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M.S. Silberberg, Chemistry, The Molecular Nature of Matter and Change, 3rd Ed., Chapter 3 P. Atkins & L. Jones, Molecules, Matter and Change, 3rd Ed., Chapter 2 – PowerPoint PPT presentation

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Title: JF Tutorial: Mole Calculations


1
JF Tutorial Mole Calculations
  • Shane Plunkett
  • plunkes_at_tcd.ie
  • Some Mathematical Functions
  • What is a mole?
  • Avogadros Number
  • Converting between moles and mass
  • Calculating mass from a chemical formula
  • Determining empirical and molecular formulae from
    mass
  • Recommended reading
  • T.R. Dickson, Introduction to Chemistry, 8th
    Ed., Wiley, Chapters 2 4
  • M.S. Silberberg, Chemistry, The Molecular Nature
    of Matter and Change,
  • 3rd Ed., Chapter 3
  • P. Atkins L. Jones, Molecules, Matter and
    Change, 3rd Ed., Chapter 2
  • Multiple choice tests http//www.mhhe.com/silber
    berg3

2
Carrying out Calculations
  • In chemistry, must deal with several mathematical
    functions.
  • Scientific Notation
  • Makes it easier to deal with large numbers,
    especially concentrations
  • Written as A 10b, where A is a decimal number
    and b is a whole number

Example Avogadros number 602 213 670 000 000
000 000 000 It is very inconvenient to write
this. Instead, use scientific notation 6.022
1023
  • Calculators
  • Sharp Casio Type in 6.022
  • Press the exponential function EXP
  • Key in 23

3
Questions
How would you write the following
7.84 108 2.3 10-4 9.22 106 1.5 10-8
  • (a) 784000000
  • (b) 0.00023
  • (c) 9220000
  • (d) 0.000000015

Calculate the following
1.13 1011
(a) (1.38 104) (8.21 106)
2.05 10-3
(b) (8.56 10-8) (2.39 104)
4
Common Decimal Prefixes
Prefix Symbol Number Word Exponential Notation
tera T 1,000,000,000,000 trillion 1012
giga G 1,000,000,000 billion 109
Mega M 1,000,000 million 106
kilo k 1,000 thousand 103
hecto h 100 hundred 102
deca da 10 ten 101
deci d 0.1 tenth 10-1
centi c 0.01 hundredth 10-2
milli m 0.001 thousandth 10-3
micro ? 0.000001 millionth 10-6
nano n 0.000000001 billionth 10-9
pico p 0.000000000001 trillionth 10-12
femto f 0.000000000000001 quadrillionth 10-15
5
2. Logarithms
  • Makes dealing with a wide range of numbers more
    convenient, especially pH
  • Two types common logarithms and natural
    logarithms
  • Common Logarithms
  • Common log of x is denoted log x
  • gives the power to which 10 must be raised to
    equal x
  • 10n x
  • written as log10x n (base 10 is not always
    specified)

Example The common log of 1000 is 3, i.e. 10
must be raised to the power of 3 to get
1000 Written as log101000 3
103 1000
6
Calculators
  • Sharp Press the LOG function
  • Type the number
  • Hit answer
  • Casio Key in the number
  • Press the LOG function

Questions
Calculate the common logarithms of the following
  1. 10
  2. 1,000,000
  3. 0.001
  4. 853

1 6 -3 2.931
log 10 log 1000000 log 0.001 log 853
7
Natural logarithms
  • Natural log of x is denoted ln x
  • the difference here is, instead of base 10, we
    have base e (where e 2.71828)
  • Gives the power to which e must be raised to
    equal x
  • lnx or logex n or en x

Example The natural log of 10 is 2.303, i.e. e
must be raised to the power of 2.303 to get 10
  • Calculators
  • Sharp Press the ln function
  • Enter the number and hit answer
  • Casio Enter the number
  • Press the ln function

