General Chemistry: Atoms, Molecules, Stoichiometry

• The notes are for General Chemistry, revision list for Phase Test 1 (Formal)
• Revision 0, February 2025
• You MUST be able to perform the following in order to maximise the marks you can attain in the Phase Test 1 (Formal):

Topic 1: Atoms, Molecules and Stoichiometry

• Relative Isotopic Mass is the mass of an atom of the isotope compared to 1/12th the mass of a carbon-12 isotope.
• The isotopic mass of chlorine-35 isotope is 35
• The isotopic mass of chlorine-37 is 37
• Relative Atomic Mass (Ar) is the smaller value that is seen by the elemental symbol in the Periodic Table.
• Relative Atomic Mass (Ar) is the average mass of an atom compared to 1/12th the mass of a carbon-12 isotope.
• Relative Atomic Mass (Ar) has units of g/mol.
• Relative Atomic Mass (Ar) can only be calculated using: Ar = ∑(% abundance)x(isotopic mass) / 100, OR Ar = ∑(relative abundance)x(isotopic mass)
• Chlorine has two isotopes, given the % abundances:
• Isotope 35Cl has an Isotopic Mass of 35 and % abundance of 75%.
• Isotope 37Cl has an Isotopic Mass of 37 and % abundance of 25%.
• Ar of Cl = (75x35)+(25x37) / 100 = 35.5 g/mol
• Relative Molecular Mass (Mr) is the larger value that is seen by the elemental symbol in the Periodic Table.
• Relative Molecular Mass (Mr) is the average mass of one molecule compared to 1/12th the mass of a carbon-12 isotope.
• Relative Molecular Mass (Mr) can be calculated by adding all of the Ar of the atoms in the molecule.
• Relative Molecular Mass (Mr) has units of g/mol
• Example: Cl2 molecule, Mr of Cl2 = 2 x 35.5 = 71.0 g/mol
• Carbon-12 isotope does not undergo radioactive decay, this means carbon-12 is able to retain its number of protons and neutrons (no losses).
• Carbon-12 has been used as a "standard" when measuring relative masses, because of its stability
• The mole quantifies the amount of any substance in Chemistry.
• 1 mole of anything is equal to Avogadro's constant and 6.02 x 10^23 of anything (particles, atoms, ions, molecules, etc)

Mole Equations Formulas

• No of moles, n = mass (in g) / molar mass (in g/mol)
• mass (in g) / molar mass (in g/mol) = *Ar if atom/ion and *Mr
• At room temperature and pressure (25°C, 1atm) for gases ONLY: Volume of gas (in dm³) = moles of gas x 24 dm³/mol, v = n x 24
• At standard temperature and pressure (0°C, 1atm) for gases ONLY: Volume of gas (in dm³) = moles of gas x 22.4 dm³/mol, v = n x 22.4
• When dealing with liquids: Molarity = no. of moles (in mol) / volume (in dm³), M = n/V
• Concentration = mass (in g) / volume (in dm³), c = m/V
• Concentration (in g/dm³) = Molarity (in mol/dm³) x Mr (g/mol), c = M x Mr

Empirical and Molecular Formula Definitions

• Empirical formula shows the simplest ratio of the atoms of each element in one molecule.
• Steps to calculate empirical formula (assuming one compound has a 100g): Convert grams to moles -> Divide each by the smallest number of moles -> Convert to whole numbers
• For Elements Phosphorous and oxygen, grams are converted to moles by dividing the grams (43.7g and 56.3g respectively), by the relative atomic mass (31 and 16 respectively) giving 1.42 and 3.52 respectively.
• 1.42 and 3.52 are divided by the smallest number of moles 1.42, giving 1 and 2.5 respectively.
• 1 and 2.5 are mulitplied by 2 to convert to whole numbers of 2 and 5
• Giving the empirical formula P2O5
• Molecular formula shows the actual number of the atoms of each element in one molecule.
• Molecular Formula = n(empirical formula), n = scaling factor
• Usually, the M, of the actual molecule is given in the question. Mr of the empirical formula needs to be calculated using the Periodic Table.
• Find the molecular formula, given the empirical formula is P2O5 and the Mr of the actual molecule is 284.0 g/mol.
• Mr of empirical formula = (2 x 31.0) + (5x16.0) = 142.0 g/mol
• Scaling factor, n = (Mr of actual molecule) / (Mr of empirical formula)

Balancing Chemical Equations

• Write the unbalanced equation.
• Check the formula for each molecule that appear in the equation.
• Count the number of atoms of each element on both the reactants side and the products side.
• Balance the elements by starting with the element that appears least frequently on either side.
• Add stoichiometric coefficients (numbers in front of the formulae) to balance each element, one at a time.
• After balancing each element, re-count the atoms to ensure the equation is balanced.
• Verify that the equation is finally balanced.
• Only stoichiometric coefficients can be changed; do NOT change the numbers inside the formula itself.
• H2O stays as H2O, don't change the subscripted number.
• But you can change the number in front of H2O, e.g. 3H2O, 4H2O, etc.

