1. The Mole—Quantifying Equations
Stoichiometry
The Mole—Quantifying Equations

2. Atomic Mass
The mass of a single atom is far too small in grams to use conveniently.
Chemists use the unit called the unified atomic mass unit (amu) or Dalton (Da).
Definition of amu is exactly 1/12 the mass of an atom of 12C
Amu (Da) = x 10-24g

3. Atomic Mass Unit Mass of one 12C atom = 12 amu (exactly)
1 amu approximates the mass of one proton or neutron.
Mass of electron is neglible in comparison.
Particle
Mass
Charge
grams
amu
coulombs e
Electron
x 10-28
x 10-4
x 10-19 -1
Proton
x 10-24
x 10-19 1
Neutron
x 10-24

4. Mass Number Mass Number (A) = Z + N
Elements differ in the number of protons in their atoms.
The atomic number Z
All atoms of a given element have the same number of protons.
Number of electrons equals protons.
Number of neutrons = N
Mass Number (A) = Z + N
Mass number is the total number of nucleons.

5. Mass for Specific Elements
Why do all element not have atomic mass number listed in the periodic table that is not a whole number or very close to it?
Are all atoms of an element the same?

6. Isotopes
Isotopes are atoms with the same atomic number, but different mass number.
The larger mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12).  This explains how Lithium can have an atomic mass of Da. 

7. Calculating Average Atomic Mass
The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature, therefore, more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as:
m(E) = ∑(m(In)*p(In)) 
where ∑ represents a n-times summation over all isotopes In of element E, and p(I) represents the relative abundance of the isotope I.

8. Atomic mass based on isotopes
Find the average atomic mass of Boron
Mass and abundance of Boron isotopes
n  isotope In    mass m (Da) isotopic abundance p
1 10B
2 11B
Solution: The average mass of Boron is:
m(B) = ( Da)(.199) + ( Da)(.801) = 1.99 Da Da = Da

9. Molecular Mass
Molecular mass: sum of atomic masses of all atoms in a molecule
Formula mass: sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.

10. Mole Puns can you dig it?
What is a mole's favorite movie?
The Green Mole
What do you get when you have a bunch of moles acting like idiots?
Moleasses
What line from Shakespeare do high school moles have to memorize?
“To mole or not to mole, that is the question.”
How much does Avogadro exaggerate?
He makes mountains out of molehills.
What element do moles love to study in chemistry?
Moleybdenum

11. Calculate the Molecular Mass
Copper (II) Nitrate Cu(NO3)2
[(14 + {3 x 16}) x 2] = 187.5g
Ca3(PO4)2
3 moles of Ca
2 moles of P
2 x 4 moles of O.
1 mole of Ca is 40.08g, so 3 moles are    g
1 mole of P is g, so 2 moles are g
1 mole of O is g, so 8 moles are g
1 mole of Ca3(PO4)2 is    g

12. Molecular Mass Ratio
From a balanced equation, the coefficients define the ratio of reactants needed for the products that result from the reaction.
Counting atoms is impractical.
Use a mass ratio:
to predict mass of products in ideal conditions.
to calculate the percentage yield in actual conditions.
to obtain the mass of each reactant needed.

13. Example C2H4 + HCl C2H5Cl 1 : 1 yields 1 for ratio of molecules
Balanced equation:
C2H4 + HCl C2H5Cl
1 : yields for ratio of molecules
: yields for mass ratio
Ethylene: Atomic mass of 2C = 2 x 12.0amu = 24.0amu
Atomic mass of 4H = 4 x 1.0amu = 4.0amu
Molecular mass of C2H = 28.0amu
Hydrogen chloride: at. mass of H = 1.0amu
at. Mass of Cl = 35.5amu
Molecular mass of HCl = 36.5amu
Ethyl chloride: at. mass of 2C = 2 x 12.0amu = 24.0amu
at. mass of 5H = 5 x 1.0amu = 5.0amu
at. mass of Cl = 35.5amu = 35.5amu
Molecular mass of C2H5Cl = 64.5amu

14. Try these: 1. sodium fluoride 2. potassium hydroxide
3. copper (I) chloride
4. manganese (IV) oxide
5. calcium sulfate
6. magnesium phosphate

15. Amadeo Avogadro
Amadeo Avogadro was an Italian physics professor who proposed in 1811 that equal volumes of different gases at the same temperature contain equal numbers of molecules.

