Transcript stioch1-3

Copyright Sautter 2003
STIOCHIOMETRY
“Measuring elements”
Determing the Results of
A Chemical Reaction
PREDICTING HOW MUCH OF A SUBSTANCE CAN BE
MADE BY A CHEMICAL REACTION BEFORE IT IS
CARRIED OUT!!
• STEP I
• ALWAYS WRITE THE
EQUATION USING
CHEMICAL FORMULAE
AND BALANCE IT.
• (REMEMBER TO BE SURE
YOU USE THE CORRECT
SUBSCRIPTS FOR THE
FORMULAE AND CHANGE
ONLY THE COEFFICIENTS
WHEN BALANCING THE
EQUATION)
• EXAMPLE:
•
•
•
•
•
•
•
•
•
•
CALCIUM CARBONATE
(LIMESTONE) WHEN HEATED
GIVES CALCIUM OXIDE (LIME)
AND CARBON DIOXIDE
CALCIUM = Ca (+2)
CARBONATE = CO3 (-2)
CALCIUM CARBONATE = CaCO3
OXIDE = O (-2)
CALCIUM OXIDE = CaO
CARBON DIOXIDE = CO2
CaCO3(S)  CaO(s) + CO2(g)
THERE IS ONE Ca ON EACH SIDE OF
THE EQUATION, ONE C ON EACH
SIDE AND THREE O ON EACH SIDE.
THE EQUATION IS BALANCED!
IN ORDER TO PREDICT THE RESULTS OF A CHEMICAL
REACTION WE MUST BE GIVEN THE QUANTITIES OF
MATERIALS THAT ARE TO BE USED IN THE REACTION
SUPPOSE WE ARE GIVEN 200 GRAMS OF CALCIUM CARBONATE
AND ASKED TO DECIDE HOW MUCH CALCIUM OXIDE AND
CARBON DIOXIDE CAN BE MADE??
WE ALREADY HAVE THE BALANCED EQUATION BUT WHAT DO WE
DO NEXT??
WELL, SINCE BALANCED EQUATIONS SHOW THE NUMBER OF
ATOMS AND MOLECULES INVOLVED, WE MUST WORK WITH
NUMBERS OF ATOMS AND MOLECULES.
REMEMBER WE COUNT ATOMS AND MOLECULES WITH MOLES.
(6.02 X 1023 = 1 MOLE)
**REMEMBER **
TO CONVERT GRAMS (MASS) TO MOLES WE DIVIDE
THE WEIGHT OF ONE MOLE FROM THE PERIODIC
TABLE INTO THE GIVEN NUMBER OF GRAMS
• FOR EXAMPLE:
• SINCE CaCO3 CONTAINS
• 1Ca FROM THE PERIODIC
TABLE WE USE 40grams x 1
• 1 C FROM THE PERIODIC
TABLE WE USE 12 grams x1
• 3 O FROM THE PERIODIC
•
•
TABLE WE USE 16 grams x 3
(1x 40) + (1x 12) + (3 x 16) = 100
THE MASS OF ONE MOLE OF
CALCIUM CARBONATE IS
100 grams
• IN OUR PROBLEM WE
ARE USING 200 grams of
CaCO3
• DIVIDING 200grams by
100grams per mole of CaCO3
WE FIND THAT WE HAVE
EXACTLY 2 MOLES OF
THE CALCIUM
CARBONATE REACTANT.
.
STEP II IN SOLVING STIOCHIOMETRY PROBLEMS
THEN IS TO CONVERT THE GIVEN NUMBER OF
GRAMS TO MOLES.
CaCO3(S)  CaO(s) + CO2(g)
• THE EQUATION SAYS THAT ONE CALCIUM CARBONATE
MAKES ONE CALCIUM OXIDE. HOW MANY CALCIUM
OXIDES WOULD TEN CALCIUM CARBONATES MAKES?
• HOW ABOUT TEN?
• WHAT ABOUT ONE HUNDRED CALCIUM CARBONATES?
• WOULDN’T THEY MAKE ONE HUNDRED CALCIUM
OXIDES?
• OF COURSE !!
• THEN WHAT ABOUT A MOLE OF CALCIUM CARBONATE?
WOULDN’T THEY BE EXPECTED TO MAKE A MOLE OF
CALCIUM OXIDE?
• AND OF COURSE THEN TWO MOLES WOULD MAKE TWO
MOLES!
• STEP III IS PROBABLY THE MOST DIFFICULT ONE!
• IT REQUIRES US TO PUT TOGETHER THE
BALANCED EQUATION FROM STEP I AND THE
MOLES THAT WE CALCULATED IN STEP II !!
• ** REMEMBER **
• HERE’S OUR BALANCED EQUATION
• CaCO3(S)  CaO(s) + CO2(g)
• AND HERE’S THE MOLES WE CALCULATED
• DIVIDING 200grams by 100grams per mole of CaCO3
WE FIND THAT WE HAVE EXACTLY 2 MOLES OF
THE CALCIUM CARBONATE REACTANT
CaCO3(S)  CaO(s) + CO2(g)
WHAT ABOUT THE CARBON DIOXIDE?
