
The Materials:


Pick up a packet and assessment plan.
Paper, pencil, scientific calculator, periodic table

The Plan:


Learn about 3 definitions of a mole
Solve dimensional analysis problems

HOMEWORK:

See your calendar!




The mole is a unit of measurement used in chemistry.
The unit can be defined in multiple ways.
(We’ll learn 3 today.)
In its simplest terms, it represents a specific number.


Dozen = what number?
Pair = what number?


Baker’s dozen = what number?
Mole = 6.022 x 10 23


Just like a dozen means 12 of anything...
6.022x10
23 of anything equals of mole.

Consider the size of 6.022x10
23 . Is it large or small?

Would you commonly use 6.022x10
23 with large things or small things?


In chemistry, we are often dealing with VERY
SMALL things.

Atoms are SUBmicroscopic. In order to have an amount large enough with which to really interact, we need quite a few atoms.
6.022x10
23 is the number that Amadeo
Avogadro chose in his lab using carbon.


6.022x10
23

My pet mole is named Avogadro.


Counting

Weighing

Amount of Space Needed

Let’s represent our 3 mole definitions in a graphic organizer.


A compound is a collection of atoms.

To calculate the mass of one mole of a compound, you’d need to add up the mass of all the atoms. This is called the
MOLAR MASS .
= ______ g CH
4

Example: 1 mole CH
4

1 C = 12.011 g C

4 H = 4(1.0079 g H)

Total = 16.04 g/mol


Calculate the molar mass of sulfur dioxide, a gas produced when sulfurcontaining fuels are burned.

SO

2
S = 32.07 g


O = 2(16.00 g)
Total = 64.07 g/mol

Can also be expressed as 1 mol SO
2
= 64.07 g

POLYVINYL CHLORIDE, CALLED PVC,
WHICH IS WIDELY USED FOR FLOOR
COVERINGS (“VINYL”) AS WELL AS FOR
PLASTIC PIPES IN PLUMBING SYSTEMS,
IS MADE FORM A MOLECULE WITH THE
FORMULA C
2
H
3
CL. CALCULATE THE
MOLAR MASS OF THIS SUBSTANCE.
RECORD THE ANSWER TO TWO
DECIMAL PLACES.


Polyvinyl chloride, called PVC, which is widely used for floor coverings (“vinyl”) as well as for plastic pipes in plumbing systems, is made form a molecule with the formula
C
2
H
3
Cl. Calculate the molar mass of this substance.

62.49 g/mol


Using the Mole Concept to Calculate

Page 182-7

EX 6.3, EX 6.4, EX. 6.6



Aluminum (Al), a metal with a high strengthto-weight ratio and a high resistance to corrosion, is often used for structures such as high-quality bicycle frames. Compute both the number of moles of atoms and the number of atoms in a 10.0-g sample of aluminum.
How many calculations is this prompt asking me to carry out?


Aluminum (Al), a metal with a high strengthto-weight ratio and a high resistance to corrosion, is often used for structures such as high-quality bicycle frames. Compute both the number of moles of atoms and the
number of atoms in a 10.0-g sample of aluminum.

Concentrate on one calculation at a time

A.) 3.36X10
-25
B.) 1.22X10
23
C.) 1.22X10
20
D.) none of these


Calcium carbonate, CaCO
3
(also called calcite), is the principal mineral found in limestone, marble, chalk, pearls, and the shells of marine animals.
a) b)
Calculate the molar mass of calcium carbonate.
A certain sample of calcium carbonate contains 4.86 mol. What is the mass in grams of this sample?

A.) 2.112
B.) 42,600
C.) 2.520
D.) none of these



If converting between the units of moles and liters of a gas, what conversion factor is needed?


1 mole = ________________
1 mole = 22.4 Liters
Example: During cellular respiration, a cell releases 2.1 mol of O
2 gas. What volume is needed to hold that gas?


Juglone, a dye known for centuries, is produced from the husks of black walnuts.
The formula for juglone is C
10
H
6
O
3
.

A sample of 1.56 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent?

How many atoms of carbon are present in the sample?


On a clean sheet of paper, show your work for
10-2 Practice Problems (1-3, 12-14, 17, 18, 23-
25).


Last week, you created a poster using chalk.



How many grams of chalk did you use on the poster?
How many moles of chalk did you use?
How many formula units of chalk did you use?

How many atoms of Ca did you use?


How many atoms of C did you use?
How many atoms of O did you use?


I marked mistakes on your papers, but I did not correct them.

Correct and complete 1-8 tonight for homework.

Let’s solve 9 and 10 together now.


