Download Chapter 6 - Chemical Calculations

B) Formula Weights
Chapter 6
Used for ionic substances
- consists of formula units,
NOT molecules
Composition and Stoichiometry
I) Molecular & Formula Weights
Formula wt.
A) Molecular Weights
Sum of the atomic weights of the
atoms in the molecule
Sum of the atomic weights of the
atoms as given in the formula
1) Ex 1: Find the M.W. of the ethyl
alcohol (ethanol), C2H6O
1) Ex 1: What is the F.W. of
(NH4)2CO3 ?
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II) Percent Composition
B) Ex 2: A 40.00 mg sample of a cmpd.
was found to contain 38.55 mg
bromine & 1.45 mg carbon. What is
the mass % composition?
A) Ex 1: Determine the mass % comp.of
copper (II) nitrate.
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III) Avogadro’s Number & the Mole
A) The Mole
How do you weigh out the
same number of items?
If we weigh quantities in ratios of
weights of individual items, we obtain
equal numbers of items.
A.W. of H
1.008 amu
A.W. of C
12.011 amu
1 atom of H
1.008 amu
1 atom of C
12.011 amu
10 atoms of H
10.08 amu
10 atoms of C
120.11 amu
X atoms of H
1.008 g
Y atoms of C
12.011 g
X = Y = 6.022 x 1023 atoms
Avogadro’s Number, NA
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The unit for a very large number of
particles is
Mole
b) Apply to molecules & formula units
1mol C2H6O = 6.02 x 1023 molecules C2H6O
= 46.08 g C2H6O
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1 mole = 6.02 x 10 particles
1 mol (NH4)2CO3 = 6.02 x 1023 (NH4)2CO3 f.u.
1 mole C = 6.02 x 1023 C atoms =12.011g C
= 96.11 g (NH4)2CO3
1) Molar Mass
Mass in grams numerically equal
to A.W., M.W., or F.W.
Note: 1 mol of (NH4)2CO3 contains:
2 x (1 mol) NH4+ ions
a) A given AW tells you:
1) avg. mass of a single atom; amu
2 x (6.02 x 1023) NH4+ ions
2) mass of a mole of atoms;
grams / mole
8 mol H atoms
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IV) Calculations
B) Ex 2: How many grams of (NH4)2S are
required to obtain 0.50 mol of NH4+?
A) Ex 1: How many moles of He are in
40.0 g of He?
a) Need FW
How many atoms are there?
__ N x 14.01 amu =
1 mol He = 6.02 x 1023 He atoms = 4.00 g He
__ H x 1.01 amu =
__ S x 32.06 amu = ___________
1) # moles He
1 mol (NH4)2S =
1 mol (NH4)2S =
2) # atoms He
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C) Ex 3: Typically smog contains about
0.040 g CO per m3 of air.
How many molecules of CO are in a
m3 of air?
MW of CO =
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V) Empirical & Molecular Formulas
C) Procedure for Determining E.F.
A) Molecular Formula
Actual numbers and kinds of atoms in
a molecule
C6H6
C2H5OH
1) Express composition in grams.
If % comp. given,
assume 100 g sample
Benzene
Ethanol
2) Determine # moles of each element
B) Empirical Formula
Relative number of atoms of
each kind in a molecule
- smallest whole-number
ratio of atoms
C1H1
3) Divide by smallest # moles to
obtain mole ratio - this is also the
atom ratio
4) If needed: Multiply by simplest
factor to get whole numbers
Benzene or acetylene
5) Write the formula
Subscripts in a molecular formula are
always some integer multiple of
subscripts in empirical formula
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6) Ex 1: A 10.45 g sample of Bi
combines w. oxygen to produce
11.65 g of a bismuth oxide.
c) Step 3: Determine mole ratio
Divide by smallest # moles
Determine the E.F. of the oxide.
a) Step 1: determine mass of oxygen
d) Step 4: Multiply by factor to get
whole numbers
Bi:
b) Step 2: Convert to moles
O:
e) Step 5: Write formula
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D) Molecular Formula Determination
Molecular formula is always some
integer multiple of the E.F.
Benzene
Acetylene
EF
CH
CH
1) Ex : Analysis of an unknown cmpd.
gave 39.72% C, 1.67% H, 58.61% Cl.
The MW was found to be 181.4 amu.
Determine the molecular formula.
MF
C6H6
C2H2
a) Determine Emp. Formula
C:
MF = (CH)n
n = multiplying factor
H:
Cl:
Find MW experimentally
Benzene
78.1
amu
______________
n=
=6
13.0 amu
C:
Acetylene
26.0 amu
_____________
n=
=2
13.0 amu
H:
Cl:
E.F. = C H Cl
E.F.W. =
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VI) Stoichiometry & the Balanced Eqn.
b) Determine Molecular Formula
Stoichiometry / determination of
quantities of reactants & products
involved in chem. rx’s.
