chemistry 100 (1cph)

8/8/2019 Chemistry 100 (1CPH)

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Science is a systematic or organized body of

knowledge obtained through observation andexperimentation.

Chemistry is a branch of science that dealswith the study of matter(properties, structure,composition, changes and principles/laws thatgovern changes in matter).



M easurement SI Units commonly used

Length M eter (m) M eter (m)

M ass Kilogram (kg) Gram (g)

Volume Cubic meter(m 3) Liter (l) , ml.

Amount of Substance

M ole(mol) M ole(mol)

Temperature Kelvin (K) Celsius( C)

1ml=1cm 3 (or cc.);1Liter = 1000 ml.=1000 cm 3 (or cc.) = 1dm 3

* 273 K = 0 C = 32 F



P refix Symbol M eaning

Tera T 10 12

Giga G 10 9

M ega M 10 6

Kilo k 10 3

Deci d 10 -1 or 1/10

Centi c 10 -2 or 1/100M illi m 10 -3 or 1/1000M icro u 10 -6 1/1000,000

Nano n 10 -9

P ico p 10 -12



Temperature:

F = 9 C + 32;5

C = 5 ( F - 32)9

K= 273 + C



Density:D(g/ml) = M ass/Volume ; D (g/L ) = M / V

for solids and liquids for gas

Accuracytells how close a measurement is to the truevalue.

P recisiontells how close two or more measurements agree

with one another.



Accuracy ± how close a measurement is to the true valueP recision ± how close a set of measurements are to eachother

Accurate&P recise

P recise butnot accurate

not precise &

not accurate



P roperties of M atter

P roperties ± qualities or characteristics bywhich a given substance is identified.

P hysical P ropertiesqualities that can be observed with thesenses,without changing the composition of

the susbstance. Ex. colors, shapes, size



Chemical P roperties

qualities that can be observed after alteration of composition of the substance

Ex: iron rust, hydrogen is flammable



Extrinsic or Extensive P roperties

are properties that depend on the amount of thesubstance in a sampleEx: volume, length, mass

Intrinsic or Intensive P ropertiesdo not depend on the amount of substance ex.

density, boiling pt. melting pt



Changes of M atter

P hysical Changechange in size, shape, physical state,no change in nature or compositon.

Ex: freezing of water,pulverizing salt

distillation of wine



Chemical Changechange in the compositon of the

substance

a new substance is formed.ex iron nail (Fe) rust (Fe 2O3)

charcoal (C) CO 2

Fermentation of fruit juice



Molecules = aggregate of two or more atomsheld by a chemical bond.

I. Diatomic molecule is composed of twoatoms.

a. with two similar atoms (element) ex:H

2, N

2, Cl

2, F

2b. with two different atoms (compound)

ex. HI, HF, CO, HCl2. P olyatomic molecule is composed of three

or more atoms.a. with three similar atoms ex: O 3

b. with three different atoms ex: HCN,

NH3, C 6H12 O 6



Ion = an atom or a group of atoms that has apositive(+) or negative(-) charge.

a. Cations are positively charged ions ( formedwhen there is a loss of one or more

electrons) ex: Na + , Ca +2 , Al+3

Na atom 11p +11 e- = O; Na +cation 11p + 0 e- = +1

b. Anions are negatively charged ions (formed

when there is a gain of one or moreelectrons) ex: Cl - , O -2

Cl atom 17p + 17 e- = o; Cl- anion 17p + 18e- = -1



M ass number ( A) = the total number of protons +number of neutrons

Atomic number (Z) = the number of protons in thenucleus of an atom .

