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Chemistry 100 Chapter 3 Balancing Chemical Equations and Stoichiometry

Chemistry 100 Chapter 3

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Chemistry 100 Chapter 3. Balancing Chemical Equations and Stoichiometry. The Chemical Reaction. What happens in a chemical reaction? Example 2 H 2 + O 2  2H 2 O Starting materials (reactants) are converted into different chemical substance(s) (the product(s)). - PowerPoint PPT Presentation

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Page 1: Chemistry 100 Chapter 3

Chemistry 100 Chapter 3

Balancing Chemical Equations and Stoichiometry

Page 2: Chemistry 100 Chapter 3

The Chemical Reaction

What happens in a chemical reaction? Example

2 H2 + O2 2H2O

Starting materials (reactants) are converted into different chemical substance(s) (the product(s)).

Described by ‘standard shorthand’ - chemical equation

Page 3: Chemistry 100 Chapter 3

Balancing Equations

Balanced equations have the same number of atoms of a given element on the LHS and the RHS Law of conservation of mass.

All the reactants and products must be identified!

Chemical equations report the results of experimentation!

Page 4: Chemistry 100 Chapter 3

A General Method of Balancing Chemical Equations.

Write the basic, or ‘skeletal’ equation, showing the formula of all reactants and products.

Balance the equation according to the law of conservation of mass. Balance by adjusting the coefficients in

front of the chemical formulas, never by adjusting the subscripts.

Best to start with an atoms that appears only once on the left and right hand side of an equation.

Page 5: Chemistry 100 Chapter 3

A Balanced Chemical Equation

Page 6: Chemistry 100 Chapter 3

Chemical Reaction Types

Combustion reaction can be written for any compound containing C, H, O or C, H, S etc. O2 (g) is present as a reactant and is usually in excess.

We can also have combination and decomposition reactions

Combination reaction two or more substances combine to form one product

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

Page 7: Chemistry 100 Chapter 3

Reactions Types (II)

Decomposition reaction one substance breaks down to form two or more different substances MgCO3(s) MgO(s) + CO2(g) NH4Cl(s) NH3(g) + HCl(g)

Decomposition of sodium azide, NaN3

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

Page 8: Chemistry 100 Chapter 3

Molecular Masses

How do we use the atomic masses of elements to determine the molecular masses of molecules?

We simply add the masses of the constituent elements in the molecule.

Page 9: Chemistry 100 Chapter 3

The Mole and Avogadro’s Number

NOTE: 1 amu = 1.6605 x 10-24 g; this is a very small mass!!

It is not a very convenient unit of measurement on the laboratory scale.

In chemistry, we use a special unit (the mole) when dealing with atoms, molecules, and ions.

Page 10: Chemistry 100 Chapter 3

The Definition of a Mole

1 mole (SI definition) the amount of substance that contains as

many elementary particles (atoms, molecules, ions) as there are atoms in 12 grams (exactly) of carbon-12.

This number is called Avogadro's number (it is an experimentally determined quantity).

Page 11: Chemistry 100 Chapter 3

The Molar Mass

1 mole = 6.022 x 1023 particles. 1 mole of carbon-12 has a mass of

12.00000 .... g exactly! The molar mass of any element in

g/mole is the same numerically as its atomic mass in amu’s!

Page 12: Chemistry 100 Chapter 3

The molar mass of a compound is the mass (in grams) of 1 mole of a compound.

Numbers in periodic table e.g., Cl = 35.453 g/mole = 35.453 amu/atom

Page 13: Chemistry 100 Chapter 3

Converting Between Moles and Molecules

Page 14: Chemistry 100 Chapter 3

Masses of Anions and Cations

Electron mass is small mass of Na+ mass of Na mass of Cl- mass of Cl mass of O2- mass of O atom!!

Page 15: Chemistry 100 Chapter 3

Percent Composition

Percent composition is the percent by mass of each element in a compound.

Q: If we have the percent composition of the compound, can we calculate its empirical formula?

Page 16: Chemistry 100 Chapter 3

A Schematic for Doing Empirical Formula Calculations

Page 17: Chemistry 100 Chapter 3

Chemical Analysis Use a variety of techniques to obtain

the identity and the % by mass of each element in a sample. Combustion analysis a sample

containing C, H, or C,H, and O is combusted (burned). (C,H,O) CO2 + H2O all the C in original compound gets converted

to CO2

all the H in original compound is converted to H2O

Page 18: Chemistry 100 Chapter 3

Combustion Analysis

Page 19: Chemistry 100 Chapter 3

Quantitative Information from Chemical Reactions

Example A + B C + D

Normally, we start out with certain quantities of reactants. How much product can we expect? How much reactant would we need to

obtain a specified amount of product?

Page 20: Chemistry 100 Chapter 3

The Mole Method

Stoichiometric coefficients in the balanced chemical equation represent the number of moles of reactants and products. certain mass of reactants ? mass of

products initial volume of reactants ? Volume

of products

Page 21: Chemistry 100 Chapter 3

ExampleN2 (g) + 3 H2 (g) 2 NH3 (g)1 mole N2 3 moles H2 2 moles NH3

the symbol means stoichiometrically equivalent

Page 22: Chemistry 100 Chapter 3

A Schematic of the Mole Method

Page 23: Chemistry 100 Chapter 3

Limiting Reagent

Chemical equations give the molecular or molar ratios of reactants needed and the products obtained. N2 (g) + 3 H2 (g) 2 NH3 (g) We assumed 100% complete conversion

of of the N2 (g) and H2 (g) to NH3 (g). Normally, the reactants will react in the

proper ratio until one of them is consumed completely.

Page 24: Chemistry 100 Chapter 3

The Definition of the Limiting Reagent

The amount of product formed is limited by the reactant that is completely consumed the limiting reagent.

The amount of product obtained assuming complete consumption of the limiting reagent theoretical yield.

The reactant that is left over the excess reagent.

Page 25: Chemistry 100 Chapter 3

Limiting Reagents

Page 26: Chemistry 100 Chapter 3

The Limiting Reagent Schematic

Use amount of Limiting Reagent

Mass of A and B

Use Molar Masses of A, B

Moles of A, B Determine Limiting Reagent

Obtain Theor. Yield

Use Coefficients from balanced Equation