Download LECTURE_pptnotes Fipps Stochiometry

Chemical Quantities & Stoichiometry



By mass
By volume
By counting the # of atoms/molecules
◦ We can use a word like a “dozen” to specify a certain
quantity.




Mole (mol): SI unit for measuring the amount of
a substance.
1 mol = 6.02 x 1023 representative particles
Avogadro’s Number: 6.02 x 1023
Representative Particle: smallest unit that has
all the characteristics of that substance.

Atom
Element (ex. Cu): ___________________
◦ Exception: The representative particle of the
7 diatomic elements is a molecule. (ex. H2)

Molecule
Covalent compound (ex. H2O): ____________

Ionic Compound (ex. NaCl):
Formula Unit
_____________
(Molecule)

4 moles Ca =
4 moles Ca
atoms Ca.
6.02 x 1023 atoms Ca
1 mole Ca
= 2.41 x 1024 atoms Ca

5 x 1018 atoms Cu =
Cu.
5 x 1018 atoms Cu
moles
1 mole Cu
6.02 x 1023 atoms Cu
= 8.3 x 10-6 moles Cu

9.2 moles F2 =
9.2 moles
molecules F2?
6.02 x 1023 molecules F2
1 mole
= 5.5 x 1024 molecules F2

9.2 moles F2 =
9.2 moles F2
6.02 x 1023 molecules F2
1 mole
= 1.1 x 1025 atoms F
atoms F?
2 atoms F
1 molecule F2

3.4 moles C2H4 =
atoms?
3.4 moles C2H4
6.02 x 1023 molecules C2H4
1 mole C2H4
= 1.22 x 1025 atoms
total
6 atoms
1 molecules C2H4

Molar Mass: The mass of one mole of an element
or compound.
◦ Molar mass of a compound = the sum of the masses of
the atoms in the formula
◦ Use the atomic masses in grams on the periodic table.
Find the molar mass:
1.
Sr
= 87.62 grams/mol
2.
MgBr2
3.
Ba3(PO4)2
24.3 + (2x 79.9) =
Ba =
P
=
O =+
=
184.1 grams/mol
3 x 137.38 g
2 x 30.97 g
8x
16 g
602.08 grams/mol

3.4 moles C2H4 =
atoms?
3.4 moles C2H4
6.02 x 1023 molecules C2H4
1 mole C2H4
= 1.22 x 1025 atoms
total
6 atoms
1 molecules C2H4
1 mol = molar mass (in grams)

68 grams F2 =
68 grams F2
moles F2?
1 mole F2
38 grams
68 / 38 = 1.8 moles F2

Standard Temperature and Pressure (STP):
0oC, 1 atm
◦ See Reference Tables

Avogadro’s Hypothesis: equal volumes of gases
at the same temperature and pressure contain
equal numbers of particles.
◦ At STP, 1 mole of any gas occupies a volume of 22.4 L.
1 mol = 22.4 L at STP (gases only!!!)

5.4 moles He =
5.4 moles He
L He at STP?
22.4 L He
1 mole He
5.4 x 22.4 = 120.96 L He

5.4 moles CH4 =
5.4 moles CH4
L CH4 gas at STP?
22.4 L CH4
1 mole CH4
5.4 x 22.4 = 120.96 L CH4

560 L SO3 =
560 L SO3
mol SO3
1 mole SO3
22.4 L SO3
560 / 22.4 = 25 mole SO3
22.4 L
at STP
(gases only)
1 mole
Molar Mass
6.02 x 1023 particles

How would you do this???
 grams (Molar Mass)
(Density) grams
liters
mole
Example:
 A gaseous compound composed of sulfur and oxygen
has a density of 3.58 g/L at STP. What is the molar mass
of this gas?
3.58 g
L
22.4 L
1 mole
3.58 x 22.4 = 80.3 g/mole

What is the density of krypton gas at STP?
83.8 grams Kr
mole
1 mole
22.4 Liters
83.8 / 22.4 = 3.74 g/L Kr


Law of Definite Proportions:
In samples of any chemical compound, the
masses of the elements are always in the
same proportion.
=> Allows us to write chemical formulas.
Percent Composition by Mass
◦ Worksheet


Percent Composition - % by mass of each
element in a compound
Percent =
Part x 100
Whole


Al:
S:
O:
Percent Comp = Mass of 1 element
x 100
Mass of compound
Example: Find the mass percent composition
of Al2(SO4)3
2 x 27 = 54
3 x 32 = 96
12 x 16 = 192
54 x 100= 15.8%
% Al:
342
% S:
96 x 100 = 28.1%
342
%O:
192 x 100 = 56.1%
342
Total Comp. = 342
Percent
Composition
18.8% Na
# grams in
100 grams
How many moles
of each element?
Divide by smallest
#moles
29.0% Cl
52.2% O
Empirical Formula: ___________________

Empirical Formula: lowest whole-number ratio.
◦ The formula for an ionic compound will
always be the empirical formula.
◦ The formula for a covalent compound will not
always be the empirical formula.

