UNIT 2 – QUANTITATIVE CHEMISTRY (I.E. IN COMES THE MATHS!! BRING A CALCULATOR!!)
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UNIT 2 – QUANTITATIVE CHEMISTRY(I.E. IN COMES THE MATHS!! BRING A CALCULATOR!!)
Scientific Notation… A (very) brief introduction
Stoichiometry The study of the relationships between the
quantities of reactants and products involved in chemical reactions.
Stoikheion = element metron = measure
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.
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.
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.
Atomic mass unit
Now we know that it weighs
1 u = 1.660 540 2 x 10-27 kg.
This is the standard used world wide
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 12.01 u.
Isotopic abundance
mc = (98.89/100 x 12) + (1.11/100 x 13) =
mc = 11.8668 + 0 .1443 = 12.01 u
No atom will contain 12.01 u. it is an average.
Example
Cl-35 = 75.53% Cl-37 = 24.47%
(35 x .7553) + (37 x .2447) = 35.49 u
Therefore the average atomic mass is 35.49u
Page 81 # 1
Atomic Mass and Molecular Mass Atomic 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.
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 17.04
units.
Formula unit of mass
The mass of one formula unit of an ionic compound, expressed in atomic mass units, u
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 58.44 units.
Page 81 # 1-3
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.
The Mole
It's not a spy, a machine for digging tunnels, a burrowing animal, or a spot of skin pigmentation - it's
- a unit of measurement containing about
6.02 x 106.02 x 102323
particles.
Chemists created the mole as their own counting unit.
1 mole of atoms is
602, 214,199, 000, 000, 000, 000, 000
atoms!
(This is the number of Carbon atoms there are in 12.01 g or Carbon)
Avogadro’s Number
6.02 x 1023 particles / mole
When you say you have a mole, it’s kind of like saying you have ‘a dozen’ of something…
A dozen donuts = 12 donutsA dozen apples = 12 apples
A mole of CO2 molecules = 6.02 x 1023 moleculesA mole of Na atoms = 6.02 x 1023 atoms
Just How Big is a Mole?Just How Big is a Mole?
Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.
If you had Avogadro's number of unpopped
popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.
If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.
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.
Steps to calculating molar mass
1. Write the chemical formula of the substance2. Determine the number of atoms or ions of each
element in one formula unit of the substance. 3. Us the atomic molar masses from the periodic table
and the amounts in moles to determine the molar mass of the chemical
4. Write the molar mass in units of grams per mol (g/mol).
Example
Molar mass of copper(Cu) = mass of 1 mol of Cu atoms M = 63.546 g/mol Cu M = mass of 6.022 x 10 23 atoms
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 12.01 (C)) + (12 x 1.01(H)) + (6x 16.00
(O)) ) / mol M = ( (72.06 g) + (12.12 g) + (96.00g) ) / mol M = ( 180.18 g) / mol M = 180.18 g/mol Therefore, 1 mol glucose has a mass of 180.18 g
and contains 6.022 x 1023 glucose molecules(C6H12O6)
Important things to note Specify what is being measured Ex. molar mass of oxygen atom = 16.00 g/mol
Molar mass of oxygen molecules = 32.00 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.
Moles of Elements Why that number?
1 mole element = (atomic mass) in grams of that element
1 mole C = 12.011 g C
1 mole B = 10.81 g B
1 mole Cu = 63.55 g Cu
i.e. atoms are teeny, teeny, tiny, teeny, itty, bitty things and it is not reasonable to talk about them in their atomic weights, but if scale them up to moles, then working with balanced chemical equations etc. becomes easier.
Determine the mass one mole of the following in grams:
a. CO2
b. NaCl
c. K2CrO4
d. Fe(NO3)3
PAGE 85 # 4
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 = 6.022 x 1023 oxygen atoms
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
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.
Example Lets say that we are given 22 grams of graphite.
How many moles do we have?
Lets say we want to know how much CaCl2 we need for a reaction that asks for 13.52 moles
You are given an unknown sample that has a mass of 32.04 grams, 2.000 moles. Find M.
Practice:
1. Determine the number of mol in 13.7 g of CO
2. What is the mass of 2.75 mol of C3H8
Homework
Page 86 # 5-7 Isotopic Abundance and Mole calculations Worksheet
Calculating Number of Entities
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.
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
N = nNA
Number of entities = (# of mols) x(Avogadro's #)
Example
Find the number of atoms in 0.004 moles of water.
Find the number of molecules in 30.0 g of Cl2 (g)
Determining the number of particles in a mole
6.02 x 1023 particles 1 mole
or
1 mole
6.02 x 1023 particles
Note that a particle could be an atom OR a molecule!
Avogadro’s Number as Avogadro’s Number as Conversion FactorConversion Factor
Equation to find number of particles!
number of moles x = number of particles
Example: How many particles of sucrose are in 3.50 moles?
