Transcript Chemical Quantities

Starter S-89
1. List the elements and how many there
are of each in Iron (III) oxide.
2. Multiply the number of each element by
the average atomic mass of each.
3. Add these numbers together, this is
called the molar mass.
Starter S-90
1. What is the molar mass of H2S?
2. How many moles of H2S would be found
in 100 g?
3. How many formula units (particles) of
H2S would be in 100g?
Remember the lab from Tuesday
Chapter 10
Chemical Quantities
Chapter 10
10.1 The Mole: A Measurement of Matter
10.1 The Mole
Quantitative – yield a number value
Most common methods
count
mass
volume
10.1 The Mole
Words often mean a quantity
pair – 2
dozen – 12
mole?
10.1 The
The Mole
10.1
Mole
Mole – means a specific number of particles
6.022 x10
23
This is called Avogadro’s Number
The unit is used for
atoms
molecules (covalent compounds)
formula units (ionic compounds)
10.1 The
The Mole
10.1
Mole
To convert numbers of particles to moles we need
an equality
1mol  6.022 x10 23 atoms
This could also be
1mol  6.022 x10 23 molecules
Or
1mol  6.022 x1023 formula.units
Depending on what type of particle you are trying to
convert
10.1 The
The Mole
10.1
Mole
The rest is just the conversion we have always done
remember – the unit you have goes on the
bottom of the fraction
-the unit you are converting to goes on the top
So to convert 2500 atom of C to moles
1molC


2500atomC

23
 6.022 x10 atomsC 
4.2 x10 21 molC
10.1 The
The Mole
10.1
Mole
Converting from moles to partciles follows the same
process
How many molecules of Carbon Dioxide are in 1.55
mol?
 6.022 x1023 moleculesC O2 

1.55molCO2 
1molCO2


9.33x1023 moleculesCO2
10.1 The
The Mole
10.1
Mole
Convert the following quantities to moles
1. 9,487,212 formula units of AgNO3
1.575 x10
17
molAgNO3
2. 5.78x1023 atoms of Nitrogen
0.960molN
3. 4.1x1024 molecules of Carbon Monoxide
6.8molCO
10.1 The
The Mole
10.1
Mole
Convert the following to the correct type of particle
1. 95 moles of CCl4
25
5.7 x10 moleculesC Cl4
2. 7.211x10-3 moles of CuCO3
21
4.342 x10 formula.unitsCuCO3
3. 0.08 moles of Helium
22
5 x10 atomsHe
10.1 The
The Mole
10.1
Mole
By definition, the atomic mass of an element in
grams is the mass of one mole of the element.
This is called the molar mass
10.1 The
The Mole
10.1
Mole
For a compound we must calculate the molar mass
1. Write down what type of atoms, and how many
of each are present
CH4
C 1
H 4
2. Multiply by the molar mass of each element
10.1 The
The Mole
10.1
Mole
For a compound we must calculate the molar mass
1. Write down what type of atoms, and how many
of each are present
CH4
C 1 x 12.0107g =
H 4 x 1.00794g =
2. Multiply by the molar mass of each element
10.1 The
The Mole
10.1
Mole
For a compound we must calculate the molar mass
1. Write down what type of atoms, and how many
of each are present
CH4
C 1 x 12.0107g = 12.0107g
H 4 x 1.00794g = 4.03176g
2. Multiply by the molar mass of each element
3. Add to get a total
10.1 The
The Mole
10.1
Mole
For a compound we must calculate the molar mass
1. Write down what type of atoms, and how many
of each are present
CH4
C 1 x 12.0107g = 12.0107g unit
H 4 x 1.00794g = 4.03176g
2. Multiply by the molar mass of each element
3. Add to get a total 16.0425g CH4
Starter S-92
1. What is the molar mass of CO?
2. How many moles of CO would
be found in 0.56 g?
3. How many molecules of
CO would be found in
3.51 moles?
Chapter 10
10.2 Mole-Mass and Mole-Volume
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
We don’t convert from particles to moles nearly as
often as we do from moles to grams.
The reason is that we usually measure the amount
of a substance on the balance
We need to know numbers of
particles so that we can
compare ratios of atom or
compounds
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
This is just another conversion problem
The equality is
Value Molar Mass = 1 mole
Again, the quantity you have goes on the bottom,
the quantity you want goes on top
So if you have 9.5g of Carbon
 1molC 
  0.79molC
9.5 gC 
 12.0107 gC 
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
If you have 2.2 moles of Silver Nitrate convert to
mass
First we need to know the formula of Silver Nitrate
AgNO3
Then the molar mass
169.8731g
And finally we can convert
 169.8731gAgNO3 
  370 gAgNO3
2.2molAgNO3 
 1molAgNO3 
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
Try the following example: How many moles is 8.2g
of Copper (II) Chloride
Formula
CuCl2
Molar mass
134.451g
Moles
 1molCuCl2 
  0.061molCuCl2
8.2 gCuCl2 
 134.451gCuCl2 
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
And another one: How many grams is 2.4 mol of
Iron (III) Sulfate
Formula
Fe2 SO4 3
Molar mass
399.881g
Moles
 399.881gFe2 SO4 3 
  960 gFe2 SO4 3
2.4molFe2 SO4 3 



