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Chapter 10
What are practical ways to measure a
container full of sand??
Mass
 Volume
 How many (Count)

Matter is generally measured in one
of three ways:

By count:
 1 dozen apples = 12 apples

By mass:
 1 dozen apples = 2.0 kg of apples

By volume:
 1 dozen apples = .20 bushels of apples
Converting units

If you know how count, mass, and
volume relate to each other, you can do
conversions between them.

EX: How many bushels of apples are in
32.7 kg of apples?
Numbers as Symbols

It is not practical to individually count out
small objects like sand grains, so
instead we use a value that represents a
larger quantity
 Ex: 120 eggs = 10 dozen eggs
Chemical Numbers

Atoms, molecules, and ions are extremely
small, so chemists use a unit similar to a
“dozen” to measure a specific number of
particles.
The MOLE

A MOLE is 6.02 x 1023 representative
particles of whatever is being talked about
○ Avogadro's number = 6.02x1023
○ Representative particles refer to the type of
substance that you are talking about
○ Abbreviated mol
You talking
about me?
Examples of Representative Particles
Substance
A balloon full of Neon
A cup full of Water
A spoon full of NaCl
Representative Particle
Neon Atoms
Water Molecules
NaCl Formula Units
Number of Iron ions in Water
Fe Ions
How much oxygen is in the air
O2 molecules
Measuring by Count

If you know the number of moles of
something or how many particles of
something there are, you can convert to
the other.
 Ex: How many moles of magnesium are
there in 1.23 x 1023 atoms of magnesium?
Measuring by Count
 Ex: How many atoms are in 1.22 mol of
Iron?
*Measuring by Count
 Ex: How many atoms of carbon are in 3.21
mol of C8H18?
Homework

Pg. 296
 # 10, 13, 14,

Pg. 315
 # 49, 52,
Measuring by Mass

The mass of a mole of atoms of a given
element will always be the same.
 It equals the atomic mass expressed in
grams

Molar Mass: the mass of one mole of a
given particle (atom, compound, etc.)
Finding Molar Mass
(Important notes below…write this down)
Find the element;
 Find the molar mass;
 Multiply by the number of atoms of
that element;
 Write this number down;
 Repeat for each element;
 Take the sum.

Finding Molar Mass

C:
 C: 1 x 12.00g = 12.00 g/mol C

O2:
 O: 2 x 16.00 g = 32.00 g/mol O2

H2O:
 H: 2 x 1.01 g = 2.02 g H
 O: 1 x 16.00 g = 16.00 g O
18.02 g/mol H2O
Finding Molar Mass

C12H22O11:
 C: 12 x 12.01g = 144.12 g C
 H: 22 x 1.01g = 22.22 g H
 O: 11 x 16.00g = 176.00 g O
342.34 g/mol C12H22O11
Measuring by Mass

If you know how many moles of a
substance you have, you can find its
total mass using the molar mass of that
substance.
 Ex: What is the mass of 2.42 mol of H2O?
= 43.6 g H2O
Your turn!

Measuring by Mass

If you are given the mass, you can also
find the number of moles
 Ex: How many moles are there in 227.4 g of
Na2CO3?
106.0 g/mol Na2CO3
Measuring by Volume
The volume of 1 mole of a solid and
liquid are very different depending on
the substance
 However, the volume of one mole of a
gas is equal for every gas at the same
physical conditions

Avogadro’s Hypothesis

Equal volumes of gases at the same
temperature and pressure contain equal
number of particles.
 Even though particles are different sizes,
they still occupy the same amount of space
because there is a great deal of space
between gas particles
Standard Temperature and
Pressure

In order to compare the volume of a gas
to one mole of the gas, we must define
the temperature and pressure
 As Temp goes up, Volume goes up
 As Volume goes down, Pressure goes up
Standard Temperature and
Pressure

Standard Temperature and Pressure
(STP): temperature of 0 ˚C and a
pressure of 1 atm
 STP is used as a universal standard at
which we compare things
Measuring by Volume

At STP one mole of a gas occupies
22.4 L of space
 Ex: How many moles of oxygen (O2) would
be present in a 63.2 L container at STP?
Practice

How many moles are present in a
sample of 4.02 x 1024molecules of NH3?