THE BALANCED EQUATION ALSO SAYS THAT ONE
CALCIUM CARBONATE MAKES ONE CARBON
DIOXIDE TOO.
• SO USING THE SAME LOGICAL THAT WE APPLIED
TO THE CALCIUM OXIDE, IT IS OBVIOUS THAT
TWO MOLES OF CALCIUM CARBONATE WILL
PRODUCE EXACTLY TWO MOLES OF CARBON
DIOXIDE ALSO.
• STEP III – USING THE BALANCED EQUATION
RATIOS FOUND IN STEP I AND THE MOLES
DETERMINED IN STEP II, FIND THE MOLES OF
EACH PRODUCT MATERIAL FORMED.
STEP IV – CONVERT THE MOLES FOUND IN STEP III
TO GRAMS (MASS)
• ** REMEMBER** TO CONVERT MOLES TO GRAMS, MULTIPLY THE
MASS OF ONE MOLE FROM THE PERIODIC TABLE BY THE NUMBER
OF MOLES.
• EXAMPLE:
• CaO CONTAINS 1 Ca (1 X 40grams from the Periodic Table) and 1 O (1 x 16
from the Periodic Table)
• THE MASS OF ONE MOLE OF CaO IS (1 x 40) + (1 x 16) = 56 grams per
mole
• TWO MOLES OF CaO ARE FORMED THEREFORE, 2 MOLES x 56 grams
per mole = 112 gram of CaO ARE FORMED IN THE REACTION
• CO2 CONTAINS 1 C (1 x 12grams from the Periodic Table) and 2 O (2 x 16
grams from the Periodic Table)
• THE MASS OF ONE MOLE OF CO2 IS (1 x 12) + (2 x 16) = 44 grams per
moles
• TWO MOLES OF CO2 ARE FORMED THEREFORE, 2 MOLES x 44 grams
per mole = 88 grams of CO2 ARE FORMED IN THE REACTION
CaCO3(S)  CaO(s) +
200 grams
2 moles
0 grams
0 moles
CO2(g)
0 grams
0 moles
before reaction occurs
0 grams
0 moles
112 grams
2moles
after reaction occurs
88 grams
2moles
STARTING TOTAL MASS = 200 +0 +0 = 200 GRAMS
FINAL TOTAL MASS = 0 + 112 + 88 = 200 GRAMS
(CONSERVATION OF MASS)
• QUESTIONS ?????
NOW FOR A HARDER ONE!
• HYDROGEN GAS REACTS WITH OXYGEN GAS TO
FORM WATER. HOW MUCH HYDROGEN AND OXYGEN
MUST BE COMBINED TO MAKE 45 GRAMS OF WATER?
•
•
•
•
•
•
STEP I – WRITE AND BALANCE THE EQUATION
HYDROGEN = H2 (DIATOMIC ELEMENT)
OXYGEN = O2 (DIATOMIC ELEMENT)
WATER = H2O
2 H2(g)
+ 1 O2(g)  2 H2O(g)
THERE ARE 4 ATOMS OF HYDROGEN ON EACH SIDE
AND TWO ATOMS OF OXYGEN ON EACH SIDE.
• THE EQUATION IS BALANCED !!
STEP II – CONVERT THE GIVEN GRAMS TO MOLES
• WATER HAS A MOLAR MASS (MASS OF ONE
MOLE) FROM THE PERIODIC TABLE IS 18 GRAMS.
• { (2 x 1) FOR HYDROGEN AND (1 x 16) FOR
OXYGEN }
• 45 grams of water DIVIDED BY 18 grams per mole = 2.5
moles
• We want to make 2.5 moles of water in our reaction.
STEP III – USE THE BALANCED EQUATION TO FIND
THE MOLES OF REACTANT OR PRODUCT REQUIRED
• Here’s our balanced equation from STEP I
• 2 H2(g)
+
1 O2(g)  2 H2O(g)
• TWO MOLES OF WATER ARE MADE FROM TWO
MOLES OF HYDROGEN. HOW MANY MOLES OF
HYDROGEN WOULD BE NEEDED TO MAKE 2.5
MOLES OF WATER? (A ONE FOR ONE RATIO)
• HOW ABOUT 2.5 ??
• NOW FOR THE OXYGEN !
• IT TAKES ONLY 1 MOLE OF OXYGEN TO MAKE 2
MOLES OF WATER, HALF AS MUCH ! (A TWO FOR
ONE RATIO)
• THEREFORE TO MAKE 2.5 MOLES OF WATER
NEEDS ONLY ½ OF 2.5 MOLES OF OXYGEN!
THEREFORE 1.25 MOLES OF OXYGEN IS
REQUIRED.
STEP IV – CONVERT MOLES OF UNKNOWN
TO GRAMS
• SINCE WE NOW KNOW THAT 2.5 MOLES OF
HYDROGEN IS REQUIRED, WE CAN MULTIPLY 2.5
TIMES 2.0 GRAMS, THE MOLAR MASS OF
HYDROGEN TO GET 5.0 GRAMS OF H2(g) NEEDED!