AX

Z

12

6 C17

9 Cl



Isotopes of Hydrogen

2

1 H1

1 H3

1 H

P rotons = _

Neutrons=_

P rotons = _

Neutrons=_

P rotons = _

Neutrons =_



Isotopes of Hydrogen

2

1 H1

1 H3

1 H

P rotons = 1

Neutrons= 0

P rotons = 1

Neutrons= 1

P rotons = 1

Neutrons = 2



Isotopes of Uranium

235

92 U234

92 U238

92 U

P rotons = 92

Neutrons=

P rotons = 92

Neutrons=

P rotons = 92

Neutrons =



Isotopes of Carbon and % Abundance

13

6C12

6 C

P rotons = 6

Neutrons= 6

P rotons = 6

Neutrons= 7

% abundance =98.89%% Abundance = 1.11%



Computations:

12 x 0.9889 =11.866813 x 0.0111 = 0.1443

12.0111 atomic weight

12. atomic mass (mass no.)M ass no.- the whole no. nearest to the exact

atomic weight



Atomic Structure : Atom - basic unit of an element that can enter into achemical reaction, retains the identity of element

Electron (discovered by Joseph John Thomson) is

negatively charge(-) particle

Neutron (discovered by James Chadwick) iselectrically neutral particle.

P roton and atomic nucleus (by Ernest Rutherford)positively charged(+) particle .



Chemical formula represents the composition of one molecule or ionic compound in terms of chemical symbols.

a.M

olecular formula shows the actual number of atoms of elements in amolecule.ex: C 6H 12 O 6 ( sucrose), C 6H6 (benzene ) ,C 2H2 (acetylene) .

b.Empirical formula gives the simplest whole number ratio of atoms of elements in a molecule ex: C H 2 O ( sucrose), C H (benzene ) , C H(acetylene)

c. Lewis formula shows the valence electrons of atoms of elements in amolecule.

.. .. ..ex: H: O : O: H ( hydrogen peroxide) H: O :H (water)

.. .. ..d. Structural formula shows the binding of atoms of elements in a molecule.ex: H-O ±O -H ( hydrogen peroxide) H- O -H (water)



Electronic Structure of Atoms

Aufbau P rinciple Atomic orbitalsElectronic configurationOctet RuleHund¶s Rule



Reduction Oxidation Reaction

Molecular Equation (Valence Change Method) NaNO 2+ K 2Cr 2O7 + H 2SO 4 = NaNO 3 +K 2SO 4 + Cr 2 (SO 4) 3+ H 2O

Ionic Equation (Valence Change Method) NO 2 + Cr 2O7 + H + = NO 3

-1 + Cr + H 2O

Half Reaction or Ion-electron method NO

2

-1 = NO3

-1

Cr 2O7-2 = Cr +3

+3-2-1



Redox

B alance the eq. using Valence change or Ionelectron method . Answer #1-10MnO 4

-1 + ClO 2-1 +H 2O = MnO 2 + ClO 4

-1 + OH -

___1. Oxidation no. of Mn in MnO 4 -1

___2. Oxidation no. of Cl in ClO 4-1

___3. electron lost/mole ___9. coefficient of ClO 2-1

___4. electron gained/mole ___10. coefficient of MnO 2

___5 Identify the OA ___6. Which is the RA? ___7. Which is the reduction product? ___8. Which is the oxidation product?



Concentration Units

M olarity ± moles of solute per liter of solutionM = moles solute moles = g solute

liter of solution molar mass

g soluteM = molar mass

liter solution



Normality ± no of equivalents of solute per liter of solution

N = equivalents of solute

liter of solution

g soluteN = eq wt

liter of solution



molality - moles of solute per kg solventg solute

m = molar masskg solvent



% by mass = mass solute X 100mass of solution

M ole fraction solute (X 2) = mole solute

mole solution

M ole fraction solvent (X 1) = mole solvent

mole solution



Concentration Units

1.Calculate the % by mass of the solute in:a. 5.50 g NaBr in 78.2 g solutionb. 31.0 g KCl in 152.0 g water

2.A solution is prepared by mixing 62.5 mlC6H6 (MW 78g/mol) with 80.3 ml toluene( MW 92.0g/mol)

density C 6H6 0.867 g/mldensity C 7H8 0.87 g/ml.

chemistry 100 (1cph)

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