The relative amounts of each element in a compound are expressed in percent composition. AKA: percent by mass of each element

% of element = grams of element X 100 grams of compound


Carvone is a substance that occurs in two forms, both of which have the same molecular formula (C
10
H
14
O) and molar mass.
One type of carvone give caraway seeds their characteristic smell; the other is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone.


Individually, calculate the mass percents of the first three compounds on the worksheet.

Let’s check your work.


Mass % is easily related to chemical formulas

The relationship of moles to chemical formulas requires a little more thought...


Empirical formula: lowest whole number ratio of the atoms of the elements in a compound

Ionic compounds are criss-crossed and then reduced. Ionic compound formulas are nearly always empirical formulas.

Empirical formula doesn’t have to be the same as the actual molecular formula of the compound.

Remember, “molecule” means covalent compound. Covalent compounds weren’t always in a reduced form.




CH

O
Calculate the mass percent of each element.

2
C = 40%


H = 6.7%
O = 53.3%
Isn’t this the reduced formula for: C
C




O H , & C O
3
H
6
O
3
,
4
H
8 4
, C
5 10
O
5 6
H
12 6
Row 1: Calculate the mass percent of C
Row 2: Calculate the mass percent of C
4
Row 3: Calculate the mass percent of C
5
Row 4: Calculate the mass percent of C
3
6
H
H
8
6
O
H
10
H
12
O
O
4
3
O
.
.
The percentages (ratio) of the elements is the same with all of these formulas because they share the same reduced form.
5
.
6
.


Molecular formula: actual formula for the compound which gives the composition of the molecule

Glucose shares an empirical formula with many compounds, but it has its molecular formula all to itself.


6(CH
2
O) = glucose
C
6
H
12
O
6




When an unknown compound is found, instruments can tell scientists the mass percent composition of the compound.
Calculations are required to convert that series of percentages into a chemical formula.
We start by converting to the empirical formula.


White powder found in the hallway

Mass spectroscopy instrument used to analyze the unknown powder

Data from the instrument:




40.9 % Carbon
4.58% Hydrogen
54.5% Oxygen
Molar mass of 180 grams/mole

Use these %s, formula knowledge, & mole knowledge to figure out the chemical formula of the powder


Use a simple rhyme!




% to gram
Gram to mole
Divide by the smallest
Multiply ‘til whole.


An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with
3.692 g of oxygen. Calculate the empirical formula for this compound.



% to gram = IS DONE FOR YOU
Gram to mole = dimensional analysis
Let’s carry out the calculation on the board.


A sample of lead arsenate, an insecticide used against the potato beetle, contains
1.3813 g of lead, 0.00672 g of hydrogen,
0.4995 g of arsenic, and 0.4267 g of oxygen.
Calculate the empirical formula for lead arsenate.


The most common form of nylon is 63.68% carbon, 12.38% nitrogen, 9.80% hydrogen, and 14.4% oxygen. Calculate the empirical formula for nylon.


Molecular Formula= n(empirical formula)

(Remember 6(CH
2
O) = glucose)
We know how to calculate the empirical formula, but how do we know what number to multiply it by?
n= molecular formula mass/molar mass of empirical


A white powder is analyzed and found to have an empirical formula of P
2
O
5
. The compound has a molar mass of 283.88 g/mol.
What is the compound’s molecular formula?


A compound used as an additive for gasoline to help percent engine knock shows the following percentage composition:

71.65% Cl

24.27% C

4.07% H
The molar mass is known to be 98.96 g. Determine the empirical formula and the molecular formula for this compound.


Page 208 4-6


On a clean sheet of paper, work the following problems individually :



Calculate the number of oxygen atoms in 3.5 g aluminum sulfate.
Calculate the number of molecules of O in 2.5 L of O
2
.
2 gas
Calculate the number of grams of iron that contain the same number of atoms as 2.24 g of cobalt. (pg 214 #57)

Quiz answers are coming.


The final product in protein metabolism is urea. Urea contains 20.00% C, 6.73% H,
46.65% N, and 26.64% O. The molar mass of urea is 60.07g/mol. Calculate the empirical formula and molecular formula.


You’ll have to READ (not skim) the lab to be successful.

Steps 1-5 are probably unnecessary.

Steps 6-11 are vital. Substitute hot plate for
Bunsen burner.

Notice the “Observations and Data” area on the back. You’ll need those measurements.

The calculations are described to you in each question. Just follow the directions.

You’ll need to answer the “Questions for
Discussion,” too.

1)
2)
3)
4)
5)
Subtract.
% water lost =[water lost/hydrate mass]100
% water in hydrate =
[5(water molar mass)/total molar mass] 100
Subtract. %water in hydrate - %water lost from your hydrate
% error = [#4 answer/#3 answer]100