Chemical Quantities Lecture

```CHEMICAL QUANTITIES
THE MOLE CONCEPT AND ITS APPLICATIONS
CHEMICAL QUANTITIES
OBJECTIVES
1.
STUDENTS WILL BE ABLE TO
 UNDERSTAND AND KNOW
 THE DEFINITIONS OF THE MOLE AND AVOGADRO’S
NUMBER
 RELATIONSHIP BETWEEN MOLES AND MASS
 RELATIONSHIP BETWEEN MOLES AND PARTICLES

CALCULATE THE FORMULA MASS OF A COMPOUND GIVEN THE
PERIODIC TABLE

CALCULATE THE MASS PERCENT COMPOSITION OF A COMPOUND
GIVEN THE FORMULA, FORMULA MASS, AND PERIODIC TABLE.
DRILL & OBJECTIVES
How many eggs are there in a dozen?
1.
If I bought 3 dozen eggs, how many eggs do I have?
1.
What are some ways that you can measure things? (such as
the amount of a substance)
2.
OBJECTIVES
1.

STUDENTS WILL BE ABLE TO
UNDERSTAND AND KNOW
 THE DEFINITIONS OF THE MOLE AND AVOGADRO’S
NUMBER
 RELATIONSHIP BETWEEN MOLES AND MASS
 RELATIONSHIP BETWEEN MOLES AND PARTICLES

CALCULATE THE FORMULA MASS OF A COMPOUND GIVEN THE PERIODIC
TABLE

CALCULATE THE MASS PERCENT COMPOSITION OF A COMPOUND GIVEN
THE FORMULA, FORMULA MASS, AND PERIODIC TABLE.
CHEMICAL QUANTITIES

How many eggs are there in a dozen?
12!
CHEMICAL QUANTITIES

How many roses are there in a dozen roses?
12!
CHEMICAL QUANTITIES

Does it matter what the _____________ is?
CHEMICAL QUANTITIES

Does it matter what the _____________ is?


NO! A Dozen is a dozen! No matter if it’s flowers, eggs, bagels,
ect.
It represents a number of _______
CHEMICAL QUANTITIES


In Chemistry the same concept is valid,
Let’s talk about “The Mole” and how it relates to
Chemistry…
THE MOLE?
The Mole?
THE MOLE?

“The Mole” or “A Mole” is an amount of a substance


There are 6.02 x1023 atoms (representative particles of a
substance) in 1 Mole (of that substance).
This number (6.02 x1023) is called Avogadro’s number

Named after Amedeo Avogadro di Quarenga (1776 – 1856), an
Italian scientist.
THE MOLE!

Let’s put this into perspective

How much is 1 mol of a substance?
6.023 x
23
10 !!
THE MOLE!

IF you had 6.02 X 1023 Watermelon Seeds…
THE MOLE!

IF you had 6.02 X 1023 Watermelon Seeds…
it would be found inside a watermelon slightly larger than
the moon!
THE MOLE!

If you had 6.02 X 1023 Grains of Sand…
THE MOLE!

IF you had 6.02 X 1023 Grains of Sand…
it would be more than ALL the sand on Miami Beach.
THE MOLE?


A mole is the amount of substance or representative
particles of any substance.
Therefore, as we have just seen


1 mol of N2 there are 6.02x1023 MOLECULES of N2
1 mol of C12H22O11 has how many molecules?
6.02x1023 can be used for “atoms” if they are the same
element
OR
It can be used for “molecules” for many elements that are
together, via compounds.
CHEMICAL QUANTITIES

How many moles of Magnesium is 3.01x1022 atoms of
magnesium?
Step 1 - What do we know?
We know that there are 3.01x1022 atoms of Mg
Step 2 - What do we want know?
We want to know the number of moles of Mg
Step 3 – What do we know that can be used for these conversions?
We know that there are 6.02x1023 atoms in 1 mol
Step 4 – Solve the problem
CHEMICAL QUANTITIES

How many moles of Magnesium is 3.01x1022 atoms of
magnesium?
