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Chapter 3. Stoichiometry. Stoichiometry. The study of quantities of materials consumed and produced in chemical reactions. 3.1 Chemical equations. The '+' is read as 'reacts with' and the arrow '' means 'produces'.

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Chapter 3

Chapter 3

Stoichiometry


Stoichiometry
Stoichiometry

  • The study of quantities of materials consumed and produced in chemical reactions.


3 1 chemical equations
3.1 Chemical equations

  • The '+' is read as 'reacts with' and the arrow '' means 'produces'


Because atoms are neither created nor destroyed in a reaction, a chemical equation must have an equal number of atoms of each element on each side of the arrow (i.e. the equation is said to be 'balanced').


Balancing equations
Balancing Equations reaction,

  • Write 'un-balanced' equation using formulas of reactants and products

  • Write 'balanced' equation by determining coefficients that provide equal numbers of each type of atom on each side of the equation (generally, whole number values)

  • Note! Subscripts should never be changed when trying to balance a chemical equation. Changing a subscript changes the actual identity of a product or reactant. Balancing a chemical equation only involves changing the relative amounts of each product or reactant.


Examples
Examples reaction,


We seem to be o.k. with our number of reaction, carbon atoms in both the reactants and products, but we have only half the hydrogen in our products as in our reactants. We can fix this by doubling the relative number of water molecules in the list of products:


Note that while this has balanced our reaction, carbon and hydrogen atoms, we now have 4 oxygen atoms in our products, and only have 2 in our reactants. We can balance our oxygen atoms by doubling the number of oxygen atoms in our reactants:



Balance the following equations
Balance the following equations using the symbols (

C2H5OH (aq) + O2 (g)  CO2 (g) + H2O (g)

Fe (s) + O2 (g)  FeO2 (s)

C2H4 (g)+ O2 (g)  CO2 (g) + H2O (l)


Answer
Answer using the symbols (

2Na (s) + 2H2O (l)  2NaOH (aq) + H2(g)

Fe (s) + O2 (g)  FeO2 (s)

C2H4 (g)+ 3O2 (g)  2CO2 (g) + 2H2O (l)


Which box represents the reaction between NO and O using the symbols (2 to produce NO2.


Homework
Homework using the symbols (

Balancing / Writing Rxn wks


Classifying reactions movie
Classifying reactions movie using the symbols (


Chemical reactivity
Chemical Reactivity using the symbols (

Combination/Synthesis Reaction:

2 or more substances react to form one new product

A + B  C

+ 


solid magnesium and oxygen gas react to produce solid magnesium oxide

2Mg (s) + O2(g)  2MgO (s)

Metal nonmetal ionic compound Diatomic

2+ 2-


Decomposition rxn
Decomposition Rxn magnesium oxide

  • One substance undergoes a reaction to produce two or more substances.

  • Typically occurs when things are heated.

    AX  A + X

     +


Solid calcium carbonate reacts to produce solid calcium oxide and carbon dioxide gas
Solid calcium carbonate reacts to produce solid calcium oxide and carbon dioxide gas

CaCO3 (s)  CaO (s) + CO2 (g)

2+ (2-) 2+ 2- 4+ 2(2-)


Single displacement
Single displacement oxide and carbon dioxide gas

  • One element replaces a similar element in a compound

    A + BX  AX + B

    BX + Y  BY + X

    +  +


Solid copper is dissolved in aqueous silver nitrate to produce solid silver and aqueous copper II nitrate.

Cu(s) + AgNO3 (aq) Ag(s) + Cu(NO3)2 (aq)

Write the sentence for this reaction:

Fe (s) + Cu(NO3)2 (aq)Fe(NO3)2 (aq)+ Cu (s)


Activity series
Activity Series produce solid silver and aqueous copper II nitrate.

  • We need to know what metals are most likely to oxidize others.

  • Example: We can’t store nickel nitrate in an iron container because the solution would eat through the container.


Activity series1
Activity Series produce solid silver and aqueous copper II nitrate.

