Stoichiometry Stoichiometry: Calculations with Chemical Formulas and Equations.

Stoichiometry

Stoichiometry:Calculations with Chemical

Formulas and Equations

Stoichiometry

EVIDENCE OF A CHEMICAL REACTION

•A gas is released.•An insoluble substance (called a precipitate) is produced.•A permanent color change is observed.

•Changes in energy. Heat, light, sound, and electrical.

Exothermic reaction: releases heat,light and sound.Endothermic reaction: absorbs heat and light.

•pH change •Odor given off

Stoichiometry

EVIDENCE OF A CHEMICAL REACTION USING YOUR SENSES:

Sight- Change in color, formation of solid, formation of gasses (bubbles), light emission

Hearing- reaction leads to a popping noise, fizzes

Smell- pungent odor, change in smell

Touch- heat absorption and emission

Stoichiometry

Law of Conservation of Mass

“We may lay it down as an incontestable axiom that, in all

the operations of art and nature, nothing is created; an equal

amount of matter exists both before and after the experiment. Upon this principle, the whole art

of performing chemical experiments depends.”

--Antoine Lavoisier, 1789

Stoichiometry

Chemical Equations

Concise representations of chemical reactions

Stoichiometry

Anatomy of a Chemical Equation

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry


Reactants appear on the left side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry


Products appear on the right side of the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry


The states of the reactants and products are written in parentheses to the right of each compound.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry


Coefficients are inserted to balance the equation.

CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

Stoichiometry

Subscripts and Coefficients Give Different Information

• Subscripts tell the number of atoms of each element in a molecule

• Coefficients tell the number of molecules

Stoichiometry

Coefficients in Chemical Reactions

• Coefficients tell how many molecules or atoms there are.

• Coefficients are numbers placed in front of each molecule or atom and are used to balance equations.

Stoichiometry

Conservation of Mass and Atoms

• Law of Conservation of Matter: • Mass is neither created or destroyed

in chemical reactions. – Lavoisier (1743-1794) proved this.

• The number and kinds of atoms present in the products is the same number and kind of atoms present in the reactants. – The atoms just rearrange.

Stoichiometry

Writing and Balancing Equations

• Determine the reactants and products.

• Write the equation.

– Use the correct formulas for each compound—watch for the diatomics; those that end in –ine and –gen. (H2, N2, O2, F2, Cl2, Br2, I2, At2).

• Balance the equation (same # and kind of elements on both sides of the arrow).

Stoichiometry

Balancing Examples

• Solid copper(II) oxide reacts with hydrogen gas to form solid copper and liquid water.

CuO (s) + H2 (g) ---> Cu (s) + H2O (l)

• Aluminum metal reacts with oxygen gas to form solid aluminum oxide.

Al (s) + O2 (g) ---> Al2O3 (s)

Stoichiometry

Balancing Examples• When balancing an equation, ONLY the

coefficients can be changed.

• NEVER change the subscripts.

• For example: 3H2O 3 is the coefficient. 2 and 1 are the subscripts.

• Changing the subscripts changes the compound.

H2O2 is not water but hydrogen peroxide.

Stoichiometry

HELPFUL HINTS

• Keep polyatomic ions together as a unit if they show up on both sides of the equation.

-- NO3-1 , C2H3O2

-1 , NH4+1 , ….

-- Ca(NO3)2 + NaCl ---> CaCl2 + NaNO3

• Save H and O until last.

-- They will tend to balance out together with the H2O.

Stoichiometry

Example

Aluminum sulfide reacts with water to produce aluminum hydroxide and hydrogen sulfide.

Al2S3 + H2O --> Al(OH)3 + H2S What should we look for immediately? Save O and H for last - there is water.

Al2S3 + H2O ---> 2 Al(OH)3 + H2S

Al2S3 + H2O ---> 2 Al(OH)3 + 3 H2S The Al and S are balanced - now H & O.

Al2S3 + 6 H2O ------> 2 Al(OH)3 + 3 H2S

Stoichiometry

Balance The Following Equations

• Al + S8 Al2S3

• 16 Al + 3 S8 8 Al2S3 • CO + O2 CO2 • 2 CO + O2 2 CO2 • C2H4 + O2 CO2 + H2O • C2H4 + 3 O2 2 CO2 + 2 H2O • Pb(NO3)2 + NaCl PbCl2 + NaNO3 • Pb(NO3)2 + 2 NaCl PbCl2 + 2 NaNO3 • Mg2C3 + H2O Mg(OH)2 + C3H4 • Mg2C3 + 4 H2O 2 Mg(OH)2 + C3H4

• Ca(OH)2 + HBr CaBr2 + H2O • Ca(OH)2 + 2 HBr CaBr2 + 2 H2O

Stoichiometry

Reaction Types

Stoichiometry

Combination or (Synthesis) Reactions

• Examples:N2 (g) + 3 H2 (g) 2 NH3 (g)

C3H6 (g) + Br2 (l) C3H6Br2 (l)

2 Mg (s) + O2 (g) 2 MgO (s)

• Two or more substances react to form one product

Stoichiometry

2 Mg (s) + O2 (g) 2 MgO (s)

Stoichiometry

Decomposition Reactions

• Examples:CaCO3 (s) CaO (s) + CO2 (g)

