1. Moles and Formula Mass

2. The Mole 1 dozen = 12 1 gross = 144 1 ream = 500 1 mole = 6.022 x 1023
There are exactly 12 grams of carbon-12 in one mole of carbon-12.

3. I didn’t discover it. Its just named after me!
Avogadro’s Number
6.022 x 1023 is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ).
I didn’t discover it. Its just named after me!
Amadeo Avogadro

4. Calculations with Moles: Converting moles to grams
How many grams of lithium are in 3.50 moles of lithium?
3.50 mol Li
6.94 g Li
= g Li
45.1
1 mol Li

5. Calculations with Moles: Converting grams to moles
How many moles of lithium are in 18.2 grams of lithium?
18.2 g Li
1 mol Li
= mol Li
2.62
6.94 g Li

6. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 3.50 moles of lithium?
3.50 mol
6.02 x 1023 atoms
= atoms
2.07 x 1024
1 mol

7. Calculations with Moles: Using Avogadro’s Number
How many atoms of lithium are in 18.2 g of
lithium?
18.2 g Li
1 mol Li
6.022 x 1023 atoms Li
6.94 g Li
1 mol Li
(18.2)(6.022 x 1023)/6.94
= atoms Li
1.58 x 1024

8. Calculating Formula Mass
Calculate the formula mass of magnesium carbonate, MgCO3.
24.31 g g + 3(16.00 g) =
84.32 g

9. Calculating Percentage Composition
Calculate the percentage composition of magnesium carbonate, MgCO3.
From previous slide:
24.31 g g + 3(16.00 g) = g
100.00

10. Formulas
Empirical formula: the lowest whole number ratio of atoms in a compound.
Molecular formula: the true number of atoms of each element in the formula of a compound.
molecular formula = (empirical formula)n [n = integer]
molecular formula = C6H6 = (CH)6
empirical formula = CH

11. Formulas (continued)
Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).
Examples:
NaCl
MgCl2
Al2(SO4)3
K2CO3

12. Formulas (continued)
Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11

13. Empirical Formula Determination
Base calculation on 100 grams of compound.
Determine moles of each element in 100 grams of compound.
Divide each value of moles by the smallest of the values.
Multiply each number by an integer to obtain all whole numbers.

14. Empirical Formula Determination
Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

15. Empirical Formula Determination (part 2)
Divide each value of moles by the smallest of the values.
Carbon:
Hydrogen:
Oxygen:

16. Empirical Formula Determination (part 3)
Multiply each number by an integer to obtain all whole numbers.
Carbon: 1.50
Hydrogen: 2.50
Oxygen: 1.00
x 2
x 2
x 2 3 5 2
Empirical formula:
C3H5O2

17. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
1. Find the formula mass of C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = g

18. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
2. Divide the molecular mass by the mass given by the emipirical formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g

19. Finding the Molecular Formula
The empirical formula for adipic acid is C3H5O2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?
3. Multiply the empirical formula by this number to get the molecular formula.
3(12.01 g) + 5(1.01) + 2(16.00) = g
(C3H5O2) x 2 =
C6H10O4

20. Chemical Analysis Law of Definite Proportions:
Different samples of the same compound always contain its constituent elements in the same proportion by mass.
This means that the ratio of atoms in a compound must be constant.
Example: We can analyze CO2 from different sources and find that each sample has the same ratio by mass of carbon to oxygen.

21. Chemical Analysis Law of Multiple Proportions:
If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in ratios of small numbers
Different compounds made up of the same elements differ in the number of atoms of each kind that combine.
Example: Carbon can form two different compounds with oxygen, CO and CO2. The ration of oxygen in each compound is 1:2

22. Chemical Analysis
Because of these two laws AND the Law of conservation of mass…..
You can determine the empirical formula from percent composition.
We can analyze percent composition of a sample by various methods:
Mass Spectrometer
Burning in a combustion train

23. Pure O2 in
CO2 is absorbed
Sample is burned completely to form CO2 and H2O
H2O is absorbed

24. Burning a g sample of a carbon–hydrogen–oxygen compound in oxygen yields g CO2 and g H2O. Determine the percent composition of the compound.
The amount of CO2 tells us how much carbon was in the sample.
Carbon:

25. Burning a g sample of a carbon–hydrogen–oxygen compound in oxygen yields g CO2 and g H2O. Determine the percent composition of the compound.
The amount of H2O tells us how much hydrogen was in the sample.
Hydrogen:

26. Burning a g sample of a carbon–hydrogen–oxygen compound in oxygen yields g CO2 and g H2O. Calculate the percent composition of the sample.
Carbon:
Hydrogen:

27. Burning a g sample of a carbon–hydrogen–oxygen compound in oxygen yields g CO2 and g H2O. Calculate the percent composition of the sample.
Oxygen:
So our sample has a percent composition of:
53.30% C, 11.19% H and 35.51% O.
Now we can calculate empirical formula!!!!! YIPEE!!!!

28. Hydrates
Compounds that have a specific number of water molecules attached to them.
Copper sulfate in it’s normal state has the five water molecules associated with it.
The formula is CuSO4 ‧ 5H2O
The name is copper (II) sulfate pentahydrate

29. Hydrates
When hydrates are heated the water molecules are evaporated off
What is left is CuSO4.
We call this anhydrous copper (II) sulfate
Anhydrous means the water is gone!