  • A list of metals arranged in order of decreasing ease of oxidation.

  • Page 139 table


Using activity series
Using activity series produce solid silver and aqueous copper II nitrate.

  • Any metal on the list can be oxidized by the metal below it.

  • Give: FeCl2 + Mg

  • Find: will iron oxidize

    Magnesium metal?

  • I finger on Fe

  • 1 finger on Mg

  • Is the bound chemical below

  • Yes Fe is below Mg.

  • Then complete the reaction


  • Give: NaCl + Mg produce solid silver and aqueous copper II nitrate.

  • Find: will sodium oxidize

    Magnesium metal?

  • I finger on Na

  • 1 finger on Mg

  • Is the bound chemical below

  • no

  • Then the reaction is not possible


What if you don t have an activity series table
What if you don’t have an Activity Series table? produce solid silver and aqueous copper II nitrate.

  • Down Group 1 (I) the "Alkali Metals" the activity increases Cs > Rb > K > Na > Li

  • Down Group 2 (II) the activity increases e.g. Ca > Mg

  • On the same period, the Group 1 metal is more reactive than the group 2 metal

  • the group 2 metal is more reactive than the Group 3 metal,

  • All three are more reactive than the "Transition Metals". e.g. Na > Mg > Al (on Period 3) and K > Ca > Ga > Fe/Cu/Zn etc. (on Period 4)


Double replacement rxn metathesis
Double Replacement Rxn/ Metathesis produce solid silver and aqueous copper II nitrate.

  • The ions of two compounds exchange places in an aqueous solution to form two new compounds.

    AX + BY  AY + BX

  • One of the compounds formed is usually a

    precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, usually water.


Double replacement rxn metathesis1
Double Replacement Rxn/ Metathesis produce solid silver and aqueous copper II nitrate.

AX + BY  AY + BX

+  +


Write the sentence for these double replacement reactions produce solid silver and aqueous copper II nitrate.

KOH (aq) + H2SO4 (aq) K2SO4 (aq) + H2O (l)

FeS (aq) + HCl (aq) FeCl2 (aq) + H2S (aq)


Combustion reaction
Combustion Reaction produce solid silver and aqueous copper II nitrate.

A substance combines with oxygen, releasing a large amount of energy in the form of light and heat.

C3H8 (g)+5O2 (g)  3CO2 (g) + H2O (g)

Usually CO2 (carbon dioxide) / CO (carbon monoxide) and water are produced.


  • Reactive elements combine with oxygen produce solid silver and aqueous copper II nitrate.

    P4(s) + 5O2(g)  P4O10 (s)

    (This is also a synthesis reaction)

  • The burning of natural gas, wood, gasoline

    C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)


Homework1
Homework produce solid silver and aqueous copper II nitrate.

  • Classifying types of Rxns worksheet


3 3 formula weights
3.3 Formula Weights produce solid silver and aqueous copper II nitrate.

  • Although we can’t “count atoms” in a molecule directly, we can count them indirectly if we know their masses.


Formula weights molecular weight
Formula Weights/ Molecular Weight produce solid silver and aqueous copper II nitrate.

  • Sum of atomic masses of each atom in a molecule.

    F.W of H2SO4 = 2(H) + S + 4(O)

    2(1) + 32 + 4(16) = 98amu

    98 g/mol


Percent composition

We can describe composition in two ways produce solid silver and aqueous copper II nitrate.1. number of atoms (amu)

2. % (by mass) of its elements.

Percent Composition


Percent composition1
Percent Composition produce solid silver and aqueous copper II nitrate.

We can find % mass of an atom in a compound from formula mass, by comparing each element present in 1 mole of compound to the total mass of 1 mole of compound


Percent composition equation
Percent Composition Equation produce solid silver and aqueous copper II nitrate.

% element = # of atoms element (atomic weight of the element ) * 100

Formula Weight


Example
Example produce solid silver and aqueous copper II nitrate.