• THE MOLAR MASS OF OXYGEN IS 32 GRAMS.
MULTIPLYING 32 TIMES 1.25 MOLES GIVES 40
GRAMS OF OXYGEN ARE NEEDED.
THEN, IN ORDER TO MAKE 45.0 GRAMS OF
WATER, USING HYDROGEN GAS AND
OXYGEN GAS WE HAVE CALCULATED THAT:
• THE MOLES OF HYDROGEN NEEDED ARE 2.5
MOLES OR 5.0 GRAMS AND
• THE MOLES OF OXYGEN NEEDED ARE 1.25 MOLES
OR 40.0 GRAMS AND
• ALL OF THIS HAS BEEN DETERMINED WITHOUT
EVER PICKING UP A TEST TUBE BY USING
STOICHIOMETRY!!!
Limiting Reactant Calculations
• In the reactions previously discussed, an amount of only
one of the reactants was given. We assumed that we
could use as much of the other reactant as we needed.
Unfortunately, this is not always the case.
• Situations where specific amounts of both of the
reactants are given are called “limiting reactant”
problems.
• The limiting reactant is the one that runs out first!
• In limiting reactant problems three possibilities exist.
Possibility one is that the first reactant is used up first.
Possibility two is that the second reactant runs out first
and possibility three is that both reactants run out at the
same time.
Containers of nuts and bolts are to be threaded together.
One nut threaded on one bolt. How many combinations
Can be made ?
bolts
nuts
Only four nut – bolt
combinations can
be made. The bolts
have run out.
The bolts are the
limiting factor
Containers of nuts and bolts are to be threaded together.
Two nuts threaded on one bolt. How many combinations
Can be made ?
bolts
nuts
The one with the smallest number is
not always the limiting factor.
It depends on the ratio of combinations!
Only three nut – bolt
combinations can
be made. The nuts
have run out.
The nuts are the
limiting factor
Limiting Reactant
• Now, let’s try a limiting factor (reactant) problem using
a chemical reaction! Remember, numbers of atoms and
molecules are measured in moles.
• The balanced equation tells us the ratio of
combination of the atoms and molecules that are used
to make the products.
• In the reaction: H2 + I2  2 HI, one molecule of
hydrogen is combined with one molecule of iodine to
give two molecules of hydrogen iodide.
• It is also true to say that one mole of hydrogen is
combined with one mole of iodine to give two moles of
hydrogen iodide. All we have really done is multiply the
entire equation through by 6.02 x 1023 (1 mole).
Limiting Reactant
• H2 + I2  2 HI
• Suppose that exactly one mole of H2 and exactly one mole
of I2 are available. In this case we can make exactly two
moles of HI and no H2 or I2 will be left.
• Now, suppose that we have one mole of H2 and two moles
of I2. Once the H2 is used up, no more I2 can be reacted.
One mole of H2will use exactly one mole of I2 leaving an
extra mole of I2 unused. The limiting reactant is H2. It ran
out first. The excess reactant is I2. We have extra I2.
• Only two moles of HI can be made. Accordingly to our
balanced equation, for each H2 used, two HI are formed.
Only one mole of H2 was used so only two moles of HI are
produced.
The quantity of products formed is based on the limiting
reactant!
Limiting Reactant
• Problem: Given the reaction: 1Ca + 1Cl2  1CaCl2. If we mix
120 grams of calcium and 71 grams of chlorine, which reactant is the
limiting factor? How many grams of CaCl2 can be made?
• Solution: Remember that balanced equations are based on moles. We
must first convert the given grams to moles.
• Moles for Ca = 120 / 40 = 3.0,
• Moles for Cl2 = 71 / (2 x 35.5) = 1.0
• From the balanced equation, 1 Ca requires exactly 1 Cl2. Since only
1.0 mole of Cl2 is available only 1.0 mole of Ca can be consumed and
2.0 moles of Ca remain unused.
• Chlorine (Cl2) is the limiting reactant.
• The amount of product that can be formed is based on the limiting
reactant. The equation tells that for each Cl2 used, one CaCl2 is made.
Since 1.0 moles Cl2 are used, 1.0 moles of CaCl2 are produced.
• Grams CaCl2 = 1.0 moles x (40 +2(35.5)) grams per mole = 111 g
Limiting Reactant
• Problem: Given the equation: 2Na + Cl2  2NaCl. How
many grams of NaCl can be made by reacting 69 grams of
Na with 5.0 moles of Cl2 ?
• Solution: Again we must work in moles. Cl2 is already
moles but Na must be converted (69 / 23 = 3.0 moles of Na)
• From the equation, 2Na needs 1Cl2 (half the moles of Na)
so 3 Na needs 1.5 moles of Cl2.
• We have 5.0 moles of Cl2, more than enough. Therefore all
of the Na is used and Na is the limiting reactant!
• The amount of product is based on the limiting reactant.
Since 2Na make 2NaCl, 3.0 Na will make 3.0 NaCl
• Grams of NaCl = 3.0 moles x (23 + 35.5) gram per mole of
NaCl = 175.5 grams of NaCl are formed.
STOICHIOMETRY