moles and molar mass

1. Y11
STOICHIOMETRIC
RELATIONSHIPS
REACTING MASSES AND VOLUMES

2. ESSENTIAL IDEA
Mole ratios in chemical equations
can be used to calculate reacting
ratios by mass and gas volume.
NATURE OF SCIENCE (1.8)
Making careful observations and obtaining
evidence for scientific theories – Avogadro’s
initial hypothesis.

3. INTERNATIONAL-MINDEDNESS
The SI unit of pressure is the Pascal (Pa, N/m2
),
but many other units remain in common usage in
different countries. These include atmosphere
(atm), millimeters of mercury (mm Hg), Torr, bar
and pounds per square inch (psi). The bar (105
Pa) is now widely used as a convenient unit, as it
is very close to 1atm. The SI unit for volume is
m3
, although liter is a commonly used unit.
.

4. STOICHIOMETRY
For any stoichiometry problem, you must
know what unit you are given and what
unit you are looking for.
Step one is to ALWAYS convert to moles
unless you are already in moles.
Step two is to multiply by the mole ratio
with the unknown on top.
Step three is to convert to the unit you are
looking for using the mole conversion
chart.

5. Stoich Problems – Mole to Mole
Given moles – looking for moles
Step 1: Label your equation
How many moles of H2
O are produced from 6.0 moles of oxygen?
Step 2: Write down the given and multiply by the mole ratio with the
unknown on top.
2H2
+ O2
2H2
O

6. MOLE TO MOLE EXAMPLES
1. N2
+ 3H2
2 NH3
How many moles of N2
are needed to make 12.2 moles of NH3
?
2. 2C2
H2
+ 5O2
4CO2
+ 2H2
O
How many moles of CO2
are produced from .80 moles of O2
?
3. 2H2
S + 3O2
2SO2
+ 2H2
O
How many moles of H2
S must react with .68 moles O2
?

7. PRACTICE - MOLE TO MOLE
2C2
H2
+ 5O2
4CO2
+ 2H2
O
1. How many moles of CO2
are produced when
14.3 moles of O2
are burned?
2. How many moles of water are produced when
.8432 moles of CO2
are produced?
3. How many moles of oxygen are needed to
react with 11.44 moles of C2
H2
?

8. Stoich Problems – Mole to Mass
Given moles – looking for grams
2H2
+ O2
2H2
O
How many grams of water are formed from 6.0 moles of H2
?
Step 1: Label the equation correctly.
Step 2: Write down the given then multiply by the mole ratio –
unknown on top.
Step 3: Multiply by molar mass of unknown to convert from
moles to grams.

9. MOLE TO MASS EXAMPLES
2H2
+ O2
2H2
O
1. How many grams of water are
produced when 4.84 mol hydrogen
reacts?
2. How many grams of oxygen are
need to react with 54.2 mol of
hydrogen?
3. How many grams of hydrogen are
needed to produce 1.634 mol of
water?

10. PRACTICE – MOLE TO MASS
Ca3
(PO4
)2
+ 3SiO2
+ 5C 3CaSiO3
+ 5CO +
2P
1. How many grams of SiO2
are needed to
produce 10.0 mol CO?
2. How many grams of carbon are needed to
react with 3.0 mol of Ca3
(PO4
)2
?
3. 42.0 mol of C produces how many grams of
CaSiO3
?

11. Stoich Problems – Mass to Mole
Given grams – looking for moles
2H2
+ O2
2H2
O
How many moles of water are formed from 16.0 grams of H2
?
Step 1: Label the equation correctly.
Step 2: Put down the given and divide by molar mass of the given to
get moles.
Step 3: Multiply by mole ratio – unknown on top – to get moles.

12. MASS TO MOLE EXAMPLES
2H2
+ O2
2H2
O
1. How many moles of water are
produced when 4.84 grams of hydrogen
reacts?
2. How many moles of oxygen are need
to react with 54.2 grams of hydrogen?
3. How many moles of hydrogen are
needed to produce 1.634 grams of water?

13. PRACTICE – MASS TO MOLE
Ca3
(PO4
)2
+ 3SiO2
+ 5C 3CaSiO3
+ 5CO +
2P
1. How many moles of SiO2
are needed to
produce 420 g of CO?
2. How many moles of CaSiO3
are produced
when 100.0 g of Ca3
(PO4
)2
are reacted?
3. How many moles of Ca3
(PO4
)2
are needed to
produce 280.0 g of CO?

