The document provides a comprehensive overview of key concepts in stoichiometry relevant to industrial technology, including atomic mass, the mole concept, percent composition, and chemical reactions. It explains how to calculate average atomic mass, percent composition of compounds, and empirical and molecular formulas through various examples. The content also covers chemical equations and their balancing, along with mass relationships in chemical reactions.

1. STOICHIOMETRY
CHEMISTRY FOR INDUSTRIAL TECHNOLOGIST

2. a. Atomic mass
b. The Mole Concept and Molar Mass
c. Percent Composition and Chemical
Formulas
d. Chemical Reactions and Chemical
Equations
e. Mass Relationships in Chemical
Reactions

3. Atomic Mass

4. Ms. Lilia sells shelled peanuts in a
store. But she meets customers asking
for 150 peanuts, another for 750
peanuts, and another for 2,000
peanuts. Obviously, it will take Ms.
Lilia a very long time to count the
peanuts. What would be another way
to count them?

5. Ms. Lilia takes 20 peanuts and weighs them.
She finds out that 20 peanuts weigh 32 g.
How much then will each peanut weigh?
32 𝑔
20 𝑝𝑒𝑎𝑛𝑢𝑡𝑠
=
1.6 𝑔
𝑝𝑒𝑎𝑛𝑢𝑡𝑠
Hence the weight of 150 peanuts would be:
1.6 𝑔
𝑝𝑒𝑎𝑛𝑢𝑡𝑠
x 150 peanuts = 240 g
It will be easier to weigh the peanuts than to
count them.

6. Now, 960 g is appropriately
how many peanuts?
150 peanuts x
1 𝑝𝑒𝑎𝑛𝑢𝑡
1.6 𝑔
= 600 peanuts
Is it possible to count objects
by weighing?

7. Whether it is peanuts or mongo beans or
candies or atoms, the procedure should
be the same. The problem, however, is
atoms are very, very small and it is not
possible to see them and count them
individually to get the average mass. We
need to look for another way to get the
average mass of the atom.

8. What is average atomic mass?
Average Atomic Mass
Average atomic mass is the weighted
average of the atomic masses of the
naturally occurring isotopes of an element.

9. Example
Calculating a Weighted Average
A box contains two size of marbles. If
25.0% have masses of 2.00 g and 75.0%
have masses of 3.00 g. What is the
weighted average?
(.250)(2.00)+(.750)(3.00)=.500+2.25= 2.75g

10. Example
Calculating a Average Atomic Mass
Boron has two isotopes:
B-10 (mass 10.013 amu) 19.8% abundance
B-11 (mass 11.009 amu) 80.2% abundance
Calculate the average atomic mass.
(.198)(10.013)+(.802)(11.009) =
1.98 amu + 8.83 amu = 10.81 amu

11. Calculate the average atomic mass of Mg
Calculating a Average Atomic Mass
Isotope 1 – 23.985 amu (78.99%)
Isotope 2 – 24.986 amu (10.00%)
Isotope 3 – 25.982 amu (11.01%)
(23.985)(.7899)+(24.986)(.1000)+(25.982)
(.1101)
18.95+2.499+2.861 = 24.31 amu

12. The Mole
Concept and
Molar Mass

13. Percent
Composition and
Chemical Formula

14. the percent by mass of each element in a compound
Percent Composition
Percent =
Part
Whole
x 100%
=
Mass of element in 1 mol
Mass of 1 mol
x 100%
Percent
composition of a
compound or
molecule

15. Example
What is the percent composition of Potassium
permanganate (KMnO₄)?
Percent Composition
Molar Mass of KMnO₄
K = 1(39.1) = 39.1
Mn = 1(54.9) = 54.9
O = 4(16.0) = 64.0
MM = 158 g/mol

16. Determine the percentage composition of sodium
carbonate (Na₂CO₃)?
Percent Composition
Molar Mass of Na₂CO₃
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) = 48.0
MM = 106 g/mol
Percent Composition
% Na =
46.0 g
106 g
x 100% =43.4 %
% C =
12.0 g
106 g
x 100% = 11.3 %
% O = 48.0 g
106 g
x 100% =45.3 %

17. Determine the percentage composition of
ethanol (C2H5OH)?
Percent Composition
Determine the percentage composition of
sodium oxalate (Na2C2O4)?

18. Calculate the mass of bromine in 50.0 g of Potassium
bromide.
Percent Composition
1. Molar Mass of KBr
K = 1(39.10) = 39.10
Br =1(79.90) =79.90
MM = 119.0
79.90 g
___________
119.0 g
= 0.6714
3. 0.6714 x 50.0g = 33.6 g Br
2.

