1. Amounts in Chemistry
Unit 2
Week 5 Tuesday

2. Stoichiometry
The study of the relationships between the quantities of reactants and products involved in chemical reactions.
Stoikheion = element
metron = measure

3. Proportions in compounds
The composition of a compound depends on two things
1. The elements they are composed of
2. The quantities of each element.

4. Law of definite proportions
A specific compound always contains the same elements in definite proportions by mass.
This is always constant even if produced other ways.

5. Relative atomic mass and isotopic abundance (p80)
Relative atomic mass = the mass of an element that would react with a fixed mass of a standard element, currently carbon-12.
One atomic mass unit (u) is described as 1/12 the mass of a carbon-12 atom.

6. Atomic mass unit Now we know that it weighs
1 u = x kg.
This is the standard used world wide

7. Isotopic abundance
Some elements have isotopes. In calculating the relative atomic mass of an element with isotopes, the relative mass and proportion of each is taken into account. For example, naturally occurring carbon consists of atoms of relative isotopic masses C-12 (98.89%) and-C 13 (1.11%). Its relative atomic mass is u.

8. Isotopic abundance mc = (98.89/100 x 12) + (1.11/100 x 13) =
mc = = u
No atom will contain u. it is an average.

9. Example
Cl-35 = 75.53%
Cl-37 = 24.47%
(35 x .7553) + (37 x .2447) = u
Therefore the average atomic mass is 35.49u

10. Page 81 # 1

11. Atomic Mass and Molecular Mass
The mass of one atom of an element expressed in atomic units, u.
Molecular mass
The mass of one molecule, expressed in atomic mass units, u.

12. Example MNH3 = 1(mN) + 3(mH) = 1(14.01 u) + 3(1.01 u) = 17.04 u .
Therefore, one molecule of ammonia has a mass of units.

13. Formula unit of mass
The mass of one formula unit of an ionic compound, expressed in atomic mass units, u

14. Example NaCl = 1(mNa) + 1(mCl) = 1 (22.99 u) + 1(35.45 u) = 58.44 u
Therefore, one formula unit of sodium chloride has a mass of units.

15. Page 81 # 1-3

16. The Periodic Table and the Mass of Elements
The atomic number of a substance is written near the top of the symbol. This number tells the number of protons in the element.
The Atomic mass is usually written near the bottom. This number is given as a numerical number of the elements mass. The mass of one mole of the element measured in grams.

17. Molar mass (g/mol) = M
the mass, in grams per mole, of one mole of a substance. The SI unit for a mol is g/mol
Molar mass is conventionally abbreviated with a capital M. For elements it is numerically equal to the atomic mass and is often called atomic mass units.

18. Steps to calculating molar mass
Write the chemical formula of the substance
Determine the number of atoms or ions of each element in one formula unit of the substance.
Us the atomic molar masses from the periodic table and the amounts in moles to determine the molar mass of the chemical
Write the molar mass in units of grams per mol (g/mol).

19. Example Molar mass of copper(Cu) = mass of 1 mol of Cu atoms
M = g/mol Cu
M = mass of x atoms

20. Example 2
Molar mass of glucose ( C6H12O6) = mass of exactly 1 mol of glucose molecules
M = (6 x C) + (12 x H) + ( 6 x O)
M = ((6 x (C)) + (12 x 1.01(H)) + (6x (O)) ) / mol
M = ( (72.06 g) + (12.12 g) + (96.00g) ) / mol
M = ( g) / mol
M = g/mol
Therefore, 1 mol glucose has a mass of g and contains x 1023 glucose molecules(C6H12O6)

21. Important things to note
Specify what is being measured
Ex molar mass of oxygen atom = g/mol
Molar mass of oxygen molecules = g/mol
When talking about oxygen we are normally referring to the diatomic molecule
Since electrons have negligible mass compared to protons and neutrons, ions will essentially have the same mass as the atoms.