14. Stoich Problems – Mass to Mass
Given grams – looking for grams
2H2
+ O2
2H2
O
How many grams of O2
are needed to react with 20.0 g of H2
?
Step 1: Label the equation.
Step 2: Write down the given and divide by the molar mass of
the given.
Step 3: Multiply by the mole ratio to find moles of unknown.
Step 4: Multiply by molar mass of unknown to find grams of
unknown.

15. MASS TO MASS EXAMPLES
2H2
+ O2
2H2
O
1. How many grams of water are
produced when 4.84 grams of hydrogen
reacts?
2. How many grams of oxygen are need
to react with 54.2 grams of hydrogen?
3. How many grams of hydrogen are
needed to produce 1.634 grams of water?

16. PRACTICE – MASS TO MASS
N2
+ 3H2
2NH3
1. How many grams of ammonia are formed
from 15.0 g N2
?
2. How many grams of H2
are needed to react
with 48.3 g N2
?
3. How many grams of H2
are needed to produce
.914 g NH3
?

17. UNDERSTANDINGS/KEY IDEA
1.3.A
Reactants can be either limiting or
excess.

18. APPLICATION/SKILLS
Be able to solve problems relating
to reacting quantities, limiting and
excess reactants, theoretical,
experimental and percentage
yields.

19. Limiting Reactant
The reactant that will run out first in a
reaction.
The excess reactant is the reactant that
is not used up completely in reaction.

20. Limiting Reactant
You will recognize a limiting reactant
problem because there will be 2 “givens”
in the problem.
2H2
+ O2
2H2
O
If 4 moles of H2
react with 8 moles of O2
,
how much water will be formed?

21. Limiting Reactant - Calculations
Step 1: Write down and convert both givens to
moles (if not already in moles).
Step 2: Set up two stoichiometric problems with the opposite
reactant as the mole ratio.
Step 3: Interpret the equations by crossing and comparing to
determine the limiting reactant.
4 mols H2 given
X = 2 mols O2needed
8 mols O2 given
X = 16 mols H2needed
1 mol O2
2 mol H2
2 mol H2
1 mol O2
You need 2 moles of O2
and you have 8 mols O2
given
You need 16 moles of H2
, but you were only given 4 moles H2
.
Since you do not have enough H2
given, it will run out first.
Therefore, H2
is the limiting reactant.
Step 4: Use the limiting reactant, calculate the answer.

22. 2H2(g)
+ O2(g)
🡪 2H2
O(l)
How many grams of water are formed when 12.0 g of
hydrogen reacts with 17.0 g of oxygen?
● First of all, I recognize that I am given an amount of hydrogen
and an amount of oxygen so I need to figure out which one is
going to run out first and stop the reaction. This one is called the
limiting reactant. The other reactant will be the excess reactant.
● 12.0 g H2
x mol = 5.94 mol H2
given
2.02g
● 17.0 g O2
x mol = .531 mol O2
given
32.0g
● Notice that my first step was to convert to moles and to actually
stop after the conversion to moles and write the word “given”.

23. 2H2(g)
+ O2(g)
🡪 2H2
O(l)
● The next step is to find out how much of each reactant is needed to
react with each other. You do this by multiplying each reactant by the
mole ratio with each other.
12.0 g H2
x mol = 5.94 mol H2
given x 1 mol O2
= 2.97 mol O2
needed
2.02g 2 mol H2
17.0 g O2
x mol = .531 mol O2
given x 2 mol H2
= 1.06 mol H2
needed
32.0g 1 mol O2
● Now cross compare to find the reactant that you do not have enough
of. You need 1.06 mol H2
and you are given 5.94 mol H2
so you have
more than enough. You need 2.97 mol O2
and are only given .531
mol O2
so you do not have nearly enough. This means that oxygen is
your limiting reactant; therefore, hydrogen is your excess reactant.

24. 2H2(g)
+ O2(g)
🡪 2H2
O(l)
To solve the problem, you use the limiting reactant.
(Important! Be sure to use the given moles, not the
needed moles.)
The problem asked you for the mass of water produced.
You now have a mole – mass problem.
.531 mol O2
x 2 mol H2
O x 18.02g = 19.1 g H2
O
1 mol O2
mol
The answer is your theoretical yield.
If you were asked how much excess reactant remained,
simply subtract the moles H2
needed by the moles given.
5.94 mol H2
given – 1.06 mol H2
needed = 4.88 mol H2
in
excess.