  • Calculate the percentage of nitrogen in Ca(NO3)2


Think
Think: produce solid silver and aqueous copper II nitrate.

% N = # N atoms (m.w N) X 100

m. w Ca(NO3)2


Answer1
Answer produce solid silver and aqueous copper II nitrate.

% N = 2(14.02 N amu) X 100

164.12 Ca(NO3)2amu

= 17%


Question
Question produce solid silver and aqueous copper II nitrate.

  • Calculate the percent composition of each element in C12H22O11


Homework2
Homework produce solid silver and aqueous copper II nitrate.

Molar mass wks

Percent composition wks


3 4 the mole
3.4 The Mole!!!!! produce solid silver and aqueous copper II nitrate.

  • The unit for dealing with, atoms, molecules, ions

  • Abbreviation = mol

    (oh the time you will save!)


History
History produce solid silver and aqueous copper II nitrate.

  • Avogadro

  • Italian

  • 1776-1856


Molar mass
Molar Mass produce solid silver and aqueous copper II nitrate.

  • A dozen eggs = 12

  • A dozen elephants = 12

  • But 12 eggs has a different weight than 12 elephants


  • Thus 1 mole of carbon is 6.02 x 10 produce solid silver and aqueous copper II nitrate.23 molecules but weighs 12 grams

  • 1 mole of sodium is 6.02 x 1023 molecules but weighs 23 g



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.




Grams mass units (amu)

Moles

Molecules

Atoms


Application
Application mass units (amu)

How many moles of water are in 5.380g of water?

  • Molar mass of water = 2(1) + 1(16) = 18g/mol

    5.380g H2O x 1 mol = 0.2989 moles H2O

    18g H2O


Using moles in calculations
Using Moles in calculations mass units (amu)

How many oxygen atoms are present in 4.20 grams NaHCO3?


4.20 g NaHCO mass units (amu) 3 x(1mole NaHCO3) x (6.02e 23molec)3 Oxygen atoms

84 g NaHCO3 1 mol 1 molec NaHCO3

= 9.03 x 10 22 atoms of Oxygen in 4.20 grams NaHCO3


Using moles in calculations1
Using Moles in calculations mass units (amu)

  • How many nitrogen atoms are in 0.25 mol of Ca(NO3)2


Answer2
Answer mass units (amu)

0.25mol x 6.02 x 1023 molec Ca(NO3)2 x _2N atoms___

1 mol 1 molec Ca(NO3)2

= 3.0 x 10 23 moles


Homework3
Homework mass units (amu)

  • Molar mass worksheet

  • G-mol-molec atoms wks


3.5 Determining empirical formula from mass percent mass units (amu)

  • Recall: Empirical formula: simplest whole # ratio of atoms in a compound.

  • Recall: We can find % mass from formula mass, by comparing each element present in 1 mole of compound to the total mass of 1 mole of compound


Example1
Example mass units (amu)

Example: Vitamin C is composed of 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula?


How to attack the problem mass units (amu)

1. Convert mass % into grams (assume 100g ie: 40.92% = 40.92 g)

2. Convert grams to moles using molar mass.

3. Divide moles of each element by the smallest number of moles present. You may round to nearest whole # . (This establishes a ratio of comparison)


Answer3
Answer mass units (amu)

40.92 g C 1mol C= 3.4 moles C /3.4 = 1 C

12 g C

54.40 g O 1mol O = 3.4 moles O / 3.4 = 1 O

16g O

4.58 g H 1mol H = 4.58 moles H/ 3.4 = 1.3 = 1 H

1 g H

  • CHO


Empirical formula mass units (amu)

Mass % elements

Find:

Given:

Calculate mole ratio

Assume 100g of sample

Grams of each element

Moles of each element

Use atomic weights


Question1
Question mass units (amu)

Compound X is composed of 55.3% K, 14.6% P , and 30.1% O.

What is the empirical formula of compound X?