1
molFe
SO
2
4 3


10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
In 1811, Amedeo Avogadro proposed
Avagador’s Hypothesis – equal volumes of gases at
the same temperature and pressure contain
equal numbers of particles
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
Standard Temperature and Pressure (STP)
T – 0oC or 273K
P – 101.3 kPa, or 1 atm
At STP the volume of one mole is 22.4L
So the equality for conversion is
1mol  22.4L
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
To do these problems, the identity of the gas doesn’t
really matter.
If we have 15 L of Chlorine gas
Cl2
The number of moles would be
 1molCl2 
  0.67molCl2
15LCl2 
 22.4 LCl2 
10.2 Mole-Mass
andMole
Mole-Volume
10.1 The
If instead we calculate for a more complicated gas
such as propane
And we also have 15 L of propane gas
C3 H 8
The number of moles would be
 1molC3 H 8 
  0.67molC3 H 8
15LC3 H 8 
 22.4 LC3 H 8 
Starter S-96
1. What is the molar mass of Pb(SO4)2?
2. How many moles of Pb(SO4)2 would be
found in 250 g?
3. How many moles of H2 gas are found in
250 L?
Chapter 10
10.3 Percent Composition and Chemical Formulas
10.3 10.1
Percent
Composition
The
Mole
The relative amounts of the elements in a
compound is called the percent composition
The percent by mass of an
element is the number of
grams of the element
divided by the mass in
grams of the compound multiplied by 100%
 mass.element 
%mass  
 x100%
 mass.compound 
10.3 10.1
Percent
Composition
The
Mole
Calculating the mass percent from a formula
CH 4
1. Formula
2. Calculate the total
C  1x12.0107 g  12.0107 g
mass of each element H  4 x1.00794  4.03176 g
3. Calculate the total
CH 4  12.0107  4.03176  16.0425 g
mass of
the compound
 12.0107 
C 
4. Calculate the
 x100%  74.8680%
 16.0425 
percent by
 4.03176 
H 
mass for
 x100%  25.1317%
 16.0425 
each element
10.3 10.1
Percent
Composition
The
Mole
Calculate the mass percent
C

