Download Unit 7 (part 2) Stoich brief review notes

Topic 7 – Moles and mole calculations
Key Concept
Definitions and facts
1. Equation
shows changes that are
taking place in substances
2. Physical
equation
shows changes that do
not change compositions
of substances (physical
change)
3. Chemical
equation
Shows changes in
compositions of one or
more substances
(chemical changes)
4. Nuclear
equation
Shows changes of
nucleus contents of one
or more atoms (nuclear
change or transmutation)
5. Chemical
reactions
ways by which chemical
changes occur
6. Reac tants
Starting substances that
will be changed in
chemical reaction
7. Products
Substances remaining
after a change had
occurred in reaction
8. Coefficients
whole-numbers in front
of substances in
equations.
Indicate number of
moles (how many) of the
substance
9. Arrow
Separates Reactants from
Products, and can read as
“yields” or “produces”
Examples
H2O(l) ---- > H2O(s)
H2(g) + O2(g) ---> H2 O(l)
220 Fr
---- > 4He + 216At
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Topic 7 – Moles and mole calculations
Types of reactions
Explanations
10. Synthesis
(combination)
Two or more
reactants combined
to make one product
11.Decomposition
(Analysis)
One reactant breaks
to make two or more
products
Examples
+
2H2 + O2 -- > 2H2O
+
2H2O ----> 2H2 + O2
12. Single
Replacement
13. Double
Replacement
14. Combustion
(oxidation)
15. Law of
Conservation
One (a more
reactive) reactant
replaces a less
reactive element of a
compound
+
+
Zn + 2HCl --->ZnCl2 +H2
Ions from
compound reactants
+
+
switche with each
other to form
NaCl + AgNO3 ---->NaNO3 + AgCl
different compounds
a carbon compound
burns in the present
of oxygen
CH4 + O2 --- > CO2 + H2O
)
During
chemical reactions: neither atoms, mass,
charge or energy is created nor destroyed.
The total number of atoms, charge, and energy are
the same before and after a reaction (conserved)
16. A balanced
chemical
equation
An equation that shows conservation
N2 + H2 ---- > NH3
allows
2N
2N
this equation to be
6H
6H
balanced.
atoms of
atoms of
=
reactants
2
products
Topic 7 – Moles and mole calculations
Key concept
17.
36.Law of conservation
of mass
Example
During a chemical reaction, mass of
substances before and after the reaction is
the same.
3Fe
20.9 g
+
2O2 ------- > Fe3O4 :
8.0 g
=
28.9 g
NOTE: The total mass of Fe and O2
(reactants) is the same as the total mass of
Fe3O4 (product)
18. Balancing equation
37.
An equation
Balancing
is balanced
equationwhen
usingit acontains the correct combination of
smallest
RAPwhole-number
table.
coefficients. The coefficients allow number of each
atomThe
on both
use ofsides
a Table
of the
is optional
equation to be the same.
To balance an equation:
One or more coefficients in front of the substances must be changed in the
equation.
The right combination of coefficients will make the number of atoms on
both sides equal.
Suggestion to balancing equations
. Make a table (optional) to keep track of number of atoms as coefficients
are changed
. Try balancing one atom at a time
. Every time a coefficient is changed, RECOUNT the number of each
atom affected by the change. NOTE changes on the table. (be sure to
count atoms correctly)
. Always change coefficients of free elements (Na, Cl2) last
. Always put parenthesis around polyatomic ions and count it as one unit
. Be sure that coefficients are in smallest whole-number ratio
19. Sums of all coefficient: After you had balance an equation, you
may be asked for the sum of all coefficients.
(which may not be written in front of the substances)
Examples of balanced equations and sums of coefficients are given on the
next page.
.
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Topic 7 – Moles and mole calculations
20. Balanced equations and sum of coefficients
Li3N
Unbalanced
------> Li
+ N2
Balanced
2Li3N -------> 6 Li
+ N2
Sum of all coefficients
2
+ 1
3Ca(OH)2 + 2H3PO4 ---- > Ca3(PO4)2 + 6H2O
Sum of coefficients 3
Unbalanced
+ 2
C3H4 +
+
O2
Balanced
C3H4 + 4O2
Sum of coefficients
1
4
= 9
Ca(OH)2 + H3PO4 ----- > Ca3(PO4)2 + H2O
Unbalanced
Balanced
+ 6
+ 4
1
----- > CO2
+ 6=
+
------ > 3CO2 +
+
3
+
12
H2O
2H2O
2 = 9
Topic 7 – Moles and mole calculations
Key concept
21. Mole
Definition and fact
A unit of quantity
equals to 6.02 x 1023
(Avogadro’s number)
How many atoms
22. Moles
interpretation in
formula
Examples
1 dozen = 12
1 mole = 6.02 x 1023
(602000000000000000000000)
In 1 mole of H2O there are:
2 moles of H atoms
1 mole of O atom
In 2 moles of H2O there are:
4 moles of H atoms
2 moles of O atoms
23. Molar mass
The mass of 1 mole of Molar mass of H2O is 18 g
(18g/mole)
a substance
Molar mass of NaCl is 58 g
( 58g/mole)
24. Gram-atomic
mass
The mass of 1 mole of Gram-atomic mass of sodium
(Na) is 22.989 g/mole.
an element.
