Transcript COMPOUNDS AND MOLECULES

STOICHIOMETRY
Chapter 3
- the study of the
quantitative
aspects of
chemical
reactions.
Counting Atoms
Chemistry is a quantitative
science—we need a
“counting unit.”
MOLE
1 mole is the amount of
substance that contains as
many particles (atoms,
molecules) as C atoms in
12.0 g of 12C.
Particles in a Mole
Avogadro’s Number
Amedeo Avogadro
1776-1856
6.02214199 x
23
10
There is Avogadro’s number of
particles in a mole of any substance.
Molar Mass
1 mol of 12C
= 12.00 g of C
= 6.022 x 1023 atoms
of C
12.00 g of 12C is its
MOLAR MASS
Taking into account all
of the isotopes of C,
the molar mass of C is
12.011 g/mol
One-mole Amounts
PROBLEM: What amount of Mg
is represented by 0.200 g? How
many atoms?
Mg has a molar mass of 24.3050 g/mol.
1 mol
0.200 g •
= 8.23 x 10-3 mol
24.31 g
How many atoms in this piece of Mg?
23
6.022
x
10
atoms
8.23 x 10-3 mol •
1 mol
= 4.95 x 1021 atoms Mg
MOLECULAR WEIGHT
AND MOLAR MASS
Molecular weight = sum of the
atomic weights of all atoms in the
molecule.
Molar mass = molecular weight in
grams per mol.
What is the molar
mass of ethanol,
C2H6O?
1 mol contains
2 moles of C (12.01 g C/1 mol) = 24.02 g C
6 moles of H (1.01 g H/1 mol) = 6.06 g H
1 mol of O (16.00 g O/1 mol) = 16.00 g O
TOTAL = molar mass = 46.08 g/mol
How many moles of alcohol (C2H6O)
are there in a “standard” can of beer
if there are 21.3 g of C2H6O?
(a) Molar mass of C2H6O = 46.08 g/mol
(b) Calc. moles of alcohol
1 mol
21.3 g •
= 0.462 mol
46.08 g
How many molecules of alcohol are
there in a “standard” can of beer if
there are 21.3 g of C2H6O?
We know there are 0.462 mol of C2H6O.
6.022 x 1023 molecules
0.462 mol •
1 mol
= 2.78 x 1023 molecules
How many atoms of C are there in
a “standard” can of beer if there are
21.3 g of C2H6O?
There are 2.78 x 1023 molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
2 C atoms
23
2.78 x 10 molecules •
1 molecule
= 5.57 x 1023 C atoms
Empirical & Molecular
Formulas
A pure compound always consists of the
same elements combined in the same
proportions by weight.
Therefore, we can express the molecular
composition as PERCENT BY
WEIGHT
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
Percent Composition
Consider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt. % N =
14.0 g N
• 100% = 30.4 %
46.0 g NO2
Wt. % O  2 (16 .0 g O per mole ) x 100 %  69 .6%
46 .0 g
What are the weight percentages of N and O in NO?
Determining Formulas
In
chemical analysis we determine
the % by weight of each element in a given
amount of pure compound and derive the
EMPIRICAL or SIMPLEST formula.
PROBLEM:
A compound of B and H
is 81.10% B. What is its empirical
formula?
A compound of B and H is 81.10% B. What is
its empirical formula?
• Because it contains only B and H, it
must contain 18.90% H.
• In 100.0 g of the compound there are
81.10 g of B and 18.90 g of H.
• Calculate the number of moles of each
constitutent.
A compound of B and H is 81.10% B. What is
its empirical formula?
Calculate the number of moles of each
element in 100.0 g of sample.
1 mol
81.10 g B •
= 7.502 mol B
10.81 g
1 mol
18.90 g H •
= 18.75 mol H
1.008 g
A compound of B and H is 81.10% B. What is
its empirical formula?
Now, recognize that atoms combine in the
ratio of small whole numbers
Find the ratio of moles of elements in
the compound.
A compound of B and H is 81.10% B. What is
its empirical formula?
Take the ratio of moles of B and H. Always
divide by the smaller number.