55.3 g K mass units (amu) 1mol K= 1.4 moles K /.47 = 3 K

39 g K

14.6 g P 1mol P = 0.47 moles P/ .47 = 1 P

31 g P

30.1 g O 1mol O = 1.9 moles O / .47 = 4 O

16g O

  • K3PO4


Determine Molecular formula from Empirical Formula mass units (amu)

Recall:

Molecular formula: the exact formula of a molecules, giving types of atoms and the number of each type.

1. Using mass % and molar mass, determine mass of each element in 1 mole of compound (same)

2. Determine number of moles of each element in 1 mole of compound. (same)

3. The integers from the previous step represent the subscripts in the molecular formula ( you just don’t divide by smallest mole ratio)


Let’s look back at our work mass units (amu)

40.92 g C1mol C= 3.4 moles C /3.4 = 1 C

12 g C

54.40 g O1mol O = 3.4 moles O / 3.4 = 1 O

16g O

4.48 g H1mol H = 4.48 moles H/ 3.4 = 1 H

1 g H

C3H4O3= molecular formula


Shortcut mass units (amu)

n = Molecular Formula Weight

Empirical Formula Weight

(where n = # of atoms)


Try it ….. mass units (amu)

The molecular weight of butyric acid is

88 amu. If the empirical formula is C2H4O. What is the molecular formula?


  • The empirical formula was given, us it to find the empirical formula weight.

  • C2H4O = 12 + 12+ 1+1+1+1+16 = 44 amu

  • 2. The molecular formula weight was given (88amu) plug it into the short cut formula.

  • n = 88 amu = 2

  • 44

  • 3. Apply the new number of atoms.

  • Molecular formula = (empirical) n

  • (C2H4O)2 = Molecular Formula = C4H8O2


Homework4
Homework formula weight.

  • Chan pg 108 43, 44, 45, 49, 50, 52, 53

  • BLPg 106-107: 37a, 39, 42, 44, 45, 46


3 6 qualitative information from balanced equations
3.6 Qualitative information from balanced equations. formula weight.

  • Stoichiometry: mixing exactly enough chemical so that all is used

    Mass-Mass problems a new highway!!!!!!

    g given  mol given  mol required  g required

    Think: (grams to moles to moles to grams)


Time out for mole ratios
Time out for mole ratios formula weight.

2H2 (g) + O2 (g)  2H2O (l)

Coefficients tell us that 2 molecules of H2 react with each molecule of O2 to form 2 molecules of H2O.

(recall: 6.02 x 1023 molecule = 1 mol)

These coefficients can be used to convert between quantities of reactants and products.


Example of mole ratios
Example of mole ratios formula weight.

  • Calculate the number of moles of H2O produced from 1.57 mol of O2?

    2H2 (g) + O2 (g)  2H2O (l)


2H formula weight.2 (g) + O2 (g)  2H2O (l)

1.57 mol O2 ( 2 mol H2O) =

1 mol O2

= 3.14 mole H2O


Silicon carbide is made by heating silicon dioxide to high temperatures.

SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)

How many grams of CO are formed by complete rxn of 5.00 g SiO2?

HINT: always make sure your equation is balanced first or mole ratios will be wrong.


Grams temperatures.substance A

Grams of substance B

Find:

Given:

Use molar mass of B

Use molar mass of A

Moles of substance A

Moles of substance B

Use coefficients of A and B from balanced equation

Every line of dimensional analysis should have a unit AND a chemical formula!!!!!


Given: temperatures.5.00 g SiO2

Find : CO g

SiO2 (s) + 3C (s)  SiC(s) + 2CO (g)

think: grams to moles to moles to grams

5.00 g SiO21mol SiO22 mol CO28 g CO = 4.67 g CO

60 g SiO21 mol SiO21 mol CO

Mole ratio


How many moles of temperatures.sulfuric acid would be needed to produce 4.80 moles of molecular iodine (I2) according to the following balanced equation.

10HI + 2KMnO4 + 3H2SO4 5I2 + 2MnSO4 + K2SO4 + 8H2O


4.80 mol I temperatures.23 mol H2SO4= 2.88 mol H2SO4

5 mol I2


Homework5
Homework temperatures.

