3,stoichiometry

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Stoichiometry: the study of quantities of materials consumed and produced in chemical reactions.

Stoichiometry derived from the Greek ward

stoichion ≡ element , metron ≡ measure

So stoichiometry used to describe the quantitative aspects of chemical composition and reactions

To understand chemical stoichiometry, you must first understand the concept of relative atomic masses.

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Atomic Mass� Atoms are so small, it is difficult to discuss how much they weigh in

grams.� An atom consists of a small dense nucleus containing +vely charged

protons (p) and neutral neutrons (n). The nucleus surrounded by –vely charged electrons.

� In neutral atoms: +ve charge of nucleus balanced by

-ve charged electrons (e)� number of (p) = number of (e), called atomic number (Z)

The identity of an atom can be determine from its atomic number, .. e.g. the atomic number of carbon is 6. This means that each C-atom has 6p and 6n or every atom in the universe that contain 6p and 6n is correctly named carbon

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� number of (p) + number of (n), called mass number (A)

mass no. = number of (p) + number of (n)

= atomic no. + number of (n)

A = Z + number of (n)

number of (n) = A - Z� Each (p) and (n) contribute one unit to the mass no.

� C-atom contain 6 (p) and 6 (n)

A = 12 , Z = 6 126C

U-atom contain 92 (p) and 146 (n)

A = 238 , Z = 92 U�

23892

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� The mass of an atom is measured relative to the mass of chosen atomic standard which is C-12, which is define exactly 12 atomic mass unit (amu).

(each proton and neutron consider as 1 amu), C-12 contain 6 protons and 6 neutrons so its mass=12amu

� an atomic mass unit (amu) for any element = (1/12) of the mass of a carbon-12 atom

� This gives us a basis for comparison� 1H= 1amu, 12C has almost 12 times the mass of 1H

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� The decimal numbers on the periodic table are atomic masses in amu

� They are not whole numbers because all atoms of the same element identical in atomic number but not in mass number.

Example: atomic mass of carbon is not 12.0 amu but 12.01 amu.� Because they are based on averages of atoms and of isotopes.� Isotopes: atoms that have different number of neutrons, and

therefore different mass number� can figure out the average atomic mass from the mass of the

isotopes and their relative abundance.� add up the percent as decimals times the masses of the isotopes as

shown in the following examples.

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Example (1):There are two isotopes of carbon C with a mass of 12.00000 amu(98.892%), and 13C with a mass of 13.00335 amu (1.108%), what is the average atomic mass of carbon.

Average atomic mass = (atomic mass x %abundance) first isotope

+ (atomic mass x %abundance) second isotope

= 12.00 x + 13.00335 x

= 12.011 amu

98.892100

1.108100

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Example (2):� There are two isotopes of lithium (Li) , one with an atomic mass of

6.0151 amu and one with a mass of 7.0160 amu. What is the percent abundance of each if the average atomic mass of lithium 6.941 amu

Average atomic mass = (atomic mass x %abundance) first isotope

+ (atomic mass x %abundance) second isotope

If the %abundance for the first isotope = χ

Then %abundance for the second isotope = 1 – χ

6.941 = 6.0151 (χ) + 7.0160 (1 – χ)

6.941 = 6.0151 (χ) + 7.0160 – 7.0160 χ

χ = 0.075 (%abundance for the first isotope = 0.075 x 100 = 7.5%)

%abundance for the second isotope = 100 – 7.5 = 92.5%

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The Mole� Because atoms have small masses, no usable scale can be devised

to weigh them. In any real situation, we deal with macroscopic samples containing enormous number of atoms.

� The mole has been established to count atoms� Mole define as the number equal to the number of carbon atoms

in exactly 12 g of pure C-12 which is equal to 6.022 x 1023

� This number is called Avogadro’s number� It is a very large number, but still, just a number� 1.0 mol of something consist of 6.022 x 1023 unit of that thing.