8
Questions
  • What is the natural log of
  1. 50
  2. 1.25 105
  3. 2.36 10-3
  4. 8.98 1013

ln 50 ln 1.25x105 ln 2.36x10-3 ln 8.98x1013
3.91 11.74 -6.05 32.13
9
3. Graphs
  • Experimental data often represented in graph
    form, especially in straight lines
  • Equation of straight line given by
  • y mx c
  • where x and y are the axes values
  • m is the slope of the graph
  • c is the intercept of the plot

y- axis
Slope
Intercept
x-axis
10
  • Sign of slope tells you the direction of the line
  • Magnitude of slope tells you steepness of line
  • Slope found by taking two x values and the two
    corresponding y values and substituting these
    into the following relation

Example Given the (x, y) coordinates (2, 4) and
(5, 9), find the slope of the line containing
these two points.
x1 2 y1 4
x2 5 y2 9
Sub into above relation m 9 4 5
1.67 5 2 3
11
4. Quadratic Equations
  • May be encountered when dealing with
    concentrations
  • Involve x2 (x-squared terms)
  • Take the form ax2 bx c 0
  • Can be solved by

  • this expression finds the roots or the solution
    for x of the quadratic equation

12
Find the roots of the equation x2 6x 8 0
Example
ax2 bx c 0
a 1
b -6
c 8
x
x
x

2
or
4
Therefore, x
13
Question
  • You have been asked to calculate the
    concentration of H3O ions in a
  • chemical reaction.

x H3O
The following quadratic equation has been given
Solve for x. 2.4x2 1.5x 3.6 0

14
Therefore
or
x 0.95 or x -1.58
Because we are dealing with concentrations, a
negative value will not make sense. Therefore,
we report the positive x value, 0.95, as our
answer. Never round up this number!
15
Important formulae so far
  • y mx c

Graphs..
ax2 bx c 0
Quadratic equations..
16
Calculations The Mole
  • Stoichiometry is the study of quantitative
    aspects of chemical
  • formulas and reactions
  • Mole SI unit of the amount of a substance
  • Definition A mole is the number of atoms in
    exactly 12g of the
  • carbon-12 isotope
  • This number is called Avogadros number and is
    given by 6.022 1023
  • The mole is NOT just a counting unit, like the
    dozen, which specifies only the number of
    objects. The definition of a mole specifies the
    number of objects in a fixed mass of substance.
  • Mass spectrometry tells us that the mass of a
    carbon-12 atom is
  • 1.992610-23g.

No. of carbon-12 atoms atomic mass (g)
mass of one atom (g)
12g _ 1.992610-23g
6.022 1023 atoms
17
Other definitions of the Mole
  • One mole contains Avogadros Number (6.022 x
    1023)
  • A mole is the amount of a substance of a system
    which contains as many elementary entities as
    there are atoms in 0.012kg (or12g) of Carbon-12
  • A mole is that quantity of a substance whose mass
    in grams is the same as its formula weight
  • E.g. Fe55.85
  • Iron has an atomic mass or 55.85g mol-1, so one
    mole of iron has a mass or 55.85g

18
One mole of any object always means 6.022 1023
units of those objects.For example, 1 mol of H2O
contains 6.022 1023 molecules 1 mol of NaCl
contains 6.022 1023 formula units
Calculating the number of particles
  • Avogadros number is used to convert between the
    number of
  • moles and the number of atoms, ions or molecules.

Example 0.450mol of iron contains how many atoms?
Number of atoms number of moles Avogadros
number (NA)
Therefore No. of atoms (0.450mol) (6.022
1023) 2.7 1023 atoms
19
Example
  • How many molecules are there in 4 moles of
    hydrogen peroxide (H2O2)?

No. of molecules no. of moles Avogadros
number (NA)
4mol (6.022 1023 mol-1)
24 1023 molecules
2.4 1024 molecules
Questions
How many atoms are there in 7.2 moles of gold
(Au)?
Answer 4.3 1024 atoms
The visible universe is estimated to contain 1022
stars. How many moles of stars are there?
Answer 1022 stars 1022 0.17
mol. 6.0221023
20
Calculating the mass of one molecule
  • Example What is the mass of one molecule of
    water?

