1. Chemical Formulas and equations as recipes
Stoichiometry
Chemical Formulas and equations as recipes

2. Section 1
“Stoichiometry”: quant-itative information available from chemical formulas and equations
Chemical Formulas show number composition of chemical compounds.
Chemical Equations are recipes for reactions…if you know how to read them.

3. Section 1
“Stoichiometry”: quant-itative information available from chemical formulas and equations
Ex: Hydrogen is a flammable gas. It reacts with oxygen in the air to produce water vapor.
If we react 4 grams of hydrogen,
how many grams of oxygen would react?
And how much water would be produced?

4. Step 1 Generate the ‘recipe’ using chemical formulas:
Ex: Hydrogen is a combustible gas. It reacts with oxygen in the air to produce water vapor.
If we reacted 4 grams of hydrogen,
how many grams of oxygen would react?
And how much water would be produced?
Step 1 Generate the ‘recipe’
using chemical formulas:
hydrogen + oxygen  water
H2 + O2  H2O
This is called
a chemical “equation”

5. Ex: H2 + O2  H2O Notice that our equation is an atom/molecule recipe?
Step 2: Balance the Equation
Ex: H2 + O2  H2O
Notice that our equation is an atom/molecule recipe?

6. Step 2: Balance the Equation
Ex: H2 + O2  H2O
2 O atoms vs. 1 O atom?
to start afterward
Lets make two waters
Now 2 O atoms and 2 O atoms
to start afterward
Since no atoms can be created or destroyed, we must first balance our recipe.

7. Ex: H2 + O2  H2O Step 2: Balance the Equation
Now, 2 H atoms vs H atoms?
to start afterward
Lets start with two H2 molecules
Now, 4 H atoms and H atoms
to start afterward

8. To finish we put in the coefficients
Step 2: Balance the Equation
Ex: H2 + O2  H2O 2 2
To finish we put in the coefficients
and the equation (recipe) is balanced!

9. Now, In order to weigh out chemicals in the lab we need to interpret our atom/molecule recipe into a mass recipe.
Ex: H2 + O2  H2O 2 2
H2 = 2 (1 amu )
2 H2 = 2(2) = 4 amu
4 grams H2
We find the mass ratios using the atomic masses:
atomic
mass 

10. Now, In order to weigh out chemicals in the lab we need to interpret our atom/molecule recipe into a mass recipe.
Ex: H2 + O2  H2O 2
4 grams H2
+ 32 grams O2
O2 = 2(16 amu)
O2 = 32 amu
32 grams O2
atomic
mass 

11. Now, In order to weigh out chemicals in the lab we need to interpret our atom/molecule recipe into a mass recipe.
Ex: H2 + O2  H2O 2
4 grams H2
+ 32 grams O2
 36 grams H2O
H2O = 2(1) + 16
2 H2O = 2(18 amu) = 36 amu
36 grams H2O

12. So our recipe tells us to weigh out 4 mass units of Hydrogen
Now, In order to weigh out chemicals in the lab we need to interpret our atom/molecule recipe into a mass recipe.
Ex: H2 + O2  H2O 2 2
4 grams H2
+ 32 grams O2
 36 grams H2O
So our recipe tells us to weigh out
4 mass units of Hydrogen
+ 32 mass units Oxygen
to creates 36 mass units of H2O

13. Ex: H2 + O2  H2O 2 2
4 grams H g O2 = g H2O
Is mass conserved?
Notice the use of “gram” atomic masses?
A gram atomic mass is the atomic mass of an element with its units as grams

14. Notice the use of “gram” atomic masses?
Ex: H2 + O2  H2O 2 2
4 grams H g O2 = g H2O
Is mass conserved?
Notice the use of “gram” atomic masses?
This creates a chemical unit called the “mole”
A mole is a “gross” quantity used in chemistry

15. Quantities of Measure Quantities can be represented by
mass (weight) ex: 3 lbs of coffee
volume ½ gallon of milk
number 24 eggs
gross units 2 dozen (2 units of 12)
Gross units are developed for convenience of grouping small items into larger packages

16. Chemical Quantities Chemical quantities can be represented by
mass grams
volume liters or milliliters
number particles (atoms, molecules, etc.)
Atoms and molecules of course are so tiny that it is impossible to count them out.

