Unit 2/3: Atom, Elements, and Stoichiometry

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Unit 2/3 Atom, Elements, and Stoichiometry

• Calculations with Chemical Formulas and Equations
• By Ms. Buroker
• Scott High School

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Isotopes

• Isotopes atoms of the same element with
  different masses
• The different masses come from different numbers
  of neutrons
• The number of protons always remains the same,
  since those give us the identity of the element
• The atomic mass shown on the PT is the weighted
  average of all the isotopes of the element

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Natural lithium is 7.42 6Li (6.015 amu) 92.58
7Li (7.016 amu)
Average atomic mass of lithium
6.941 amu
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So is there anything else that can change?

• The protons of an element do not change
• The neutrons of an element can change (isotopes)
• The electrons of an element can also change
• Ions atoms of an element with a different amount
  of electrons than protons

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Charged Atoms

• When electrons change, they create a charged
  atom, since the numbers of protons and electrons
  are now different
• More electrons negative ion (called an anion)
• Less electrons positive ion (called a cation)

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Micro World atoms molecules
Macro World grams
Atomic mass is the mass of an atom in atomic mass
units (amu)
By definition 1 atom 12C weighs 12 amu
On this scale 1H 1.008 amu 16O 16.00 amu
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First Things First!!

• The Law of Conservation of Mass!!
• Mass is neither created nor destroyed
  therefore, chemical equations must be balanced in
  order to show equal numbers of moles of elements
  on both sides of the reaction!

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A process in which one or more substances is
changed into one or more new substances is a
chemical reaction
A chemical equation uses chemical symbols to show
what happens during a chemical reaction
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How to Read Chemical Equations
2 Mg O2 2 MgO
2 atoms Mg 1 molecule O2 makes 2 formula units
MgO 2 moles Mg 1 mole O2 makes 2 moles MgO 48.6
grams Mg 32.0 grams O2 makes 80.6 g MgO
2 grams Mg 1 gram O2 makes 2 g MgO
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Balancing Equations
A chemical equation is made up of reactants on
the left And the products on the right. An arrow
signifies reactants going to products
Why?-----Atoms are conserved in chemical reactions
The equation is balanced only by adjusting
the coefficients of the formulas as necessary
to get whole number coefficients. NEVER
introduce extraneous formulas NEVER change
subscripts of the formulas
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Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on
   the left side and the correct formula(s) for the
   product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide
and water

1. Change the numbers in front of the formulas
   (coefficients) to make the number of atoms of
   each element the same on both sides of the
   equation. Do not change the subscripts.

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Balancing Chemical Equations

1. Start by balancing those elements that appear in
   only one reactant and one product.

start with C or H but not O
multiply CO2 by 2
multiply H2O by 3
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Balancing Chemical Equations

1. Balance those elements that appear in two or more
   reactants or products.

7 oxygen on right
remove fraction multiply both sides by 2
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Balancing Chemical Equations

1. Check to make sure that you have the same number
   of each type of atom on both sides of the
   equation.

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Lets Take a Little Pre-Test
Get a Separate Piece of Paper and do the Best you
can!
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What is Chemical Stoichiometry?

• Stoichiometry is simply the study of the
  quantities of materials consumed and produced in
  chemical reactions.

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Counting by Weighing
Suppose you work in a candy store that sells
jelly beans by the bean. Customers come in and
order 50, 200, or 500 jelly beans. Counting
these out by hand would be horrible! You (being
the smart people that you are) decide to buy a
scale and count the jelly beans by weighing them.

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• Atoms are very small therefore, we deal with
  samples of matter that contain huge numbers of
  atoms. So, we determine the number of atoms in a
  given sample by finding its mass.
• We need to know that average mass of the
  particular atom!

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Atomic Mass

• The modern system for atomic mass is based on 12C
  
• 12C is assigned a mass of exactly 12 atomic mass
  units (amu)

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Mass Spectrometer
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Average Atomic Mass

• The atomic mass reflected on the periodic table
  is the average mass of each naturally occurring
  isotope of that elements and its abundance in
  nature.
• Another way

Sample of Ne
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The Mole

• We use the mole as our unit of measurement in
  determining the number of atoms
• One mole (abbreviated mol) is the number of
  carbon atoms in exactly 12g of pure 12C 6.02 X
  1023
• Avogadros Number

So . One mole of any element (6.02 X 1023 atoms)
is equal to its atomic mass!!!!
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Sample Problem

• Cobalt (Co) is a metal that is added to steel
  to improve its resistance to corrosion.
  Calculate both the number of moles in a sample of
  cobalt containing 5.00 x 1020 atoms and the
  mass of the sample.

