Ch 5.11 & Ch 6

On the first right side page (after Ch 5 notes),

• Create a Cover page for Chapter 6. (see next slide for

ideas)

• THEME: The Mole, Avogadro’s #, Mole conversions

• Make it the whole page

• Include color

On the back side of cover page (left side),

• Paste the Ch 6 Learning Objectives in.

TN Ch 6.1 Date

Title and

Highlight

Topic:

EQ:

NOTES:

Write out the notes from my website.

Use different types of note-taking

methods to help you recall info (different

color pens/highlighters, bullets, etc)

When I lecture we will add more info, so

leave spaces in your notes

Summary Questions:

2-3 sentences… What did you learn

today from the notes?

THESE ARE AT THE VERY END OF

NOTES

Right Side – NOTES ONLY

Reflect

Question:

Reflect on

the

material by

asking a

question

(its not

suppose to

be

answered

from

notes)

TN Ch 6.1

Title and

Highlight

DRAW ANY PICTURES, FIGURES,

AND WRITE OUT ANY PRACTICE

PROBLEMS/QUESTIONS.

WE WILL ANSWER THEM TOGETHER.

LEAVE SPACES SO WE CAN ANSWER

QUES.

LEFT Side – PICTURES, PRACTICE PROBLEMS, ETC

READ Ch 5.11 (pg. 148-149)

For any compound, the formula weight is

the sum of the atomic masses of all the

atoms in its chemical formula

◦ (units = amu – atomic mass unit):

What is the formula weight for CaCl2

Ca 1 x 40.078 amu = 40.078 amu

+ Cl 2 x 35.453 amu = 70.906 amu

= 110.984 amu CaCl2

Formula Weight Rules

Find the formula weights for the following

ionic compounds

a. MgSO4

b. KCN

c. Ca(OH)2

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.1-6.3 (pg. 165-168 stop at “converting grams ad moles)

6.02 x 1023

What is a conversion factor?

◦ It relates how an object compares to another.

◦ Usually one item is set to equal 1.

Examples:

1 min = 60 sec

1 year = 365 days

1 doz eggs= 12 eggs

1 pound = 16 ounces

With atoms, we must use their mass as a way to count them.

Atoms - too small & too many to count individually.

We need a larger # because atoms are so small.

Chemist’s use mole (mol).

Avogadro’s number = 602 billion trillion

602,000,000,000,000,000,000,000

6.02 X 1023 (in scientific notation)

This number is named in honor of Amedeo Avogadro (1776 – 1856),

1 mol = 6.02 1023 items (units change)

A large amount!!!!

How big is a mole? (ed.ted.com)

One mole of marbles = 6.02 × 1023 marbles.

One mole of sand grains = 6.02 × 1023 sand grains.

The value of the mole is defined as being equal to the number of atoms in exactly 12 g of pure carbon-12. (Very important)

This definition of the mole establishes a relationship between mass (grams of carbon) and number of atoms (Avogadro’s number).

• 1 mole of hockey pucks would equal

the mass of the moon!

(Moon mass 7.3 x 1022 kg)

• 1 mole of basketballs would fill a bag

the size of the earth!

Enough soft drink cans to cover the

surface of the earth to a depth of

over 200 miles.

If you had Avogadro's number of

unpopped popcorn kernels, and

spread them across the United

States of America, the country would

be covered in popcorn to a depth of

over 9 miles.

If we were able to count atoms at the

rate of 10 million per second, it

would take about 2 billion years to

count the atoms in one mole.

= 6.02 x 1023 C atoms

= 6.02 x 1023 H2O molecules

= 6.02 x 1023 NaCl formula units

1 mole C

1 mole H2O

1 mole NaCl

Representative Particles:

Atom = 1 element

Formula unit (f.u.) = Ionic Compound

Molecule = Molecular Compound

Ion = charge atom

How many molecules are in 2.50 moles of C12H22O11?

Left side

a. Number of atoms in 0.500 mole of Al

b.Number of moles of S in 1.8 x 1024 atoms S

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.3 (pg. 168-170)

6.02 x 1023

The molar mass of any element—the mass of 1 mol of that element. (Cool part is that it’s the same calculation as gram formula mass – Ch 5.11)

Need to see the difference: 1 Cu atom has atomic weight = 63.55 amu. 1 mol of Cu atoms has mass of 63.55 g. ……Do you see the difference????

Molar Mass - 1 mole equals the sum of

the atomic masses in grams ( not amu). ◦ Need Periodic Table!!!!

molar mass

Mass (g) Moles (mol)

1 step conversion!!

