Download Les Brown

Shoot for the moon. Even if you miss, you'll land
among the stars (space debris).
- Les Brown -
LET’S REVIEW WHAT FORMULAS TELL US. IF
WE WRITE THE FOLLOWING:
2NaCO3
THE 2 TELLS US THAT WE HAVE 2 FORMULA
UNITS OF SODIUM CARBONATE. EACH FORMULA
UNIT CONTAINS 1 ATOM OF SODIUM, 1 ATOM
OF CARBON, AND 3 ATOMS OF OXYGEN.
IN A LABORATORY, WE WANT TO BE ABLE TO
DETERMINE THE FORMULAS OF COMPOUNDS AND
DETERMNE HOW MANY FORMULA UNITS OR
MOLECULES WE ARE DEALING WITH.
ATOMS ARE TOO TINY TO BE ABLE TO
DETERMINE THEIR MASS EASILY. EVEN
MOLECULES (COMBNATIONS OF SEVERAL
ATOMS ARE TOO SMALL)
1 ATOMIC MASS UNIT (AMU) = 1.6 X 10-24 g
THE BEST LAB BALANCES WILL ONLY
MEASURE MASS TO 0.000001 g (1 microgram).
WE NEED SOME OTHER WAY.
YOU CAN COUNT OBJECTS BY WEIGHING.
AN EXAMPLE THAT IS SOMETIMES USED IS SELLING JELLY
BEANS. IF YOU NEEDED 1000 JELLY BEANS, AND EACH
JELLY BEAN WEIGHED 5 GRAMS, IT WOULD BE MUCH EASIER
TO WEIGHT 5000 GRAMS RATHER THAN COUNTING 1000.
THIS IS BASICALLY WHAT WE DO IN CHEMISTRY. WE WANT
TO WORK WITH EQUAL NUMBERS OF ATOMS OR WHOLE
NUMBER RATIOS OF ATOMS, SO WE USE THE MOLE
CONCEPT.
IN THIS CONCEPT, 12 GRAMS OF CARBON 12 CONTAIN 6.02 X
1023 ATOMS OF CARBON (ONE MOLE).
IF WE TAKE THE ATOMIC WEIGHT OF ANY ELEMENT IN GRAMS
(THE GRAM ATOMIC WEIGHT), IT CONTAINS 6.02 X 1023 ATOMS.
OR, IF WE TAKE THE MOLAR MASS IN GRAMS OF A
COMPOUND, WE HAVE 6.02 X 1023 MOLECULES.
6.02 X 1023 IS AVOGADRO’S NUMBER, NAMED IN HONOR OF
AVOGADRO, AND WE USE THE SYMBOL N.
N = 6.02 X 1023 = ONE MOLE
CONVERSIONS WE CAN DO:
1)THERE ARE 6.02 X 1023 PARTICLES IN ONE MOLE. IF WE
KNOW THE NUMBER OF MOLES, WE CAN CALCULATE THE
NUMBER OF PARTICLES – MOLECULES, ATOMS, OR FORMULA
UNITS – OR VICE VERSA.
2)THE NUMBER OF MOLES OF A SUBSTANCE IN A GIVEN
MASS CAN BE DETERMINED BY DIVIDING THE MASS IN
GRAMS BY THE FORMULA MASS.
EXAMPLES:
1)CONVERT 60.0 g of NaOH TO MOLES.
2)CONVERT 3.00 MOLES OF ZnCl2 TO GRAMS.
3)HOW MANY MOLECULES ARE IN 0.0045 MOLES OF
Al2(CO3)3?
4)HOW MANY MOLES ARE PRESENT IN 1.00 X 1025
MOLECULES OF H2?
5)HOW MANY MOLECULES (ATOMS) ARE PRESENT IN 8.00 g
He?
6)HOW MANY GRAMS OF AlPO4 ARE IN 1.00 X 1023
MOLECULES?
PERCENT COMPOSITION IS IMPORTANT IN CHEMISTRY.
WE HAD AN EXAMPLE IN NUCLEAR CHEMISTRY. IF YOU ARE
MINING FOR URANIUM, AN ORE THAT CONTAINED 1%
URANIUM WOULD BE FAR MORE VALUABLE THAN A ORE
CONTAINING ONLY 0.01%.
THEN, ONCE YOU PURIFY THE URANIUM, IF IT CONTAINED
1% U-235, IT WOULD BE FAR MORE VALUABLE THAN IF IT
CONTAINED 0.5%. THE NATURAL AVERAGE IS 0.7%.