Gram atomic mass of
Gram-atomic mass of chlorine
elements can be found
(Cl) is 35.5 g /mole
on the Periodic Table
25. Gram- formula The mass of one mole
(molecular )
of a compound
mass
The sum of all the
atomic masses in a
formula
The gram-formula mass of
NaCl is 58.4 g/mole.
It is the sum of the mass of
1 Na (22.9 g ) + the mass of
1 Cl (35.5 g )
The gram-formula mass of
H2O is 18 g
It is the sum of the mass of
2 H (2 g ) + the mass of
1 O (18 g)
26. Percent
composition
Portion of a mass of a % by mass of Na in NaCl = 39.3%
substance that is due to
the mass of individual % by mass of Cl in NaCl = 60.7 %
element in its formula
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Topic 7 – Moles and mole calculations
Concept
27. Hydrates
Facts and definitions
Examples
Ionic substances
CuSO4 .5H2O (copper
containing water in their sulfate pentahydrate)
crystalline structure
CuSO4
H 2O
A diagram of the hydrate
CuSO4.5H2O showing
5 water molecules attached
to the crystalline structure
of CuSO4
28. Anhydrous
a hydrate with the
water removed
In the hydrate (copper
sulfate pentahydrate)
.
The hydrate is CuSO4 5H2O
The anhydrous is CuSO4
29. Formula mass The sum of the mass
of a hydrate
of the anhydrous +
the mass of water
30. Percent
composition of
a hydrate
Portion of a hydrate’s
mass that is due to
the mass of the water.
The formula mass of
CuSO4 .5H2O is 250 g.
This is the sum of the mass
of 1CuSO4 ( 160 g) + the
mass of 5H2O (90 g)
% of water in the hydrate
CuSO4.5H2O is 36 %
This is the portion of the
mass of CuSO4.5H2O
(250 g) that is due to the
mass of 5H2O (90 g)
In the next few pages, you will learn step-by-step how to calculate
number of moles of atom, mass, mole, and percent composition of
a given formula.
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Topic 7 – Moles and mole calculations
Calculations with formula
31. Determining moles of
atoms
Given moles x how many atom
Examples
How many moles of oxygen are in
mole of Ca(ClO )
x 6 = 6 moles of oxygen
How many moles of oxygen are in
mole of Ca(ClO3)2?
x 6 = 3 moles of oxygen
What is the total number of moles of atoms
Total moles of atoms =
sum of all atoms in a formula in mole of Ca(ClO3)2?
Ca = 1 mole
Total moles of atoms
Cl = 2 moles 1 + 2 + 6 = 9 moles
O = 6 moles
32. Determining
Gram-Atomic mass
What is the Gram-Atomic mass of
silver (Ag)? 107.868 g
107.868
Ag
47
LOOK on Periodic Table
-
33. Calculating
7 Gram-Formula mass.
What is the Gram-Formula mass of
. count each atom in formula Al2(SO4)3
. multiply how many by
Atoms Atomic How
Total
atomic mass of the element
mass x many = mass
. add up all products (total
27 g
2
54 g
mass of each element) to get Al
formula mass
S
32 g
3
96 g
Gram-formula mass =
sum of all masses
O
16 g
12
Gram- Formula mass =
192 g
342 g
of Al2(SO4)3
.
7
Topic 7 – Moles and mole calculations
Calculation with formulas
34. Calculating Gram-Formula
mass of hydrates
NOTE: when an atom appears on
both sides of the (.) in a hydrate, you
must add them up.
Examples
What is the Gram-Formula mass of
the hydrate CaCO3 4H2O ?
Atoms
Atomic How
Total
mass x many =mass
Ca
40
1
40 g
In example to the right, there are:
C
12
1
12 g
3 O in CaCO3 (left of .) and
O
16
7
112 g
4 O in 4H2O (right of .)
H
1
8
8g
Total # of O atoms = 3 + 4 = 7
Gram-Formula mass = 172 g
of
35. Calculating mass from moles
What is the mass of
Al2(SO4)3 ?
Mass = moles x formula mass
Mass = 2.5 x 342 = 855 g
What is the mass of 0.3 mole of water,
H2O ?
Formula mass of H2O:
2 H 2 (1) = 2 g
18 g
1 O 1 (16) = 16 g
Mass = mole x formula mass
Mass = 0.3 x 18 = 5.4 g
36. Calculating moles from
mass
Given mass
Mole =
Formula mass
Use this Table T equation to
calculate mass and mole
8
How many moles are represents in
200 grams of Al2(SO4)3 ?
200
moles =
= 0.58 mol
342
How many moles of H2O are in 54
grams of the substance?
