Topic B – Atoms, Atomic Masses and Moles

Chapter 1

Chemistry 10 TBD01

TA Ch1 Mr. Martin Brakke

Topic B: Atoms, Atomic Masses

1.1 Atoms and Molecules 1.2 Comparing the masses of atoms (Mass

spectrometry) 1.3 Relative atomic masses – the relative atomic

mass scale 1.4.1 - Avogadro's Number

TA Ch1 Mr. Martin Brakke

1.1 – The Atom

Atoms have two main parts, the nucleus and the electron shells. The atom is composed of three sub-atomic particles

Protons (nucleus) Electrons (electron shells) Neutrons (nucleus)

The model of the atom we generally draw is known as the Bohr model

TA Ch1 Mr. Martin Brakke

1.1 – Atomic Theory Why do we believe what we believe about the atom?

John Dalton (1766-1844) Proportions – Billiard Model

J.J. Thomson (1856-1940) Cathode Ray Tubes (CRT) Plum Pudding Model

Ernest Rutherford (1871-1937) Gold Foil Experiment – Empty Space Model

Niels Bohr (1885-1962) Spectra of atoms – Planetary Model

Erwin Schrodinger (1887-1961) & Werner Heisenberg (1901-1976) Waves & Uncertainty – Quantum Model

TA Ch1 Mr. Martin Brakke

1.1 – Relative Mass and Size

Neutrons help stabilize the nucleus, separating

protons. (if too many/few n, radioactive) Electrons control the chemical properties of the

elements Protons identify an atom as a specific element

Sub-atomic Particle

Symbol Relative Mass Relative Charge

Proton p 1 +1

Neutron n 1 0

Electron e 5x10-4 -1

TA Ch1 Mr. Martin Brakke

1.1 – Atoms vs Molecules

The term atom was first used by John Dalton to mean the smallest particle of an element He also stated the law of definite proportions

which states that atoms can combine in whole number ratios to form molecules or compounds

TA Ch1 Mr. Martin Brakke

1.2 – Operation of Mass Spec

What’s it for? A mass spectrometer allows chemists to determine: Relative atomic masses of atoms Relative molecular masses of compounds Structure of molecules

How? By splitting up atoms, isotopes or molecules by their mass to charge ratio

TA Ch1 Mr. Martin Brakke

1.2 – Steps of a Mass Spec

Neon-20

Neon-21 Neon-22

Sample

1. Vaporization

2. Ionization

3. Acceleration 4. Deflection

5. Detection

TA Ch1 Mr. Martin Brakke

1.3 – What can we use MS data for?

Elements in the Periodic table have: Chemical Symbol Atomic Number (protons) Relative Atomic Mass (Ar is NOT mass number)

The mass number is the number of protons and

neutrons in a single atom The relative atomic mass is the weighted average

of all the mass numbers from all the isotopes in abundance

10

Ne 20.18

TA Ch1 Mr. Martin Brakke

1.3 – Relative Atomic Mass

The weighted average (Ar) is only given a value compared to the mass of the carbon-12 atom 1/12th of carbon 12 would be one unit! So every mass is based on 1 amu

Use chlorine as an example:

Chlorine-35 = 75% Chlorine-37 = 25% Before we calculate, what number (35,37) should

the weighted average be closer to?

TA Ch1 Mr. Martin Brakke

1.3 – Simple Ar Calculation

Ar Cl = (0.75 x 35amu) + (0.25 x 37amu) Ar

Cl = 35.5 g/mol What is an amu?

Atomic Mass Units (1/12th of the carbon mass) The units are grams per mole (g/mol)

Now try the ones on your notes…..

TA Ch1 Mr. Martin Brakke

1.4 - Classification of Matter

Matter – anything that has mass and takes up space. Anything! Chemistry – the study of matter and the

changes it undergoes. All matter can exist in three (3) states:

Solid Liquid Gas Plasma

TA Ch1 Mr. Martin Brakke

1.4 - Matter Flow Chart MATTER

Can it be physically separated?

Homogeneous Mixture

(solution)

Heterogeneous Mixture

Compound

Element

MIXTURE PURE SUBSTANCE

yes no

Can it be chemically decomposed?

no yes Is the composition uniform?

no yes

Colloids Suspensions

TA Ch1 Mr. Martin Brakke

1.4 - The Mole Chemical reactions involve atoms and molecules. The ratios with which elements combine depend on

the number of atoms not on their mass. The masses of atoms or molecules depend on the

substance. Individual atoms and molecules are extremely

small. Hence a larger unit is appropriate for measuring quantities of matter. A mole is equal to exactly the number of atoms

in exactly 12.0000 grams of carbon 12. This number is known as Avogadro’s number. 1 mole is equal to 6.02 x 1023 particles.