18.75 mol H
2.499 mol H
2.5 mol H
=
=
7.502 mol B
1.000 mol B
1.0 mol B
But we need a whole number ratio.
2.5 mol H/1.0 mol B = 5 mol H to 2 mol B
EMPIRICAL FORMULA = B2H5
A compound of B and H is 81.10% B.
Its empirical formula is B2H5. What is
its molecular formula ?
Is the molecular formula B2H5, B4H10,
B6H15, B8H20, etc.?
B 2H 6
B2H6 is one example of this class of compounds.
How to Determine the molar mass?
Mass spectrometer
A compound of B and H is 81.10% B. Its empirical
formula is B2H5. What is its molecular formula?
We need to do an EXPERIMENT to find
the MOLAR MASS.
Here experiment gives 53.3 g/mol
Compare with the mass of B2H5
= 26.66
g/unit
Find the ratio of these masses.
53.3 g/mol
2 units of B2H5
=
26.66 g/unit of B2H5
1 mol
Molecular formula = B4H10
Law of conservation of mass [Lavoisier, 1743-1794]
Mass is neither created nor destroyed
in a chemical reaction
Law of definite proportion [Joseph Proust, 1754-1826]
A given compound always contains the same
proportion of elements by mass
Law of multiple proportions [John Dalton, 1766-1844]
When 2 elements form multiple compounds, the mass
of the second element per gram of the first one can
always be reduced to small whole numbers
ELEMENTS are composed of identical particles, atoms
CHEMICAL COMPOUNDS are formed when atoms of different
elements combine with each other: A given compound always has
the same relative numbers and types of atoms
Chemical Equations
• Because the same atoms are
present in a reaction at the
beginning and at the end, the
amount of matter in a system
does not change.
• The Law of the
Conservation of Matter
Demo of conservation of matter, See
Screen 4.3.
Chemical Equations
Because of the principle of the
conservation of matter,
an equation
must be
balanced.
It must have the same
number of atoms of the
same kind on both
sides.
Lavoisier, 1788
Balancing
Equations
___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s)
Chemical Equations
Depict the kind of reactants and products and their
relative amounts in a reaction.
2 Al(s) + 3 Br2(g) ---> Al2Br6(s)
The numbers in the front are called
stoichiometric coefficients
The letters (s), (g), and (l) are the physical states of
compounds.
Chemical Equations
4 Al(s) + 3 O2(g)
---> 2 Al2O3(s)
This equation means
4 Al atoms + 3 O2 molecules
---give--->
2 molecules of Al2O3
4 moles of Al + 3 moles of O2
---give--->
2 moles of Al2O3
STOICHIOMETRY
It rests on the principle of the conservation of matter.
2 Al(s) + 3 Br2(liq) ------> Al2Br6(s)
PROBLEM:
If 454 g of NH4NO3 decomposes, how much N2O and
H2O are formed? What is the theoretical yield of
products?
STEP 1
Write the balanced chemical
equation
NH4NO3 ---> N2O + 2 H2O
454 g of NH4NO3 --> N2O + 2 H2O
STEP 2 Convert mass reactant
(454 g) --> moles
1 mol
454 g •
= 5.68 mol NH4NO3
80.04 g
STEP 3 Convert moles reactant
(5.68 mol) --> moles product
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3 Convert moles reactant --> moles
product
Relate moles NH4NO3 to moles product
expected.
1 mol NH4NO3 --> 2 mol H2O
Express this relation as the
STOICHIOMETRIC FACTOR.
2 mol H2 O produced
1 mol NH4NO3 used
454 g of NH4NO3 --> N2O + 2 H2O
STEP 3 Convert moles reactant (5.68
mol) --> moles product
2 mol H2O produced
5.68 mol NH4NO3 •
1 mol NH4NO3 used
= 11.4 mol H2O produced
454 g of NH4NO3 --> N2O + 2 H2O
STEP 4 Convert moles product
(11.4 mol) --> mass product
Called the THEORETICAL YIELD
18.02 g
11.4 mol H2O •
= 204 g H2O
1 mol
ALWAYS FOLLOW THESE STEPS IN
SOLVING STOICHIOMETRY PROBLEMS!