3.01x1022
atoms Mg
X
1 mol Mg
6.02x1023 atoms of Mg
= 5.00x10-2 mol Mg
CHEMICAL QUANTITIES

Individual Problems




How many moles are 1.20x1025 atoms of phosphorus?
How many atoms are in 0.750 mol of Zinc?
How many molecules are there in 4 mol of glucose, C6H12O6?
How many molecules are there in 0.44 mol N2O5
CHEMICAL QUANTITIES




Now there is a difference between asking how many
atoms are there in the entire compound and how many
atoms of one of the compound there are.
We said that there are the same number or atoms in a
compound or a diatomic molecule. However, there is a
difference between these questions…
How many atoms are there in aluminum fluoride?
VS.
How many fluoride ions are there in aluminum fluoride?
CHEMICAL QUANTITIES

So let’s answer the question.


How many fluoride ions are there in 1.46 mol of aluminum
fluoride?
If we follow the four steps that we went through prior
CHEMICAL QUANTITIES
How many fluoride ions are there in 1.46 mol of aluminum
fluoride?
Step 1 - What do we know?
We know that there are 1.46 mol of aluminum fluoride
Step 2 - What do we want know?
We want to know the number of F-
Step 3 – What do we know that can be used for these conversions?
NO!
We know that there are 6.02x1023 atoms in 1 mol
BUT is that enough information??
CHEMICAL QUANTITIES
How many fluoride ions are there in 1.46 mol of aluminum
fluoride?
We want to know the number of FBut how can we find out the number of fluoride ions
we have from what is given?
We can get the information we need by writing out the
chemical formula of the compound.
AlF3
CHEMICAL QUANTITIES
How many fluoride ions are there in 1.46 mol of aluminum
fluoride?
1.46 mol AlF3
X
6.02x1023 Formula units of AlF3
3 F- ions
1 formula unit AlF3
1 mol AlF3
X
= 26.3676 x 1023 F- ions
INDIVIDUAL WORK

How many ammonium ions are in 0.036 mol ammonium
phosphate, (NH4)3PO4?
DRILL

What is Avogadro's number?

How many particles are there in 1 mol of a substance?

How many atoms are there in 1.45 mol of Na?
CHEMICAL QUANTITIES
THE GRAM FORMULA MASS
CHEMICAL QUANTITIES

Chemists have defined the gram atomic mass as the
number of grams of an element that is numerically equal
to the atomic mass in amu



The Gram atomic mass is the mass of one mole of atoms of a
monatomic element
This can also be described as MOLAR MASS – in place of
gram formula mass to refer to the mass of a mole of any
element or compound.
I will be using the term Molar Mass more frequently
CHEMICAL QUANTITIES
Atomic Mass Number
CHEMICAL QUANTITIES
This number tells you how many grams of this
element there are in 1 mol of the element
Therefore,
there are 47.88 grams/mol of Ti
Or there are 47.88 grams of Ti IN 1 mol of Ti
CHEMICAL QUANTITIES
B
C
N
O
F
10.811g
12.0107g
14.0067g
15.9994g
18.998g
The molar mass in grams/mol
CHEMICAL QUANTITIES


You can calculate the total molecular weight of a
molecule by adding up the molar masses of each element.
The molar mass of the molecule SO4 is equal to
S = 32.065 g/mol x 1 S = 32.065 g/mol
O = 15.9994 g/mol x 4 O = 63.9976 g/mol
SO4 = 96.0626 g/mol

I will specify the amount of significant figures needed in
INDIVIDUAL WORK

What is the molar mass of the following compounds?
1.
2.
3.
4.
5.
PCl3
Sodium Carbonate
C8H18
Aluminum Sulfate
(NH4)2CO3
CHEMICAL QUANTITIES

Why is this important?



This is important because it will allow us to covert from moles
to grams and grams to moles, allowing for quantitative
experimentation.
Let’s do a calculation…
How many grams are in 7.20 mol of dinitrogen trioxide?