17. Chemical Quantities Chemical quantities can be represented by
mass grams
volume liters or milliliters
number particles (atoms, molecules, etc.)
gross units moles of particles
(gram-relative masses)
Moles are the gross unit used for convenience in chemistry = 6 x 1023 particles
This is called Avogadro’s number
One mole contains Avogadro's number of particles

18. A.k.a. the gram formula mass
1 Helium atom
Weights 4 amu’s
1 mole He atoms
(6 x atoms)
Weighs 4 grams
Notice?
Gram relative masses:
1 mole = 6.02 x 1023
A.k.a. the gram formula mass

19. A.k.a. the gram formula mass
1 mole S atoms
(6 x atoms)
32 grams
1 Sulfur atom
32 amu’s
Notice?
Gram relative masses:
1 mole = 6.02 x 1023
A.k.a. the gram formula mass

20. A.k.a. the gram formula mass
1 mole H2 molecules
(6 x molecules)
2 grams
1 Hydrogen
molecule
2 amu’s
Notice?
Gram relative masses:
1 mole = 6.02 x 1023
A.k.a. the gram formula mass

21. A.k.a. the gram formula mass
1 water molecule
18 amu’s
1 mole H2O molecules
(6 x molecules)
18 grams
Notice?
Gram relative masses:
1 mole = 6.02 x 1023
A.k.a. the gram formula mass

22. Formula Mass: the mass of one
Atomic mass: Mass of one atom
ex: H = 1.0 C = 12.0
Na = Cl = 35.5
Molecular mass or Formula mass (ionic)
ex: H2 = 2(1.0) NaCl =
= =
Ex: CuSO4 = (16.0)
64.0
= 159.5

23. Practice problems
Calculate the formula mass of each:
HNO3
2. Cu(NO3)2

24. Practice problems = 63  = 188  Calculate the formula mass of each:
HNO3
2. Cu(NO3)2
= 63 
= 188 

25. Gram-formula mass: The mass of one Mole
1 gram = 6 x 1023 atomic mass units
H = So, 6 x 1023 H atoms = 6 x 1023 amu = 1 g
He = x 1023 He atoms = 4(6 x 1023) = 4 g
H2 = x 1023 H2 molecules = 2(6 x 1023) = 2 g
H2O = 8 6 x 1023 H2 molecules = 18 g

26. Gram-formula mass: The mass of one Mole
1 gram = 6 x 1023 atomic mass units
H = So, 6 x 1023 H atoms = 6 x 1023 amu = 1 g
He = x 1023 He atoms = 4(6 x 1023) = 4 g
H2 = x 1023 H2 molecules = 2(6 x 1023) = 2 g
H2O = 2(1) + 16 = 18 “ g
Chemicals are packaged into units called a “mole”
1 mole = 6 x 1023 particles (atoms, molecules, etc.)
1 mole = same mass value as one particle, but unit is … grams

27. Gram-formula mass: The mass of one Mole
1 gram = 6 x 1023 atomic mass units
H = So, 6 x 1023 H atoms = 6 x 1023 amu = 1 g
He = x 1023 He atoms = 4(6 x 1023) = 4 g
H2 = x 1023 H2 molecules = 2(6 x 1023) = 2 g
H2O = 2(1) + 16 = 18 “ g
Formula mass: mass of a formula, the mass of one
Gram formula mass: mass of a formula in grams
the mass of one mole
mass of 6 x = one mole

28. “Simple” mole problems32
What is the atomic mass of sulfur?
…the gram atomic mass?
So, What is the mass of one mole of S (molar mass)?
How many atoms are in one mole of sulfur?
How many atoms are in 32 g of S?
How many atoms are in 64 g?
How many moles is 64 g of S?
How many moles is 16 grams of S
How many atoms in 16 g of S?
What is the mass of 18 x 1023 S atoms?
32 g
32 g
6 x 1023 atoms
6 x 1023 atoms
Finish these:

29. More “simple” mole problems
What is the formula mass of H2O?
the gram formula mass?
So, What is the mass of one mole of H2O?
How many molecules are in 18 g of H2O?
How many molecules are in 36 g?
How many moles is 36 g of H2O?
How many moles is 9 grams of H2O
What is the mass of 18 x 1023 H2O molecules?
2(1) + 16 = 18
Now you’re thinking in moles!
18 g
18 g
Finish these