What is the mass of 2.14mol of C2H4O2?

How many molecules can be found in
3.42mol of BaSO4?
Practice makes perfect

Under STP, what volume does 4.95mol
of CH4 take up?

If a gas takes up 34.71L of space under
STP, how many moles of that gas is
present?

How many moles are present in a 134g
sample of HBr?
Mo’ Practice = Mo’ Perfect

How many cookies are in 24.1moles of
Oreo’s?

How many/much _________ can be
found in __________ while under STP?
Quiz Practice

HCl
CO2
Quiz Practice

How many moles are in 4.03 𝑥 1022
molecules of silver (I) oxide?
22

4.03 𝑥 10 𝑎𝑡𝑜𝑚𝑠 𝐴𝑔2 𝑂

=0.0669 𝑚𝑜𝑙𝐴𝑔2 𝑂
1𝑚𝑜𝑙
6.02 𝑥 1023
=
Quiz Practice

How much mass does 3.79 moles of
𝐵𝐹3 have?
B: 10.8 x 1=10.8g
 F: 19.0 x 3= 57.0g
 Total: 67.8g

67.8𝑔
1𝑚𝑜𝑙

3.79 𝑚𝑜𝑙

= 257.0g 𝐵𝐹3
=
Quiz Practice
Quiz Practice

How much volume in liters does 6.307
moles of NaCl fill at STP?
22.4𝐿
1𝑚𝑜𝑙

6.307𝑚𝑜𝑙

141𝐿 NaCl
=
Quiz Practice
How much mass does 1.805 moles of
Lithium Bromide (LiBr) have?
 Li: 6.9 x 1= 6.9g
 Br: 79.9 x 1= 79.9g
 Total= 86.8g



1.805𝑚𝑜𝑙
=
86.8𝑔
1𝑚𝑜𝑙
156.7𝑔
𝐿𝑖𝐵𝑟
1 𝑚𝑜𝑙
=
Mixed Mole Conversions

We now know:
1 mole = 6.02 x 1023 particles = molar mass = 22.4 L of gas at STP
 Because
of this, we can convert from
count, mass or volume to count, mass or
volume.
 BUT, you must first convert to moles
Mixed Mole Conversions
Practice!

44.01g
CO2
Practice, part dos!

Ugh, More? 

180.16
Practice??

60.05
Even more practice???

32.00
(mind=blown)
And I thought my mind was
already blown…
Practice Extraordinaire!

What is the mass of 3.72 x 1024 molecules
of C3H8?
3.72 x 1024
molecules
=273g C3H8
1mol
44.11g
6.02 x 1023
molecules
1mol
To practice or not to prac…
well let’s just practice

What is the volume of 57.9g of methane
CH4 gas at STP?
57.9g
= 80.8L CH4
1mol
22.4L
16.05g
1mol
Oh the practice!

What is the mass of 1.5 x 1025
molecules of CO2?
1.5 x 1025
molecules
1mol
44.01g
6.02 x 1023
molecules
1mol
= 1097g CO2 = 1.10 x 103g CO2
Percentages Matter!

Fertilizers
 Higher nitrogen content  greener grass
 Higher potassium content  stronger roots

A man walks into a restaurant….
Percent Composition
The percent by mass of each element in
a compound
 Usually used to determine the chemical
experimentally
 If the formula is known, the percent can
be easily determined from each atom’s
molar mass

Percent Composition

Ex: What is the percent composition of H2O
H: 2 x 1.01 g/mol = 2.02 g/mol
O: 1 x 16.00 g/mol = 16.00 g/mol
18.02 g/mol H2O
% H = (2.02 g/mol) / (18.02 g/mol) x 100 = 11.2 % H
% O = (16.00 g/mol) / (18.02 g/mol) x 100 = 88.79% O
Percent Composition