19. Calculate the mass of nitrogen in 85.0 mg of the amino acid
lysine, C6H14N2O2.
1. Molar Mass of C6H14N2O2
C = 6(12.01) = 72.06
H =14(1.01) = 14.14
MM = 146.2
28.02 g
___________
146.2 g
= 0.192
3. 0.192 x 85.0 mg = 16.3 mg N
2.
N = 2(14.01) = 28.02
O = 2(16.00) = 32.00
Percent Composition

20. Percent composition allow you to calculate the simplest
ratio among the atoms found in compound.
Empirical Formula – formula of a compound that expresses
lowest whole number ratio of atoms.
Molecular Formula – actual formula of a compound
showing the number of atoms present
Formulas
Examples:
C4H10 - molecular
C2H5 - empirical
C6H12O6 - molecular
CH2O - empirical

21. An oxide of aluminum is formed by the reaction of 4.151 g of
aluminum with 3.692 g of oxygen. Calculate the empirical formula.
Calculating Empirical Formula
1. Determine the number of grams of each element in the
compound.
4.151 g Al and 3.692 g O
2. Convert masses to moles.
4.151 g Al 1 mol Al
26.98 g Al
= 0.1539 mol Al
= 0.2308 mol O
3.692 g O 1 mol O
16.00 g O

22. Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of 4.151 g of
aluminum with 3.692 g of oxygen. Calculate the empirical formula.
3. Find ratio by dividing each element by smallest amount of moles.
0.1539 moles Al
0.1539
= 1.000 mol Al
0.2308 moles O
0.1539
= 1.500 mol O
4. Multiply by common factor to get whole number. (cannot have
fractions of atoms in compounds)
O = 1.500 x 2 = 3
Al = 1.000 x 2 = 2
therefore, Al2O3

23. Calculating Empirical Formula
When a 2.000 g sample of iron metal is heated in air, it reacts with
oxygen to achieve a final mass of 2.573 g. Determine the empirical
formula.
2.000 g Fe 1 mol Fe
55.85 g Fe
= 0.03581 mol Fe
0.573 g O 1 mol O
16.00 g
= 0.03581 mol Fe
Fe = 2.000 g O = 2.573 g – 2.000 g = 0.5730 g
1 : 1
FeO

24. Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato
beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of
arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for
lead arsenate.
1.3813 g Pb 1 mol Pb
207.2 g Pb
= 0.006667 mol Pb
0.00672 g H 1 mol H
1.008 g H
= 0.00667 mol H
0.4995 g As 1 mol As
74.92 g As
= 0.006667 mol As
0.4267g Fe 1 mol O
16.00 g O
= 0.02667 mol O

25. Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato
beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of
arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for
lead arsenate.
0.006667 mol Pb
0.00667 mol H
0.006667 mol As
0.02667 mol O
0.006667
0.006667
0.006667
0.006667
= 1.000 mol Pb
= 1.00 mol H
= 1.000 mol As
= 4.000 mol O
PbHAsO4

26. Calculating Molecular Formula
A white powder is analyzed and found to have an empirical formula
of P2O5. The compound has a molar mass of 283.88g. What is the
compound’s molecular formula?
Step 1: Molar Mass
P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g = 80.00 g
141.94 g
Step 2: Divide MM by
Empirical Formula Mass
238.88 g
141.94g
= 2
Step 3: Multiply
(P2O5)2 =
P4O10

27. Calculating Molecular Formula
A compound has an experimental molar mass of 78 g/mol. Its
empirical formula is CH. What is its molecular formula?
C = 12.01 g
H = 1.01 g
13.01 g
78 g/mol
13.01 g/mol
= 6
(CH)6 =
C6H6

28. CHEMICAL
REACTION AND
EQUATION

29. 3 ways of representing the reaction of H2 with O2 to form H2O
A process in which one or more substances is changed into one
or more new substances is a chemical reaction
A chemical equation uses chemical symbols to show what
happens during a chemical reaction
reactants products

30. Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left
side and the correct formula(s) for the product(s) on the
right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. Change the numbers in front of the formulas (coefficients)
to make the number of atoms of each element the same on
both sides of the equation. Do not change the subscripts.
2C2H6 NOT C4H12

31. Balancing Chemical Equations
3. Start by balancing those elements that appear in only
one reactant and one product.
C2H6 + O2 CO2 + H2O start with C or H but not O
2 carbon
on left
1 carbon
on right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O

32. Balancing Chemical Equations
4. Balance those elements that appear in two or more
reactants or products.
2 oxygen
on left
4 oxygen
(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen
(3x1)
multiply O2 by
7
2
= 7 oxygen
on right
C2H6 + O2 2CO2 + 3H2O
7
2
remove fraction
multiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O

33. Balancing Chemical Equations
5. Check to make sure that you have the same number
of each type of atom on both sides of the equation.
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C
12 H
14 O
4 C
12 H
14 O
4 C (2 x 2) 4 C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)

34. MASS
RELATIONSHIPS IN
CHEMICAL
REACTIONS

35. Mass Changes in Chemical Reactions
1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to calculate the number
of moles of the sought quantity
4. Convert moles of sought quantity into desired units

36. Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion,
what mass of water is produced?
grams CH3OH moles CH3OH moles H2O grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
209 g CH3OH 1 mol CH3OH
32.0 g CH3OH
x 4 mol H2O
2 mol CH3OH
x
18.0 g H2O
1 mol H2O
x =
235 g H2O