Topic 2: Atomic Structure

• Structure of an atom:
• Proton has a relative mass of 1 and Relative Charges of +1
• Neutron has a relative mass of 1 and Relative Charges of 0
• Electron has a relative mass of ~0 and Relative Charges of -1
• Mass number = no. of protons + no. of neutrons
• Proton number = atomic number (Z)
• Difference between atom, isotope, cation and anion.
• Inside a shell, there is/are subshell(s), and Inside a subshell, there is/are orbital(s)
• Shell 1 has Subshell 1s and Orbitals 1s
• Shell 2 has Subshell 2s and Orbitals 2s. Shell 2 has Subshell 2p and Orbitals 2px, 2py, 2pz
• Shell 3 has Subshell 3s and Orbitals 3s Shell 3 has Subshell 3p and Orbitals 3px, 3py, 3pz
• 3d Not required
• 4s is a Subshell, containing Orbitals of 4s
• 4p is a Subshell, containing Orbitals of 4px, 4py, 4pz
• 4d and 4f are Not Required
• Type of subshell s has 1 orbital (n) and a Max no. of e- in the subshell (2 x n) of 2 x 1 = 2 e-
• Type of subshell p has 3 orbitals (n) and a Max no. of e- in the subshell (2 x n) of 2 x 3 = 6 e-
• Type of subshell d has 5 orbitals (n) and a Max no. of e- in the subshell (2 x n) of 2 x 5 = 10 e-
• Type of subshell f has 7 orbitals (n) and a Max no. of e- in the subshell (2 x n) of 2 x 7 = 14 e-
• s-orbital is spherical, and p-orbital is dumbbell-shaped.

Electron Configuration

• Know THREE (03) forms of electron configuration of atoms (and ions)
• Orbital Diagram (or electron-in-box diagram)
• Spdf notation
• Noble Gas Core
• Example for oxygen (O) atom:
• Orbital Diagram has 1 box as 1 orbital, and the box is filled in the following order, 1s, 2s and 2p
• Spdf notation is 1s2 2s2 2p4
• Noble Gas Core is [He] 2s2 2p4
• Know how to write electron configurations of atoms: Fill up electrons from the lowest-energy orbital (Aufbau Principle)
• Each orbital can be filled with max. 2 electrons with opposite spins↑↓ (Pauli Exclusion Principle)
• For subshells with more than 1 orbital (like p-subshell or d-subshell), fill the orbitals singly first before pairing up electrons (Hund's Rule)
• Know the arrangement of subshells in order of increasing energy: 1s 2s 2p 3s 3p 4s 3d (yes, 4s comes first!)
• When writing down electron configuration of cations, remove from highest numbered SHELL!
• Zinc atom: 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10
• NOTE: Remove e- from 4s first as it is in the 4th SHELL!
• Zinc 2+ ion: 1s^2 2s^2 2p^6 3s^2 3p^6 3d^10
• When writing down electron configuration of anions: Follow steps as you do for atoms.

Ionisation Energy Definitions

• 1st Ionisation Energy is the energy required to remove one electron from each atom of 1 mole of gaseous atoms to form 1 mole of gaseous +1 cations. M (g) → M+ (g) + e-
• 2nd Ionisation Energy is the energy required to remove one electron from each ion of 1 mole of gaseous +1 cations to form 1 mole of gaseous +2 cations. M+ (g) → M2+ (g) + e-
• 1st ionisation energy: There is a general increase across each Period.
• For each Period, electron is added onto the same shell.
• All valence electrons experience similar shielding from inner shell.
• Number of protons increases across the Period: Increased nuclear charge
• Valence electrons are pulled closer to the positive nucleus causing Atomic radius to decrease.
• More energy is needed to remove the valence electron as they get much closer to the nucleus, therefore 1st IE increases across the Period.

Ionisation Energy Continued

• 1st IE of Group 3 element is LOWER than 1st IE of Group 2 element (e.g. B < Be & Al < Mg): Draw out the orbital diagram (or electron-in-box diagram) to visualise your answer
• For example, Be is 1s^2 2s^2, valence electron of Be is in a 2s subshell. B is 1s^2 2s^2 2p^1, valence electron of B is in a 2p subshell.
• Valence electron of B is much further from the nucleus than that of Be. Therefore, Less energy is required to remove the valence electron of B.
• 1st IE of Group 6 element is LOWER than 1st IE of Group 5 element (e.g. O < N & S < P): Draw out the orbital diagram (or electron-in-box diagram) to visualise your answer
• All of the orbitals in the 2p subshell of N are singly-filled. N is 1s^2 2s^2 2p^3
• Meanwhile, there is a pair of electrons in one of the orbitals in the 2p subshell in O. These electrons experience electron pair repulsion. Oxygen 1s^2 2s^2 2p^4 This makes it much easier to remove the valence electron of O.
• 1st Ionisation Energy decreased down a Group, the number of shells increase down each Group.
• Valence electron gets further from the nucleus making it easier to remove, the number of inner shells also increases down a Group.
• Valence electron experiences more shielding effect from the inner shells, resulting in Less energy is required to remove valence electron going down the Group.
• It is understandable successive ionisation energy by removing electrons until there is no electron left.
• Can be used to identify the Group in which the element belongs.
• For example, if there is a huge increase from the 3rd Ionisation Energy to the 4th Ionisation Energy
• It means that the fourth electron is removed from an "inner shell" which is much closer to the nucleus, hence, this element belongs in Group 3.