Ex: CH2Cl2
C: 1 x 12.01 g/mol = 12.01 g/mol
H: 2 x 1.01 g/mol = 2.02 g/mol
Cl: 2 x 35.45 g/mol = 70.90 g/mol
84.93 g/mol CH2Cl2
% C = (12.01g/mol) / (84.93g/mol) x 100 = 14.14 % C
% H = (2.02g/mol) / (84.93g/mol) x 100 = 2.38 % H
% Cl = (70.90g/mol) / (84.93g/mol) x 100 = 83.48 % Cl
More Practice?!?!
What is the percent composition of ZnBr2?
 Zn: 65.39g/mol x 1 = 65.39g/mol
 Br: 79.90g/mol x 2 = 159.8g/mol
 Total: 225.2g/mol

Zn: (65.39g/mol) / (225.2g/mol) x 100 = 29.04%
 Br: (159.8g/mol) / (225.2g/mol) x 100 = 70.96%

“You’re killing me smalls”

If a nickel ore sample that had a mass of
159.62g was found to contain 48.97g of
nickel, what is the percent composition of
nickel in the ore sample?

(48.97g Ni) / (159.62g ore) x 100 = 30.68% Ni
Ah-Yea-ah!!

What is the percent composition of
𝐶𝑎(𝑁𝑂3 )2 ?
Ca: 40.08g/mol x 1 = 40.08g/mol
 N: 14.01g/mol x 2 = 28.02g/mol
 O: 16.00g/mol x 6 = 96.00g/mol
 Total: 164.10g




Ca: (40.08g/mol) / (164.10g/mol) x 100 = 24.42%
N: (28.02g/mol) / (164.10g/mol) x 100 = 17.07%
O: (96.00g/mol) / (164.10g/mol) x 100 = 58.50%
Yes-sir-ee Bob!

What is the percent composition of
NaOH?
Na: 22.99g/mol x 1 = 22.99g/mol
 O: 16.00g/mol x 1 = 16.00g/mol
 H: 1x 1 = 1.01g/mol
 Total: 40.00g/mol




Na: (22.99g/mol) / (40.00g/mol) x 100 = 57.48%
O: (16.00g/mol) / (40.00g/mol) x 100 = 40.00%
H: (1.01g/mol) / (40.00g/mol) x 100 = 2.53%
Bubblegum and Magic Cards!
Empirical Formula

The ratio of atoms in a compound
 NOT the chemical formula!
Calculating the Empirical Formula from
Percent Composition
1.
2.
3.
4.
5.
Assume you have 100 g of the compound.
Change “%” to “g”
Convert grams to moles for each element
Divide each mole amount by the smallest
mole amount, these numbers are the
coefficients for the E.F.
If the numbers from step 4 are not all whole
numbers, multiply the step 4 numbers by a
whole number so all step 4 numbers are
whole numbers
.5 moles  x 2
.333 moles  x 3
.25 moles  x 4
.20 moles  x 5
Empirical Formulas

Ex: A compound is determined to be
26.56% K, 35.41% Cr, and 38.08 % O.
What is the empirical formula of the
compound?
Molecular Formula of a
Compound

Molecules can have formulas that are
not the simplest ratio. They can be
whole number multiples of the Empirical
Formula
Calculating Molecular Formula
1.
2.
3.
Start with the E.F.
Determine molar mass of E.F.
Divide the molar mass of the unknown
molecule by the molar mass of the E.F.
 This tells how many E.F.’s are in the M.F
4.
Multiply the subscripts of the E.F. by
the number found in step 3
Molecular Formulas

Ex: What is the molecular formula of a
compound that has an empirical formula
of P2O5 and a molar mass of 283.89 g?
P: 2 x 31.0g = 62.0 g
O: 5 x 16.0g = 80.0 g
142.0g P2O5
283.89g =1.999 ≈ 2 therefore, MF=P4O10
142.0g