  • Chan pg: pg 109 46, 66,67,68,69

  • BLPg: 107 #’s 51, 53, 55, 57, 59, 60

    Note: you must be able to pass a pop quiz with 80% accuracy to be able to move on to the next section. If not additional problems and help will be mandatory.


3 7 limiting reactant
3.7 Limiting Reactant temperatures.

The number of products that can form is limited by the amount of reactant present.

The limiting reactant is the one that gives the least amount of product.

Reactants  Products



If you have 10 slices of bread and 7 slices of ham, how many sandwiches will you be able to make according to the previous reaction?

2Bd + 1Hm  Bd2Hm

What is your limiting reagent?


Extra Tasty Ham Slabs sandwiches will you be able to make according to the previous reaction?


What is the limiting reactant blue or red
What is the limiting reactant, Blue or Red? sandwiches will you be able to make according to the previous reaction?


Attacking limiting reactant problems
Attacking limiting reactant problems sandwiches will you be able to make according to the previous reaction?

  • Using the grams of each reactant and their mole ratios calculate how many grams of product will be formed.

  • Compare the amount of product formed by each reactant.

  • The reactant that give the lesser amount of product is the limiting reactant.


Try one
Try One…. sandwiches will you be able to make according to the previous reaction?

When a mixture of silver and sulfur is heated, silver sulfide is formed:

16 Ag (s) + S8 (s)  8 Ag2S (s)

What mass of Ag2S is produced from a mixture of 2.0 g of Ag and 2.0 g of S8?


2 g Ag sandwiches will you be able to make according to the previous reaction? 1 mol Ag 8 mol Ag2S 248 g Ag2S

108 g Ag 16 mol Ag 1 mol Ag2S

= 2.3g Ag2S can be formed from 2 g Ag

2 g S81 mol S88 mol Ag2S 248 g Ag2S

256 g S8 1 mol S8 1 mol Ag2S

= 15.5 g Ag2S can be formed from 2.0 g S8


= 15.5 g Ag sandwiches will you be able to make according to the previous reaction? 2S can be formed from 2.0 g S8

Most amount of product can be formed (aka: ham slabs!!!)

= 2.3g Ag2S can be formed from 2 g Ag

Least amount of product can be formed Thus Ag is the limiting reactant!!!! (aka: bread)


Try one again man these are fun
Try one again…. (man these are fun!!!) sandwiches will you be able to make according to the previous reaction?

How many grams of urea can be produced from 10.0 g of NH3 and 10.0 g of CO2?

2NH3 + CO2 CH4N2O + H2O

(Urea)


Answer4
Answer sandwiches will you be able to make according to the previous reaction?

17.6 g of Urea from 10g NH3

13.6 g of urea from 10 g CO2 LIMITNG REACANT


Theoretical yield
Theoretical Yield sandwiches will you be able to make according to the previous reaction?

  • The amount of product calculated based on the limiting reactant.

    % yield = collected yield X 100

    Predicted yield

    Actual/collected = what you really get

    Theoretical/Predicted = What you might get or predict you will get.


Try one1
Try one! sandwiches will you be able to make according to the previous reaction?

In a chemical reaction the theoretical yield is 145 g. If the percent yield was 92.0%, what was the actual yield?


Answer5
Answer sandwiches will you be able to make according to the previous reaction?

92.0 = actual x 100

145

= 133 g


Calculating theoretical yield
Calculating Theoretical Yield sandwiches will you be able to make according to the previous reaction?

When you are doing a limiting reactant equation,

(g – mol-mol-g) you are calculating the theoretical yield. It is not until you actually run the experiment in the lab that you will get your actual yield.

Some times not all the reactants react, or they may react in a way different than you desired. Chemistry is not perfect ( unlike you guys)


Homework 3 7
Homework 3.7 sandwiches will you be able to make according to the previous reaction?

Chang: pg 107-108 # 63, 67, 69, 70, 77, 78

BL: Page


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