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mass of 1 atom C = 12 amu (it is called atomic mass)

mass of 2 atoms C = 2 X 12 amu

mass of 3 atoms C = 3 x 12 amu

mass of 6.022 x 1023 atoms C = 6.022 x 1023 x 12 amu

1 mole C atoms = 12 g of C atoms (it is called atomic mass)

in the same manner :

1mole of O atoms = 16 g of O atoms

1 mole of Fe atoms = 55.85 g Fe atoms

so:

1 mole of any element = molar mass = 6.022 x1023atoms

1mole of H = 1.008 g = 6.022 x1023atoms

1 mole of Ag = 107.9 g = 6.022 x1023atoms

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� Converting mass to molesHow many moles are in 0.551 g of potassium (K)

Calculate the number of atoms of Cu in 3.5 g of Cu

no. of moles of K = 0.551 g K x

•Converting mass to atoms

39.1 g K1 mol K = 0.0141 mol K no. of moles of K = 0.551 g K x

39.1 g K

no. of Cu atoms = 3.5 g Cu x63.5 g Cu

6.022 x 1023 atoms Cu

= 3.32 x 1022 atoms Cu

1 mole K = 39.1 g (molar mass mass) = 6.022 x 1023 atoms

1 mole Cu = 63.5 g (molar mass mass)= 6.022 x 1023 atoms

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� Converting moles to mass: how many grams are present in 0.35 moles Na

mass of Na = 0.35 mol Na x1 mol Na23.5 g Na

= 8.225 g Na

•Converting moles to atoms:How many oxygen atoms are in 0.27 moles

no. of O atoms = 0.27 mol O x 6.022 x 1023 atoms O

1 mol O= 1.63 X 1023 atoms O

1 mole O = 16 g O = 6.022 x 1023 atoms O

1mol Na = 23.5 g Na = 6.022 x 1023 atoms Na

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� Converting atoms to moles:How many moles are in 500 atoms Fe.

1 mol Fe = 55.85 g Fe = 6.022 x 1023 atoms Fe

no. of mol Fe = 500 atoms Fe x 1 mol Fe6.022 x 1023 atoms Fe

= 8.3 x 10-22 mol Fe

•Converting atoms to mass (g):What is the mass of 1.5 x 1020 atoms Cl

1 mol Cl = 35.5 g Cl = 6.022 x 1023 atoms Cl

mass of Cl (g) = 1.5 x 1022 atoms Cl x 35.5 g Cl6.022 x 1023 atoms Cl

= 0.884 g Cl

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molecular mass� A chemical compound is a collection of atoms.� H2O contain:

2 atoms H + 1 atom O2 (1.008 amu H) + 1 ( 16.00 amu O)

2.016 amu + 16.00 amu = 18.016 amuThis called the molecular mass of H2O

If we use more than one molecule of H2O

2 x 2 atoms H + 2 x 1 atoms O 2 x 6.022 x 1023 atoms H + 1 x 6.022x 1023 atoms O (2 x 1 mol H + 1 mol O)

2 X 1 g H + 1 x 16 g OSo: 1 mol H2O contain 2 mol H and 1 mol O

1 mol H2O = 2 (mass of H) + 1 (mass of O)

= 2 x 1 g + 1 x 16 g = 18 g15

� Mass in gram of 1 mole of a substance often called molar mass or molar weight.

� To determine the molar mass of a compound, add up the atomic masses of the elements that make it up.