Find the molar mass from PT

(DON’T ROUND!!!)

A.1 mole of Br atoms

B.1 mole of Sn atoms

= 79.904 g

= 118.710 g

How many moles of carbon are in 26 g of carbon?

Left side

How many grams of Al are in 3.00 moles

of Al?

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.3 (pg. 171-172)

6.02 x 1023

molar

mass

MASS

IN

GRAMS

MOLES

NUMBER

OF

PARTICLES

6.02 1023

Everything must go through Moles!!!

How many atoms of Cu are present in

35.4 g of Cu?

Remember… 6.02 X 1023 atoms = 1 mole = molar mass (g)

Left side

How many atoms of K are present in

78.4 g of K?

Left side

Remember… 6.02 X 1023 atoms = 1 mole = molar mass (g)

How many grams of Ca are

present in 5.23 x 1024 atoms Ca?

Left side

Remember… 6.02 X 1023 atoms = 1 mole = molar mass (g)

READ Ch 6.4 (pg. 172-174)

6.02 x 1023

Molar mass is the mass of 1 mol of molecules

(units used for only molecules) or formula units (units used for only ionic compounds) of that compound.

Just like Ch 5.11!!! Just different units because of moles

1 mole of CaCl2

Ca 1 x 40.078 g = 40.078 g

+ Cl 2 x 35.453 g = 70.906 g

= 110.984 g/mol CaCl2

molar

mass

(g/mol)

MASS

IN

GRAMS

MOLES

NUMBER

OF

PARTICLES

6.02 1023

(particles/mol)

Everything must go through Moles!!!

Find the mass of 2.1 1024 formula units of NaHCO3.

Left side

Find the formula units of 51.5g of NaHCO3.

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.5 (pg. 175-179)

6.02 x 1023

The relationships in a chemical formula allow us to convert between moles of the compound and moles of an element (and vice versa).

Left side

Find the moles of sodium in 15 g of NaCl.

Left side

How many grams of O are present in 78.1 g

CO2?

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.6-6.7 (pg. 180-182)

6.02 x 1023

Show the conversion through Avogadro

10.0g of water = _______________ H atoms

Left side

Left side – Draw and color

Left side – Draw and color

Left side – Draw and color

The mass percent composition, or percent composition, of an element is the element’s percentage of the total mass of the compound.

◦ You use this calculation everyday….called your grades!!!

Step 1: Calculate the molar mass of the compound.

Step 2: Divide each elements total mass by the

molar mass of the compound.

Step 3: Convert the decimal to a % by multiplying

by 100.

Step 4: Al ways check your work!!!! All element

%’s should equal 100%.

Find the % composition of each element in Na3PO4

Left side

Find the % composition of CARBON in C4H10

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.8 (pg. 183-185)

6.02 x 1023

Empirical and Molecular Formulas give the numbers of atoms or moles of each element always in a whole number ratio (the law of definite proportions).

1. Empirical Formula: The formula of a compound with the smallest whole number ratio of the atoms present.

1. Determine the mass in grams of each element present.

If %’s are given, check to see if %’s equal 100%. If yes, then the 100%=100g.

So, change %’s to grams.

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 (**See bottom of page 185 for help with what # to multiply with to get whole subscript number)

* Be careful! Do not round off numbers prematurely

A sample of a brown gas, a major air pollutant,

is found to contain 2.34 g N and 5.34g O.

Determine the empirical formula for this

substance.

Left side

A substance has the following composition by

mass: 60.80 % Na ; 28.60 % B ; 10.60 % H

What is the empirical formula of the substance?

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES

READ Ch 6.8 (pg. 186-187)

6.02 x 1023

2. Molecular Formula: The formula that states the actual number of each kind of atom found in one molecule of the compound. (biggest ratio)

1. Need Empirical Formula first (look at rules from Ch 6.8)…Solve for it!!

2. Calculate the molar mass of E.F.

3. Book MUST give you the molar mass of

Molecular Formula.

4. To find M.F…..

molar mass of M.F. = whole number ( E.F. ) = M.F. molar mass of E.F.

A compound has an empirical formula

of NO2. The colorless liquid, used in

rocket engines has a molar mass of

92.0 g. What is the molecular formula

of this substance?

Left side

Let’s reflect…. What did you learn today from the notes? (2-3 sentences)

RIGHT side - THESE ARE AT THE VERY END OF NOTES