2) Anion: Write nonmetal name with –ide OR the polyatomic anion name. (–ate, –ite) Iron(II) sulfide, FeS Magnesium sulfate, MgSO4
Common Polyatomic Ions * these 12 will be on Quiz 1 - all 20 Polyatomic Ions will be on Quiz 2 WS 2d
Name Symbol Charge *ammonium NH4+ 1+ *acetate C2H3O2– 1– – (ethanoate) (CH3COO ) *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 1– HCO3– (hydrogen carbonate) *carbonate CO32– 2– *sulfate SO42– 2– 2– sulfite SO3 2– 2– *chromate CrO4 2– dichromate Cr2O72– 2– *phosphate PO43– 3–
“Oxyanion” Names (elbO’s) C Si
N O P S As Se Te
perchlorate chlorate chlorite hypochlorite
ClO4– ClO3– ClO2– ClO–
nitrate nitrite
NO3– NO2–
In Out 4 –
sulfate sulfite
SO42– SO32–
phosphate
PO43–
3 2 1
4 3 –
F Cl Br I
Ion Name
per-___-ate ___-ate ___-ite hypo-___-ite
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Naming Acids Ion
add H+
Acid
In Out Ion Name Acid Name 4 – per-___-ate per-___-ic acid 3 4 ___-ate ___-ic acid 2 3 ___-ite ___-ous acid 1 – hypo-___-ite hypo-___-ous acid perchlorate Name Acids chlorate from these chlorite oxyanions: hypochlorite WS 2e
ClO4– ClO3– ClO2– ClO–
nitrate NO3– nitrite NO2– sulfate SO42– sulfite SO32–
Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Reactants appear on the left side of the equation.
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Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Products appear on the right side of the equation.
Anatomy of a Chemical Equation CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
States (s, l, g, aq) written in parentheses next to each compound
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 CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(g)
Coefficients show the amount of each particle and are inserted to balance the equation.
Reaction Types
Combination A + B → AB
Demo: MgO
2 Mg(s)
+
2→1
O2(g) → 2 MgO(s)
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Decomposition
1→2
AB → A + B
(50 milliseconds!)
2 NaN3(s) 2 Na(s) + 3 N2(g)
Replacement Reactions (or “Displacement”)
Single Replacement AB + C → A + CB (aq)
+
(s) →
(s)
video clip
+
(aq)
Double Replacement AB + CD → AD + CB
Pb(NO3)2(aq) + KI(aq) → PbI2(s) + KNO3(aq) Demo:
Combustion CxHy + _O2 _CO2 + _H2O • Often involve hydrocarbons reacting with oxygen in the air WS 4a
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
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Formula Weights
Formula Weight (FW)
Molecular Weight (MW)
• Sum of the atomic weights for the atoms in a chemical formula • Formula Weight of calcium chloride, CaCl2, is…
• Sum of the atomic weights for the atoms in a molecule or compound • Molecular Weight of ethane, C2H6, is…
Ca: 1(40.08 amu) + Cl: 2(35.45 amu) 110.98 amu
C: 2(12.01 amu) + H: 6(1.008 amu) 30.07 amu
Percent Composition % element =
(# of atoms)(AW) x 100 (FW)
One can find the percent by mass of a compound of each element in the compound by using this equation.
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Percent Composition So the percentage of carbon in ethane (C2H6) is… %C = =
(2)(12.01) (30.07) 24.02 x 100 30.07
= 79.88% C
Moles
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 ?