1
x
12.0107
g

12.0107
g
CH 3OH
1. Formula
2. Mass of each element H  4 x1.00794  4.03176 g
O  1x15.9994  15.9994 g
3. Total mass
CH 3OH  12.0107  4.03176  15.9994  32.0419 g
4. Mass Percent
 12.0107 
C 
 x100%  37.4844%
 32.0419 
 4.03176 
H 
 x100%  12.5828%
 32.0419 
 15.9994 
O 
 x100%  49.9327%
 32.0419 
Starter S-95
What is the percent by mass of the all the
elements in Cu(NO3)2.
10.3 10.1
Percent
Composition
The
Mole
Empirical Formula – smallest whole number ratio of
the elements in a
Compound Empirical
compound
Formula
The empirical formula
H2O
H2O
can be calculated
CH3COOH
CH2O
from the percent
composition
CH2O
CH2O
C6H12O6
CH2O
S8
S
10.3 10.1
Percent
Composition
The
Mole
To calculate the empirical formula
N  25.9%
1. List the elements and their
O  74.1%
percent compositions
 1molN 
N  25.9 gN 
 1.85molN

2. Convert the percent
 14.0 gN 
compositions to moles O  74.1gO  1molO   4.63molO


16.0
gO


3. Calculate the mole ratio (divide N   1.85mol   1.00
N
2mol 
1.85
by the smallest number of moles)
4.63
mol 

O
5
O 
  2.50
4. Smallest Whole Number ratio
 1.85mol 
5. Write the Formula
2 5
NO
10.3 10.1
Percent
Composition
The
Mole
To calculate the empirical formula
1. Elements
2. Convert to Moles
Hg  67.6%
S  10.8%
O  21.6%
 1molHg 
Hg  67.6 gHg 
  0.337 molHg
200.59
gNg


 1molS 
S  10.8 gS 
  0.337 molS
32.06
gS


 1molO 
O  21.6 gO 
  1.35molO
16.0
gO


3. Mole ratio
4. Whole Number ratio
5. Write the Formula
HgSO4
 0.337 mol 
Hg  
  1.00
Hg 0.337
 1 mol 
 0.337 mol 
S S  1
  1.00
 0.337 mol 
1.35
mol
O
4

O


  4.01
 0.337 mol 
Starter S-97
What is the empirical formula if
Lead is 59.7%
Hydrogen is 2.9%
Arsenic is 21.6%
Oxygen is 18.4%
10.3 10.1
Percent
Composition
The
Mole
The molecular formula can be calculated from the
empirical formula and the molar mass
Comparison of Empirical and Molecular Formulas
Formula
Classification
Molar Mass
CH
Empirical
13
C2H2
Molecular
26 (2x13)
C6H6
Molecular
78 (6x13)
CH2O
Empirical
30
C2H4O2
Molecular
60 (2x30)
C6H12O6
Molecular
180 (6x30)
10.3 10.1
Percent
Composition
The
Mole
Steps in calculations (if mass=60.0g)
4
1. Determine the empirical formula
C  1x12.0  12.0 gC
2. Calculate the mass
H  4 x1.01  4.04 gH
of the empirical formula
4
CH N
CH N  30.0 g
N  1x14.0  14.0 gN
3. Divide the actual molar mass by
this number
4. Multiply the empirical formula
60.0 g
 2.00
30.0 g
C2 H 8 N 2
Starter S-99
What is the empirical formula if
Silver – 63.5%
Nitrogen – 8.2 %
Oxygen – 28.2%
If the formula mass is 170g, what is the
formula of this compound?
Practice Problem 1
A. How many moles of AgNO3 are found in
125 g?
B. How many grams of AgNO3 are found in
2.99 moles?
Practice Problem 2
A. How many molecules are found in 19.5
moles of CH2O?
B. How many grams of CH2O are found in
1.8x1024 molecules?
Practice Problem 3
A. What is the empirical formula of the
following compound
Na-43.4%
C – 11.3%
O – 45.3%
Practice Problem 4
A. What is the empirical formula
C – 3.2g
H – 0.53g
O – 4.3 g
B. If the molar mass is 330, what is the
molecular formula?
Starter S-102
Test Day
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Starter S-105
What is the empirical formula if
Carbon – 49.5%
Hydrogen – 5.2%
Nitrogen – 28.9%
Oxygen – 16.5%
What if the molecular formula, if the molar
mass is 194g?