2 KClO3 (s) 2 KCl (s) + O2 (g)

2 NaN3 (s) 2 Na (s) + 3 N2 (g)

• One substance breaks down into two or more substances

Stoichiometry

Combustion Reactions

• Examples:CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

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

• Rapid reactions that produce a flame

• Most often involve hydrocarbons reacting with oxygen in the air

Stoichiometry

Formula Weights

Stoichiometry

Formula Weight (FW)• Sum of the atomic weights for the atoms

in a chemical formula• So, the formula weight of calcium

chloride, CaCl2, would be Ca: 1(40.1 amu) + Cl: 2(35.5 amu)

111.1 amu

• These are generally reported for ionic compounds

Stoichiometry

Molecular Weight (MW)

• Sum of the atomic weights of the atoms in a molecule

• For the molecule ethane, C2H6, the molecular weight would be

C: 2(12.0 amu)+ H: 6(1.0 amu)

30.0 amu

Stoichiometry

Percent Composition

One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:

% element =(number of atoms)(atomic weight)

(FW of the compound)x 100

Stoichiometry

Percent Composition

So the percentage of carbon in ethane is…

%C =(2)(12.0 amu)

(30.0 amu)

24.0 amu

30.0 amu= x 100

= 80.0%

Stoichiometry

Moles

Stoichiometry

Avogadro’s Number

• 6.02 x 1023

• 1 mole of 12C has a mass of 12 g

Stoichiometry

The Mole—How big is 6.022 x 1023;

Avogadro ’s number?

602,200,000,000,000,000,000,000

Stoichiometry

It would take 10 billion chickens laying 10 eggs per day more than 10 billion years to lay a

mole of eggs.

Stoichiometry

Avogadro's number of inches is 1,616,434 light years, or

across our galaxy and back 8 times.

Stoichiometry

Avogadro's number of seconds is about 19 quadrillion years!

That is 4,240,666 times the age of the earth, or 954,150 times the age of the universe itself.

Stoichiometry

Avogadro's number of kilograms is just over 20 times

the mass of the earth.

Stoichiometry

Avogadro's number of cents could repay the

United States National Debt 86 million times.

Stoichiometry

Molar Mass

• By definition, these are the mass of 1 mol of a substance (i.e., g/mol)– The molar mass of an element is the mass

number for the element that we find on the periodic table

– The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

Stoichiometry

Using Moles

Moles provide a bridge from the molecular scale to the real-world scale

Stoichiometry

Mole Relationships

• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles

• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

Stoichiometry

Finding Empirical Formulas

Stoichiometry

Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition

Stoichiometry


The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

Stoichiometry


Assuming 100.00 g of para-aminobenzoic acid,

C: 61.31 g x = 5.105 mol C

H: 5.14 g x = 5.09 mol H

N: 10.21 g x = 0.7288 mol N

O: 23.33 g x = 1.456 mol O

1 mol12.01 g

1 mol14.01 g

1 mol1.01 g

1 mol16.00 g

Stoichiometry

Calculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:

C: = 7.005 7

H: = 6.984 7

N: = 1.000

O: = 2.001 2

5.105 mol0.7288 mol

5.09 mol0.7288 mol

0.7288 mol0.7288 mol

1.458 mol0.7288 mol

Stoichiometry


These are the subscripts for the empirical formula:

C7H7NO2

Stoichiometry

Combustion Analysis

• Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this– C is determined from the mass of CO2 produced

– H is determined from the mass of H2O produced

– O is determined by difference after the C and H have been determined

Stoichiometry

Elemental Analyses

Compounds containing other elements are analyzed using methods analogous to those used for C, H and O

Stoichiometry

Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products

Stoichiometry

Stoichiometric CalculationsFrom the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

Stoichiometry

Stoichiometric Calculations

Starting with 1.00 g of C6H12O6…

we calculate the moles of C6H12O6…

use the coefficients to find the moles of H2O…

and then turn the moles of water to grams

C6H12O6 + 6 O2 6 CO2 + 6 H2O

Stoichiometry

Limiting Reactants

Stoichiometry

How Many Cookies Can I Make?

• You can make cookies until you run out of one of the ingredients

• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat)

Stoichiometry

How Many Cookies Can I Make?

• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make

Stoichiometry

Limiting Reactants

The limiting reactant is the reactant present in the smallest stoichiometric amount

Stoichiometry

Limiting Reactants

• The limiting reactant is the reactant present in the smallest stoichiometric amount– In other words, it’s the reactant you’ll run out of first (in

this case, the H2)

Stoichiometry

Limiting Reactants

In the example below, the O2 would be the excess reagent

Stoichiometry

Theoretical Yield

• The theoretical yield is the amount of product that can be made– In other words it’s the amount of product

possible as calculated through the stoichiometry problem

• This is different from the actual yield, the amount one actually produces and measures

Stoichiometry

Percent Yield

A comparison of the amount actually obtained to the amount it was possible to make

Actual YieldTheoretical YieldPercent Yield = x 100

Stoichiometry Stoichiometry: Calculations with Chemical Formulas and Equations.

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