Step 1 Calculate the molar mass of water
Molar mass of water (2 atomic mass H) (1
atomic mass O)
Molar mass H2O (2 1.008g mol-1) (1 16.000g
mol-1) 18.00 g mol-1
Step 2 Employ Avogadros number
Mass of one molecule Molar mass
Avogadros no.
18.00g mol-1 6.0221023mol-1
2.99210-23g
Note Always check the units you have in your
answer to ensure you are correct
21
Example
  • Calculate the mass of one molecule of ammonium
    carbonate (NH4)2CO3

Step 1 Calculate the molar mass
2 14.01gmol-1 8 1.008 gmol-1 1 12.01gmol-1 3
16.00 gmol-1
28.02 gmol-1 8.064 gmol-1 12.01 gmol-1
48.00 gmol-1
2 Nitrogen atoms 8 Hydrogen atoms 1 Carbon atom 3
Oxygen atoms
Total 96.09 gmol-1
Step 2 Employ Avogadros Number, NA
Mass of one molecule
96.09 gmol-1 . 6.0221023mol-1
1.59 10-22g
Questions
Calculate the mass of one molecule of
  1. Ethanoic acid (CH3COOH)
  2. Methane (CH4)
  3. Potassium dichromate (K2Cr2O7)

9.96 10-23 g
2.66 10-23 g
4.89 10-22 g
22
Converting between mass and moles
  • In the lab, we measure the mass of our reactants
    in grams using a
  • balance. However, when these react they do so in
    a ratio of moles.
  • Therefore, we need to convert between the mass we
    measure and the
  • number of moles we require.

The expression relating mass and number of moles
is
Mass of sample (g) no. of moles (mol) molar
mass (gmol-1)
Example
Calculate the mass in grams in 0.75mol of sodium
hydroxide, NaOH
Step 1 Find the molar mass of the compound
Na 22.99 gmol-1
Mr 40.00 gmol-1
O 16.00 gmol-1
H 1.008 gmol-1
Step 2 Substitute into the above expression
Mass of sample
0.75mol 40.00 gmol-1
30g
23
Questions
  • Calculate the mass in grams present in

(a) 0.57mol of potassium permanganate (KMnO4)
Answer Molar mass KMnO4 158.03 gmol-1
Mass in grams 0.57mol 158.03 gmol-1
90.07 g
(b) 1.16mol of oxalic acid (H2C2O4)
Answer Molar mass H2C2O4 90.04 gmol-1
Mass in grams 1.16mol 90.04 gmol-1
104.44 g
(c) 2.36mol of calcium hydroxide (Ca(OH)2)
Answer Molar mass Ca(OH)2 74.1 gmol-1
Mass in grams 2.36mol 74.1 gmol-1 174.87
g
24
Converting between moles and mass
  • Number of moles mass of sample (g)
  • molar mass (gmol-1)

Example
Convert 25.0g of KMnO4 to moles
Step 1 Calculate the molar mass
K Mn O
1 39.10 gmol-1 1 54.93 gmol-1 4 16.00 gmol-1
39.10 gmol-1 54.93 gmol-1 64.00 gmol-1
Mr 158.03 gmol-1
Step 2 Substitute into above expression
25.0g . 158.03gmol-1
No. of moles
0.158 mol
25
Questions
  • Calculate the number of moles in

(a) 1.00g of water (H2O)
Answer Molar mass water 18.02 gmol-1
1.00g H2O 0.055mol
(b) 3.0g of carbon dioxide (CO2)
Answer Molar mass carbon dioxide 44 gmol-1
3.0g CO2 0.068mol
(c) 500g of sucrose (C12H22O11)
Answer Molar mass sucrose 342.30 gmol-1
500g C12H22O11 1.46mol
(d) 2.00g of silver chloride (AgCl)
Answer Molar mass silver chloride 143.38 gmol-1
2.00g AgCl 0.014mol
26
Important formulae so far.
Defining the mole
No. of carbon-12 atoms atomic mass (g)
mass of one atom (g)
Calculating the number of atoms or molecules,
given the number of moles
No. of atoms No. of moles Avogadros number
(NA)
No. of molecules No. of moles Avogadros
number (NA)
Calculating the mass of an individual molecule
Mass of one molecule Molar mass
Avogadros no.
Most important equation
Mass of sample (g) no. of moles (mol) molar
mass (gmol-1)
  • Number of moles mass of sample (g)
  • molar mass (gmol-1)

27
Calculating mass percentage from a chemical
formula
  • Many of the elements in the periodic table of the
    elements occur in
  • combination with other elements to form compounds.