� Example:

mass of 1 mol CH4 = mass of 1 mol C + 4 x mass of H

= 1 x 12 g + 4 x 1 g

= 16 g

So:

1 mol of a compound = molar mass = 6.022 x 1023 molecules

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� Calculate the molar mass for each of the following compounds:

1) Mg3P2

molar mass Mg3P2 = 3 x molar mass Mg + 2 molar mass P

= 3 x 24.3 g + 2 x 31 g = 134.9 g

2) Ca(NO3)2

molar mass Ca(NO3)2 = 40 g + 2 {14 + (3 x 16)} = 164 g

3) Al2(Cr2O7)3

molar mass Al2(Cr2O7)3 = (2 x 27) g + 3 {(2 x 52) +(7 x 16)}

= 702 g 4) CaSO4· 2H2O

molar mass CaSO4· 2H2O = 40 g + 32g + 4 x 16g + 2 x 18g

= 172 g

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Example (1):

a) how many moles of Ca(NO3)2 are present in 3.5 g Ca(NO3)2

b) how many molecules Ca(NO3)2 present in 3.5 g Ca(NO3)2 1 mol Ca(NO3)2 = molar mass = Avogadro’s no. 1 mol Ca(NO3)2 = 164 g Ca(NO3)2 = 6.022 x 1023 molecules

a) no. of mol of Ca(NO3)2 = 3.5 g Ca(NO3)2 x1 mol Ca(NO3)2 164 g Ca(NO3)2

= 0.0213 mol Ca(NO3)2

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b) no. of Ca(NO3)2 molecules =

3.5 g Ca(NO3)2 x6.022 x 1023 molecules Ca(NO3)2

164 g Ca(NO3)2

= 1.29 x 1022 molecules Ca(NO3)2

Example (2):

a) what is the mass of 0.75 mol Mg3P2 b) how many molecules of Mg3P2 present in

0.75 mol Mg3P2 c) how many P atoms present in 0.75 mol Mg3P2

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a) mass of Mg3P2 =

1 mol Mg3P2 = 134.9 g Mg3P2 = 6.022 x 1023 molecules

0.75 mol Mg3P2 x 134.9 g Mg3P2

1 mol Mg3P2

= 101.175 g Mg3P2

b) no. of Mg3P2 molecules =0.75 mol Mg3P2 x 1 mol Mg3P2

6.022 x 1023 molecules

= 4.52 x 1023 molecules

c) 1 mol Mg3P2 contain 2 mol P

mol of P = 0.75 mol Mg3P2 x2 mol P

1 mol Mg3P2= 1.5 mol P

no. of P atoms = 1.5 mol P x 6.022 x 1023 molecules1 mol P

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• Example: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 H = 6.022 x 1023 atoms H

5.82 x 1024 atoms H

1 mol C3H8O molecules = 8 mol H atoms

72.5 g C3H8O1 mol C3H8O60 g C3H8O

x8 mol H atoms1 mol C3H8O

x6.022 x 1023 H atoms

1 mol H atomsx =

� Example:

How many H-atoms are present in 25.6g urea, (NH2)2CO, (molar mass = 60.06 g)

grams of urea moles of urea moles of H atoms of H

no. of H-atoms = 1.03 x 1024

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25.6 g urea xno. of H-atoms = 1 mol urea60.06 g urea

x 4 mol H1 mol urea

x 6.022 x 1023 atoms

1 mol H

Percentage composition of a compound� Two ways to describe the composition of a compound:

a) in terms of the number of its constituent atoms.

e.g.: CH4 1 C-atom : 4 H-atoms

H2O 2 H-atoms : 1 O-atom

b) in terms of the percentage (by mass) of its elements

Find the mass of each element, divide by the total mass of all elements, multiply by a 100.