25. LIMITING REACTANT
EXAMPLE
If you have 6.70 mol Na reacting with 3.20
mol Cl2
, what is your limiting reactant, how
many moles of product will be formed and
how much excess reactant remains?
2Na + Cl2
2NaCl

26. UNDERSTANDINGS/KEY IDEA
1.3.B
The experimental yield can be
different from the theoretical yield.

27. Theoretical Yield –
the maximum amount of product that can
be produced from a given amount of
reactant (found by using stoichiometry).
Experimental Yield –
the measured amount of product
obtained from a reaction (what you got in
the lab)

28. Percent Yield

29. You will recognize this type of
problem when they ask you to find the
percent yield.
The experimental value will be given
in the problem and you have to
calculate the theoretical yield by using
stoichiometry.

30. EXAMPLE
2H2(g)
+ O2(g)
🡪 2H2
O(l)
If 17.0 g of oxygen reacts to form 18.7 g of
water, what is the percent yield?
● First of all, you recognize that this is a percent yield
problem. Circle 18.7 g and label it “experimental”. Next
work the problem to solve for the theoretical yield.
17.0 g O2
x mol x 2 mol H2
O x 18.02 g = 19.1 g
32.0 g 1 mol O2
mol
% yield = exp x 100% = 18.7g x 100% = 97.9%
theo 19.1g

31. UNDERSTANDINGS/KEY IDEA
1.3.C
Avogadro’s law enables the mole
ratio of reacting gases to be
determined from volumes of the
gases.

32. Avogadro’s hypothesis states that
equal volumes of different gases
contain equal numbers of particles at
the same temperature and pressure.
This means that if you are given gas
volumes and asked for gas volumes at
the same conditions, you can find your
answer using the mole ratios without
doing any conversions.

33. APPLICATION/SKILLS
Be able to calculate reacting
volumes of gases using Avogadro’s
law.

34. Example Problem
40 cm3
of carbon monoxide is reacted with 40
cm3
of oxygen in the following reaction.
2CO + O2
→ 2CO2
What volume of carbon dioxide is produced?
This is actually a limiting reactant “volumes of
gases” problem and is quite often assessed on
the IB exam.

35. 40 cm3
CO given x 1 mol O2
= 20 cm3
O2
needed
2 mol CO
40 cm3
O2
given x 2 mol CO = 80 cm3
CO needed
1 mol O2
Notice you can use the mole ratio directly
without converting cm3
to moles first.
You need 80 cm3
CO and only have 40cm3
so CO is the limiting reactant.
40 cm3
CO given x 2 mol CO2
= 40 cm3
CO2
produced
2 mol CO
Oxygen is in excess by 20 cm3
.

36. UNDERSTANDINGS/KEY IDEA
1.3.E
The molar concentration of a
solution is determined by the
amount of solute and the volume of
solution.

37. A solution is a homogeneous mixture of the
solvent with the solute.
● The solute is the substance being dissolved in the
solvent and it is usually the less abundant component.
● The solvent is the substance doing the dissolving. It
is usually a liquid and is the more abundant
component.
An aqueous solution is a solution with water as
the solvent.
Solutes can be solids, liquids or gases, but the
solvent is generally a liquid.

38. The concentration is the composition of the
solution expressed in moles or grams solute
over volume of the total solution.
When no more solute can be dissolved in a
solution, the solution is saturated.
Square brackets [ ] are used to represent
concentration or molarity.
Concentration [ ] = moles of solute/volume of
solution
The unit of volume in [conc] is usually dm3
.
***To solve for moles, multiply the concentration
by the volume given. This is a very common
calculation.***

39. If you are given a volume of cm3
and
asked to find moles with a concentration
using mol/dm3
, you must divide by 1000.
1 dm3
= 1000 cm3

40. GUIDANCE
Units of concentration include:
g/dm3
, mol/dm3
and parts per
million (ppm).

41. GUIDANCE
Use square brackets to denote
molar concentration.

42. UNDERSTANDINGS/KEY IDEA
1.3.F
A standard solution is one of known
concentration.

43. UNDERSTANDINGS/KEY IDEA
Strong electrolytes are assumed to
completely break down into ions in
solution.

44. ELECTROLYTES
Electrolytes are substances that when in
solution break into ions and conduct
electricity due to the presence of said ions.
All ionic compounds are electrolytes
meaning that when in solution, they
dissociate into the ions making up the
compound.
Strong acids and strong bases dissociate
completely into their ions.