A chemical formula of a compound tells you the
composition of that compound in terms of the
number of atoms of each element present.
The mass percentage composition allows you to
determine the fraction of the total mass each
element contributes to the compound.
Example
Ammonium nitrate (NH4NO3) is an important
compound in the fertiliser industry. What is the
mass composition of ammonium nitrate?
Step 1 Calculate the molar mass of ammonium
nitrate
Two N atoms 28.016 gmol-1
Four H atoms 4.032 gmol-1
Three O atoms 48.00 gmol-1
Molar mass NH4NO3 80.05 gmol-1
28
Step 2 Determine the mass composition for each
element
  • Nitrogen 28.016g N in one mol of ammonium nitrate

Mass fraction of N 28.016g 80.05g
Mass composition of N 28.016g 100
80.05g 34.99 35
Hydrogen 4.032g H in one mol of ammonium nitrate
Mass fraction of N 4.032g 80.05g
Mass composition of H 4.032g
100 80.05g 5.04 5
Oxygen 48.00g O in one mol of ammonium nitrate
As above, the mass composition of O is found to
be 60
29
Therefore, the mass composition of ammonium
nitrate (NH4NO3) is
  • Nitrogen 35
  • Hydrogen 5
  • Oxygen 60

To check your answer, make sure it adds up to 100
Question
What is the mass composition of C12H22O11?
Answer Carbon 42.1 Hydrogen 6.5
Oxygen 51.4
30
Determining empirical formula from mass
  • The empirical formula of a compound tells you the
    relative number of
  • atoms of each element present in that compound.
    It gives you the
  • simplest ratio of the elements in the compound.

For example, the empirical formula of glucose
(C6H12O6) is CH2O, giving the CHO ratio of 121
If you know the mass composition and the molar
mass of elements present in a compound, you can
work out the empirical formula
Example
What is the empirical formula of a compound which
has a mass composition of 50.05 S and 49.95 O?
Step 1 Find the atomic masses of the elements
present
Sulfur (S) 32.066 gmol-1
Oxygen (O) 16.000 gmol-1
31
Step 2 Determine the number of moles of each
element present
  • Since we are dealing with percentages, we can
    express the mass as
  • grams if we assume we have 100g of the compound.

Therefore, 100g of our compound contains 50.05g
of sulfur and 49.95g of oxygen.
Convert number of grams to number of moles
Number of mol Sulfur mass of sulfur in sample
(g) atomic mass of sulfur (gmol-1)
50.05g . 32.066 gmol-1
1.56 mol
Similarly, the no. of mol of Oxygen is found to
be 3.12mol
Step 3 Determining the ratios of elements
Sulfur 1.56mol Oxygen 3.12mol
Ratio 1.56 3.12
Ratio must be in whole numbers. Here we must
divide across by 1.56
Therefore, we have a ratio of 12 giving an
empirical formula of SO2
32
Question
  • Determine the empirical formula of a compound
    that contains 27.3
  • mass Carbon and 72.7 mass Oxygen.

Answer No. of mol Carbon 2.27mol No. of mol
Oxygen 4.54mol Ratio 12 Empirical formula CO2
Monosodium glutamate (MSG) has the following mass
percentage composition 35.51 C, 4.77 H,
37.85 O, 8.29 N, and 13.60 Na. What is its
molecular formula if its molar mass is 169
gmol-1? 
Answer C5H8O4NNa
33
Important calculations
  • Calculating mass percentage from a chemical
    formula
  • Step 1 Calculate the molar mass
  • Step 2 Determine the mass composition for each
    element
  • Determining empirical formula from mass
  • Step 1 Find the atomic masses of the elements
    present
  • Step 2 Determine the number of moles of each
    element present
  • Step 3 Determining the ratios of elements