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� Easiest if you use a mole of the compound.• Percent composition is the atomic mass for each element divided by

the molar mass of the compound multiplied by 100:

Example (1): Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

% element = (no. of atoms x atomic mass) elementmolar mass of the compound

100 %

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% Ag = 29.0 g33.3 g

x 100% = 87.09 %

% S = 4.30 g33.3 g

x 100% = 12.91 %100 %

� Example (2): calculate the mass percent of each element in the following compounds:

a) C2H6 : molecular mass C2H6 = 2 x 12 + 6 x 1 = 30 g

% C = 2 x 12 g

30 gx 100 % = 80 %

% H = 6 x 1 g30 g

x 100 % = 20%100 %

b) C3H4O2 : molar mass = 3 x 12 + 4 x 1 + 2 x 16 = 72 g

% C = 3 x 12 g72 g

x 100 % = 50.01 %

% H = 4 x 1 g72 g

x 100 % = 5.57 %

% O = 2 x 16 g72 g

x 100 % = 44.42 %

100 %

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� Example (3):Aspartame, an artificial sweetener, is 9.52% nitrogen. There are two nitrogen atoms per molecule. What is the molecular mass of aspartame

mass % N =no. of N atoms x molar mass Nmolar mass of aspartame

x 100 %

molar mass of aspartame =no. of N atoms x atomic mass N

mass % Nx 100 %

2 x 14 g9.52 %

= x 100 %

= 294.12 g/mol

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Determining the formula of compound If a new compound is prepared, the first thing to

do is to determine its formula.This is done by:� Taking a weighed sample and decompose it into its

component elements. Determine the mass of each element or the mass % of each one.

� Reacting a weighed sample with O2 to produce CO2, H2O, N2, which are then collected and weighed.

from the mass of these compounds, determine the mass of each element in the compound.

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Then find the number of moles of each element

Find the lowest mole ratio of atoms in the

Compound which is called the Empirical Formula

The molecular formula: the actual ratio of elements in a compound. The two formulas can be the same.

CH2 : empirical formula

C2H4 : is molecular formula , empirical formula is: CH2

C3H6 :is molecular formula , empirical formula is: CH2

H2O: both

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To obtain an Empirical Formula1. 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.

4. If whole numbers are not obtained* in step 3), multiply through by the smallest number that will give all whole numbers

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you can determine the empirical formula from:

1) The mass of each element Example: A sample is found to contain 2.34 g N and

5.34g O. Determine a empirical formula for this substance.

mol of N = 2.34 g N x 1 mol N14 g N

= 0.167 mol N

mol of O = 5.34 g O x 1 mol O16 g O = 0.334 mol O

mol N : mol O

0.3340.167 :0.167 0.1671 : 2

Empirical formula is: NO2

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2) From percent composition,

Example: A sample is 57.53% C, 5.5%H, 9.59%N, and 27.40%O what is its empirical formula.

Consider a sample size of 100 grams. This will contain:

57.53 g C, 5.50 g H, 9.59 g N

mol of C = 57.53 g C x 1 mol C 12 g C

= 4.79 mol C

mol of H = 5.38 g H x 1 mol H1.008 g H = 5.46 mol H

mol of N = 9.59 g N x 1 mol N14 g N

= 0.67 mol N

mol of O = 27.40 g O x 1 mol O16 g O = 1.71 mol O

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mol C : mol H : mol N : mol O

4.79 : 5.46 : 0.67 : 1.710.67 0.67 0.67 0.67

7 : 8 : 1 : 2.5

14 : 16 : 2 : 5

So the empirical formula : C14H16N2O5

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3) Reacting a weighed sample with O2 to produce CO2, H2O, N2

….ect

Example: A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O2 . 0.2998 g of CO2 and 0.0819 g of H2O are produced. What is the empirical formula of the compound

1 mol CO2 = 44 g 1 mol H2O = 18 g

mol CO2 = 0.2998 g CO2 x 1 mol CO244 g CO2

= 6.81 X 10-3 mol CO2

1 mol CO2 contain 1 mol C mol C = 6.81 X 10-3 mol

mass of C = 6.81 X 10-3 mol C x1 mol C12 g C

= 0.0818 g C

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mol H2O = 0.0819 g H2O x 1 mol H2O18 g H2O

