the study of chemical change is the heart of chemistry
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The study of chemical change is the heart of chemistry. Law of Conservation of Mass. - PowerPoint PPT PresentationTRANSCRIPT
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The study of chemical change is the heart of chemistry.
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LAW OF CONSERVATION OF MASS“We may lay it down as an
incontestable axiom that, in all the operations of art and nature, nothing is
created; an equal amount of matter exists both before and after the
experiment. Upon this principle, the whole art of performing chemical
experiments depends.”--Antoine Lavoisier, 1789
“Atoms are neither created nor destroyed during any chemical reaction.”
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3.1 CHEMICAL EQUATIONS Consider a simple chemical equation:
2 H2 + O2 → 2 H2O + : “reacts with”, → : “produces”
•reactants and products•coefficients•balanced
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3.1 CHEMICAL EQUATIONS
The difference between a subscript and a coefficient Subscripts should not be change when balancing
equation
BALANCING EQUATIONS
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3.1 CHEMICAL EQUATIONS
Consider a chemical reaction:
BALANCING EQUATIONS
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3.1 CHEMICAL EQUATIONS
Express the chemical reaction
Balance carbon and hydrogen
Complete the chemical equation by balancing oxygen
BALANCING EQUATIONS
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3.1 CHEMICAL EQUATIONS
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3.1 CHEMICAL EQUATIONS INDICATING THE STATES OF
REACTANTS AND PRODUCTS
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3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYCOMBINATION AND DECOMPOSITION REACTIONS
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3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYCOMBINATION AND DECOMPOSITION REACTIONS
Figure 3.7
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3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY
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3.2 SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITYCOMBUSTION IN AIR
When hydrocarbons are combusted in air, they react with O2 to form CO2 and H2O.
Combustion of hydrocarbon derivatives•CH3OH, C2H5OH•Glucose (C6H12O6) – oxidation reaction
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3.3 FORMULA WEIGHTS How do we relate the numbers of atoms or molecules to
the amounts we measure in the laboratory?
Although we can not directly count atoms or molecules, we can indirectly determine their numbers if we know their masses.
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3.3 FORMULA WEIGHTSFORMULA AND MOLECULAR WEIGHTS
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3.3 FORMULA WEIGHTSPERCENTAGE COMPOSITION FROM FORMULAS
Needed for identifying unknown samples
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3.4 AVOGADRO’S NUMBER & THE MOLE The definition of a mole
•The amount of matter that contains as many objects (atoms, molecules, etc) as the number of atoms in exactly 12 g of isotopically pure 12C.
Avogadro’s number•The number of objects in 1 mole of matter.
•6.0221421 × 1023
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3.4 AVOGADRO’S NUMBER & THE MOLE
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3.4 AVOGADRO’S NUMBER & THE MOLEMOLAR MASS
The mass of a single atom of an element (in amu) is numerically equal to the mass (in grams) of 1mol of that element.
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3.4 AVOGADRO’S NUMBER & THE MOLEMOLAR MASS
The molar mass of a substance is the mass in grams of one mole of the substance.
Figure 3.10
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3.4 AVOGADRO’S NUMBER & THE MOLE
0.02989 mol
6.05 mol
180.0 g/mol
84.0 g/mol
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3.4 AVOGADRO’S NUMBER & THE MOLE
71.1 g
(a) 532 g, (b) 0.0029 g
164.1 g/mol
84.0 g/mol98.1 g/mol
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3.4 AVOGADRO’S NUMBER & THE MOLE
Molecules C6H12O6
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3.5 EMPIRICAL FORMULAS FROM ANALYSES The ratio of the number of moles of each element in a
compound gives the subscript in a compound’s empirical formula.
Consider a compound containing Hg & Cl (MWHg 200.6, MWCl 35.5) (73.9% Hg, 26.1% Cl by mass) How to get the empirical formula for the compound?
HgCl2
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Procedure for calculating an empirical formula from percentage composition.
3.5 EMPIRICAL FORMULAS FROM ANALYSES
40.92 g C, 4.58 g H, and 54.50 g O.
C:H:O = 3(1:1.33:1) = 3:4:3C3H4O3
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3.5 EMPIRICAL FORMULAS FROM ANALYSES
The subscripts in the molecular formula of a substance are always a whole-number multiple of the corresponding subscripts in its empirical formula.
MOLECULAR FORMULA FROM EMPIRICAL FORMULA
The formula weight of the empirical formula C3H4 is 3(12.0 amu) + 4(1.0 amu) = 40.0 amu
the molecular formula: C9H12
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3.5 EMPIRICAL FORMULAS FROM ANALYSES
A common technique used for the determination of the
empirical formula for compounds containing principally
carbon and hydrogen.
COMBUSTION ANALYSIS
Fig 3.14 Apparatus for combustion analysis.
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3.5 EMPIRICAL FORMULAS FROM ANALYSES COMBUSTION ANALYSIS
Mass of O = mass of sample - (mass of C + mass of H) = 0.255 g - (0.153 g + 0.0343 g) = 0.068 g O
C3H8O
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3.6 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS
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3.6 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS
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3.6 QUANTITATIVE INFORMATION FROM BALANCED EQUATIONS Procedure for calculating amounts of reactants consumed or
products formed in a reaction
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KClO3 122.55, KCl 74.5, O2 32.00 Answer: 1.76 g
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MWLiOH 23.95
2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l)
Answer: 3.64 g
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3.7 LIMITING REACTANTSLet’s consider a sandwich-making process:
If you have 10 slices of bread and 7 slices of cheese,
You will have 5 sandwiches and 2 slices of cheese leftover. In this case,
•Limiting reactant (limiting reagent) : Bd•Excess reactant (excess reagent) : Ch
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3.7 LIMITING REACTANTS
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Sample Exercise 3.18 Calculating the Amount of Product Formed from a Limiting Reactant
The most important commercial process for converting N2 from the air into nitrogen-
containing compounds is based on the reaction of N2 and H2 to form ammonia (NH3):
N2(g) + 3 H2(g) → 2 NH3(g)
How many moles of NH3 can be formed from 3.0 mol of N2 and 6.0 mol of H2?
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Answer: (a) AgNO3, (b) 1.59 g, (c) 1.39 g, (d) 1.52 g Zn
Zn: 65.39 g/molAgNO3: 169.87 g/molAg: 107.9 g/molZn(NO3)2: 189.39 g/mol
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3.7 LIMITING REACTANTS THEORETICAL YIELD The quantity of product that is calculated to form when all of
the limiting reactant reacts
(a) The theoretical yield is
146.14 g/mol84.16 g/mol
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3.7 LIMITING REACTANTS THEORETICAL YIELD The quantity of product that is calculated to form when all of
the limiting reactant reacts
146.14 g/mol84.16 g/mol
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EXERCISES
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EXERCISES
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EXERCISES3.94
3.94
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EXERCISES
Atomic number 57, Lanthanum
3.96
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EXERCISES
3.101
3.101