topic 1:stoichiometry (12.5 hours) 1.1 the mole concept and avogardo’s number 1.2 formulas 1.3...
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Topic 1:Stoichiometry(12.5 hours)
1.1 The mole concept and Avogardo’s number1.2 Formulas
1.3 Chemical equations1.4 Mass and gas vol. relationships in chemical reactions
1.1 The mole concept and Avogadro’s constant1.2 Formulas
1.1.1 Apply the mole concept to substances1.1.2 Determine the number of particles and the amount of
substance (in moles)
1.2.1 Define the terms relative atomic mass (Ar) and relative molecular mass (Mr)
1.2.2 Calculate the mass of one of mole of species from its formula
1.2.3 Solve problems involving the relationships between the amount of substance in moles, mass, and molar mass
Quick Review:
Atoms are the smallest unit The same atoms make up elements (Na or
Cl2)
Different whole combination of atoms make up compounds (NaCl) or molecules (CH4)
Atomic mass
Atoms contain different numbers of protons and neutrons, so they have different masses (isotopes)
Carbon-12 (6 p+, 6 no)has a mass 12 times greater than hydrogen-1(1 p+, 0 no)
It’s impossible to measure the mass of individual atoms on a balance, because it is so small
Chemists developed their own unit, called the mole
The Mole
Similar to a dozen, except instead of 12, it’s 602,000,000,000,000,000,000,000
6.02 X 1023 mol-1(in scientific notation) It applies to all kinds of particles: atoms,
particles, molecules, ions, electrons, formula units depending the way the question is asked, so be careful.
This number is named in honor of Amedeo Avogadro (1776 – 1856)
Just how big is Avogadro's constant?
The earth is estimated to be 4.54 billion years old, that is approximately 1.43 x 1017 s. Still much smaller than a mole.
If you were to count every grain of sand in the Sahara desert, assuming its 4000 km by 1500 km, and its depth is 10 m, and assuming that 1 cm3 contains 1000 grains of sand. It would contain 6 x 1022 grains of sand. Only one tenth the size of a mole!!!
Relative atomic mass (Ar)
As the weighted mean mass of all the naturally occurring isotopes of an element relative to one twelfth of the mass of a carbon-12 atom.
Not whole numbers No mass unit Also applies to relative molecular mass (Mr)
The Mass of 1 mole (in grams)
Equal to the numerical value of the average atomic
mass (get from periodic table), or add the atoms
together for a molecule
1 mole of C atoms = 12.0 g
1 mole of Mg atoms = 24.3 g
1 mole of O2 molecules = 32.0 g
Molar Mass (or atomic mass)
Molar Mass of Compounds
The molar mass (MM) of a compound is determined the same way, except now you add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride =
40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl2
20
Ca 40.08 17
Cl 35.45
Atoms or Molecules
Moles
Mass (grams)
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Multiply by atomic/molar mass from periodic tableDivide by
atomic/molar mass from periodic table
Practice
Calculate the Molar Mass of calcium phosphate Formula = Masses elements:
Ca: 3 Ca’s X 40.1 = P: 2 P’s X 31.0 = O: 8 O’s X 16.0 =
Molar Mass =
Ca3(PO4)2
120.3 g
62.0 g
128.0 g
120.3g + 62.0g +128.0g310.3 g/mol
molar mass Avogadro’s number Grams Moles particles
Everything must go through Moles!!!
Calculations
Atoms/Molecules and Grams
How many moles of Cu are present in 35.4 g of Cu? How many atoms?
35.4 g x 1 mol = 0.557 mol Cu
63.55 g
0.557 mol Cu x 6.02x1023 atoms = 3.35x1023 atoms
1 mol
On your own!
How many moles of Fe are present in 102.4 g of Fe? How many atoms?
102.4 g x 1 mol = 1.83 mol Fe55.85 g
1.83 mol Fe x 6.02x1023 atoms = 1.10x1024 atoms 1 mol
Work backwards!
What is the mass (in grams) of 1.20x1024 molecules of glucose (C6H12O6)?