= 4.55 X 10-3 mol H2O

1 mol H2O = 2 mol H

mol H = 4.55 x 10-3 mol H2O x 2 mol H1 mol H2O

= 9.1 x 10-3 mol H

mass H = 9.1 x 10-3 mol H x 1.008 g H1 mol H

= 9.173 x 10-3 g H

Mass of sample = mass C + mass H + mass O

0.2000 g = 0.0818 g + 9.173 x 10-3 g + mass O

mass of O = 0.109 g

mol O = 0.109 g O x 1 mol O16 g O

= 6.18 x 10-3 mol O

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To find the empirical formula:

mol C : mol H : mol O

6.81 X 10-3 : 9.1 x 10-3 : 6.18 x 10-3

6.18 x 10-3 6.18 x 10-3 6.18 x 10-3

1 : 1.47 : 1

2 : 3 : 2

Must be whole number

Empirical formula is : C2H3O2

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Empirical To Molecular Formulas� Empirical is lowest ratio� Molecular is actual molecule� Need Molar mass� Ratio of empirical to molar mass will tell you the

molecular formula� Must be a whole number because...

Molecular formula = x ( empirical formula)

X =Molar mass of molecular formulaMolar mass of empirical formula

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Example:A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its empirical and molecular formula? %N = 100% - 69.6% = 30.4%

mol of S = 69.6 g S x 1mol S32 g S

= 2.175 mol S

mol of N = 30.4 g N x 1 mol N14 g N

= 2.172 mol Nmol S : mol N

2.175 : 2.172

1 : 1

Empirical formula : NS X = 184/46 = 4Molecular formula : N4S4

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Chemical equationsChemical changes involves reorganization of atoms in one or more substances.

When CH4 (natural gas) combine with O2, CO2 and H2O are formed.

This process is represented by chemical equation

A chemical equation uses chemical symbols to show what happens during a chemical reaction

All chemical equations have two parts� Reactants: the substances you start with� products: the substances you end up with.

in the chemical equation, the reactants are written on the left side of an arrow and the product on the right side.

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CH4 + O2 CO2 + H2O

reactants products

Notice that in chemical reactions:� Bonds have been broken, new bonds have bee formed.� Atoms neither created nor destroyed. All atoms present in

reactants must be present in products (law of conservation of mass).

(number of each type of atoms in reactants = number of each type atoms in products)

This is done by balancing a chemical equation

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Balancing equations

CH4 + O2 → CO2 + H2O

ReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

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CH4 + 2 O2 → CO2 + 2 H2O

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Chemical equation gives two important types of information:� The physical nature of reactants and products, if it is

liquid (ℓ) , solid (s) , gas (g) , aqueous (aq)� Relative number of reactants and products which indicated

by the coefficients in the balance equation

e.g. : CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

1 molecule 2 molecules 1 molecules 2 molecules

6.022x1023 2x6.022x1023 6.022x1023 2x6.022x1023

1 mol 2 mol 1 mol 2 mol

16 g 2 x 32 g 44 g 2 x 18 g

80 g 80 g

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Balancing Chemical Equations1. 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

C2H6 + O2 CO2 + H2O

2. 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.

2C2H6NOT C4H12

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

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right

multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

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

2 oxygenon left

4 oxygen(2 x 2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3 x 1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

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

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products

4 C12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)

14 O (7 x 2) 14 O (4 x 2 + 6)

Practice� 3 Ca(OH)2 +2 H3PO4 →6 H2O + Ca3(PO4)2

� Fe2O3(s) + 2Al(s) → 2Fe(s) + Al2O3(s)

� Fe2S3(g) + 6 HCl(g) → 2 FeCl3(s) + 3 H2S(g)

� C4H9OH(ℓ) +6 O2(g) → 4CO2(g) +5 H2O(g)

� C8H18(ℓ) +(25/2) O2(g) → 8 CO2(g) + 9 H2O(g)

2 C8H18(ℓ) +25 O2(g) → 16 CO2(g) + 18 H2O(g) 46

Stoichiometric calculations: amount of reactants and products

A basic question raised in the chemical lab. is ‘‘how much product will be formed from specific amount of starting materials reactants’’ or ‘‘how starting material must be used to obtain a specific amount of product’’. To interpret a reaction quantitatively, apply the concepts of molar masses and the mole concept.

Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

We use moles to calculate the amount of product formed in a reaction. This approach is called the mole method, which means that the stoichiometric coefficients in a chemical equation can be interpreted as a number of moles of each substance

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� The coefficient in chemical equation represent the number of molecules, not masses of molecules

� To develop the principles for dealing with the stoichiometry of reactions, consider the reaction of C3H8 with O2, what mass of O2 will react with 96.1 g of C3H8

� In doing stoichiometry, follow the following steps:1) write the balance chemical equation for the reaction

C3H8(g) + 5O2(g) → 3CO2(g) + 4 H2O(g)

1 molecule 5 molecules 3 molecules 4 molecules

1 mol 5 moles 3 moles 4 moles

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2) Convert the known masses of reactants and products to moles

molar mass C3H8 = 3 x 12.01 + 8 x 1.008 = 44.1g

Moles of C3H8 = 96.1 g C3H3 x1 mol C3H8

44.1 g C3H8= 2.18 mol C3H8

3) Use the balance equation to set up the appropriate mole ratio and calculate the number of moles of the desired reactant or product.

1mol C3H8 react with 5 mol O2

mol of O2 = 2.18 mol C3H8 x 5 mol O2

1 mol C3H8= 10.9 mol O2

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4) Convert from moles back to grams if required by the problem

Mass O2 = 10.9 mol O2 x32 g O2

1 mol O2

= 349 g O2

What mass of CO2 produce when 96.1g C3H8 is combusted with O2 1 mol C3H8 produce 3 mol CO2

Mol of CO2 = 2.18 mol C3H8 x3 mol CO2

1 mol C3H8

= 6.54 mol CO2

Molar mass CO2 = 44 g CO2

molar mass O2 = 32 g O2

Mass of CO2 = 6.54 mol CO2 x44 g CO2

1 mol CO2

= 288 g CO2

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Use the molar mass

of (A)

Moles of compound

(A)

Moles of compound

(B)

Mass (g) of

compound (B)

Use the molar mass

of (B)

Stoiochiometry of Balanced

Equation

Mass(g) of

compound (A)

•Decide where to start based on the units you are given •and stop based on what unit you are asked for

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Examples (1)The decomposition of KClO3 (molar mass = 122.5 g/mol) occurs according to the following equation:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

A 25.5 g sample of Potassium chlorate is decomposed.

1.How many moles of O2(g) are produced?

2.How many grams of O2(g) (molar mass = 32g/mol)?

3.How many grams of KCl (molar mass = 74.5g/mol) is produce?

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mol of KClO3 = 25.5 g KClO3 x 1 mol KClO3

122.5 g KClO3

= 0.208 mol KClO3

1) 2 mol KClO3 form 3 mol O2

mol O2 = 0.208 mol KClO3 x 3 mol O2

2 mol KClO3 = 0.312 mol O2

2) mass of O2 = 0.312 mol O2 x

1 mol O2

32 g O2= 9.98 g O2

3) 2 mol KClO3 form 2 mol KCl

mol KCl = 0.208 mol KClO3 x 2 mol KClO3 2 mol KCl = 0.208 mol KCl

mass of KCl = 0.208 mol KCl x 1 mol KCl74.5 g KCl = 15.50 g KCl

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Example (2)For the following reaction:

K2PtCl4(aq) + 2 NH3(aq) → Pt(NH3)2Cl2(s)+ 2KCl(aq)

1) What is the mass of NH3 (molar mass = 17g/mol) needed to react completely with 65 g K2PtCl4 (molar mass = 415g/mol) ?