30. 45 grams _______ 18 g/mol More complex problems:
How many molecules are in 45 grams of water?
H2O = 18 grams per mole
So 45 grams is two and some fraction more of a mole?
45 grams
_______
18 g/mol

31. Notice that moles are the bridge between all the units in chemistry
ROADMAP
Notice that moles are the bridge between all the units in chemistry
= 2.5 moles
X 6x1023
molecules/ mol
45 grams
= 15 x 10 23
molecules
__________
18 g/mol

32. Mole conversion map multiply divide
 Notice: 1 mole of a gas = 22.4 liters
at STP (Standard Temperature and Pressure)
multiply
divide

33. Conversions (dimensional analysis)
1 mole = 6 x 1023 Particles
1 mole = gram formula mass
1 mole = 22.4 liter of gas (at STP)
What is the mass of 2.5 moles of H2O?
Unit Plan:
moles  grams
1 mole H2O = gram formula mass = 2(1) + 16 = 18 grams 18
_________ 1
2.5 moles
x ___gram
mole
= 45 grams 1
Starting Unit on bottom
to cancel out
Ending unit on top

34. problems What is the mass of 3.5 moles of Cu?
How many moles are 16.8 g of CO2 gas ?(at STP)
3. How many moles are in 66 grams of CO2?
3.5 moles Cu
x 63.5 grams Cu
1 mole Cu
= 220 grams Cu
= grams 
= 1.5 moles 

35. Try these: (write the setups and solve)
He = 4 g/mol
How many moles is 32 g of He?
What is the mass of moles of O2 gas?
What is the volume of moles of O2 gas? (at STP)
What is the mass of 67.2 liters of O2 gas? (at STP)
32 g x 1 mol
4 g
= 8 moles
= 0.8 grams 
= 0.56 liters 
= 96 grams 

40. Section 2
Stoichiometry II

41. Gas Density
22.4 liters = 1 mole = molar mass (gram formula mass)
What is the density of H2 gas (at STP) in g/L?
Density = Mass
Volume
g/L
Density = grams = liters

42. 1) What is the mass of 50 liters of H2 gas at STP?
H2 gas density = g/L
grams = 1 liter
50 liters H2
grams
x ____________
= 4.5 grams H2
1 liter H2
Or D = M M = DV V
M = ( g/L)(50 L)
M =4.465 g

43. 2) What is the molar mass of a gas with a density of 1.43 g/L?
(Hint: what is the mass of 22.4 liters?)
1.43 grams
1 liter
22.4
x liters
mole
= 32 g/mole 1
Or D = M M = DV V
M = (1.4 g/L)(22.4 L)

46. Percent-age (per hundred) composition
Composition of a compound as the Part of out of a the whole
Ex: How can we express the composition of SO2 gas?
What part is Sulfur compared to the whole SO2?
What fraction?
1 mole SO2 = 32 g s + 2(16)= 32 g of O = 64g total
Sulfur = 32 g out of 64 g = 0.50 (half),
Sulfur = 0.50 x 100 = 50 parts per hundred or 50 per-”cent” (50%)

47. From chemical formulas:
what is the % composition of H2O?
1 mole H2O = 2(1.00g ) = 18.0 grams total
% = Part x 100
whole
% H = 2.00 g x 100 % O = g x 100
18.0 g g
% H = 11.1 % % O = 88.9 %

48. Try some: 5.9%H, 94.1%O 27.3%C, 72.7%O 82.3%N, 17.6%H
Calculate the % composition of the following:
H2O2
CO2
NH3
5.9%H, 94.1%O
27.3%C, 72.7%O
82.3%N, 17.6%H

49. Percents What percent is Oxygen? 100% – 11.1% = 88.9% O
Percentage = Part x 100
whole
From mass data:
Ex: What is the percent composition of water if 200. grams of water decomposes into 22.2 grams of hydrogen and 178 grams of oxygen?
% = Part x 100
% = g H x 100
200. g total
%= x 100
% = 11.1 % H
What percent is Oxygen?
100% – 11.1% = 88.9% O

50. Try one:
150 grams of potassium chlorate is heated, giving off oxygen gas. 91 grams of potassium chlorate remains.
What is the mass of the oxygen gas released?
What percentage of the original potassium chlorate is oxygen?
59 g
39 %

