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An in-depth exploration of atomic mass units, atomic mass, natural abundance, and the concept of averaging atomic masses considering relative abundances. It also covers the calculation of molecular and formula masses, as well as the use of moles to avoid dealing with huge numbers of atoms or molecules.
What you will learn
Typology: Exercises
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1 amu = 1.66054 x 10 -24^ 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
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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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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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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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5.2: How to Avoid Huge Numbers
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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5.2: A mole is 6.02 × 10^23 of anything
The NUMBER is constant, NOT the MASS
1 mole = 6.02 × 10^23
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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
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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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
23 things