2) What mass of Pt(NH3)2Cl2 (molar mass = 297g/mol) can be produced from 65g of K2PtCl4

3) What is the mass of KCl (molar mass= 74.5g/mol) will be produced?

4) What is the mass of Pt(NH3)2Cl2 produce from the reaction of 65g of NH3?

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mol of K2PtCl4 = 65 g K2PtCl4 x1 mol K2PtCl4 415g K2PtCl4

= 0.157 mol of K2PtCl4

I mol K2PtCl4 react with 2 mol NH3

mol of NH3 = 0.157 mol of K2PtCl4 x 2 mol NH3 I mol K2PtCl4

= 0.314 mol of NH3

Mass of NH3 = 0.314 mol of NH3 x 1 mol NH3 17 g NH3 = 5.338 g NH3

I mol K2PtCl4 produce 1 mol Pt(NH3)2Cl2

mol Pt(NH3)2Cl2 = 0.157 mol of K2PtCl4 x I mol K2PtCl4

1 mol Pt(NH3)2Cl2 = 0.157 mol

Mass of Pt(NH3)2Cl2 = 0.157 mol Pt(NH3)2Cl2 x1 mol Pt(NH3)2Cl2

297g Pt(NH3)2Cl2

= 46.63 g Pt(NH3)2Cl2

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3) 1 mol K2PtCl4 produce 2 mol KCl

mol of KCl =0.157 mol of K2PtCl4 x 2 mol KCl

1 mol K2PtCl4 = 0.317 mol of KCl

mass of KCl = 0.317 mol of KCl x1 mol of KCl74.5 g KCl = 23.62 g KCl

4) 2 mol NH3 produce 1 mol Pt(NH3)2Cl2

mass of Pt(NH3)2Cl2 = 65g of NH3 x

1 mol NH3 17 g NH3

x2 mol NH3

1 mol Pt(NH3)2Cl2

x1 mol Pt(NH3)2Cl2

297g Pt(NH3)2Cl2

= 567.79 g Pt(NH3)2Cl2

1)massto moles

2)Stoichiometric factor

3) Moles to mass

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Calculation involving a limiting Reagent

When chemicals are mixed to react, they are often mixed in stoichiometric quantities; that is in exactly the proportions indicated by the balance equation so that all reactants are used up at the same time.We can explain what we mean by stoichiometric quantities in the following example.

CH4(g) + H2O(g) → 3 H2(g) + CO(g)

If 2.5g CH4(molar mass=16g/mol react with H2O (molar mass = 18g/mol) …..calculate the mass of H2O required to react completely with CH4 .

mass of H2O = 2.5 g CH4 x 16 g CH4

1 mol CH4 x 1 mol CH4

1 mol H2O x1 mol H2O 18 g H2O

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mass H2O = 2.81g

Thus if 2.5 g CH4 mixed with 2.18 g H2O both reactants will run out as the same time:

reactants mixed in stoichiometric amounts

If 2.5 g CH4 mixed with 3.0 g H2O , CH4 will be consumed before H2O run out (some of H2O remain unreacted):

H2O present in excess

Amount CH4 of limit the amount of product that can be formed. Once the CH4 is consumed no more product can be formed .

CH4 is called the limiting reactant

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The limiting reactant is:� Reactant that determines the amount of product formed. � the reactant used up first in the reaction.� Makes the least product.

The excess reactant is:� the one left over.

Determining the Limiting reagent� To determine the limiting reagent requires that you do two

stoichiometry problems.� Figure out how much product each reactant makes.� The one that makes the least is the limiting reagent.

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Example (1)Ammonia is produced by the followin reaction

N2 + 3 H2 → 2 NH3

1) What mass of NH3 (molar mass = 17g/mol) can be produced from a mixture of 500. g N2(molar mass =28g/mol) and 100. g H2 (2 g/mol)?

2) How much unreacted material remains?