51. Part = % x whole 100 Part from percentage and whole:
How many grams of hydrogen are in 150. grams of water?
(water is 11.1% Hydrogen)
% = Part x 100
whole
X = % x g water
100
X = 16.7 grams H
Part = % x whole
100

52. Try this:
Copper sulfate hydrate is 36% water. What mass of water will be lost by heating 2.50 grams to the hydrate, to drive off the water?
0.90 g

56. Empirical and molecular formulas
What is the empirical formula for each molecular formula?
C2H8
Na2O2
C12H22O11
Give a possible molecular formula for each empirical formula: HO
CH3
NH2

57. Calculating empirical (ratio) formulas
 Find the number ratio (moles) from mass data
Problem 1: What is the empirical formula of a compound which contains 0.56 grams of Hydrogen and 4.44 grams of oxygen?
Solution: convert masses into moles
0.56 g g O
1 g/mol g/mol
0.56 mol H mol O
Divide larger by smaller / = 2.02
2:1 ratio
H2O = H2O

58. Problem 2: What is the empirical formula of a compound which is 52
Problem 2: What is the empirical formula of a compound which is 52.9% Al and 47.1% O?
Solution: convert percent (mass) into moles
52.9 g Al x 1 mole 47.1 g O x 1 mol
27 g g
1.96 mol Al mol O
2.94 / = : 1.5 ratio?
= Al1 O1.5
= Al2O3

59. Problem 3: What is the name of the compound which has 0
Problem 3: What is the name of the compound which has 0.35 moles of tin combined with 1.40 moles of chlorine? Solution?
Find the empirical formula, then name it.
SnCl4
Tin (IV) Chloride

62. Molecular formulas From empirical formula
and the Molar (molecular) mass
Problem: What is the empirical formula of a compound with 5.88% H and 94.1% O?
In chemistry we think in two worlds:
Big/outside world  measurements are by mass
weight/ mass / percent's

63. Molecular formulas From empirical formula
and the Molar (molecular) mass
Problem: What is the empirical formula of a compound with 5.88% H and 94.1% O?
The Atomic/particle world
 measurements are by number
Atom ratios / chemical formulas ex: H2O
…and Moles!

64. Molecular formulas Which world are we in now? From empirical formula
and the Molar (molecular) mass
Problem: What is the empirical formula of a compound with 5.88% H and 94.1% O?
Which world are we in now?

65. Molecular formulas
Problem: What is the empirical formula of a compound with
5.88% H and 94.1% O?
Step 1: Big world to particle world
Convert percent's (grams) to moles:
5.88 g H x 1 mol = mol H g O x 1 mol = mol O
1 g g
Empirical formula = HO

66. Molecular formulas
Problem: What is the empirical formula of a compound with
5.88% H and 94.1% O?
Empirical formula = HO
Step 2: Change from empirical (lowest ratio) formula
to Molecular (actual) formula
Calculate the EFM: empirical formula mass
HO = = 17

67. Molecular formulas
Empirical formula = HO
Step 2: Change from empirical (lowest ratio) formula
to Molecular (actual) formula
Calculate the EFM: empirical formula mass
34 / 17 = 2
Multiply the empirical formula: 2 (HO)
Molecular formula = H2O2

68. Find these molecular formulas
Empirical formula = CH2O
molar mass = 180 g/mol
Empirical formula = C3H8
molar mass = 44 g/mol
C6H12O6
C3H8

69. 1) Calculate its empirical formula first:
Problem: Octene is an organic monomer used in the production of polyethylene plastic. Analysis show octene is 85.7% C and 14.3%H. Its molar mass is 112.
1) Calculate its empirical formula first:
85.7 g C / 12.0 g/mol = mol C
14.3 g H / 1.0 g/mol = mol H
14.3 / 7.14 = 2 1:2 ratio = CH2

70. 2) Calculate its molecular formula second:
Problem: Octene is an organic monomer used in the production of polyethylene plastic. Analysis show octene is 85.7% C and 14.3%H. Its molar mass is 112.
2) Calculate its molecular formula second:
Molecular formula:
EFM: CH2 = (1) = 14
Molar mass / EFM  112 / 14 = 8 (molecule is 8 x larger)
8 x CH2 = C8H16

72. Regents questions