# Atomic structure for SQA National 5 Chemistry

1. Subatomic particles
2. Relative atomic mass

Each element has a characteristic number of protons, called the atomic number, which can be found in the periodic table. In atoms the number of electrons is the same as the number of protons. Since they have opposite charges, this ensures that every atom is neutral.

In atoms:
number of protons = atomic number
number of electrons = atomic number
number of neutrons = mass number − atomic number

For charged ions, the number of negatively charged electrons changes. Negative ions will have a higher number of electrons since they have a negative charge that comes from the negative electrons. Thus, positive ions will have the usual number of electrons that belongs to the atoms minus their charge which indicates the number of electrons lost.

Relative atomic mass is a weighted average mass of all the isotopes of an element. Weighted means that it takes into account the natural abundance of each isotope.

To find the relative atomic mass from given atomic masses of isotopes and their abundances, multiply corresponding atomic masses and their abundances in decimals and add the multiplication products together. If you use abundances as percentages, divide your final result by 100%.

# 1

Identify the element and ionic charge from the diagram representing an ion.

atomic number = number of protons = 4
Beryllium has an atomic number of 4 and as an atom has 4 electrons.
2 electrons are missing, thus, this is an ion with a charge of 2+.

beryllium, 2+

# 2

What is the relative atomic mass of platinum correct to 3 decimal places? The data in the table has been adapted for the question.

abundance of ¹⁹²Pt (a₁) = 0.782% = 0.00782
abundance of ¹⁹⁴Pt (a₂) = 32.864% = 0.32864
abundance of ¹⁹⁵Pt (a₃) = 33.787% = 0.33787
abundance of ¹⁹⁶Pt (a₄) = 25.211% = 0.25211
abundance of ¹⁹⁸Pt (a₅) = 7.356% = 0.07356
RAM = m₁a₁ + m₂a₂ + m₃a₃ + m₄a₄ + m₅a₅
RAM = 192 × 0.00782 + 194 × 0.32864 + 195 × 0.33787 + 196 × 0.25211 + 198 × 0.07356 = 195.121

195.121

# 3

Boron has two main isotopes, ¹⁰B (20%) and ¹¹B (80%). What is the relative atomic mass of boron? The data provided has been adjusted for the question.

abundance of ¹⁰B (a₁) = 20% = 0.2
abundance of ¹¹B (a₂) = 80% = 0.8
RAM = m₁a₁ + m₂a₂ = 10 × 0.2 + 11 × 0.8 = 10.8

10.8

# 4

Bromine has two stable isotopes, ⁷⁹Br and ⁸¹Br with their abundances respectively being 51% and 49%. What is the relative atomic mass of bromine?

abundance of ⁷⁹Br (a₁) = 51% = 0.51
abundance of ⁸¹Br (a₂) = 49% = 0.49
RAM = m₁a₁ + m₂a₂ = 79 × 0.51 + 81 × 0.49 = 79.98

79.98

# 5

How many protons, neutrons and electrons does a F⁻ ion have?

number of protons = atomic number = 9
number of neutrons = atomic mass − atomic number = 19 − 9 = 10
number of electrons in a F atom = atomic number = 9
− charge in the ion means that one electron has been gained.
number of electrons in F⁻ = 9 + 1 = 10

9 protons, 10 neutrons, 10 electrons

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