1) a)calculate the mass NH3 of from the mass of N2

mass NH3 = 500 g N2 x

1 mol N2

28 g N2

x1 mol N2

2 mol NH3 x1 mol NH3

17g NH3

= 607.14 g NH3

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b)calculate the mass NH3 of from the mass of H2

mass NH3 = 100 g H2 x1 mol H2

2 g H2

x 2 mol NH3

3 mol H2

x17g NH3

1 mol NH3

= 566.67 g NH3

Smaller amount

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Example:A 2.00 g sample of NH3(molar mass = 17 g/mol) is mixed

with 4.00 g of O2.(molar mass= 32 g/mol)

4 NH3(g) + 5 O2(g)→4 NO(g) + 6 H2O(g)

1) Which is the limiting reactant

2) What is the mass of H2O produce

3) how much excess reactant remains after the reaction has stopped?

mol of NH3 = 2.00 g NH3 x1) 1 mol NH3 17 g NH3

= 0.1176 mol NH3

mol of O2 = 4.00 g O2 x1 mol O2

32 g O2 = 0.125 mol O2

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4 NH3 5 O2

Moles: 0.1176 0.125(0.125/5)(0.1176/4)

0.0250.0294

Amount of the products depends on the amount of O2

2) mass of H2O = 0.125 mol O2 x

5 mol O2

6 mol H2O x1mol H2O18 g H2O

= 2.7 g H2O Convert mass to mol

Use stoichiometry from the equation

Convert mol to mass

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3)(Amount of excess = (total amount of - ( amount reactedreactant unreacted) excess reactant) of excess reactant)

mol of NH3 reactedwith all O2

= = 0.125 mol O2 x 4 mol NH3 5 mol O2

=0.10 mol NH3

mass of NH3 reacted =1 mol O2 17 g NH3

mass of unreacted NH3 = 2.00 g – 1.7 g = 0.3 g

0.10 mol NH3 x = 1.7 g NH3

Total amounts Reacted amounts

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Percent Yield� The theoretical yield what the balanced equation tells you,

the amount of product formed when the limiting reactant is completely convert to product.

� The actual yield is what you make in the lab., obtain experimentally.

� Percent yield tells us how “efficient” a reaction is.� The actual yield always less than the theoritical yield� Percent yield can not be bigger than 100 %.

Percent yield =actual yield

Theoritical yieldx 100%

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Example(1)

A 6.0 g of Al (atomic mass = 27g/mol) reacts with excess Br2. 50.3 g of AlBr3 (molar mass = 267g/mol) are produced. What is the percent yield of AlBr3.

2Al(s) + 3 Br2(ℓ) → 2AlBr3(s)

mass of AlBr3 = 6.0 g Al x 27g Al

1 mol Al x2 mol Al2 mol AlBr3 x

1 mol AlBr3

267g AlBr3

= 59.33g AlBr3

Theortical yield Percent yield AlBr3 =50.3 g59.33g x 100%

= 84.78%

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Example(2):

Years of experience have proven that the percent yield for the following reaction is 74.3%

Hg + Br2 → HgBr2

If 10.0 g of Hg (atomic mass =200.5g/mol) and 9.00 g of Br2 (molar mass = 160g/mol) are reacted, how much HgBr2 (molar mass= 360.5g/mol) will be produced?

mol of Hg = 10.0 g x 1 mol Hg200.5g Hg

= 0.05 mol Hg

mol Br2 = 9 g Br2 x1 mol Br2 160g Br2

= 0.056 mol Br2

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1 Hg 1 Br2

mol 0.05 0.056 Hg is the limiting reagent

Mass of HgBr2 = 0.05 mol Hg x 1 mol Hg1 mol HgBr2 x 1 mol HgBr2

360.5g HgBr2

= 18.025 g HgBr2 Theortical yield

74.3% =Percent yield =Actual yield

18.025x 100%

Actual yield = 18.025 x 74.3100

= 13.39 g

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