6.02 x 1023 particles
1 mole
Equation to find number of moles!
number of particles x = number of moles
Example: How many moles are contained in 4.50 x 1024 atoms of zinc?
1 mole
6.02 x 1023 particles
1. Number of atoms in 0.500 mole of Al
2.Number of moles of S in 1.8 x 1024 S atoms
Learning CheckLearning Check
Molar Mass
The Mass of 1 mole (in grams) Molar mass is equal to the atomic mass (look on
the periodic table)
1 mole of C atoms = 12.01 g
1 mole of Au atoms = 196.97g
1 mole of Cu atoms = 63.55 g Always round to the hundredths place.
Molar MassMolar Mass
Mole to Mass Conversion
Number of moles x number of grams = mass 1 mole
Example: Calculate the mass in grams of 0.0450 moles of chromium.
0.0450 moles Cr x 52.0 g Cr = 2.34 g Cr 1 mol Cr
Where will you find the number of
grams per mole?
The Periodic Table
Look at your Periodic Table
Mass to Mole Conversion
Mass x 1 mole = number of moles number of grams
Example: How many moles of calcium are there in 525 grams of calcium?
525 g Ca x 1 mol Ca = 13.1 mol Ca 40.1 g Ca
Where will you find the number of
grams per mole?
The Periodic Table
Look at your Periodic Table
Mass to Atoms Conversion
Two steps:
1. Use the mass to moles conversion
2. Now calculate the number of atoms using the moles to particles equation.
Example:
How many atoms of gold are in a pure gold nugget having a mass of 25.0 g?
Atoms/Molecules and GramsAtoms/Molecules and Grams
How many atoms of Cu are present in 35.4 g of Cu?
35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu 63.5 g Cu 1 mol Cu
= 3.4 X 1023 atoms Cu
Learning Check!Learning Check!
How many atoms of K are present in 78.4 g of K?
78.4 g K 1 mol K 6.02 X 1023 atoms K 39.1 g K 1 mol K
= 1.20 X 1024 atoms K
Empirical and Molecular Formulas11.4
Percent CompositionMass of element x 100 = percent by massMass of compound
Example: Determine the percent composition of NaHCO3.
Determine the mass of each element first!
%mass element = mass of element in 1 mol compound x 100 molar mass of compound
Percent Na = 22.99 g Na x 100 = 27.37% Na
84.01 g NaHCO3
You have to complete this step for each element in the compound!!!!
Divide the mass of each element by the molar mass of the compound and multiply by 100%
52.14%C100%46.07g24.02g
13.13%H100%46.07g6.048g
34.73%O100%46.07g16.00g
Determine the Percent Composition from the Formula C2H5OH
Types of Formulas
Empirical FormulaEmpirical FormulaThe formula of a compound that expresses the The formula of a compound that expresses the smallest whole number ratiosmallest whole number ratio of the atoms of the atoms present.present.
Molecular FormulaMolecular FormulaThe formula that states the The formula that states the actualactual number of each number of each kind of atom found in kind of atom found in one moleculeone molecule of the of the compound.compound.
Convert the percentages to grams by assuming you have 100 g of the compound– Step can be skipped if given masses
47gC100g47gC100g
47gO100g47gO100g 6.0gH
100g6.0gH100g
Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Convert the grams to moles
C mol 3.9 12.01g
C mol 1C 47g
O mol 2.9 16.00g
O mol 1O g 47
Hmol 6.0 1.008g
Hmol 1 Hg 6.0
Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Divide each by the smallest number of moles
1.3 2.9 C mol 3.9
1 2.9 O mol 2.9
2 2.9 H mol 6.0
Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
If any of the ratios is not a whole number, multiply all the ratios by a factor to make it a whole number– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then
multiply by 3; if ?.25 or ?.75 then multiply by 4
4 3x 1.3 2.9 C mol 3.9
3 3x 1 2.9 O mol 2.9
6 3x 2 2.9 H mol 6.0 Multiply all theRatios by 3 Because C is 1.3
Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
° Use the ratios as the subscripts in the empirical formula
4 3x 1.3 2.9 C mol 3.9
3 3x 1 2.9 O mol 2.9
6 3x 2 2.9 H mol 6.0 C4H6O3
Determine the Empirical Formula ofAcetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Molecular Formulas• The molecular formula is a multiple of
the empirical formula• To determine the molecular formula you
need to know the empirical formula and the molar mass of the compound
Divide the given molar mass of the compound by the molar mass of the empirical formula– Round to the nearest whole number
407.63
252 gg
Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an
empirical formula of C5H3
Multiply the empirical formula by the calculated factor to give the molecular formula
(C5H3)4 = C20H12
Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an
empirical formula of C5H3
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