1 UNIT 6 STOICHIOMETRY
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UNIT 6 STOICHIOMETRY
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There are three ways to measure matter
– count (number of particles – representative particles)
– mass (grams)
– volume (Liters)
Mole – unit for amt of matter relating these quantities
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Representative Particles (rp) The type of particle depends on the type
of substance – elements – atoms
– molecular cmpds (2 nonmetals) – molecules
– ionic cmpds (metal + nonmetal) – formula units (fu)
Avogadro determined the # of any type of particle in one mole is constant
Avogadro’s # = 6.02 x 1023
There are ALWAYS 6.02 x 1023 particles in 1 mole of anything
6.02 x 1023 rp = 1 mole
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Conversion factor – any fraction made up of equivalent measures
– conversion factors = 1
– used to change from one unit to another
Factor–Label Method – problem solving process that uses conversion factors
– Problems are set up so that units cancel one another out
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To Convert from moles to rp or rp to moles: rp’s are: atoms, molecules or fu
6.02 x 1023 rp 1 mole
1 mole 6.02 x 1023 rp
Perform the following conversions:
25 moles Ag = ? atoms Ag
25 moles 6.02 x 1023 atoms
1 mole
= 1.51 x 1025 atoms
9.2 x 1023 fu NaCl = ? moles
9.2 x 1023 fu
6.02 x 1023 fu
1 mole = 1.53 moles
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Molar Mass # of grams per mole of any substance
determined using the atomic mass in the PT
Add the masses of all the elements in a cmpd
Used to convert: mass ↔ moles
Au 197.0 g
1 mole
NaCl
Na = 23.0
Cl = 35.5
58.5 g
1 mole
Cu(NO3)2
1 Cu x 63.5 =
2 N x 14.0 =
6 0 x 16.0 =
187.5 g
1 mole
63.5
28.0
96.0
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mass (g) ↔ moles
3.2 moles Ca(OH)2 = ? grams
1 Ca
2 O
2 H
x 40.1
x 16.0
x 1.0
= 40.1
= 32.0
= 2.0
74.1 g
1 mole
3.2 moles
1 mole
74.1 g = 237.1 g
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mass (g) ↔ moles
250 g H3PO4 = ? moles
3 H
1 P
4 O
x 1.0
x 31.0
x 16.0
= 3.0
= 31.0
= 64.0
98.0 g
1 mole
250 g
98.0 g
1 mole = 2.6 mol
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Molarity (M) Unit of concentration for solutions
# moles
1 L
Used to convert from moles to L or L to moles for solutions
EX: How many moles in 1.5 L of a 0.1 M solution of CaBr2?
Rewrite: 0.1 moles
1 L
1.5 L
1 L
0.1 moles = 0.15 moles
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EX: You need 15 moles of H2SO4 in a 6 M solution. How many liters of solution will you have?
Rewrite: 6 moles 1L
15 moles
6 moles
1 L = 2.5 L
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Molar Volume the volume of 1 mole of a gas at STP
STP = Standard Temperature and Pressure
0 C and 1 atm 22.4 L 1 mole
EX: How many moles of H2 gas in a 3.4 L balloon?
3.4 L
22.4 L
1 mole = 0.15 moles
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Mole
Representative particles (rp)
atoms, molecules, formula units
6.02 x 1023 rp
1 mole
mass (g)
molar mass
(PT)
# g
1 mole
Volume (L)
gas
@ STP
22.4 L
1 mole
Liquid Solutions
Molarity (M)
# moles
1 L
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Empirical Formula
tells the relative number of atoms of each
element that a cmpd contains
simplest formula
smallest whole # ratio of the atoms in a cmpd
does NOT necessarily indicate the actual #’s of
atoms present in each molecule
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Percent to mass,
mass to mole,
divide by small,
multiply ‘til whole
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Example: p. 90-91 73.9% Hg and 26.1% Cl
73.9 g Hg
26.1 g Cl
1 mol Hg
200.6 g Hg
1 mol Cl
35.5 g Cl
= 0.368 mol Hg
= 0.735 mol Cl
/0.368
/0.368
= 1
= 2
HgCl2
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Sample 3.13: p. 92
40.92 g C
4.58 g H
54.50 g O
1 mol C
12.01 g C
1 mol H
1.008 g H
1 mol O
16.00 g O
= 3.407 mol C
= 4.54 mol H
= 3.406 mol O
/3.406
/3.406
/3.406
= 1
= 1.333
= 1
x 3 = 3
x 3 = 4
x 3 = 3
C3H4O3
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Molecular Formula actual formula for a cmpd
whole number mulitple of the empirical formula
To determine the whole number multiple:
compare the empirical molar mass with the
experimental molar mass
Experimental molar mass
Empirical molar mass
multiply the empirical formula by the whole
number to get the molecular formula
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From sample exercise 3.13, the empirical formula
for a cmpd is C3H4O3. The experimentally
determined molecular mass is 176 amu.
Determine the molecular formula.