GENERAL PLAN FOR
STOICHIOMETRY
CALCULATIONS
Mass
product
Mass
reactant
Moles
reactant
Stoichiometric
factor
Moles
product
454 g of NH4NO3 --> N2O + 2 H2O
STEP 5 How much N2O is formed?
Total mass of reactants = total mass of products
454 g NH4NO3 = ___ g N2O + 204 g H2O
mass of N2O = 250. g
454 g of NH4NO3 --> N2O + 2 H2O
STEP 6 Calculate the percent yield
If you isolated only 131 g of N2O, what is the
percent yield?
This compares the theoretical (250. g) and
actual (131 g) yields.
454 g of NH4NO3 --> N2O + 2 H2O
STEP 6 Calculate the percent yield
actual yield
% yield =
• 100%
theoretical yield
131 g
% yield =
• 100% = 52.4%
250. g
PROBLEM: Using 5.00 g of
H2O2, what mass of O2 and of
H2O can be obtained?
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Reaction is catalyzed by MnO2
Step 1: moles of H2O2
Step 2: use STOICHIOMETRIC FACTOR to
calculate moles of O2
Step 3: mass of O2
Reactions Involving a
LIMITING REACTANT
• In a given reaction, there is not enough of
one reagent to use up the other reagent
completely.
• The reagent in short supply LIMITS the
quantity of product that can be formed.
LIMITING REACTANTS
Reactants
2 NO(g) + O2 (g)
Products
2 NO2(g)
Limiting reactant = ___________
Excess reactant = ____________
LIMITING REACTANTS
Demo of limiting reactants on Screen 4.7
LIMITING REACTANTS
(See CD Screen 4.8)
React solid Zn with 0.100
mol HCl (aq)
Zn + 2 HCl ---> ZnCl2 + H2
1
2
3
Rxn 1: Balloon inflates fully, some Zn left
* More than enough Zn to use up the 0.100 mol HCl
Rxn 2: Balloon inflates fully, no Zn left
* Right amount of each (HCl and Zn)
Rxn 3: Balloon does not inflate fully, no Zn left.
* Not enough Zn to use up 0.100 mol HCl
LIMITING REACTANTS
React solid Zn with 0.100
mol HCl (aq)
Zn + 2 HCl ---> ZnCl2 + H2
Rxn 1
7.00
Rxn 2
3.27
Rxn 3
1.31
mass Zn (g)
mol Zn
0.107
0.050
0.020
mol HCl
0.100
0.100
0.100
mol HCl/mol Zn
0.93/1
2.00/1
5.00/1
Lim Reactant LR = HCl
no LR
LR = Zn
Reaction to be Studied
2 Al + 3 Cl2 ---> Al2Cl6
PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2.
What mass of Al2Cl6 can form?
Mass
product
Mass
reactant
Moles
reactant
Stoichiometric
factor
Moles
product
Determine the formula of a
compound of Sn and I using the
following data.
•
•
•
•
Reaction of Sn and I2 is done using excess Sn.
Mass of Sn in the beginning = 1.056 g
Mass of iodine (I2) used = 1.947 g
Mass of Sn remaining = 0.601 g
Tin and Iodine Compound
Find the mass of Sn that combined with
1.947 g I2.
Mass of Sn initially =
1.056 g
Mass of Sn recovered = 0.601 g
Mass of Sn used =
0.455 g
Find moles of Sn used:
1 mol
0.455 g Sn •
= 3.83 x 10-3 mol Sn
118.7 g
Tin and Iodine Compound
Now find the number of moles of I2 that
combined with 3.83 x 10-3 mol Sn. Mass
of I2 used was 1.947 g.
1 mol
1.947 g I2 •
= 7.671 x 10-3 mol I2
253.81 g
How many moles of iodine atoms?
-3
7.671 x 10
2 mol I atoms 
mol I2 

 1 mol I2

= 1.534 x 10-2 mol I atoms
Tin and Iodine Compound
Now find the ratio of number of moles of moles
of I and Sn that combined.
1.534 x 10-2 mol I
4.01 mol I
=
1.00 mol Sn
3.83 x 10-3 mol Sn
Empirical formula is
SnI4