Ch 7.1 Counting Units

One dozen = 12 things

We use a dozen to make it easier to count the amount of substances.

Molesthe SI base unit that describes the number of particles in a substance.

Mole is abbreviated mol

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So what does a Mole equal?Avogodro’s constant is the amount of particles in one mole of a substance.

He found that 6.02x1023 is the amountparticles are always found in one mole

This amount is exactly equal to the amount of 1.0 mole.

MolesUsed by chemist when counting large numbers of tiny particles such as atoms

One mole of anything will equal the same amount of particles.

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Ch. 7.2 Molar MassThe mass in grams of 1 molof any substance.

Using the periodic chart you can determine the molar mass of any element or compound.

g/ 1 mole

6.02 x 1023 particles

Ch. 7.3 Molar Mass Calculations

Chemists must use mole calculations in order to prepare reactions.

They act as recipes for chemists.

EXAMPLE PROBLEM #1How many grams of Calcium fluoride must a chemist prepare if the reaction requires 4.0 moles of calcium fluoride?

SolutionThe molar mass of CaF2 is 78 g/1 mol.

4.0 moles CaF2 x 78.0 grams CaF2 =

1 mole CaF2

310 g CaF2

Example Problem #2How many moles were produced if your reaction yielded 92.0 grams of sodium sulfate?

SolutionMM of Na2(SO4) is 142 g/1 mole92.0 g Na2(SO4) x 1 mole Na2(SO4) =

142.0 g Na2(SO4)

0.648 moles Na2(SO4)

Example Problem #3How many molecules were produced if your reaction yielded 12.0 grams of Carbon monoxide?

Solution

MM of CO is 28 g/1 mole12.0 g CO x 6.02x1023molecules CO

28.0 g CO

= 2.58x1023 molecules CO

Ch. 7.4 Percent Composition

A measurement that measures the amount of each element in the total compound.

Can be used to determine the amount recovered from a compound breaking down

Hydrogen

Nitrogen

Oxygen76%

22%

2%HNO3

Percent Composition

How many grams of silver can be recovered by decomposing 32 grams of silver oxide?

Calculating Empirical FormulaA compound is made from 30.5% nitrogen and 69.5% oxygen. What is the empirical formula?

Assume you have a 100 gram sample

30.5 g N/ 14.0 gram per mole = 2.2 moles N

69.5 g O/ 16.0 grams per mole = 4.3 mole O

2.2 mole N/ 2.2 mole = 1 N

4.3 mole O/2.2 mole = 2 O

Formula is NO2

Empirical Formula

• A lab determines that a compound found in water contains 66.0% calcium and 34.0 % phosphorous. What is the identity of the compound?

Ch. 7.5 Determining Molecular Formula

The empirical formula is nitrogen dioxide. Its actual molar mass is 184 g/mole. What is the molecular formula?

Molar mass of NO2 is 46 g/ mole.

184g / 46.0g = 4

NO2 x 4 = N4O8

Ch 8.1 Chemical Reactions

•When substances undergo

chemical changes, they

form new substances.

•Atoms are rearranged,

because bonds are broken

and reformed

Chemical Reactions

• Signs are:

1. evolution of heat and light

2. color change

3. gases emitted

4. a precipitate forms (solid

residue)

Equation Make-up

•Reactants

substances that

will undergo a

chemical change.

(Left side)

Equation Make-up

•Products

substances that

are formed as a

result of a

chemical change.

(Right side)

Recipes:

• 1 cup batter +1/2 cup water + 2 eggs → 4 pancake

6CO2

+6H2O→C

6H

12O

6 + 6O

2

ReactantsProducts

Ch. 8.2 Balancing

• All equations must have the

same type and number of atoms

on each side of the equation.

• Law of Conservation of Mass

• Tells you the amounts

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Beaker Violating Law of

Conservation of Mass

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Mole Ratios

• Coefficients show the amount of

moles of each substance.

• Mole ratio is the smallest

relative number of moles of the

substance involved in a

reaction.