Topic 3: Chemical Bonding

• Understand THREE (03) types of strong chemical bonding
• Ionic Bonding is between metals and non-metals involving complete transfer of electron(s) from metal to non-metal.
• Metal loses electron(s) to form cations (+), and Non-metal gains electron(s) to form anions (-).
• Know how to draw the dot-and-cross diagram of an ionic compound e.g. NaCl, MgO, etc.
• Covalent bonding is between non-metal and non-metal involving sharing of electron(s) between two (02) non-metal atoms.
• 1 single bond (-) = 1 bond pair (X) = 2 electrons
• Know how to draw the Lewis structure of covalent molecules, such as for the Hydrogen Chloride (HCl) Lewis Structure. Draw sketch then mark lone pairs
• Metallic bonding exists only in metals (Groups 1 – 3, Sn and Pb).
• Metal atoms easily “loses” their valence electrons to form a sea of delocalized electrons, hence also forming metal cations.
• Metallic bonding: Electrostatic attraction between the “sea” of delocalized (free) electrons and the metal cations.
• Strength of metallic bond depends on the number of electrons contributed into the sea of delocalised electrons, the charge of the metal cations formed, and the size of the metal cations
• Molecules containing strong chemical bonds have high melting points as a large amount of energy is needed to break the bonds.

Molecules and Bond Angles

• For covalently-bonded molecules ONLY, know how to draw shapes of molecules and predict bond angles.
• Two (02) types of electron pairs (EPs) in covalently-bonded molecules are Bonding pair (BP) and Lone Pair (LP).
• Bonding pair (BP) : pair of electrons involved in covalent bond
• Lone pair (LP): pair of electrons NOT involved in covalent bond, EP = all BP + all LP
• Shapes of molecules are determined by the repulsion of electron pairs (EPs) around the central atom.
• Since EPs are negative, EPs would like to be as far apart from each other as possible.
• LP-LP repulsion is stronger than LP-BP repulsion which is stronger than BP-BP repulsion. (VSEPR Theory = Valence Shell Electron Pair Repulsion Theory)
• The following MUST be memorised:
• EP(s) around central atom refers to the number of electron pairs around the central atom.
• BP refers to the number of bonding pairs
• LP refers to the number of lone pairs
• Linear basic shape with EP(s)/BP of 2 has a bond angle of 180 degrees
• Trigonal Planar basic shape with EP(s)/BP of 3 has a bond angle of 120 degrees
• Tetrahedral basic shape with EP(s)/BP of 4 has a bond angle of 109.5 degrees
• Trigonal bipyramidal basic shape with EP(s)/BP of 5 has bond angles between vertical and horizontal with 90 degrees, and between horizontal 120 degrees.
• Octahedral shape with EP(s)/BP of 6 has a bond angle of 90 degrees
• Bond angle: angle between two (02) BPs.

Molecules With Lone Pairs Around The Central Atom

• Derived Shapes are based on Basic Shapes
• LP present in the molecules will repel the BP more strongly, hence BPs are much closer to each other, decreasing the bond angle.
• Therefore for molecule with 4 electron pairs around the central atom, Trigonal pyramidal and V-shaped / bent are based on the basic shape “Tetrahedral".
• Though the bond angles for these DERIVED shapes are less than 109.5° because the LP-BP repulsion is much stronger, hence pushing the BPs closer together, decreasing the bond angle.
• To determine the shapes without having to draw Lewis structure: No. of electron pairs (EP) = (Group number of central atom) + (1e¯ per BP formed) / 2 Determine the number of LP: EP = BP + LP
• Methane (CH4) is calculated as having 4 electron pairs by (4 + (4 x 1)) / 2 = 4. 4= 4 + LP, therefore LP = 0
• So CH4 has 4 BP, and 0 LP, making the shape of CH4 is "tetrahedral” with a bond angle of 109.5°
• Ammonia (NH3) is calculated as having 4 electron pairs by (5 + (3 x 1)) / 2 = 4. 4 = 3 + LP, therefore LP = 1, so NH3 has 3 BP, 1 LP.
• This makes the shape of NH3 is "trigonal pyramidal" with a Bond angle of 107°

How to Draw Lewis Structure for Covalently-Bonded Molecules

• For covalently-bonded molecules with more than two atoms e.g. PF3:
• Determine the central atom, the central atom only appears once in a molecule.
• Central atom for PF3 is P
• Make one single bond with the surrounding atoms.
• Complete the octet of the surrounding atoms (For H, duet rule instead of octet rule)
• Check the structure by counting the total number of valence e
• If structure is missing e, add e onto the central atom as “lone pairs (LP)
• Analyse Lewis structure e.g. for PF3, there are 3 BP and 1 LP around the P atom.