1.20x1024 molec. X 1 mol = 1.99 mol
6.02x1023 molec.
1.99 mol x 180.12 g = 358.4 g
1 mol
From periodic table
This is as tricky as it gets!
How many atoms of carbon are found in 2.6g of glucose (C6H12O6)?
2.6 g x 1 mole= 0.014 mol C6H12O6
180.12 g
0.014 mol x 6.02 x 1023 molecules = 8.4 x 1021 moleculesmol
8.4 x 1021 molecules x 6 atoms of C = 5.06 x 1022 atoms 1 molecule C6H12O6
1.3 Chemical Equations
1.3.1 Deduce chemical equations when all reactants and products are given
1.3.2 Identify the mole ratio of any two species in a chemical equation
1.3.3 Apply the state symbols (s), (l), (g) and (aq)
Chemical equations (review)
To deduce the products of an equation we must know the type of reaction occuring.
1. Single displacement
2. Double displacement
3. Combustion
4. Synthesis
5. Decomposition
Single displacement:
either the metal changes position with the metal ion…AgNO3 (aq) + Cu(s) CuNO3(aq) + Ag(s)
or the non-metal changes position with the non-metal ion.F2(g) + NaCl(aq) NaF(aq)+ Cl2(g)
Double displacement
The cations (positively charged ions) change position with each other.
NaOH + HCl HOH + NaCl
Fe(OH)2 + 2KBr 2KOH + FeBr2
Balancing an equation means that both sides must have the same number and type of atoms present.
Coefficients are red (indicating total number of compounds), subscripts are blue (indicating number of atoms).
Organic Combustion
A hydrocarbon reacting with oxygen to produce carbon dioxide and water
CH4 (g) + 2O2 (g) 2H2O (g) + CO2 (g)
2C2H6 (g) + 7O2 (g) 6H2O (g) + 4CO2 (g)
The coefficient is also related the mole concept. For the 1st reaction, we would say that one mole of methane reacts with two moles of oxygen to produce two moles of water and one mole of carbon dioxide gas.
Synthesis
The combination of two or more particles to produce one compound.
N2 (g) + H2 (g) NH3 (l)
Use (g) for gas, (l) for liquid, (s) for solids and (aq) if it dissolves in water/aqueous.
Decomposition
A single large compound will be broken down into more simple parts.
H2O (l) H2 (g) + O2 (g)
1.4 Mass relationships in chemical equations
1.4.1 Calculate theoretical yields from chemical equations1.4.2 Determine the limiting reactant the reactant in excess
when quantities of reacting substances are given.1.4.3 Solve problems involving theoretical, experimental
and percentage yield.
Chemistry Recipes
Looking at a reaction tells us how much of something you need to react with something else to get a product
Be sure you have a balanced reaction before you start!
Example: 2 Na + Cl2 2 NaCl This reaction tells us that by mixing 2 moles of
sodium with 1 mole of chlorine we will get 2 moles of sodium chloride
What if we wanted 4 moles of NaCl? 10 moles? 50 moles?
Practice
Write the balanced reaction for hydrogen gas reacting with oxygen gas.
2 H2 + O2 2 H2O How many moles of reactants are needed? What if we wanted 4 moles of water?
What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get?
What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?
2 mol H2 1 mol O2
4 mol H2
2 mol O2
6 mol H2, 6 mol H2O
25 mol O2, 50 mol H2O
Mole Ratios
These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical
Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)?
2 Na + Cl2 2 NaCl
5 moles Na 1 mol Cl2
2 mol Na= 2.5 moles Cl2
Mole-Mole Conversions
How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?
2 Na + Cl2 2 NaCl
2.6 moles Cl2 2 mol NaCl
1 mol Cl2= 5.2 moles NaCl
You practice
Aluminum reacts with oxygen to produce aluminum oxide. If I have 2.6 mol of Al, how much Al2O3 would I produce?
