CHAPTER 2

A P CH E M

2.2 REVIEWAtomic number is the number of protonsAn atomic symbol represents the elementThe mass number is equal to the number

of protons +neutrons

19 -1 9

ISOTOPESDefinition: Two or more forms of the same

element containing different numbers of neutrons. Isotopes of an element have the same chemical properties but differing atomic mass

Examples: Carbon -12 Carbon -13Chlorine -35Chlorine - 37

PERCENT ABUNDANCEMost elements exist as atoms of different masses -

isotopes.

Examples:Carbon -12 98.90 % natural abundanceCarbon -13 1.10 % abundanceCarbon -14 less than 1/trillionth

Chlorine-35 75.77 %Chlorine 37 24.23 %

1.Low pressure gas sample ionized2.Gas bombarded by e-’s, ejecting e-’s from sample creating cations (positively charged ions). 3.Ions sorted & separated by mass and charge using magnetic field (using Inertia – heavier particles deflected less than lighter and using charges – highly charged particles interact more strongly with magnets and electric fields and are deflected more that particles of equal mass with smaller charges).4. Results measured in

detector

MASS SPECTROMETER

ATOMIC WEIGHTMass or weight???

one amu (atomic mass unit) = exactly 1/12 of the mass of a carbon-12 atom

Atomic weight is the weighted average of the masses of an elements’ isotopes

EX2.1 Boron exists in two naturally, occurring isotopes. B-10

( 10.016 amu) makes up 18.83% of each natural sample of this element. The remaining 81.17% of the sample is B-11 (11.013 amu). What atomic mass would be calculated for this mixture of isotopes?

Solve by multiplying isotope mass by %Boron -10 (10.016)x(.1883) = 1.8860Boron -11 (11.013)x(0.8117) = +8.9393 10.825 amu

Keep 5 sig figs – use percent as exact numbers

EX2.2The two natural isotopes of Lithium are 6Li (6.01512 amu)

which accounts for 7.42% of the total and 7Li which accounts for the remaining amount. If the mass of lithium is shown as 6.942 on the periodic table, what is the mass of the Li -7 isotope.

Lithium -6 (6.01512)(0.0742) = 0.4463219Lithium – 7 (Lithium 7)(.9258) = 0.9258 x

6.942 amu 0.446322 + 0.9258x = 6.942

0.9258 x = 6.4959Lithium-7 = x x = 7.016 amu

MOST ABUNDANT NATURAL ISOTOPEHydrogen is the most abundant in the

universe

2.5 ATOMS AND THE MOLEDefinition: the amount of a substance

that contains as many entities ( atoms, molecules, or other particles) as there are atoms in 0.012 kg of carbon-12 atoms.

Current accepted number1 mole = 6.022045 x 1023 particles

ANALOGYBaker’s count in dozens 1 dozen = 12

donutsChemists count atoms in Moles

http://www.youtube.com/watch?v=1R7NiIum2TI

Avogadro’s #6.022 x 1023 = 1 mole2nd DefinitionThe mass of one mole of atoms of a pure element in grams is numerically equal to the atomic weight of that element in amu

EX 2.3One mole of sulfur contains 6.022 x 10 23 atoms of

sulfur and has a mass of 32.06 grams. What is the mass of one atom of sulfur?

Sample = 1 atom sulfur

1 atom S 1 mole 32.06 grams S = 5.324 x10-23g Sulfur 6.022 x 1023 atoms 1 mole S

EX 2.4How many sulfur atoms are present in 1.00

grams of sulfur?Given: 1 mole S = 6.022 x 10 23 atoms S

Sulfur atomic weight = 32.06 grams/mole ( P.T.)

1.00 grams S 1 mole S 6.022 x 10 23 atoms = 1.88 x 10 22 atoms Sulfur 32.06 g S 1 mole

EQUIVALENCIES AND CONVERSION FACTORSFormula Formula Mass

(amu)Molar Mass (g/mol)

H 1.0079 = 1.008 1.0079 = 1.008H2 2.016 2.016H2O 18.016 18.016H2SO4 98.076 98.076

SAMPLE MOLE PROBLEMSEx2.5 How many moles are equivalent to 5.00 g CaCO3?(Add molar masses of calcium, oxygen and carbon present

to get the molar mass)5.00 grams CaCO3 1 mole CaCO3 = 0.0500 moles CaCO3

100.089 g CaCO3

Ex2.6 A microchemical experiment requires 0.0100 moles of Al(NO3)3. How many grams is this?

0.0100 moles Al(NO3)3 213.0017 g Al(NO3)3 = 2.13 g Al(NO3)3

1 mole Al(NO3)3

MORE MOLE PROBLEMSEx. 2.7 For exactly 1.000 g of carbon disulfide (CS2), how

many molecules are present? How many atoms of sulfur?

1.000 g CS2 1 mole CS2 6.022 x 1023 molecules CS2 = 76.131 g CS2 1 mole CS2

7.910 x1021 CS2

To convert to atoms (continue 3 atoms / I molecule CS2) 1.000 g CS2 1 mole CS2 6.022 x 1023 molecules CS2 2 atoms

sulfur =1.582 x 1022 76.131 g CS2 1 mole CS2 1 molecule CS2 atoms S

Ex2.8 A sample of sulfur hexafluoride (SF6) contains 1.69 x 1022 atoms of fluorine. What is the mass of the sample?

MORE MOLE PROBLEMSEx2.8 A sample of sulfur hexafluoride (SF6) contains 1.69 x

1022 atoms of fluorine. What is the mass of the sample?

1.69 x 1022atoms F 1 molecules SF6 1 mole SF6 146.054 g =

6 atoms F 6.022 x 1023 molecules SF6 1 mol

0.683 g SF6

THE PERIODIC TABLE1. History

Mendeleev, 1869 - developed the first periodic table by identifying similarities among the elementsbased on increasing atomic weights (did not know atom was divisible)

Periodic Law – The properties of elements are periodic functions of their atomic numbers

Moseley , 1913 - organized the P.T. in order of atomic number

THE PERIODIC TABLEGroups or families – columns down the P.T.Periods or series – rows across P.T.Metals - on the left side of the stair case ,

metalliods along the staricase, nonmetals on the right side

Main group elements on the left and right side of P.T.

Transistion metals – in the middleLanthanide and actinide – on th every bottomSpecific namesGroup 1 – alkali metalsGroup 2 – alkaline earth metalsGroup 17 (7A) – halogensGroup 18 (8A) – noble gases

PERIODIC TABLEhttp://www.youtube.com/watch?v=SmwlzwGMMwc

Coulomb’s Law – a helpful tool in explaining P.T. trends• An Inverse Square LawThe magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them.

If the two charges have the same sign, the electrostatic force between them is repulsive; if they have different sign, the force between them is attractive.

• Like charges repel • unlike charges attract

The Equation

1 2

1 22

( )

( ) e

e

where F is force Newtons , q and q represent charge (Coulomb's),

d is distance meters between electrons, and k is Coulomb's constant

q qF kd

Applying Coulomb’s Law to P.T.• The nucleus of the atom (with a

positive charge) is attracted to each electron (with a negative charge) and visa versa.

• The attraction increases exponentially as the distance between the e- and the nucleus decreases.

• Note: electrons repel each other

HOMEWORK CHAPTER 2HW#1 -21, 23, 25, 27, 29, 47 Composition Atoms,

IsotopesHW#2 - 31, 33, 35 Atomic Symbols WkstHW #3 – 37, 39, 59, 61, 62, 63 HW #4 - Mole Relationships Wkst.HW #5 - 43, 45, 57 PT