CHEMICAL QUANTITIES
How many grams are in 7.20 mol of dinitrogen trioxide?
 Step 1 – add up the total molar mass of the compound



If it is in word form, write the chemical formula
Step 2 – Set up the proper conversion factors
Step 3 – Solve
CHEMICAL QUANTITIES
How many grams are in 7.20 mol of dinitrogen trioxide?
 Step 1 – add up the total molar mass of the compound

If it is in word form, write the chemical formula
N2O3
N = 14.0 g/mol x 2 N = 28.0 g/mol
O = 16.0 g/mol x 3 O = 48.0 g/mol
N2O3 = 76.0 g/mol
CHEMICAL QUANTITIES
How many grams are in 7.20 mol of dinitrogen trioxide?
 Step 2 – Set up the proper conversion factors
 Step 3 – Solve
7.20 molN 2 O 3 x
76 .0 gN 2 O 3
1.00 molN 2 O 3
 547 .2 gN 2 O 3
INDIVIDUAL WORK

Find the mass of the following
1.
2.
3.

3.32 mol K
5.08 mol Ca(NO3)2
4.52x10-3 mol K2CO3
Find the number moles of the following
1.
2.
3.
0.000264g Li2HPO4
847g (NH4)2CO3
195g calcium nitrate
EXIT TICKET

Does 54.938 g of Mn have the same number of moles as
that of 112.411 g of Cd? Explain why or why not.
DRILL

What is Avogadro’s number?

How many moles are there in 245 kg of CH2COOH?

Does 54.938 g of Mn have the same number of moles as
that of 112.411 g of Cd? Explain why or why not.
Group Work

Find the mass (g) of the following:
1.
2.
3.
4.
5.
6.
10.0 mol Cr
2.20x10-3 mol Sn
0.720 mol Be
2.40 mol N2
4.52x10-3 mol C20H42
0.0112 mol Potassium Carbonate
Find the number of moles of the following:
1.
1.
2.
3.
4.
5.
72.0 g Ar
3.70x10-1 g B
333 g Tin (II) Fluoride
7.21x10-2 g He
27.4 g TiO2
CHEMICAL QUANTITIES
THE VOLUME
OF A
MOLE
OF
GAS
CHEMICAL QUANTITIES
How would we measure the amount of moles in a gas?
CHEMICAL QUANTITIES

The volume of a gas is usually measured at STP (Standard
Temperature and Pressure)

STP conditions
 Temperature
 0°C
 Pressure
 1 atmosphere (atm)
CHEMICAL QUANTITIES
At STP conditions
1 mol of ANY gas occupies a volume of 22.4 Liters (L)
22.4 Liters of a gas / 1 mol of the gas
22.4 L of a gas contains 6.02 x 1023 representative
particles of that gas
This is known as the molar volume of a gas
CHEMICAL QUANTITIES
What were the units of the molar volume of a gas?
Liters
What kind of unit is liters?
VOLUME
Therefore, 1 mol of any gas occupies the same volume
not mass
CHEMICAL QUANTITIES
He
22.4 L N2
6.02 x 1023 molecules of N2
28g of N2
Ne
22.4 L He
6.02 x 1023 molecules of He
4g
22.4 L CO2
6.02 x 1023 molecules of CO2
44g of CO2
CO2
INDIVIDUAL WORK
What is the volume at STP of these gases?
5.40 mol O2
3.20 x 10-2 mol CO2
Assuming STP conditions, how many moles are there in
these volumes
74.6 L SO2
5.78x10-2 N2
CHEMICAL QUANTITIES
Because the molar volume of a gas is a VOLUME, we can
relate density (mass over volume) of a gas to determine
the mass of the gas.
If we have a gas that has a density of 1.964 g/L we can
multiply the density with the molar volume of a gas (22.4
L) to calculate the mass of the gas.
1.964 g x 22.4 L = 44.0g
L

GROUP WORK
The densities of gases A, B, & C are 1.25 g/L, 2.86 g/L, and
0.714 g/L, respectively.