PERCENT COMPOSITION IS ALSO IMPORTANT IN DEALING
WITH COMPOUNDS. COMPOUNDS ARE COMPOSED OF TWO
OR MORE ELEMENTS CHEMICALLY COMBINED.
A GIVEN COMPOUND ALWAYS HAS THE
SAME PERCENT COMPOSITION BECAUSE
THE ELEMENTS ALWAYS COMBINE IN THE
SAME WAY – THE FORMULA DOES NOT
CHANGE.
NOTE: YOU ARE COMBINING GIVEN
NUMBERS OF WHOLE ATOMS.
THIS IS ANOTHER WAY OF STATING THE LAW
OF DEFINITE PROPORTIONS, SOMETIMES
REFERRED TO AS THE LAW OF DEFINITE
COMPOSITION.
THE LAW OF DEFINITE COMPOSITION
STATES THAT CHEMICAL COMPOUNDS ARE
COMPOSED OF A FIXED RATIO OF
ELEMENTS AS DETERMINED BY MASS.
PERCENT COMPOSITION CAN BE DETERMINED BY
EXPERIMENT, AND, IN TURN, CAN BE USED TO DETERMINE
THE MOLE RATIO (EMPIRICAL FORMULA) OF THE
COMPOUND.
FOR EXAMPLE, IN THE LAB, THE MASS OF EACH ELEMENT IN
A COMPOUND COULD BE DETERMINED BY DECOMPOSING A
COMPOUND.
EXAMPLE: LAB PROCEDURES SHOW THAT 50.0 g OF
AMMONIA YIELD 41.0 g NITROGEN AND 9.00 g HYDROGEN.
WHAT IS THE PERCENT COMPOSITION OF AMMONIA?
NOTE: EMPIRICAL FORMULAS CAN BE DETERMINED FROM
EXPERIMENTAL DATA OR FROM PERCENT COMPOSITION.
EXAMPLE: A 2.50 G SAMPLE OF A COMPOUND CONTAINS 0.900 g
CALCIUM AND 1.60 g CHLORINE. DETERMINE THE EMPIRICAL
FORMULA.
STEP 1: DETERMINE THE # OF MOLES
Ca = 0.900 g/40.1 g/mole = 0.0224 mole
Cl = 1.60 g/35.5 g/mole = 0.0451 mole
THE MOLE RATIO IS 0.0224 mole Ca TO 00451 mole Cl.
STEP 2: CALCULATE THE SIMPLEST WHOLE NUMBER RATIO.
Ca = 0.0224/0.0224 = 1
Cl = 0.0451/0.0224 = 2.01 or 2
THE EMPIRICAL FORMULA IS CaCl2
YOU COULD DO THE SAME THING FROM PERCENT
COMPOSITION.
REMEMBER: PERCENT IS PARTS PER HUNDRED.
EXAMPLE: YOU HAVE A COMPOUND WHOSE PERCENT
COMPOSITION IS 40% C, 6.71% H, AND 53.3% O. SO, IF YOU
HAD A 100 g OF THE COMPOUND, YOU WOULD HAVE 40.0 g C,
6.71 g H, and 53.3 g O.
C = 40 g/12.0 g/mole = 3.33 mole
H = 6.71 g/1.01 g/mole = 6.64 mole
O = 53.3 g/16.0 g/mole = 3.33 mole
OR C = 3.33/3.33 = 1
H = 6.64/3.33 = 2
THE EMPIRICAL FORMULA IS CH2O
O = 3.33/3.33 = 1
IF YOU KNOW THE MOLECULAR MASS, YOU CAN CONVERT
THE EMPIRICAL FORMULA TO THE MOLECULAR FORMULA.
IN THE PREVIOUS EXAMPLE, IF YOU KNEW THE
MOLECULAR MASS WAS 60:
THE EMPIRICAL FORMULA MASS IS:
C = 12.0
2H = 1.01
O = 16.0
total = 30.0
SO,
60/30 = 2
THE MOLECULAR FORMULA IS 2 TIMES
THE EMPIRICAL FORMULA OR:
C2H4O2 = MOLECULAR FORMULA
THE MOLECULAR FORMULA IS ALWAYS A WHOLE NUMBER
MULTIPLE OF THE EMPIRICAL FORMULA.