TA Ch1 Mr. Martin Brakke

1.4 - Definitions of the Mole 1 mole of a substance has a mass equal to the

formula mass in grams. Examples

1 mole H2O is the number of molecules in 18.01 g H2O 1 mole H2 is the number of molecules in 2.02 g H2. 1 mole of atoms has a mass equal to the atomic weight

in grams. 1 mole of particles = 6.02 x 1023 particles for any

substance! The Molar mass is the mass of one mole of a

substance Avogadro's number is the number of particles

(molecules) in one mole for any substance 1

TA Ch1 Mr. Martin Brakke

1.4 – Calculating Molar Mass

To find the molar mass you can simply add together the relative atomic mass of each of the atoms present in a molecule You may always round off to two unit passed the

decimal for the molar mass Oxygen = 16.00 g mol-1 Hydrogen = 1.01 g mol-1 Carbon = 12.01 g mol-1

etc

TA Ch1 Mr. Martin Brakke

1.4 – Calculating Molar Mass

Calculate the Molar Mass for glucose (C6H12O6) C – 12.01 x 6 H – 1.01 x 12 O – 16 x 6 (12.01x6) + (1.01x12) + (16x6) = 180.18 g mol-1

This value means that for every mole of glucose, it has a mass of 180.18 grams. OR that there are 180.18 grams of glucose in one mole.

TA Ch1 Mr. Martin Brakke

1.4 - Conversions

This now means that we can make simple conversions. We need to keep in mind:

UNITS!!! We are only multiplying by a value of 1 when

converting UNITS!!! We want certain units to cancel out so we can

flip the numerator/denominator as fits the problem

TA Ch1 Mr. Martin Brakke

1.4 - Simple Conversions:

80. g CuSO4

159.5 g CuSO4

1 mol CuSO4 = 0.50 mol CuSO4

0.50 mol CuSO4 159.5 g CuSO4

1 mol CuSO4 = 80. g CuSO4

- Mole / Mass Conversions -

Use the Molar Mass of a substance to convert from Moles to Mass and Mass to Moles

Mass to Moles

Moles to Mass

TA Ch1 Mr. Martin Brakke

2 mol CuSO4 6.022x1023 (mc) CuSO4

1 mol CuSO4 = 1.2x1024 (mc) CuSO4

= 2 mol CuSO4 6.022x1023 (mc) CuSO4

1 mol CuSO4 1.2x1024 (mc) CuSO4

1.4 - Simple Conversions: - Mole / Molecule Conversions -

Use Avogadro’s Number : 6.022 x 1023 molecules (mc) in one mole of the substance

Moles to (mc)

(mc) to Moles

TA Ch1 Mr. Martin Brakke

2 mol O2 22.4 L O2

1 mol O2 = 44.8 L O2

44.8 L O2

22.4 L O2

1 mol O2 = 2 mol O2

1.4 - Simple Conversions: - Mole / Volume Conversions -

At STP (standard temperature and pressure) one mole of any gas takes up 22.4 L of space

Moles to Volume

Volume to Moles

TA Ch1 Mr. Martin Brakke

Moles to Mass

(use Molar Mass)

Moles to Volume

(Molar Volume of a gas 22.4

Moles to Molecules

(use Avogadro’s Number)

TA Ch1 Mr. Martin Brakke

1.5: Empirical/Molecular Formulas

The empirical formula of a substance is the lowest whole number ratio of elements in the compound (simplified, like CH2O) The molecular formula of a substance is the

actual numbers of elements (like C6H12O6)

TA Ch1 Mr. Martin Brakke 1.5 – Using MS data to find the empirical formula

Determine the empirical formula of a compound with 79.9% Carbon and 20.1% Hydrogen 1. Assume the % to be grams (out of 100g sample) 2. Convert grams of each element to moles using

the molar mass (g/mol) (we will cover soon!) 3. Write the equation with mole ratios 4. Divide by the smallest # of moles 5. If needed, multiply until all have whole numbers

TA Ch1 Mr. Martin Brakke

1.5: % to Empirical Example

Determine the empirical formula of a compound with 79.9% Carbon and 20.1% Hydrogen

79.9% C = 79.9g C x 1𝑚𝑚𝑚 𝐶12.01 𝑔 𝐶

= 6.65 mol C

20.1% H = 20.1g H x 1 𝑚𝑚𝑚 𝐻1.01 𝑔 𝐻

= 19.9 mol H

C6.65H19.9 / 6.65 = CH2.99

Round off 2.99 to 3, so we have CH3 as our empirical formula

TA Ch1 Mr. Martin Brakke 1.5: Find % Composition from Empirical or Molecular Formulas

To backtrack, if you know the molecular formula to be C2H6, find the percentage composition of each compound. (This is theoretical, whereas Mass Spec data would be experimental)

% = #𝑚𝑚𝑚𝑚𝑚 𝑖𝑖 𝑓𝑚𝑓𝑚𝑓𝑚𝑓 𝑥 𝑀𝑚𝑚𝑓𝑓 𝑀𝑓𝑚𝑚 𝑚𝑓 𝑚𝑚𝑚𝑚𝑚𝑖𝑒𝑀𝑚𝑚𝑓𝑓 𝑀𝑓𝑚𝑚 𝑚𝑓 𝐸𝑚𝐸𝑚𝑓𝑖𝐸𝑓𝑚 𝑚𝑓 𝑀𝑚𝑚𝑚𝐸𝑓𝑚𝑓𝑓 𝐹𝑚𝑓𝑚𝑓𝑚𝑓

𝑥𝑥𝑥𝑥

%C = 2 𝑥 12.0130.08

𝑥 𝑥𝑥𝑥 = 79.9% C

%H = 6 𝑥 1.0130.08

𝑥 𝑥𝑥𝑥 = 20.1% H