Calculate the molar mass of each of these substances and
compare them to the molar mass of ammonia, sulfur
dioxide, chlorine, nitrogen, and methane
Identify the gasses A, B, & C
CHEMICAL QUANTITIES
Atoms (6.02 x 10 23)
1mol
1 mol
Molar
Mass
22.4 L
Solids and
Liquids
Gases
HOMEWORK
PROBLEMS
COPY THEM DOWN!
1. Calculate the volume of each of these gases at STP

1.
2.
9.6 mol He
4.8 mol N2
How many moles is each of the following, assuming STP
2.
1.
2.
56.o L N2O
0.224 L O2
Find each of the following quantities
3.
1.
2.
The mass of 18.0L of CH4 (STP)
The volume in liter, of 835 g of SO3 (STP)
DRILL


What are STANDARD conditions?
How many liters are there in 1 mol of gas?

Does it matter which gas it is? Why or why not?
CHEMICAL QUANTITIES
PERCENT COMPOSITION
CHEMICAL QUANTITIES- % COMPOSITION
Calculating Percent Composition
Percent Composition



The percent by mass of each element in a compound
They must add up to 100%
Ex. K2CrO4




40.3% K
26.8% Cr
32.9% O
100% Total
CHEMICAL QUANTITIES - % COMPOSITION
Percent Mass of an element in a compound is the number
of grams of the element divided by the grams of the
compound, multiplied by 100%
% mass =
Grams of element
Grams of compound
x 100%
CHEMICAL QUANTITIES- % COMPOSITION
Let’s look at our example again
K2CrO4
STEP 1
Calculate the Total Molecular Mass of K2CrO4
– Go ahead a calculate that now
K = 39 g x 2K = 78 g
Cr = 52 g
O = 16 g x 4O = 64g
K2CrO4 = 194g
CHEMICAL QUANTITIES- % COMPOSITION
Let’s look at our example again
K2CrO4
STEP 2
Using the individual molar mass of each element, divide each
element with the total molar mass
Go ahead and calculate that now with 3 Significant figures
K = 39 g x 2K = 78 g K ÷194g = .402
Cr = 52 g ÷194g = 0.268
O = 16 g x 4O = 64g ÷194g = 0.330
CHEMICAL QUANTITIES- % COMPOSITION
Let’s look at our example again
K2CrO4
STEP 3
Multiply the number calculated from Step 2 and multiply by 100%
Go ahead and calculate that now
K = 39 g x 2K = 78 g K ÷194g = .402 x 100% = 40.2% K
Cr = 52 g ÷194g = 0.268 x 100% = 26.8% Cr
O = 16 g x 4O = 64g ÷194g = 0.330 x 100% = 33% O
CHEMICAL QUANTITIES- % COMPOSITION
INDIVIDUAL WORK
Calculate the mass of carbon in 82g in C3H8
Calculate the percent composition of each of these
compounds.
1.
2.
1.
2.
3.
4.
Propane, C3H8
Sodium bisulfate, NaHSO4
Calcium acetate, Ca(C2H3O2)2
Hydrogen cyanide, HCN
CLOSURE

Take the remainder of this period and write down
in complete sentences what you learned today,
approximately 2-3 paragraphs.
DRILL


What is percent composition?
What is the formula for percent composition?
DRILL

Calculate the mass of oxygen in 142g of Sodium
bisulfate, NaHSO4
CHEMICAL QUANTITIES
EMPIRICAL FORMULAS
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
Empirical Formula
Gives the lowest whole number ratio of the elements
in a compound.
This ratio is not necessarily the molecular formula!
Calculating the empirical formula is going to require the use
of almost everything you’ve learned so far.
Each element in the compound would require different
amounts depending on the mass and % composition
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
Steps to solve for an Empirical Formula
1. Write down the base formula AxBY
2. Convert whatever you have into moles using
dimensional analysis
3. Divide each element by the lowest number of moles in
the compound
4. Round if necessary to the nearest 10ths place and/or
multiply by a number to get a whole number ratio
5. Write down the Empirical Formula
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
What is the empirical formula of a compound that is 25.9%
Nitrogen and 74.1% Oxygen?