• 2H2

+ O2

→ 2H2O = 2:1:2

Rules to Balancing

• Can only change coefficients

and never subscripts.

• Always balance H and O last if

water is in the equation.

• In a double displacement

reaction, balance the ions.

• (OH)-= H

2O

Writing Complete

Equations

• Phase identification

• s, l, g, v, aq

• Solutions and acids are mixed

with water, so are considered

aqueous.

• All ionic compounds are solids or

aqueous

• Remember which ELEMENTS are

diatomic

Ch. 8.3

Reaction Types

• 5 types of reactions

1. Synthesis Reaction

(combination RXN)

•A reaction when two

substances form at

least one new, more

complex compound.

Synthesis Reaction

• EXAMPLE:

A + B A B

2Fe + O22FeO

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• Zn + S → ZnS

• Al + Br2

→ AlBr3

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2. Decomposition Reaction

•A reaction in which

one compound breaks

into at least two

products.

Decomposition Reaction

EXAMPLE:

AB A + B

2 NI3 N

2+ 3I

2

2 H2O

2 2 H

2O + O

2

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Decomposition

of Sugar

C6H

12O

6→ C + H

2O

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3. Single displacement Reaction

•A reaction in which

atoms of one element

take the place of ions

of another

compound.

Single displacement Reaction• Example

X A + B B A + X

3CuCl2 + 2Al

2AlCl3 + 3Cu

• Fe2O

3+ 2Al → 2Fe + Al

2O

3

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4. Double Replacement Reaction

• A reaction in which the apparent exchange of ions between two compound solutions.

Double replacement reaction

AX + BY AY + BX

Pb(NO3)2

+K2(CrO

4)

→

Pb(CrO4) + 2K(NO

3)

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5. Combustion reaction

•Reaction in which an

organic compound

and oxygen burn.

•Oxygen always is a

reactant

Combustion reaction

•CO or CO2

will

always be a product

• H2O will always be a

product

Combustion reaction

C2H

50H + 3O

2

2CO2

+ 3H2O

MUPPET LABS

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Stoichiometry

Ch. 9.1

• Calculations dealing with the amounts

of substances needed or produced in

a chemical reaction.

• Mole-mole problems

• Mass-mass problems

• Percent Yield

Stoichiometry

Ch. 9.1

• Steps to solving:

1. Balance the chemical equation

2. Determine the mole ratio needed for

problem

3. Determine # of steps

4. Use Dimensional Analysis to solve

Stoichiometry

Mole-mole Problem

• How many moles of sodium nitrate

can be produced from the reaction of

0.67 moles of calcium nitrate with

excess sodium chloride?

2NaCl + Ca(NO3)2 → CaCl

2+ 2NaNO

3

0.67 moles Ca(NO3)2

x 2 mole Na(NO3)

1mole Ca(NO3)2

= 1.3 moles of Na(NO3)

Ch 9.2 Stoichiometry

Mass-mass Problem

• How many grams of

magnesium oxide could be produced

from 9.90 g of magnesium reacting

with an excess of oxygen gas?

• 2Mg + O2

→ 2MgO

9.90 grams Mg x 1 mole Mg x 2 mole MgO x 40.0 g MgO

24 g Mg 2 mole Mg 1 mole MgO

= 16.5 grams of MgO

Ch. 9.4 Stoichiometry

Percent Yield

• The actual amount of product

expressed as a percentage of the

calculated theoretical yield of that

product.

• The theoretical amount is what is

calculated using Stoichiometry

• Percent Yield = actual amount of product x 100%

theoretical amount of product

Stoichiometry

Percent Yield Problem

• What is the percent yield in the following

reaction if 5.50 grams of hydrogen reacts

with nitrogen to form 20.4 grams of

ammonia?

• N2

+ 3H2

→ 2NH3

5.50 g H2

x 1 mole H2

x 2 mole NH3

x 17 g NH3

2 g H2

3 mole H2

1 mole NH3

= 31.2 grams of NH3

• 20.4 g /31.2g x 100% = 65.4%