Molecular Formula: either the same as the
empirical formula (as for ionic compounds) or
a simple whole-number multiple of the
empirical formula.

Calculate the empirical formula of a
compound containing 0.90g Ca and 1.60g Cl.
◦ Step 1: Convert GRAMS to MOLES.
 Ca: 0.90g
1 mole
40.1 g
= 0.0224 mole Ca
 Cl: 1.60g
1 mole
35.5 g
= 0.0451 mole Cl

Step 2: DIVIDE the # of moles of each
substance by the smallest number to get the
simplest mole ratio.
Ca: 0.0224 = 1
0.0224
Cl: 0.0451 = 2.01 ~ 2
0.0224
CaCl2


Step 3: If the numbers are whole numbers,
use these as the subscripts for the formula.
If the numbers are not whole numbers,
multiply each by a factor that will make them
whole numbers.
Look for these fractions:
◦ 0.5  x 2
◦ 0.33  x 3
◦ 0.25  x 4
1.
Suppose the mass percents of a compound are 40%
carbon, 6.70% hydrogen, and 53.3% oxygen.
Determine the empirical formula for this
compound.
Since this compound is covalent, the actual formula
may not be the simplest ratio of elements. If the
molar mass of the compound is experimentally
shown to be 90.0 g/mol, what is the molecular
formula of this covalent compound?
Find the molecular formula of ethylene glycol
(CH3O) if its molar mass is 62 g/mol.
Step 1: CH3O = (12) + (3 x 1) + (16) = 31
Step 2: 62 / 31 = 2
Step 3: 2 (CH3O)  C2H6O2
The percent composition of methyl
butanoate is 58.8% C, 9.8% H, and 31.4 %
O and its molar mass is 102 g/mol.

◦
What is its empirical formula?
◦
What is its molecular formula?
58.8% C
1 mole C
12 g C
= 4.9 / 1.9 = 2.5 x 2 = 5
9.8% H
1 mole H = 9.8 / 1.9 = 5 x 2 = 10
1gH
31.4%O
1 mole O = 1.9 /1.9 = 1 x 2 = 2
16 g O
C5H10O2  (5x12) + (10x1) + (2x16) = 102 g/mol
Empirical mass = molecular mass, so molecular formula is
the same  C5H10O2


If a compound is 40% C, 7% H, and 53% O,
what is its empirical formula?
What is the molecular formula for this
element if the molecular mass is 180 g/mol?

Stoichiometry: The calculation of quantities of
substances involved in chemical reactions.
N2 (g) + 3H2 (g)  2NH3 (g)


The above equation could be read:
1 mol of N2 reacts with 3 moles of H2 to yield
2 moles of ammonia.
2A + B  3C + 7D

Given the number of moles of reactant A
(ex. 6 moles A), I can find:
◦ 1) The number of moles of reactant B needed to
react completely with 6 moles of A (all 6 moles are
used up).
◦ 2) The number of moles of product C formed.
◦ 3) The number of moles of product D formed.
N2 (g) + 3H2 (g)  2NH3 (g)
There is a 1:3:2 mole ratio
1.
If you have 2 moles of N2, how many
moles of NH3 will be produced?
2.
If you want 5 moles of product, how many
moles of hydrogen gas do you need?
3.
How many moles of nitrogen are needed to
react completely with 8 moles of hydrogen?
N2 + 3H2  2NH3
If you have 2 moles of N2, how many moles of
NH3 will be produced?
2 mol N2
2 mol NH3
1 mol N2
2 x 2 / 1 = 4 moles NH3
N2 + 3H2  2NH3
If you want 5 moles of product, how many
moles of hydrogen gas do you need?
5 mol NH3
3 mol H2
2 mol NH3
5 x 3 / 2 = 7.5 moles H2
N2 + 3H2  2NH3
How many moles of nitrogen are needed to
react completely with 8 moles of hydrogen?
8 mol H2
1 mole N2
3 mole H2
8 x 1 / 3 = 2.67 mole N2
****The only way to convert
from one compound to
something totally different in
the reaction is to use the
MOLE TO MOLE RATIO from
the coefficients!!!****
Note – If you don’t have moles already, your
first step is to convert to moles!
Mole Review:
22.4 L
at STP
1
mole
Molar
Mass
6.02 x 1023
particles
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
MOLES
(g/mol)
6.02 
1023
particles/mol
Molarity (mol/L)
LITERS
OF
SOLUTION
NUMBER
OF
PARTICLES
1.
2.
How many liters of oxygen are required to
burn 3.86 L of carbon monoxide?
2CO (g) + O2 (g)  2CO2 (g)
How many liters of PH3 are formed when
0.42 L of hydrogen reacts with phosphorus?
P4 (s) + 6 H2 (g)  4 PH3 (g)
N2 (g) + 3H2 (g)  2NH3 (g)
1.
2.
How many grams of hydrogen gas are
required for 3.75 g of nitrogen gas to react
completely?
What mass of ammonia is formed when
3.75 g of nitrogen gas react with hydrogen
gas?
N2 + 3H2  2NH3
How many grams of H2 are required to produce
5.0 grams of NH3?
Grams NH3  moles NH3  moles H2  grams H2
5.0 g NH3
1 mole NH3
17 g NH3
3 mole H2
2 g H2
2 mole NH3 1 mole H2
5.0 x 3 x 2 / 17 / 2 = 0.88 g H2
1)
The combustion of propane, C3H8, a fuel used in backyard
grills and camp stoves, produces carbon dioxide and water
vapor.
C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O (g)
What mass of carbon dioxide forms when 95.6 g of
propane burns?
2)
Solid xenon hexafluoride is prepared by allowing xenon gas
and fluorine gas to react.
Xe (g) + 3F2 (g)  XeF6 (s)
◦ How many grams of fluorine are required to produce 10.0 g of XeF6?
◦ How many grams of xenon are required to produce 10.0 g of XeF6?