STEP 1 - Write down the base formula AxBY…
NxOy
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
What is the empirical formula of a compound that is 25.9%
Nitrogen and 74.1% Oxygen?
STEP 2 - Convert whatever you have into moles using
dimensional analysis
Because this is a % composition, we know that
everything has to add up to 100%; therefore, we can
assume that if we had 100g total of the compound,
there would be 25.9g of N and 74.1g of O
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
What is the empirical formula of a compound that is 25.9%
Nitrogen and 74.1% Oxygen?
STEP 2 - Convert whatever you have into moles using
dimensional analysis
25.9g of N x
1 mol N
14.0 g N
= 1.85 mol N
74.1g of O x
1 mol O
16.0 g O
= 4.63 mol O
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
What is the empirical formula of a compound that is 25.9%
Nitrogen and 74.1% Oxygen?
STEP 3 - Divide each element by the lowest number of
moles in the compound
25.9g of N x
1 mol N
14.0 g N
= 1.85 mol N ÷ 1.85 mol = 1 mol N
74.1g of O x
1 mol O
16.0 g O
= 4.63 mol O ÷ 1.85 mol = 2.5 mol O
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
What is the empirical formula of a compound that is 25.9%
Nitrogen and 74.1% Oxygen?
STEP 4 - Round if necessary to the nearest 10ths place
and/or multiply by a number to get a whole number
ratio
2 = 2 mol N
1 mol N x ____
2 = 5 mol O
2.5 mol O x ____
CHEMICAL QUANTITIES- EMPIRICAL FORMULAS
What is the empirical formula of a compound that is
25.9% Nitrogen and 74.1% Oxygen?
STEP 5 - Write down the Empirical Formula
2 mol N
5 mol O
N2O5
CHEMICAL QUANTITIES- MOLECULAR FORMULAS
The Molecular formula
is the actual formula of a molecular compound
While the empirical formula give us the base ratio of the
compound, the molecular formula will give the actual
molecular formula by using the molar mass.
x (empirical formula) = molecular formula
Ratio of the compounds
Chemical Quantities- Molecular Formulas
 Therefore, if you have the molar mass of the compound

and divide it by the molar mass of the empirical formula,
you will get the ratio between the two.
Multiply the ratio to each element of the compound to
get the the molecular formula.
Chemical Quantities- Molecular Formulas

Ex. If the empirical formula was calculated to be CH4N and
the known molecular formula molar mass is = to 60g
 1st – Take the molar mass of the empirical formula CH4N
 In
this case it is calculated to be 30g/mol

2nd – Divide the known molar mass of 60g/mol and divide
it by 30 g/mol 60/30 = 2

3rd – Take the ratio of the two molar masses and then
multiply it by each element of the compound
C1x2H4x2N1x2= C2H8N2
Chemical Quantities- Molecular Formulas
Write the Molecular Formula for the problem below

The compound methyl butanoate has a percent composition of
58.8% C, 9.8% H, 31.4%O. If its molecular mass is 102 g/mol,
what is it’s molecular formula?
Write the Empirical Formula for the following
1. 79.8% C 20.2% H
2. 67.6% Hg 10.8% S 21.6% O
3. 27.59% C
1.15% H
16.09%N
55.17%O
4. 17.6% Na
39.7%Cr
42.7%O
DRILL

What is the difference between an empirical formula and
molecular formula?
Chemical Quantities – Review
Determine the Empirical Formula with the parameters
below
1.
71.72% Cl, 16.16% O, and 12.12%C
What is the molecular formula for the compound below?
The compound’s empirical formula and molar mass is given
Below
2.
HgCl, 472.2 g/mol
Determine the molecular formula for the compound
3.
94.1%O and 5.9% H ; molar mass = 34g
Find the empirical formula for each compound from its percent
composition
4.
5.
72.4% Fe and 27.6% O
52.8% Sn, 12.4% Fe, 16.0% C, and 18.8% N
```