45. APPLICATION/SKILLS
Be able to determine the ions
produced by an electrolyte in
aqueous solution and calculate the
concentration of those ions.

46. Sample Problems
1.0 mol/dm3
HCl breaks into 1.0 mol/dm3
H+
ions and 1.0 mol/dm3
Cl-
ions
2.0 mol/dm3
Ba(OH)2
breaks into 2.0
mol/dm3
Ba2+
ions and 4.0 mol/dm3
OH-
ions
You should be able to break any ionic
compound into its ions and be able to
determine the concentration of said ions by
the original concentration of the ionic cmpd.

47. APPLICATION/SKILLS
Be able to solve problems involving
molar concentration, amount of
solute and volume of solution.

48. EXAMPLE
Calculate the mass of copper II sulfate
pentahydrate, CuSO4
.
5H2
0, required to prepare
500 cm3
of a 0.400 mol/dm3
solution.
● First of all, they are asking for mass so you know you
need to solve for moles.
● You can solve for moles by multiplying volume by the
concentration – watch for units.
n = 500 cm3
x 1 dm3
/1000cm3
x 0.400 mol/dm3
= .200 mol
● To solve for mass when you have moles, multiply by the molar
mass.
Mass = .200 mol x 249.61g/mol = 49.9 g

49. APPLICATION/SKILLS
Be able to use the experimental
method of titration to calculate the
concentration of a solution by
reference to a standard solution.

50. https://learn360.infobase.com/p_Search.a
spx?bc=0&rd=a&q=titration
Standard solutions are used to find
concentrations of other solutions.
The volumetric technique “titration” is most
commonly used to find the concentration
of an unknown solution.

51. EXAMPLE
What volume of 2.00 mol/dm3
hydrochloric acid would have to be added to
25.0 cm3
of a 0.500 mol/dm3
sodium carbonate solution to produce a
neutral solution of sodium chloride?
● First you need a balanced equation:
2HCl + Na2
CO3
🡪 2NaCl + H2
O + CO2
● Next find moles of sodium carbonate:
25.0cm3
x 1dm3
/1000cm3
x 0.500mol/dm3
= .0125mol
● Use the mole ratio to find moles of HCl needed:
.0125 mol Na2
CO3
x 2 mol HCl = .0250 mol HCl
1 mol Na2
CO3
● Use the concentration equation to solve for volume since you have
moles and concentration.
[Conc] = mol/volume so volume = mol/[conc]
volume = .0250mol/2.00mol/dm3
= .0125dm3
x 1000cm3
/dm3
= 12.5cm3

52. EXAMPLE
Calculate the volume of carbon dioxide produced at
STP when 1.00g of calcium carbonate reacts with
25.0cm3
of 2.00 mol/dm3
hydrochloric acid.
CaCO3
+ 2HCl 🡪 CaCl2
+ H2
O + CO2
● Wow! Another limiting reactant problem with mixed units
given. Both givens have to be converted to moles:
1.00gCaCO3
x mol/100.09g = 0.0100mol CaCO3
given
25.0cm3
x 1dm3
/1000cm3
x 2.00mol/dm3
= 0.0500mol HCl
0.0100mol CaCO3
given x 2 mol HCl = 0.0200 mol HCl needed
1molCaCO3
0.0500mol HCl given x 1 mol CaCO3
= 0.025mol CaCO3
needed
2 mol HCl

53. You need 0.0250 mol CaCO3
and are only given 0.0100
mol so this is your limiting reactant.
0.0100 mol CaCO3
x 1 mol CO2
= 0.0100 mol CO2
1mol CaCO3
0.0100 mol CO2
x 22.4 dm3
/mol = .224 dm3
.224 dm3
x 1000 cm3
/dm3
= 224 cm3

54. Citations
International Baccalaureate Organization. Chemistry Guide,
First assessment 2016. Updated 2015.
Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd
ed. N.p.: Pearson Baccalaureate, 2014. Print.
ISBN 978 1 447 95975 5
eBook 978 1 447 95976 2
Most of the information found in this power point comes
directly from this textbook.
The power point has been made to directly complement the
Higher Level Chemistry textbook by Brown and Ford and is
used for direct instructional purposes only.