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Understanding Atomic and Molecular Masses through Stoichiometry, Exercises of Stoichiometry

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

  • What is the difference between molecular mass and formula mass, and when should each term be used?
  • What is the definition of an Atomic Mass Unit (amu) and how is it related to the mass of an atom?
  • How do we calculate the average atomic mass of an element considering its natural abundance?

Typology: Exercises

2021/2022

Uploaded on 09/12/2022

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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
g102.6788O -2316
8 ×=
g101.6736H -241
1×=
g103.9851U -22238
92 ×=
2
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
C
12
6
3
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 (%)
12C12.000 98.892
13C13.00335 1.108
14C14.00317 1 x 10-4
Any shovelful of Carbon from living material will have a
Naturally Occurring Abundance of
98.892% 12C, 1.108% 13C and 0.0001% 14C
pf3
pf4
pf5

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1

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

2

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

3

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

4

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

5

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

10

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

11

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 )

12

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

13

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

15

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

19

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

20

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