54
moles =
= 3.0 mol
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Topic 7 – Moles and mole calculations
Calculation with formulas
Examples
What is the percent composition of Al, S,
and O in the formula Al2(SO3)4
37. Percent composition
by mass calculation
Refer to Set 13 example for total mass
of each atom in Al2(SO4)3
See Table T for percent
composition equation
% of Al =
Total mass of an atom
%=
x 100
Formula mass
% of S =
% of O =
54
x 100 = 15.8 %
342
96
342
192
342
x 100 = 28.1 %
x 100 = 56.1 %
What is the percent of water in the hydrate
CaCO3.4H2O ?
38. Percent composition
of hydrates
Find mass of 4 H2O:
8 H = 8 (1) = 8
72 g
4 O = 4 (16) = 64
Total mass of H2O
% H2O =
x 100
Formula mass
of hydrate
72
% H2O =
= 42 %
172
Refer to Set 14 example to see how the
formula mass of CaCO3.4H2O (172 g)
was determined.
39. Percent by volume
% =
part volume
whole volume
x 100
A 350 ml alcohol solution contains
190 ml of isopropyl alcohol. What is
percent composition of the alcohol in this
solution?
% =
190
x 100 = 45 %
350
.
9
Topic 7 – Moles and mole calculations
Calculation with formulas
examples
A 3.5 grams sample of a hydrated was
heated to remove its water. The mass of
the anhydrous remaining was determined
to 2.3 grams. What is percent of water
in the hydrate
40. Percent of water
from lab data
Mass of water =
mass of hydrate – mass of anhydrous
% water =
mass of water
mass of hydrate
x 100
mass of water = 3.5 - 2.3 = 1.2 g
1.2
% water =
x 100 = 34.3 %
3.5
A student collected the following data
during a hydrate experiment:
Mass of crucible
= 15.3 g
Mass of hydrate + crucible = 19.5 g
Mass of anhydrous + crucible = 17.5 g
What is the percent of water in the hydrate
compound that was used in the experiment?
Mass of hydrate = 19.5 – 15.3 = 4.2 g
Mass of anhydrous = 17.5 – 15.3 = 2.2 g
Mass of water = 4.2 – 2.2 = 2.0 g
2.0
% water =
x 100 = 47.6 %
4.2
41. Molecular formula from
mass and empirical formula
. Determine mass of empirical
formula
. Determine unit (how many)
of empirical formula
What is the molecular formula of a
compound with a molecular mass of
56 g/mol and an empirical formula of CH2
empirical formula’s mass: 1 C = 12
2 H = 2 14 g
molecular mass = 56 =
molecular mass
4
empirical mass
14
empirical formula
. Determine molecular formula Molecular formula = 4 (CH ) = C H
2
4 8
multiply each subscript of
empirical by unit (step 2)
10
Topic 7 – Moles and mole calculations
Calculation with equations
42. A balanced
chemical equation
Remember:
Coefficients represent
number of moles.
Examples
Shows mole ratio (proportion) of substances
taking part in a chemical reaction.
mole ratio is the ratio of coefficients
(whole number in front of the substances)
In the balanced equation
4 NH3 + 5 O2 ---- > 4NO + 6 H2O
Mole ratio of NH3 to O2 is 4 : 5
Mole ratio of NH3 to NO is 1 : 1
(reduced from 4 : 4)
Mole ratio of NO to H2O is 2 : 3
(reduced from 4 : 6)
43. mole-mole
problems
Given the balanced equation below:
4 NH3 + 5 O2 ---- > 4NO + 6 H2O
when
of
4 NH3 + 5 O2 ---- > 4NO + 6 H2O
. Re-write equation
. Write
and
from question under the
correct substances.
2
Setup mole proportion
4
.
of water,
, is produced
is reacted?
X
=
6
2
. Cross multiply and
Solve for
4X
X
X
=
12
=
3 mol
.
11
Topic 7 – Moles and mole calculations
Calculations with equation
44. Volume – Volume
problem
Examples
Given the balanced equation below:
2C4H10 + 13O2 ------>8CO2 + 10H2O
is the total
combusted to produce
. Re-write equation
and
. Write
from question under the
correct substances.
. Setup volume proportion
2C4H10
2
8
=
=
=
3Cu + 2H3PO4 ----- > Cu3(PO4)2 + 3H2
is the total
by reacting
. Re-write equation
that must be
?
+ 13O2 ------>8CO2 + 10H2O
. Cross multiply and
Solve for
45. Mass – Mass
problem
of
that is produced
?
3Cu + 2H3PO4 ----- > Cu3(PO4)2 + 3H2
. Write
and under correct formula
. Setup mass proportion
Mass of 2H3PO4
6 H = 6(1) = 6 g
2 P = 2(31) = 62 g
8 O = 8(16) = 128 g
198 g
Mass of 3H2 = 6 (1) = 6g
12
mass of 2H3PO4 ==
mass of 3H2
198
==
6
198
==
294