34
Molarity
  • Some chemical reactions involve aqueous solutions
    of reactants

The concentration of a solution is the amount of
solute present in a given quantity of solvent or
solution
This concentration may be expressed in terms of
molarity (M) or molar concentration

M Molarity no. of moles volume in Litres
Molarity is the number of moles of solute in 1
Litre (L) of solution
35
What is molarity of an 85.0mL ethanol (C2H5OH)
solution containing 1.77g of ethanol?
Example
  • Step 1 Determine the number of moles of ethanol

Molar mass of ethanol, C2H5OH
2 carbon atoms
2 12.01 gmol-1
24.02 gmol-1
1 oxygen atom
1 16.00 gmol-1
16.00 gmol-1
6 hydrogen atoms
6 1.008 gmol-1
6.048 gmol-1
46.07 gmol-1
No. of moles mass in g molar mass
No. of moles ethanol 1.77g . 46.07
gmol-1
0.038 mol
36
Step 2 Convert to molarity
  • Have 85.0mL ethanol

1 L 1000mL
? Have 0.085 L of ethanol
Molarity no. of moles volume in L
0.038 mol 0.085 L
0.45 molL-1
0.45 M
Questions
Calculate the molarities of each of the following
solutions
(a) 2.357g of sodium chloride (NaCl) in 75mL
solution
Answer 0.5378 M
(b) 1.567mol of silver nitrate (AgNO3) in 250mL
solution
Answer 6.268 M
(c) 10.4g of calcium chloride (CaCl2) in 2.20
102 mL of solution
Answer 0.426 M
37
Example
  • An antacid tablet is not pure CaCO3 it contains
    starch, flavouring, etc.
  • If it takes 41.3mL of 0.206 M HCl to react with
    all the CaCO3 in one
  • tablet, how many grams of CaCO3 are in the
    tablet. You are given the
  • following balanced equation

2HCl(aq) CaCO3(s) ? CaCl2(aq) H2O(l) CO2(g)
Step 1 Determine the no. of moles of HCl that
react
Have 0.206 M HCl solution ? have 0.206 mol in one
litre
Have 41.3 mL of HCl solution ? have 0.0413 L of
HCl solution
Molarity no. of moles volume in L
? no. of moles Molarity volume in L
0.206 molL-1 0.0413L
0.0085 mol
8.5 10-3 mol HCl
38
Step 2 Determine no. of moles of CaCO3 used in
the reaction
  • 2HCl(aq) CaCO3(s) ? CaCl2(aq) H2O(l) CO2(g)

From the balanced equation, we can see that 2
moles of HCl are required to react with one mole
of CaCO3
Therefore, if 8.5 10-3 mol of HCl are present
in the reaction, we must have 4.25 10-3 mol of
CaCO3 present.
Molar mass of CaCO3
1 calcium atom
1 40.08 gmol-1
40.08 gmol-1
1 carbon atom
1 12.01 gmol-1
12.01 gmol-1
3 oxygen atoms
3 16.00 gmol-1
48.00 gmol-1
100 gmol-1
No. of mols mass in g molar mass
? Mass in g no. of mols molar mass
(4.25 10-3 mol) (100 gmol-1)
0.425 g CaCO3 present in tablet
39
Questions
  • (a) How many moles of NaCl are present in 25.00mL
    of 1.85M NaCl(aq)?

Answer 4.62 10-2 mol NaCl
(b) What volume of a 1.25 10-3 M solution of
C6H12O6(aq) contains 1.44 10-6 mol of
glucose?
Answer 1.15 mL
(c) If stomach acid, given as 0.1 M HCl, reacts
completely with an antacid tablet
containing 500mg of CaCO3, what volume of acid in
millilitres will be consumed? The balanced
equation is CaCO3(s) 2HCl(aq) ? CaCl2(aq)
CO2(g) H2O(l)
Answer 100mL acid
40
Important formulae
Calculating the number of moles
No. of moles mass in g molar mass
Calculating the molarity or concentration
Molarity no. of moles volume in L
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