How moles of CO2 are produced when 52.0 g C2H2
burns?
2C2H2 (g) + 5O2 (g)  4CO2 (g) + 2H2O (g)
How many liters of hydrogen gas are formed from 50
grams of potassium?
2K (s) + 2H2O (l)  2KOH (aq) + H2 (g)


How many molecules of oxygen are produced by
the decomposition of 6.54 g of potassium
chlorate (KClO3)?
2KClO3 (s)  2KCl (s) + 3O2 (g)
How many grams of nitrogen dioxide must react
with water to produce 5.00 x 1022 molecules of
nitrogen monoxide?
3NO2 (g) + H2O (l)  2HNO3 (aq) + NO (g)

Limiting Reagent: The reactant that limits the
amount of product that can be formed in a
reaction.
◦ The reaction will stop when all of this reactant is used
up.
◦ Determines the amount of product that is produced.

Excess Reagent: You have more than you need of
this reactant.
◦ The reaction will stop before all of this reactant is used
up. You will have some of this reactant leftover.

You have :
◦ 1 loaf of bread (containing 14 slices of bread)
◦ 4 jars of peanut butter
◦ 2 jars of jelly
A) How many peanut butter and jelly
sandwiches can you make?
B) What is the limiting reagent?
C) What are the reactants in excess?
**The amount of product
that can be formed in a
reaction is always
determined by the limiting
reactant!!**

A s’mores MUST have:
◦ 2 graham crackers
◦ 2 pieces of chocolate
◦ 1 marshmallow

If you had:
◦ 8 graham crackers
◦ 4 pieces of chocolate
◦ 6 marshmallows
How many s’mores could you make?
1) If 2.70 mol C2H4 is reacted with 6.30 mol O2,
what is the limiting reagent?
C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g)
2) Identify the limiting reagent when 6.00 g HCl reacts
with 5.00 g Mg.
Mg (s) + 2HCl (aq)  MgCl2 (aq) + H2 (g)
- How many grams of MgCl2 are produced in this
reaction? __________________
- Which reactant is in excess? _________________
- How much of your excess reagent do you have
leftover? _____________________
3) How many grams of water can be produced by
the reaction of 2.40 mol C2H2 with 7.4 mol O2?
C2H4 (g) + 3O2 (g)  2CO2 (g) + 2H2O (g)
Limiting Reagent? ______________
Excess Reagent? _______________

Theoretical Yield: the maximum amount of
product that could be formed from the given
amounts of reactants (ideal conditions).
◦ Calculated using stoichiometry.

Actual Yield: the amount of product that
actually forms in a lab.
◦ Actual yield is usually less than theoretical yield.

Percent Yield =
Actual Yield
Theoretical Yield
x 100%
1.
If 3.75 g of nitrogen completely react, what
is the theoretical yield of NH3?
N2 (g) + 3H2 (g)  2NH3 (g)
If the actual yield is 3.86 g, what is the
percent yield?

Find the percent yield if 84.8 g of iron (III)
oxide reacts with an excess of carbon
monoxide to produce 55.0 g of iron.
Fe2O3 (s) + 3CO (g)  2Fe (s) + 3CO2 (g)