MW
181.4 amu
n = ---------- = ------------------ =
EFW
Use balanced chem. eqn. - tells you not
only the reactants & products but also
how much of each is involved in the
chem. rx.
MF =
2 K (s) + 2 H2O(R)
2 KOH(aq)
+ H2(g)
2 atoms 2 molecules
2 formula
units
2 moles
2 x 56 =
112g
1 molecule
2 moles
2 x 39 =
78g
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2 moles
2 x 18 =
36g
1 mole
1x2=
2g
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A) Procedure
mole ______________________________________
moles of desired substance
=
ratio
moles of given substance
1) Calc. Number of moles of a given
substance
mole ______________________________________
coeff. of desired substance
=
ratio
coeff. of given substance
3) Convert moles of desired substance to
grams
2) Determine moles of the desired
subst.
- use coeff. in the bal. eqn.
- convert moles of given
substance to moles of
desired substance
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VII) Solving Stoichiometry Problems
A) Ex 1:
mole
given
mole
desired
How many moles of methane would be
required to produce 0.67 mole of
chloroform?
CHCl3 + 3 HCl
4) Summary
CH4 + 3 Cl2
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B) Ex 2:
mole
given
mole
desired
grams
desired
C) Ex 3:
grams
given
How many grams of chlorine are req.
to produce 0.67 mol of CHCl3?
moles
given
moles
desired
grams
desired
How many grams of KOH would be
formed by the rx. of 3.0 g of potassium
w. excess water?
2 K + 2 H2O
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2 KOH + H2
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)H corresponds to molar quantities
given in eqn. as written
VIII) Heat of Reaction (Enthalpy)
)Hrxn = HProducts & HReactants
4 H2O(R) )H = &274 kcal
4 H2(g) + 2 O2(g)
A) Exothermic Rx’s.
B) Endothermic Rx’s
HP < HR , )H < 0 , exothermic
HP > HR , )H > 0 , endothermic
heat is released
heat is absorbed
2 H2O(R) + 137 kcal
2 H2 (g) + O2(g)
137 kcal + 2 H2O(R)
2 H2(g) + O2(g)
or
or
2 H2(g) + O2(g)
2 H2O(R)
2 H2O(R)
)H = &137 kcal
2 H2(g) + O2(g)
)H = +137 kcal
reverse of previous rx.
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C) Determining )H for a Rx.
2 H2 (g)
Convenient sample sizes are reacted &
conv. factors are used to obtain the
heat energy
+ O2 (g)
1) Ex 1: When 36.0g of Al reacts w.
excess Fe2O3 how much heat is
released?
2 Al(s) + Fe2O3(s)
!137
kcal
+ 137
kcal
2 Fe(s) + Al2O3(s)
)Hrxn = !202 kcal
2 H 2O ( R )
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IX) Percent Yield
A) Ex 1:
A 15.6g sample of C6H6 is mixed w.
excess HNO3 and reacts according to
the bal. eqn. below. We obtain 18.0 g
C6H5NO2. What is the percent yield of
C6H5NO2?
Theoretical Yield
Maximum amt. product that
could be obtained
Actual Yield
C6H6 + HNO3
Amt. product actually obtained
% Yield =
C6H5NO2 + H2O
Actual Yield
x 100%
Theoretical Yield
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X) Limiting Reagent (Reactant)
B) Ex: A mixture of 1.5 mol of CH3OH
& 3.00 mol of O2 react to form CO2 &
H2O. What is the limiting reactant?
How many moles of CO2 are formed?
Nonstochiometric amounts
- one reactant is completely consumed
2 CH3OH + 3 O2
- others are in excess (left over)
2 CO2 + 4 H2O
limits the amount of product
A) Ex: A shelf requires 10 bolts and nuts.
The package came with 10 bolts and 8
nuts. What is the limiting “reagent”?
10 bolts + 10 nuts => 1 shelf
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Chapter 6 (RJO 7) - Homework
1(a,b,d), 2(a,d), 4, 5(a,d), 6(a,d),
7(a,c,d), 8(a,c,d), 9a, 10(a,d), 11c,
12(a,d), 13(a,c), 15(a,b,c), 16(a,b),
17(a,d), 18(a,b), 20(b,d), 21, 22, 24,
25a, 26b, 27(a,d), 28a, 29a, 30(a,d),
31(a,c), 34, 35, 37, 38, 40(a,b), 41,
43, 46, 48, 51, 53, 56-58, 61, 63, 65
68, 70
Computer Programs
10 - mole concept
13 - Composition
& Emp. Form.
11 - % composition
12 - Determining an
Empirical Form.
16 - Form. Wts. &
Stoichiometry
17 - Eqns. & Stoich.
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