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Topic 5: Stoichiometry - Chemical Arithmetic
Masses of some atoms:
Introducing…….the Atomic Mass Unit (amu)
1 amu = 1.66054 x 10 -24^ g
11 H =1.6736× 10 -24g 168 O =2.6788× 10 -23g
23892 U =3.9851× 10 -22 g
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5.1: Atomic Mass Unit
Atomic Mass is defined relative to Carbon -12 isotope
12 amu is the mass of the isotope of carbon
Carbon -12 atom = 12.000 amu
Hydrogen -1 atom = 1.008 amu
Oxygen -16 atom = 15.995 amu
Chlorine -35 atom = 34.969 amu
126 C
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5.1: Atomic Mass - Natural Abundance
We deal with the naturally occurring mix of isotopes , rather than pure isotopes
Carbon has three natural isotopes Isotope Mass (amu) Abundance (%)
(^12) C 12.000 98. (^13) C 13.00335 1. (^14) C 14.00317 1 x 10 -
Any shovelful of Carbon from living material will have a Naturally Occurring Abundance of 98.892%^12 C, 1.108%^13 C and 0.0001%^14 C
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5.1: Atomic Mass - Relative Abundance
How do we take into account the naturally occurring Abundances? Take the average mass of the various isotopes weighted according to their Relative Abundances
Isotope Mass (amu) Abundance (%)
(^12) C 12.000 98. (^13) C 13.00335 1. (^14) C 14.00317 1 x 10 -
Relative Abundance
1 x 10 -
N.B. The % Abundance adds up to 100 The Relative Abundance adds up to 1
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5.1: Average Atomic Mass
Isotope Mass (amu) Abundance (%)
(^12) C 12.000 98. (^13) C 13.00335 1. (^14) C 14.00317 1 x 10 -
Relative Abundance
1 x 10 -
The Average Atomic Mass is given by:
(0.98892 x 12.000 amu ) + (0.01108 x 13.00335 amu ) + (1 x 10 -6^ x 14.00317 amu ) = 12.011 amu
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5.1: Atomic and Molecular Mass
You can calculate the mass of any compound
from the sum of the atomic masses from the
periodic table.
Example: Molecular Mass of H 2 SO 4
H 2 SO 4
2 Hydrogen atoms
1 Sulfur atom
4 Oxygen atoms
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5.2: How to Avoid Huge Numbers
Recall the introduction of Atomic Mass Units
This avoids working with ridiculously small
masses
How to avoid the problem of counting huge
numbers of molecules / atoms?
1 amu = 1.66054 x 10 -24^ g
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How does a chemist say how many ATOMS or MOLECULES she reacted?
She talks of moles of Atoms or Molecules reacted
1 mole = 6.022142 x 10^23
5.2: How to Avoid Huge Numbers - Use moles!
Avogadro’s Number (N A )
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1 mole of = 6.02 x 10^23
5.2: A mole is 6.02 × 10^23 of anything
ANTS ANTS
Br 2 molecules Br 2 molecules
KMnO 4 KMnO^4
BEERS BEERS
PEANUTS PEANUTS
CH 3 COOH CH 3 COOH
The NUMBER is constant, NOT the MASS
1 mole = 6.02 × 10^23
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5.2: The mole: Where does it come from?
1 mole is the number of Carbon atoms found in EXACTLY 12.00 g of^12 C
Definition:
12 amu × 6.02 × 10^23 = 12 g
amu = g.mol -
1 mole
12 g 12 amu =
g mole
12 g
6.02 x 10^23
12 amu =
14
g mol
5.2: Molar mass
H 2 SO 4
2 Hydrogen atoms
1 Sulfur atom
4 Oxygen atoms
(2 x 1.0079 ) + (1 x 32.065 ) + (4 x 15.999 )
How to calculate the mass of 1 mole of a compound? = the molecular mass in grams
g. mol -
g mol
g mol
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Examples:
5.2: Molar mass
1 molecule of KCl has molecular mass of 74.55 amu.,
1 mol of KCl has a mass of 74.55 g.,
and contains 6.02 x 10^23 molecules of KCl.
1 mole of H 2 O has a mass of? g
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Mass of nicotine = 1.2 mg Molar mass of nicotine = 162.26 g.mol - No. of molecules 1 mole = 6.02 × 10^23 molecules.mol -
How many MOLECULES of nicotine (C 10 H 14 N 2 ) are there in an average cigarette (1.2 mg)?
5.2: Calculations of molecular amounts
no. of moles of nicotine = 1.2 × 10 -3^ g / 162.26 g.mol -
= 7.4 × 10 -6^ moles of nicotine
Step 1: Convert mass to moles using molar mass
= 1.2 × 10 -3^ g
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Mass of nicotine = 1.2 mg = 1.2 × 10 -3^ g Molar mass of nicotine = 162.26 g.mol - No. of molecules 1 mole = 6.02 × 10^23 molecules.mol -
How many MOLECULES of nicotine (C 10 H 14 N 2 ) are there in an average cigarette (1.2 mg)?
5.2: Calculations of molecular amounts
no. of molecules of nicotine = 7.4 × 10 -6^ mol × 6.02 × 10^23 molecules.mol -
= 4.4 × 10^18 molecules of nicotine
Step 2: Convert moles to molecules using Avogadro’s number
= 7.4 × 10 -4^ mol
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1 mole = 6.022142 x 10^23
AVOGADRO’S NUMBER (N a )
1 mole = 6.022142 x 10
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