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Molar mass is the mass of 1 mole of
in grams
atoms
1 mole 12C atoms 6.022 x 1023 atoms 12.00 g 1
12C atom 12.00 amu
1 mole 12C atoms 12.00 g 12C 1 mole lithium
atoms 6.941 g of Li
For any element atomic mass (amu) molar mass
(grams)
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S
C
One Mole of
Hg
Cu
Fe
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How many atoms are in 0.551 g of potassium (K) ?
1 mol K 39.10 g K
1 mol K 6.022 x 1023 atoms K
0.551 g K
8.49 x 1021 atoms K
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Molecular mass (or molecular weight) is the sum
of the atomic masses (in amu) in a molecule.
For any molecule molecular mass (amu) molar
mass (grams)
1 molecule SO2 64.07 amu 1 mole SO2 64.07 g
SO2
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How many H atoms are in 72.5 g of C3H8O ?
1 mol C3H8O (3 x 12) (8 x 1) 16 60 g C3H8O
1 mol C3H8O molecules 8 mol H atoms
1 mol H 6.022 x 1023 atoms H
72.5 g C3H8O
5.82 x 1024 atoms H
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Percent composition of an element in a compound
n is the number of moles of the element in 1 mole
of the compound
52.14 13.13 34.73 100.0
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Percent Composition

• Obtain the mass percents of the elements from
  the formula of the compound by comparing the mass
  of each element present in 1 mole of the compound
  to the total mass of 1 mole of the compound.
• C2H5OH

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Types of Formulas

• Empirical Formula
• The formula of a compound that expresses the
  smallest whole number ratio of the atoms present.
• Ionic formula are always empirical formula
• Molecular Formula
• The formula that states the actual number of
  each kind of atom found in one molecule of the
  compound.

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To obtain an Empirical Formula

• 1. Determine the mass in grams of each element
  present, if necessary.
• 2. Calculate the number of moles of each
  element.
• 3. Divide each by the smallest number of moles to
  obtain the simplest whole number ratio.
• If whole numbers are not obtained in step 3),
  multiply through by the smallest number that will
  give all whole numbers
• Be careful! Do not round off numbers
  prematurely

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• A sample of a brown gas, a major air pollutant,
  is found to contain 2.34 g N and 5.34g O.
  Determine a formula for this substance.
• require mole ratios so convert grams to moles
• moles of N 2.34g of N 0.167 moles of N
• 14.01 g/mole
• moles of O 5.34 g 0.334 moles of O
• 16.00 g/mole
• Formula

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Calculation of the Molecular Formula

• A compound has an empirical formula of NO2. The
  colourless liquid, used in rocket engines has a
  molar mass of 92.0 g/mole. What is the molecular
  formula of this substance?
• empirical formula mass 14.012 (16.00) 46.01
  g/mol
• n molar mass 92.0 g/mol emp.
  f. mass 46.01 g/mol
• n 2
• N2O4

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Empirical Formula from Composition

• A substance has the following composition by
  mass 60.80 Na 28.60 B 10.60 H
• What is the empirical formula of the substance?
• Consider a sample size of 100 grams
• This will contain 60.80 grams of Na, 28.60
  grams of B, and 10.60 grams H
• Determine the number of moles of each
• Determine the simplest whole number ratio

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Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
6.0 g C 0.5 mol C
1.5 g H 1.5 mol H
g of O g of sample (g of C g of H)
4.0 g O 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
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Mass Changes in Chemical Reactions

1. Write balanced chemical equation
2. Convert quantities of known substances into moles
3. Use coefficients in balanced equation to
   calculate the number of moles of the sought
   quantity
4. Convert moles of sought quantity into desired
   units

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Other units

• Molarity
• Moles solute / L solution
• Gases
• 22.4 L 1 mole of ANY GAS at STP

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Methanol burns in air according to the equation
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
molar mass CH3OH
molar mass H2O
coefficients chemical equation
209 g CH3OH
235 g H2O
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Limiting Reagents
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Method 1

• Pick A Product
• Try ALL the reactants
• The lowest answer will be the correct answer
• The reactant that gives the lowest answer will be
  the limiting reactant

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Limiting Reactant Method 1
LimitingReactant

• 10.0g of aluminum reacts with 35.0 grams of
  chlorine gas to produce aluminum chloride. Which
  reactant is limiting, which is in excess, and how
  much product is produced?
• 2 Al 3 Cl2 ? 2 AlCl3
• Start with Al
• Now Cl2

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g
AlCl3 27.0 g Al 2 mol Al
1 mol AlCl3
49.4g AlCl3
35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g
AlCl3 71.0 g Cl2 3 mol Cl2
1 mol AlCl3
43.9g AlCl3
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Finding Excess Practice

• 10.0g of aluminum reacts with 35.0 grams of
  chlorine gas 2 Al 3 Cl2 ? 2 AlCl3
• We found that chlorine is the limiting reactant,
  and 43.8 g of aluminum chloride are produced.

35.0 g Cl2 1 mol Cl2 2 mol Al 27.0 g Al
71 g Cl2 3 mol Cl2 1
mol Al
7.8 g Al USED!
10.0 g Al 7.8 g Al 2.2 g Al EXCESS
Given amount of excess reactant
Amount of excess reactant actually used
Note that we started with the limiting reactant!
Once you determine the LR, you should only start
with it!
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Theoretical Yield is the amount of product that
would result if all the limiting reagent reacted.
Actual Yield is the amount of product actually
obtained from a reaction.
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