Empirical molar mass:
3 C x 12 = 36
4 H x 1 = 4
3 O x 16 = 48
88 amu
Multiple:
176 = 2
88
Molecular Formula:
C3H4O3 x 2
C6H8O6
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Percent Composition Percentage by mass contributed by each element
in a substance
Can be determined experimentally or using the chemical formula
C12H22O11
12 C x 12 = 144
22 H x 1 = 22
11 O x 16 = 176
342 amu
To check: make sure that the %’s all add to 100%
/342 x 100%
/342 x 100%
/342 x 100%
= 42.1%
= 6.4%
= 51.5%
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Reaction Stoichiometry
Using a balanced eqn, if we know the
mass of one reactant or product, we can
determine the masses of every other
substance in the rxn.
The coefficients can be interpreted as the
relative numbers of moles of each
substance – mole ratio
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grams of A
use molar mass
moles
of A
use
mole
ratio
moles
of B
use molar mass
grams of B
p. 97
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EX: How many moles of water are produced from
1.57 moles of O2.
2H2 + O2 → 2H2O
1.57 mol x mol
1.57 mol O2
mole ratio
1 mol O2
2 mol H2O = 3.14 mol H2O
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EX: Calculate the mass of CO2 produced when
1.00 g of butane is burned.
C4H10 + O2 → CO2 + H2O 1.00 g x g
2 8 10 13
1.00 g C4H10
4 C = 48
10 H = 10
58 g/1mol
C4H10
1 C = 12
2 O = 32
44 g/1 mol
CO2
1 mol C4H10
58 g C4H10
8 mol CO2
2 mol C4H10
44 g CO2
1 mol CO2
3.03 g CO2
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EX: How many grams of O2 can be prepared from
the decomposition of 0.0367 moles of KClO3?
KClO3 → KCl + O2
0.037 mol x g
2 2 3
2 - O = 32 g/1 mol O2
0.0367 mol KClO3 3 mol O2
2 mol KClO3
32 g O2
1 mol O2
1.76 g O2
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Limiting Reactant (reagent) The reactant that is completely consumed
in a chem rxn
determines or limits the amt of product
formed
Excess Reactant (reagent) reactant that has some left over after the
rxn
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What are the limiting & excess
reagents?
10 bottles & 12 lids
142 seniors & 100 airplane tickets
10 slices of bread & 7 slices of cheese
2 slices of bread are required for each
sandwich
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Balanced eqn: 2 H2 + O2 → 2H2O
Moles: 2 mol 1 mol 2 mol
Molar mass: 2 g/mol 32 g/mol 18 g/mol
Mass: 4 g 32 g 36 g
Rxn according 2 mol + 1 mol → 2 mol
to bal. eqn: 4 g + 32 g → 36 g
What will happen
if 3 mol of H2 are
combined with 1
mol O2?
What will happen
if 8 g of H2 are
combined with
100 g O2?
What will happen
if 25 g of H2 are
combined with
125 g O2?
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To determine the limiting reactant, you must
compare what you have to what you need. Sample Ex 3.19, p 101
2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
3.50 g 6.40 g x g
x g – 1st
3.5 g Na3PO4
1 mol Na3PO4
164 g Na3PO4
3 mol Ba(NO3)2
2 mol Na3PO4
Since you have less than you need, Ba(NO3)2 is the
limiting reactant and must be used for calculations.
261 g Ba(NO3)2
1 mol Ba(NO3)2
= 8.36 g Ba(NO3)2 needed to
react with all the Na3PO4
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2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
6.40 g x g
6.4 g Ba(NO3)2 1 mol Ba(NO3)2
261 g Ba(NO3)2
1 mol Ba3(PO4)2
3 mol Ba(NO3)2
602 g Ba3(PO4)2
1 mol Ba3(PO4)2
= 4.92 g Ba3(PO4)2
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2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
3.50 g 6.40 g x g
6.4 g Ba(NO3)2 1 mol Ba(NO3)2
261 g Ba(NO3)2
1 mol Ba3(PO4)2
3 mol Ba(NO3)2
602 g Ba3(PO4)2
1 mol Ba3(PO4)2
= 4.92 g Ba3(PO4)2
Another way: do comparison at the end
3.5 g Na3PO4 1 mol Na3PO4
164 g Na3PO4
1 mol Ba3(PO4)2
2 mol Na3PO4
602 g Ba3(PO4)2
1 mol Ba3(PO4)2
= 6.42 g Ba3PO4
Limiting reactant: Ba(NO3)2
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Theoretical yield – the max. amt of product that
can be produced from a given amt of reactant
calculated amt of product from the limiting reactant
Actual yield – the measured amt of product
obtained from a rxn
almost always less than theoretical yield
Percent yield – ratio of actual yield to theoretical
yield expressed as a %
% yield = actual yield x 100%
theoretical yield
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Sample Ex 3.20 p 102
2C6H12 + 5O2 → 2H2C6H8O4 + 2H2O
25 g x g
25 g C6H12 1 mol C6H12
84 g C6H12
2 mol H2C6H8O4
2 mol C6H12
146 g H2C6H8O4
1 mol H2C6H8O4
Theoretical yield = 43.5 g H2C6H8O4
Actual yield = 33.5 g
% yield = 33.5 g x 100%
43.5 g = 77.0%
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