Chemistry 103 Lecture 14. Outline I. Empirical/Molecular Formulas II. Chemical Reactions - basic...
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Transcript of Chemistry 103 Lecture 14. Outline I. Empirical/Molecular Formulas II. Chemical Reactions - basic...
Outline
I. Empirical/Molecular Formulas
II. Chemical Reactions - basic symbols - balancing - classificationIII. Stoichiometry
The molecular formula Is the true or actual number of the atoms in a
moleculeThe empirical formula Is the simplest whole number ratio of the atoms
(this is the formula for ionic compounds)
H2O2 HOmolecular formula empirical formula
Types of Formulas
Empirical & Molecular Formulas Ionic Compounds - Only need Empirical
formula.
Molecules - Need information on both to determine exact make-up of your system
Molecular Formula Empirical Formula
C6H6 CH
A molecular formula Is a multiple (or equal) of its empirical
formula Has a molar mass that is the empirical
mass multiplied by a whole numbermolar mass = a whole number empirical mass
Is obtained by multiplying the empirical formula by a whole number
Relating Molecular and Empirical Formulas
Determine the molecular formula of compound that has a molar mass of 78.11 g/mole and an empirical formula
of CH.
Finding the Molecular Formula
A compound is 24.27% C, 4.07% H, and 71.65% Cl. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas?
Molecular Formula
Physical Change
In a physical change,• The identity and composition
of the substance do not change
• The state can change or the material can be torn into smaller pieces
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chemical Change
In a chemical change, • Reacting substances form
new substances with different compositions and properties
• A chemical reaction takes place
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
Chemical Reaction
In a chemical reaction, Old bonds are broken and
new bonds are formed Atoms in the reactants are
rearranged to form one or more different substances
Fe and O2 form rust (Fe2O3)
Chemical Reaction
In a chemical reaction,
• A chemical change produces one or more new substances
• There is a change in the composition of one or more substances
Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
Writing a Chemical Reaction
Chemists use a shorthand approach when writing the specifics of a chemical reaction. This approach is called the chemical equation.
Reactants -----> Products
Chemical Equations
A chemical equation,
• Gives the chemical formulas of the reactants on the left of an arrow and the products on the right
Reactants Product
C(s)
O2 (g)CO2 (g)
Symbols Used in Equations
Symbols used in chemical
equations show: The states of the reactants
The states of the products
The reaction conditions
Chemical Equations Are BalancedIn a balanced chemical reaction,
• The number of reactant atoms are equal to the number of product atoms
Chemical Equations•Chemical equations: symbolic descriptions of chemical reactions.•Two parts to an equation:
•reactants and products
H2 + O2 H2OA Chemical Equation must also be “balanced”.
2H2 + O2 --> 2H2O
Balanced Chemical Equations Chemical Equations must be balanced
There must be equal numbers of atoms of each element on both sides of the equation (both sides of the arrow) 1. Write the correct symbols and formulas for all of the
reactants and products. 2. Count the number of each type of atom on BOTH
sides of the equation. 3. Insert coefficients until there are the equal numbers
of each kind of atom on both sides of the equation.
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
H2(g) + N2(g) NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
H2(g) + N2(g) NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
H2(g) + N2(g) NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
H2(g) + N2(g) 2 NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
3 H2(g) + N2(g) 2 NH3 (g)
Example - making ammonia
Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.
BEFORE AFTER
3 H2(g) + N2(g) 2 NH3 (g)
Balancing Equations
Methane reacts with oxygen (combustion reaction) to form carbon dioxide and water.
Write a properly balanced chemical equation
Stoichiometry
Chemical Stoichiometry: using mass and quantity relationships among reactants and products in a chemical reaction to make predictions about how much product will be made.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Four molecules NH3 react with five molecules O2
to produce
four molecules NO and six molecules H2O
Quantities in a Chemical Reaction
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Four molecules NH3 react with five molecules O2
to produce
four molecules NO and six molecules H2O
and
Four mol NH3 react with five mol O2
to produce
four mol NO and six mol H2O
Quantities in a Chemical Reaction
We can read the equation in “moles” by placing the word “mole” or “mol” between each coefficient and formula.
4Fe(s) + 3O2(g) 2Fe2O3(s)
4 mol Fe + 3 mol O2 2 mol Fe2O3
Moles in Equations
Conservation of Mass
The Law of Conservation of Mass indicates:• No change in total mass occurs in a reaction• Mass of products is equal to mass of
reactants
In an ordinary chemical reaction, • Matter cannot be created or destroyed• The number of atoms of each element are equal • The mass of reactants equals the mass of products
H2(g) + Cl2(g) 2HCl(g)
2 mol H, 2 mol Cl = 2 mol H, 2 mol Cl
2(1.008) + 2(35.45) = 2(36.46) 72.92 g = 72.92 g
Law of Conservation of Mass
Conservation of Mass
2 mol Ag + 1 mol S = 1 mol Ag2S
2 (107.9 g) + 1(32.07 g) = 1 (247.9 g)
247.9 g reactants = 247.9 g product
A mole-mole factor is a ratio of the moles for two
substances in an equation.
4Fe(s) + 3O2(g) 2Fe2O3(s)
Fe and O2 4 mol Fe and 3 mol O2
3 mol O2 4 mol Fe
Fe and Fe2O3 4 mol Fe and 2 mol Fe2O3
2 mol Fe2O3 4 mol Fe
O2 and Fe2O3 3 mol O2 and 2 mol Fe2O3
2 mol Fe2O3 3 mol O2
Writing Mole-Mole Factors
How many moles of Fe2O3 can form from 6.0 mol O2?
4Fe(s) + 3O2(g) 2Fe2O3(s)
Relationship: 3 mol O2 = 2 mol Fe2O3
Write a mole-mole factor to determine the moles of Fe2O3.
6.0 mol O2 x 2 mol Fe2O3 = 4.0 mol Fe2O3
3 mol O2
Calculations with Mole Factors