16. History—or how did they find that number???
If Avogadro’s hypothesis is true, then atomic weights for gases can be derived by weighing equal volumes of different gases (Cannizzaro).
Johan Loschmidt (HS teacher) took the idea and calculated the size of a molecule of air. He developed an estimate for the number of molecules in a given volume of air.
These three ideas together lead to the number named for Avogadro. Loschmidt was the first to calculate this number.

17. Avogadro’s Number 1 mole of a substance, NA = 6.02214179(30)×1023
is known as the Avogadro constant.
For calculations please use 6.02 x 1023

18. Mole is just a number
Definition: A mole is the amount of substance that contains as many elementary particles as there are atoms in exactly 12 grams of carbon-12 (12C).
1 Mole = x 1023particles
(atoms, molecules, ions, electrons, apples, wads of gum, elephants)
= NA particles
~100 million x 100 million x 100 million

19. Conversion factors 6.022045 x 1023 whatever kind of particles per mole
One mole of common substances:
CaCO3 :100.09g
Oxygen: 32.00g
Copper: 63.55g
Water:18.02g

20. 2 H2 (g) + O2 (g) → 2 H2O(g) 2 dozen H2 molecules react with exactly
1 dozen O2 molecules to give exactly
2 dozen H2O molecules.
2 moles of H2 molecules react with exactly
1 mole of O2 molecules to give exactly
2 moles of H2O molecules.
Why do we do this? Because these last sizes are
in the gram range and easy to weigh.
Conventions: 1 mole of 12C atoms weighs 12 g exactly.
1 atom of 12C weighs 12 amu exactly.
(amu = atomic mass unit = ~mass of a proton or neutron)

21. 2 H2 (g) + O2 (g) → 2 H2O(g)
2 moles of H2 molecules react with exactly
1 mole of O2 molecules to give exactly
2 moles of H2O molecules.
2 moles of H2 molecules 4 x 1.008g = 4.032g
react with exactly
1 mole of O2 molecules 2 x g = g
to produce exactly
2 moles of H2O molecules 2(2 x 1.008g g) = 36.03g
Sum of reactants = Sum of products
Law of Conservation of Mass

22. Number – mass relationships Mole – mass relationships
12 red 7g each = 84g 12 yellow each=48g
55.85g Fe = x 1023 atoms Fe 32.07g S = x 1023 atoms S

23. Mole—Mass Relationships
Element Atomic Mass Molar Mass Number of Atoms
1 atom of H = amu g = x 1023 atoms
1 atom of S = amu g = x 1023 atoms
1 atom of O = amu g = x 1023 atoms
1 molecule O2 ( x 2) 32.00amu g = x 1023 atoms
1 molecule S8 (32.07 x 8) amu g = x 1023 atoms

24. Calculating the Number of Moles and Atoms in a Given Mass of Element – Class Problem
Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element, 3680oC. How many moles of tungsten, and atoms of the element, are contained in a 35.0 mg sample of the metal? Plan:
Convert Mass to Moles
Convert Moles to Atoms

25. Calculation Solution:
Moles of W = 35.0x10-3g W x 1 mol W = mol
183.9 g W
Moles of W = 1.90 x mol
No. of W atoms = 1.90 x mol W x x 1023 atoms
1 mole of W
= 1.15 x 1020 atoms of Tungsten

26. Molecular Mass—Molar Mass (M)
The molecular mass of a compound expressed in amu is numerically the same as the mass of one mole of the compound expressed in grams , called its molar mass.

29. Calculate the Molecular Mass of Glucose: C6H12O6
Carbon—6 x amu = amu
Hydrogen—12 x amu = amu
Oxygen—6 x amu = amu
amu

30. Calculate the Molar Mass of Glucose: C6H12O6
Carbon—6 x g/mol = g/mol
Hydrogen—12 x g/mol = g/mol
Oxygen—6 x g/mol = g/mol
g/mol

31. Conversions Go from mass (g) to moles Go from moles to particles (NA)
Go from moles to volume of gas (L)
Go from moles to mass (g)

32. Mass (g)
: g/mol
x g/mol
Volume STP
x 22.4L/mol
: 22.4L/mol
x NA/mol
: NA/mol
Number of particles (NA)

33. Percent Composition
For a compound, the percent composition for a specific element is the fraction of the compound mass that came from that element.
AnBm %A = n(A g/mol) x 100
AnBm g/mol