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
Using Moles Moles are the bridge from the particle (micro) scale to the real-world (macro) scale. macro- molar Mass
mass
bridge Moles
microAvogadro constant
Particles
1 mol #g
(groups of (atoms) 6.022x1023 (molecules) 23 particles) 6.022x10 (units) 1 mol
#g 1 mol
1 mol 6.022x1023
(grams)
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Using Moles 1.What is the mass of 1 mole of copper(II) bromide, CuBr2?(63.55) + 2(79.90) = 223.35 g = 6.022 x 1023 particles 2.How many moles are there in 112 g of copper(II) bromide, CuBr2? 1 mol CuBr2 = 0.501 mol 112 g CuBr2 x 223.35 g CuBr2 CuBr2 3.How many particles present in each of the questions #1 & #2 above? 6.022 x 1023 particles = 3.02 x 1023 0.501 mol x 1 mol particles
Stoichiometry: calculations of quantities in chemical rxns –how much reactant is consumed or –how much product is formed
•Balanced chemical equations show the amount of: atoms, molecules, moles, and mass Most important are the ratios of reactants and products in moles, or…
mol-to-mol ratios
Stoichiometric Calculations Rxn: gA
A(aq) + 2 B(aq) C(aq) + 2 D(aq) molar mass A g A 1 mol A
? 1 mol A OR ? gA ? g B 1 mol B molar gB mass B
mol A
Coefficients of balanced equation OR
mol-to-mol 2 mol B 1 mol A ratio
1 mol A 2 mol B
mol B
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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) 1 mol A x _ mol B x grams A mol A
grams A x
.
grams B = 1 mol B
1) molar mass 2) mole ratio 3) molar mass
Stoichiometric Calculations Example : g of A g of B
HW p. 114 #58
Solid magnesium is added to an aqueous solution 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) mole ratio B/A 18.0 g HCl x g of A
molar mass B
2.016 g H2 1 mol HCl x 1 mol H2 x 36.46 g HCl 2 mol HCl 1 mol H2 molar mass A
= ____ g Hg2 H2 0.498
Finding Empirical Formulas
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Types of Formulas • Empirical formulas: the lowest ratio of atoms of CH 3 each element in a compound.
C2H4O
• Molecular formulas: C2H6 C6H12O3 the total number of atoms of each element in a compound. molecular mass = emp. form. empirical mass multiple
Calculating Empirical Formulas Steps (rhyme)
from Mass % Composition
Percent to Mass assume 100 g Mass to Mole
MM from PT
÷
moles by smallest to Divide by Small get mole ratio of atoms CH4 x (if necessary) to get
Times ‘til Whole whole numbers of atoms 75 % C
75 g C
6.2 mol C
1C
25 % H
25 g H
24.8 mol H
4H
1) Percent to Mass
3) Divide by Small
2) Mass to Mole
4) Times ’til Whole
Butane is 17.34% H and 82.66% C by mass. Determine its empirical formula. 82.66 g C x 1 mol C = 6.883 mol C = 1 ≈ 1 C x 2 12.01 g C 6.883 mol =2C 17.34 g H x 1 mol H = 17.20 mol H 1.008 g H 6.883 mol
= 2.499 ≈ 2.5 H C2H5 x 2 =5H If molecular mass is 58 g·mol–1, what is the Molecular Formula? HW p. 113 #43a, 48
molecular mass empirical mass
58 =2 29.06
2 (C2H5) = C4H10
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Calculating Empirical Formulas Percent to Mass Mass to Mole Divide by Small Times ‘til Whole
Combustion Analysis
• Hydrocarbons with C and H are analyzed through combustion with O2 in a chamber. Step 1 is “combustion g H is from the g H2O produced to mass”
g C is from the g CO2 produced
g X is found by subtracting (g C + g H) from g sample
Combustion Analysis Example 1 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?
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0.579 g CO2 x ?gC
1 mol CO2 1 mol C 12.01 g C x x 44.01 g CO2 1 mol CO2 1 mol C = 0.158 g C
Step 1: “combustion to mass”
1 mol H2O x 2 mol H x1.008 g H 0.142 g H2O x 18.02 g H2O 1 mol H2O 1 mol H ?gH = 0.0159 g H 0.3000 g sample – (0.158 g C) – (0.0159 g H) = ?gO
= 0.126 g O
0.158 g C x
1 mol C = 0.0132 mol C = 1.67 C 12.01 g C 0.00788 mol x3=5C
0.0159 g H x
0.126 g O x
1 mol H = 0.0158 mol H = 2 H 1.008 g H 0.00788 mol x3=6H
1 mol O = 0.00788 mol O = 1 O 16.00 g O 0.00788 mol x3=3O
C5H6O3
Combustion Analysis Example 2 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?