4Al + 3O2 2Al2O3
2.6 mol Al x 2 mol Al2O3 = 1.3 mol Al2O3
4 mol Al
Mole-Mass Conversions
Most of the time in chemistry, the amounts are given in grams instead of moles
We still go through moles and use the mole ratio, but now we also use molar mass to get to grams
Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride?
2 Na + Cl2 2 NaCl
5.00 moles Na 1 mol Cl2 70.90g Cl2
2 mol Na 1 mol Cl2
= 177g Cl2
You Practice
Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.
2 Al + 3 I2 2 AlI3
Mass-Mole We can also start with mass and convert to moles of
product or another reactant We use molar mass and the mole ratio to get to
moles of the compound of interest Calculate the number of moles of ethane (C2H6)
needed to produce 10.0 g of water 2 C2H6 + 7 O2 4 CO2 + 6 H20
10.0 g H2O 1 mol H2O 2 mol C2H6
18.0 g H2O 6 mol H20
= 0.185 mol C2H6
Mass-Mass Conversions
Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)
Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in
Mass-Mass Conversion
Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.
N2 + 3 H2 2 NH3
2.00g N2 1 mol N2 2 mol NH3 17.06g NH3
28.02g N2 1 mol N2 1 mol NH3
= 2.4 g NH3
Practice
How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?
__Ca + __N2 __Ca3N2
Grilled Cheese Sandwich
Bread + Cheese ‘Cheese Melt’
2 B + C B2C
100 bread 30 slices ? sandwiches
Limiting Reactants
aluminum + chlorine gas aluminum chloride
Al(s) + Cl2(g) AlCl3
2 Al(s) + 3 Cl2(g) 2 AlCl3
100 g 100 g ? g
A. 200 g B. 125 g C. 667 g D. ???
Limiting Reactant
Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant.
That reactant is said to be in excess (there is too much).
The other reactant limits how much product we get. Once it runs out, the reaction
s. This is called the limiting reactant.
Limiting Reactant
To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.
The lower amount of a product is the correct answer.
The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!
Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!
Limiting Reactant: Example
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
LimitingLimitingReactantReactant
LR Example Continued
We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .
Limiting Reactant Practice
15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.
Finding the Amount of Excess
By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.
Can we find the amount of excess potassium in the previous problem?
Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2 2 KI We found that Iodine is the limiting reactant, and
19.6 g of potassium iodide are produced.
15.0 g I2 1 mol I2 2 mol K 39.1 g K
254 g I2 1 mol I2 1 mol K= 4.62 g K USED!
15.0 g K – 4.62 g K = 10.38 g K 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!
Limiting Reactant: Recap
1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.
2. Convert ALL of the reactants to the SAME product (pick any product you choose.)
3. The lowest answer is the correct answer.4. The reactant that gave you the lowest answer is the LIMITING
REACTANT.5. The other reactant(s) are in EXCESS.6. To find the amount of excess, subtract the amount used from
the given amount.7. If you have to find more than one product, be sure to start with
the limiting reactant. You don’t have to determine which is the LR over and over again!
Percent Yields
Theoretical yield: max amount of a product that is formed in a reaction.
Actual yield: amount of product that is actually obtained in a reaction Usually less than theoretical. Why?
Why?
Theoretical has assumed that all of limiting reagent has completely reacted. Many reactions do not go to completion Unexpected competing side reactions
limit the formation of products. Some reactants are lost during the
separation process (remember in the lab, pouring off water, leaving silver behind!)
Impure reactants Faulty measuring Poor experimental design or technique
How to calculate?
Percent Yield = Actual Yield x 100% Theoretical yield
Practice together:
20g of HBrO3 is reacted with excess HBr.
1. What is the theoretical yield of Br2?
2. What is the percent yield, if 47.3g is produced?
HBrO3 + 5HBr 3H2O + 3 Br2
Practice alone:
When 35g of Ba(NO3)2 is reacted with excess Na2SO4, 29.8g of BaSO4 is recovered.
Ba(NO3)2 + Na2SO4 BaSO4 + 2NaNO3
1. Calculate the theoretical yield of BaSO4
2. Calculate the percent yield of BaSO4