22. PAGE 85 # 4

23. The Mole and Molar Mass The Mole = n
Avogadro’s constant (NA) = the number of entities in one mole = 6.02 x1023
Eg. 1 dozen oxygen atoms = 12 oxygen atoms
1 mole oxygen atoms = x 1023 oxygen atoms

24. A mole is the amount of substance that contains as many elementary particles (atoms, molecules, or other particles) as there are atoms in exactly 12 g of the carbon 12 isotope. It is important to remember that one mole always contains the same number of particles, no matter what the substance. Much like one dozen eggs contains 12 eggs, one mole of a substance always contains 6.02 x 1023

25. 6.022 x 1023 particles (Avogadro’s number).
Ex 1 mole of elephants = 6.02 x 1023 elephants

26. Calculations involving Molar mass
Amount in moles (n) = mass(g)
Molar mass(g/mol)
The Magic Triangle
n = m / M
m = n x M
M = m / n

27. GRASP method 1 Given = What do you know?
2 Required = What do you want to know? 
3 Analysis = What formula can you use? 
4 Solution = Show your work 
5 Paraphrase = State the answer in a sentence.

28. Example
Lets say that we are given 22 grams of graphite. How many moles do we have? G
mC = g
MC = g / mol S
n = 22.00g / g/mol
= 1.83 moles of carbon R
n = ? P
Therefore, grams of carbon is equal to 1.83 moles A
n = m/M

29. Lets say we want to know how much CaCl2 we need for a reaction that asks for 13.52 molesG
MCaCl2 = g/mol
nCaCl2 = mol S
m = n x M
m = mol x g/mol
= 1500g R
m = ? P
Therefore, we need g of calcium chloride. A

30. You are given an unknown sample that has a mass of 32. 04 grams, 2
You are given an unknown sample that has a mass of grams, moles. Find M. G
m? = grams
n? = moles S
M = g / 2.000
M = g/mol R
M? = ? P
Therefore, the sample has a molecular mass of g/mol A
M = m / n

31. Homework
Page 86 # 5-7
Isotopic Abundance and Mole calculations Worksheet

32. Calculating Number of Entities

33. Calculating Number of Entities (N) from Mass
We are not able to count the number of atoms in reactions; we must count them indirectly by measuring out a certain mass of a substance and calculating the number of moles.
In order to estimate the number of entities in a large sample we need to take a small sample and find out the mass of a small number of entities.

34. Example
Example: You have been given a load of crushed gravel and you need to know how mainly pieces of rock you have. Instead of counting out every piece in the 2000kg load you want to do it in a faster manner. You decide to find the number of entities in 200 grams of gravel. A 200 gram sample has 40 rocks. How many rocks are in a 2000kg load?

35. G
m = 2000kg = g
Conversion factor = 40 rocks/ 200 g S
N = x (40 rocks/200 g)
= rocks R
N = ? P
Therefore, the pile of gravel has 400, 000 rocks. A
N =Total mass x conversion factor

36. Calculating Number of Entities and Avogadro’s number
There are times when need to know exactly how many atoms are in something. This is especially true when dealing with gases as they change in pressure with temperature.
To find the number of entities we use the formula

37. N = nNA
Number of entities = (# of mols) x(Avogadro's #)

38. Example Find the number of atoms in 0.004 moles of water. G
nH2O = mol
NA = 6.02 x 1023 entities / mol S
N = mol x 6.02 x 1023 entities/mol
= 2.41 x 1021 ­ molecules of water R
N = ? P
Therefore, moles of water contains 2.41 x 1021 ­ molecules of water A
N = n x NA

39. Find the number of molecules in 30.0 g of Cl2 (g)
nCl2= (30.0g / 70.9g/mol)
= mol
NA = 6.02 x 1023 entities / mol S
N = mol x 6.02 x 1023 entities/mol
= 2.53 x molecules of cl2 R
N = ? P
Therefore, moles of Cl2 contains 2.53 x ­ molecules of chlorine gas A
N = n x NA