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6.274 g CO2 x ?gC
1 mol CO2 1 mol C 12.01 g C x x 44.01 g CO2 1 mol CO2 1 mol C = 1.712 g C
Step 1: “combustion to mass”
1 mol H2O x 2 mol H x1.008 g H 3.212 g H2O x 18.02 g H2O 1 mol H2O 1 mol H ?gH = 0.3593 g H 4.599 g sample – (1.712 g C) – (0.3593 g H) = ? g Cl = 2.528 g Cl
1.712 g C x
1 mol C = 0.1425 mol C = 2 C 12.01 g C 0.07131 mol
0.3593 g H x
1 mol H = 0.3564 mol H = 5 H 1.008 g H 0.07131 mol
1 mol Cl 0.07131 mol Cl 2.528 g Cl x = 1 Cl = 35.45 g Cl 0.07131 mol
C2H5Cl If the compound has a MW of 193 g·mol–1, what MW 193 =3 is the molecular formula? EW 64.51 HW p. 114 #52b
C6H15Cl3
How Many Cookies Can I Make?
• Which ingredient will run out first? • If out of sugar, you should stop making cookies. • Sugar is the limiting ingredient, because it will limit the amount of cookies you can make.
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Before
H2
After O2
Which is limiting?
2 H2 + O2 Initial: 10 ? mol Change: –10 End: 0 mol limiting
Before
?7 mol –5 2 mol
?0 mol +10 10 mol
excess
H2
After O2
2 H2 + O2 Initial: 10 ? mol Change: –10 End: 0 mol
2 H2O
?7 mol –5 2 mol
2 H2O ?0 mol +10 10 mol
Does limiting mean smallest amount of reactant? No!
O2 is in smallest amount, but… H2 is in smallest “stoichiometric” amount
Limiting Reactant • convert
reactant A to reactant B to compare • If available < needed (limiting) • If available > needed (excess) Solid aluminum metal is reacted with aqueous copper(II) chloride in solution
2 Al + 3 CuCl2 2 AlCl3 + 3 Cu 54.0 g Al 4.50 mol CuCl2 (Which is limiting?) 54.0 g Al x 1 mol Al x 3 mol CuCl2 = 3.00 mol CuCl 2 26.98 g Al 2 mol Al (4.50 mol CuCl2) available > needed (3.00 mol CuCl2) CuCl2 is excess Al is limiting
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Theoretical Yield theoretical yield: the maximum amount of product that can be formed – calculated by stoichiometry – limited by LR (use LR only to calculate) limiting 54.0 g Al x 1 mol Al x 3 mol Cu x 63.55 g Cu = 191 g Cu 26.98 g Al 2 mol Al 1 mol Cu
produced
• different from actual yield (or experimental), amount recovered in the experiment HW p. 115 #72
Percent Yield A comparison of the amount actually obtained to the amount it was possible to make
%Yield =
Actual x 100 Theoretical
(calculate using the LR only)
NOT % Error: % Error = |Accepted – Experimental| x100 Accepted
Percent Yield 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?
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Percent Yield 4 Al + 3 O2
HW p. 116 #79, 77
2 Al2O3
1mol Al 2 mol Al2O3 101.96 g Al2O3 23.4 g Al x x x 26.98 g Al 4 mol Al 1 mol Al2O3 = 44.2 g Al2O3 39.3 g of aluminum oxide are formed. What is the percentage yield?
%Yield =
39.3 g 44.2 g
x 100
88.9 %
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