35. Mass Fraction and Mass %
3.0g/ball x 3 balls = 9g
9.0g/16.0g total = 0.56
0.56 x 100% = 56% red
2.0g/ball x 2 balls = 0.25
16g total
0.25 x 100 = 25% purple
1.0g/ball x 3 balls = 3.0g
0.19 x 100 = 19% yellow
Mass of Red Balls =
Mass Fraction Red =
Mass % Red =
Mass Fraction Purple =
Mass Fraction Yellow =
Check: 56% + 25% + 19% = 100%

36. Calculate M and % composition of NH4C2H3O2
N __mol of N x _________ = _____g N
H __mol of H x _________ = _____g H
C __mol of C x _________ = _____g C
O __mol of O x _________ = _____g O
Molar mass = M = _____g

37. Calculate M and % composition of NH4C2H3O2
N 1 mol of N x 14.01g/mol = 14.01g N
H 7 mol of H x g/mol = g H
C 2 mol of C x g/mol = g C
O 2 mol of O x g/mol = g Molar mass = M = g NH4C2H3O2
%N = g N/77.076g = 18.18%
%H = g H/77.076g = %
%C = g C/77.076g = 31.17%
%O = g O/77.076g = 41.50%
100.00%

38. Empirical Formula
Empirical Formula - A formula that gives the simplest whole-number ratio of atoms in a compound.
Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known.
Many compounds can share the same empirical formula. Alkanes are CnH2n +2

39. Determining Empirical Formula
Start with the number of grams of each element, given in the problem. 
If percentages are given, assume that the total mass is 100 grams so that  the mass of each element = the percent given.
Convert the mass of each element to moles using the molar mass from the periodic table. 
Divide each mole value by the smallest number of moles calculated. 
Round to the nearest whole number.  This is the mole ratio of the elements and is  represented by subscripts in the empirical formula. 
If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same  factor to get the lowest whole number multiple. 
e.g.  If one solution is 1.5, then multiply each solution in the problem by 2 to get 3. 
e.g.  If one solution is 1.25, then multiply each solution in the problem by 4 to get 5. 

40. Example
A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and g H.  What is the empirical formula of the compound?

41. First step
Find the moles for each element

42. Find the ratio of moles
Divide each by the smallest number of moles present. Round to nearest whole number.

43. Finding the Formula
This is the mole ratio of the elements and is represented by subscripts in the empirical formula.
Ca—1
O—2
H—2
Therefore CaO2H2 or with the correct formula, Ca(OH)2.

44. Nutrasweet
NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O.  Calculate the empirical formula of NutraSweet and find the molecular formula.  (The molar mass of NutraSweet is g/mol)

45. Start with the number of grams of each element, given in the problem.
If percentages are given, assume that the total mass is 100 grams so that the mass of each element = the percent given.
57.14% C, 6.16% H, 9.52% N, and 27.18% O.

46. Use the conversion factor: g/mol and divide or multiply by 1/g/mol.
Convert the mass of each element to moles using the molar mass from the periodic table.
Use the conversion factor: g/mol and divide or multiply by 1/g/mol.

47. Divide each mole value by the smallest number of moles calculated
Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number.
Select the smallest number of moles to divide each element (moles). Smallest one will equal 1.

48. Mole Ratio of Elements
This is the mole ratio of the elements and is represented by subscripts in the empirical formula.
If the number is too far to round (x.1 ~ x.9), then multiply each solution by the same factor to get the lowest whole number multiple.

49. Molecular Formula
Now, we can find the molecular formula by finding the mass of the empirical formula and setting up a ratio:

50. Empirical Formula Experiment
A sample of a pure oxide of nickel was analyzed by heating to drive off the oxygen. A team of students weighed an empty test tube, recording a mass of g. After adding a sample to the tube, they measured a total mass of g. The team then heated the sample in an atmosphere of natural gas reducing it to pure metal. The final mass after two heatings was g for the tube and the metal residue. Perform calculations necessary to find results below, showing all of your work.
Mass of nickel oxide sample
Mass of nickel present in sample
Mass of oxygen present in sample
Mass percent of nickel
Mass percent of oxygen
Determine the empirical formula of the oxide of nickel, showing your work clearly.
Name the compound according to IUPAC conventions.

51. Lab Setup
Place the sample of oxide of nickel in the large test tube and mount the tube at an angle.
Attach gas supply to the tubing entering the test tube.
Connect the second tube to the burner and the test tube. This produces a reducing atmosphere with very low oxygen content.

52. What are the possible formulas?
Find the possible cations that nickel forms.
Write balanced equations.
What are the molar ratios of reactant to elemental nickel that you would expect to find? (Molar conversions)
Predict which ionic unit formula will result in describing the reaction.
Calculate the formula.

53. Calculations