I.M.S. SCIENCE
MR. SCHAEFERS NAME(s):
______________ _____________________
PERIOD: ____ DATE: _______ SCORE: ____/20
KEEP EYE DROPPERS 1 CM. ABOVE !
MOLE- (6.02 X 10^ 23)
Molar
solution (Moles/Liter) is a ratio number of moles of a substance ( called
solute )
placed in enough water(called solvent)so that the total volume is one
liter.
The molecular weight (also called the formula
weight and molar mass)
of Na OH is 40 grams ( Na: 23 + O: 16 + H : 1 = 40
grams )
So to make a .5 L of a .3 M ( .3 M * .5 L = .15
) solution of:
Example: ). NaOH
= 40 grams * .15 =
6 grams
_______________________________________________________________________________________
1) K I
= K
+ I
= __ grams * .15 = ___ grams
2) Na2 ( C O 3 )
= Na2
+ C 1 + O 3
= __ grams * .15 = ___ grams
3) Pb( N O 3 )2
= Pb + N 2
+ O 6 = __ grams *
.15 = ___ grams
4) Cu( N O 3 )2
= Cu + N 2
+ O 6 = __ grams *
.15 = ___ grams
_______________________________________________________________________________________
cations & anions:
K+ I-, Na
OH , Na2
( C O 3 ) , Pb
( N O 3 )2 , Cu
( N O 3 )2
A X +
C Y
======> A Y
+ C
X
A+ B- +
C+ D- ======>
A+ D- +
C+ B-
_______________________________________________________________________________________
A TABLESPOON OF WATER - COUNT 1 MOLECULE PER SECOND= 8
X 10 ^ 6 YEARS
- AMOUNT BASED ON C-12 ,12 GRAMS =
AVOGADRO’S NUMBER
= 6.02 X 10 ^23 *See
below
- MASS OF A SUBSTANCE (6.02 X 10^ 23) IS CALLED A MOLE
- NUMERICAL PART = ATOMIC WEIGHT
SO H = 1 GRAM : C = 12 GRAMS
12: 1 RATIO OF ATOMIC WEIGHTS
- 5 MOLES OF OXYGEN = 5 X 16 GRAMS = 80 GRAMS OF OXYGEN
- 5 MOLES OF SULFUR = 5 X 32 GRAMS = 160 GRAMS OF
SULFUR
SO 80 G/ 160 G = 1/2
- IF THE QUANTITIES OF THE TWO ELEMENTS ARE IN THE SAME
RATIO AS THEIR ATOMIC WEIGHTS
THEY CONTAIN THE SAME NUMBER OF ATOMS.
- MASS OF MOLECULAR SUBSTANCE IN GRAMS IS EQUAL TO ITS
MOLECULAR WEIGHT, ITS GRAM
MOLECULAR WEIGHT /
COVALENT BONDS
- GM
MOLECULAR WEIGHT OF A DIATOMIC MOLECULAR ELEMENT IS ONE MOLE
O2
H2 CL2
- GM
MOLECULAR WEIGHT OF A MOLECULAR SUBSTANCE IS ONE MOLE OF MOLECULES OF THE
SUBSTANCE.
EXAMPLE: H2O = 18 GRAMS
CH4= 16 GRAMS
- GM MOLECULAR WEIGHT OF NACL = 58.5 GRAMS, THIS
IS A MOLE OF THIS COMPOUND.
A PRECIPITATE IS
:_______________________________________________________________
I.M.S. SCIENCE
MR. SCHAEFERS NAME(s):
______________ _____________________
MOLE- (6.02 X 10^ 23)
PERIOD:
____ DATE: _______ SCORE: ____/20
Definition:
1 mole of a substance is an amount equal to its molecular weight in grams.
Avogadro’s number is the number of molecules in a mole of any substance, and
equals
(6.02 X 10^ 23).
At standard temperature and pressure (STP),
1 mole of any gas occupies
22.414 liters of
volume.
Amadeo Avogadro (1776-1856) was an Italian physicist
who sought to reconcile findings by Dalton and Gay-Lussac and proposed that, at
equal temperatures and pressures, equal volumes of gases have the same number of
molecules.
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Zinc
Iron
Copper
Aluminum
Mass= 65.4 g
Mass= 55.8 g Mass=
63.5 g Mass=
27g
A=________________ B=_______________ C=
_______________ D=____________
1.
Determine the Identity of the element by its mass.(Do a teacher Check)
Mass=__________g
Mass=__________g
Mass=_________g Mass=__________g
2.
Calculate the mass of an Al atom.
Calculate the mass of an Zn atom.
Solution:
27.0
g = 27.0
g = ______ X 10^-23g
____ g = .0 g =
_____ X 10^-23g
1 mole 6.02 X 10^23
1 mole
6.02 X 10^2
Avogadro’s number
Avogadro’s number
3.
Calculate the volume of an Aluminum atom. Assume Al atom is cube shaped.
Volume Al
atom =(1.27 cm X1.27 cm X ____ cm)/(6.02
X 10^ -23)= _______cm^3
Volume Cu
atom =(1.27 cm X1.27 cm X ____ cm)/(6.02
X 10^ -23)= _______cm^3
Volume
Zinc atom =(1.27 cm X1.27 cm X ____ cm)/(6.02
X 10^ -23)=______cm^3
Volume Iron atom =(1.27 cm
X1.27 cm X ____ cm)/(6.02
X 10^ -23)=______cm^3
4.
Calculate the “diameter” of an Al atom
= (volume of cube just able to contain Al atom)^ 1/3
(1.6 X 10^-23)^1/3 =
_________ X 10^-8 cm.
So: the “radius” of Al atom =
(“diameter” of Al atom/
2)=_______X 10^-8cm.
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