30. Hydrates and Composition
Cupric chloride, CuCl2, when heated to 100C is dehydrated. If g of CuCl2 · x H2O gives g of CuCl2 on heating, what is the value of x?
First calculate how much H2O has been heated off:

31. Hydrates and Composition
Cupric chloride, CuCl2, when heated to 100C is dehydrated. If g of CuCl2 · x H2O gives g of CuCl2 on heating, what is the value of x?
Convert CuCl2 and H2O to moles

32. Hydrates and Composition
Cupric chloride, CuCl2, when heated to 100C is dehydrated. If g of CuCl2 · x H2O gives g of CuCl2 on heating, what is the value of x?
Now find the mole ratio of CuCl2:H2O
1:2.01
CuCl2‧2H2O

33. Mass Spectrometry
Courtesy

34. Mass Spectometry Learning Objective for EK 1.D.2:
LO 1.14 The student is able to use data from mass spectrometry to identify the
elements and the masses of individual atoms of a specific element.

35. Purpose of Mass Spectrometry
Produces spectra of masses from elements in a sample of material.
The isotopic composition of sample of an element
The relative atomic mass of an element
Produces spectra of masses from molecules in a sample of material, and fragments of the molecules.

36. Purpose of Mass Spectrometry
Used to determine
the elemental composition of a sample
the masses of particles and of molecules
potential chemical structures of molecules by analyzing the fragments
the identity of unknown compounds by determining mass and matching to known spectra
the isotopic composition of elements in a molecule

37. Mass Spectometry

38. Mass Spectometry Sample is vaporized
Passed through a beam of electrons (creates cations)
Accelerated through a magnetic field which changes their path depending on their mass
Lightest ions have greater deflection
Position on detector plate gives accurate values of their masses

39. Basics of Mass Spectrometry
20 𝑋 → 𝑋 + + 𝑒 −
Sample of element X
Shot with an electron
Detected by the spectrometer
Mass of 20 𝑋 = 𝑋 +
Appears on spectra as mass:charge (M/Z)
Relative isotopic mass of X-20 = 20
M/Z of X-20 = 20/1 = 20
X-18
X-20

40. Basics of Mass Spectrometry
Sample of element X 8
66.7% 4
33.3%
How does the height of X+ -20 compare to X +-18?
X-18
X-20

41. Basics of Mass Spectrometry
Isotopic composition of element X is
33% X-18 and 66.7% X-20 8
66.7% 4
33.3%
The relative atomic mass can be calculated :
4𝑥18 +(8𝑥20) 12 =19.3 or
33.3 𝑥 𝑥 =19.3

42. Mass Spectrum of Neon
Using the data from the mass spectrum above, calculate the average atomic mass of neon.

43. Mass Spectrum of Bromine, Br2
Bromine has two isotopes: 50.69% 79Br and 49.31% 81Br
Molecular Ion Peaks
[79Br81Br]+
[79Br79Br]+
Fragments
[81Br81Br]+
The spectra of Br (which exists as a diatomic molecule) exhibits Molecular ion peaks composed of the possible combinations of the isotopes.
79Br+
81Br+

44. Mass Spectrum of CO2 Molecular ion peak [CO2]+ = 44 Fragment Peaks

45. Stoichiometry
“In solving a problem of this sort, the grand thing is to be able to reason backward. This is a very useful accomplishment, and a very easy one, but people do not practice it much.”
Sherlock Holmes, in Sir Arthur Conan Doyle’s A Study in Scarlet

46. Review: Chemical Equations
Chemical change involves a reorganization of
the atoms in one or more substances.
C2H5OH + 3O2 ® 2CO H2O
reactants
products
When the equation is balanced it has quantitative significance:
1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

47. Calculating Masses of Reactants and Products
Balance the equation.
Convert mass or volume to moles, if necessary.
Set up mole ratios.
Use mole ratios to calculate moles of desired substituent.
Convert moles to mass or volume, if necessary.

48. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed.
1. Identify reactants and products and write the balanced equation. 4 Al + 3 O2 2
Al2O3
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

49. Working a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed?
4 Al O2  2Al2O3
6.50 x 2 x ÷ ÷ 4 =
12.3 g Al2O3
6.50 g Al
1 mol Al
2 mol Al2O3
g Al2O3 =
? g Al2O3
26.98 g Al
4 mol Al
1 mol Al2O3

50. Limiting Reactant The limiting reactant is the reactant
that is consumed first, limiting the amounts of products formed.

51. Calculating Limiting Reactants
Calculate the amount of product produced using each amount of reactant. (Yes, 2 stoich problems!)
Whichever reactant produces the LEAST amount (moles OR mass) of product is your limiting reactant. 51

52. Working a Limiting Reactant Problem
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant and how many grams of N2 will be formed?
1. Identify reactants and products and write the balanced equation.
N2(g) + 3
Cu(s) + 3
H2O(g) 2
NH3 + 3
CuO
a. Every reaction needs a yield sign!
b. What are the reactants?
c. What are the products?
d. What are the balanced coefficients?

53. Working a Limiting Reactant Problem
NH3 +
CuO
N2(g) 2 3
Cu(s)
H2O(g)
Calculate the amount of product produced using each amount of reactant.
CuO produeces the smaller amount so CuO is the limiting reactant g N2 are produced.
18.1 g NH3
1 mol NH3
1 mol N2
28.02 g N2 =
14.9 g N2
17.03 g NH3
2 mol NH3
1 mol N2 =
90.4 g CuO
10.6 g N2
1 mol CuO
79.55 g CuO
3 mol CuO
1 mol N2
28.02 g N2

54. Theoretical and Actual Yield
Theoretical yield:
The calculated amount of product produced.
Actual yield:
The amount produced by performing the experiment.
The percent yield can be calculated using the equation below: