Avogadro’s Number 6.02 X 10 23. 1 Mole of anything 6.02 X 10 23 of that thing.
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Transcript of Avogadro’s Number 6.02 X 10 23. 1 Mole of anything 6.02 X 10 23 of that thing.
![Page 1: Avogadro’s Number 6.02 X 10 23. 1 Mole of anything 6.02 X 10 23 of that thing.](https://reader035.fdocuments.net/reader035/viewer/2022062301/5697bfb51a28abf838c9de04/html5/thumbnails/1.jpg)
Avogadro’s Number
6.02 X 1023
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1 Mole of anything
6.02 X 1023 of that thing
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Can be:
- atoms- molecules- ions
Particles
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0.5 Mole
3.01 X 1023
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0.25 Mole
1.50 X 1023
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2.0 Mole
12.04 X 1023
Or1.204 X 1024
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1.0 Mole of any gas at STP
22.4 Liters
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STP
Standard Temperature = 0CStandard Pressure = 1 atm
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0.5 mole of any gas
11.2 Liters
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2.0 mole of any gas
44.8 Liters
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3.0 mole of any gas
67.2 Liters
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0.25 mole of any gas
5.6 Liters
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Formula Mass
Sum of the masses of the elements in the compound.
Expressed in atomic mass units
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Formula Mass of H2O
2 X H = 2 X 1.0 = 2.01 X O = 1 X 16.0 = 16.0
Sum = 18.0 amu’s
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Formula Mass of NH3
3 X H = 3 X 1.0 = 3.01 X N = 1 X 14.0 = 14.0
Sum = 17.0 amu
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Formula Mass of CO2
1 X C = 1 X 12.0 = 12.02 X O = 2 X 16.0 = 32.0
Sum = 44.0 amu
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Gram Formula Mass
Formula mass expressed in grams. Equals the molar mass of the compound.
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Molar Mass
Mass of one mole of the substance.
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Molar Mass
H2O = NH3 =CO2 =
18.0 grams17.0 grams44.0 grams
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Count up the atoms in (NH4)2SO4
For Paren: SubInside X Suboutside
N: 2 S: 1H: 8 O: 4
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Count up the atoms in 2Mg3(PO4)2
For Paren: SubInside X Suboutside
Coefficients X subs in formula
Mg: 6 P: 4 O: 16
The “2” applies to every element in the formula.
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# of Moles
# of Grams
# ofParticles
# ofLiters (gas)
X 6.02 X 1023
X Formula Mass
X 22.4 L/mole by 6.02 X 1023
byformula mass
by22.4
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Percent
Part X 100% Whole
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Percent H in H2O
Part X 100% = 2 X 100%
Whole 18
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Percent O in H2O
Part X 100% = 16 X 100% Whole 18
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Empirical Formula
smallest whole number ratio of the elements in a
compound
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Molecular Formula
Gives exact composition of molecule
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Covalent Compound
Formula contains all nonmetals
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Molecular (vast majority)&
Network (SiO2, SiC, Cdia, Cgraph)
Types of covalent substances
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Ionic Compound
Formula contains metal plus nonmetal
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Hydrate
Ionic compound that has H2O molecules incorporated into
its structure.
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Anhydrate
Substance that remains after the water is removed from a
hydrate.
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CuSO4•5H2O
Formula of a hydrated salt. • means “is associated with.”H2O molecules are stuffed in
the empty spaces.
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CuSO4·5H2O ?
CuSO4 + 5H2O
Heat to constant mass
Evaporates into airanhydrate
hydrate
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Formula mass of CuSO4•5H2O
Mass of CuSO4 plus mass of 5 water molecules.
249.6 grams/mole
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Percent H2O in CuSO4•5H2O
(from the formula)
Part X 100% = 90 X 100% Whole 249.6
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Metals
All elements to the left of the staircase except H
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Nonmetals
All elements to the right of the staircase plus H
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Binary Compound
Compound made from 2 elements
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Which formulas are empirical?
H2O H2O2 CH4 C2H6
C6H12O6 KCl P4O10 CaF2
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Crystal Lattices
Ionic Compounds, Metals, & Network
Solids make….
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Smallest repetitive unit in a crystal lattice
Formula Unit
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Have distinctly different properties than molecular
substances.
Substances with crystal lattices…
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Types of Substances
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Ionic, Metallic, and Network solids
What kind of substances have Crystal Lattices?
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Empirical formulas only
Substances that make crystal lattices have
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Have both empirical & molecular formulas.
The molecular formula is a whole-number multiple of the
empirical formula.
Molecular Covalent
Substances
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Given empirical formula & Formula Mass, find Molecular
Formula
1) Find empirical mass2) Divide formula mass/empirical
mass3) Multiply subscripts in empirical
formula by answer in step 2
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Empirical formula = CH & Formula Mass = 78,
find Molecular Formula
1) Empirical mass = 132) Divide formula
mass/empirical mass = 78/13 = 6
3) Multiply subscripts: C6H6
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12 grams of hydrated salt is heated to constant mass. After heating the mass is 8.0 grams. What is the percent salt & the percent H2O?
1)Mass of H2O = 12 – 8 = 4 g
2)Percent H2O = 4/12 X 100%
3)Percent salt = 8/12 X 100%
Percent water in hydrate from experimental data.
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He
1 atom of He or 1 mole of He
1 atom per molecule
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O2
1 molecule of O2 or 1 mole of O2
2 atoms per molecule
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O3
1 molecule of O3 or 1 mole of O3
3 atoms per molecule
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Percent composition to empirical formula
1.Convert to mass.2.Convert to moles.3.Divide by small.4.Multiply ‘til whole.
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Find the empirical formula:45.27% C, 9.50% H, 45.23% O
1) 45.27 g C, 9.50 g H, and 45.23 g O2) 3.773 mol C, 9.50 mol H, and 2.827 mol
O3) Divide by 2.827: C1.33H3.36O1
4) Multiply by 3: C4H10O3