For some inexplicable reasons, a small percentage of Form 6 students seemed to struggle initially with the concept of matter (The larger percentage tend to understand this intuitively) This is despite being exposed to it in SPM-level physics and chemistry. When the Berry Berry Teacher questioned her previous batches of students why they find it hard to understand the concept of it, they said that since they couldn’t understand it earlier, they have a mental blockage towards it. (Luckily, most students will eventually achieve full understanding on this subject matter.)
Everything you can see in this physical world is consist of substance, generally term as matter. (It’s not an entirely accurate description but technically applicable and correct) Berry Berry Easy advises students to stick to the terminologies used in the syllabus as different fields tend to define matter differently. So to be on the safe side, try and absorb all the terminologies and concepts presented in the notes below.
STPM Form 6 – Terminology and Concepts: Matter
Matter – anything that occupies space and has mass.
Fundamental Particles of Atoms (Historical Point of View)
John Dalton (1808) – atomic theory
- Atoms – small indivisible particles.
- Atoms – neither created nor destroyed.
- Atoms – chemical reactions result from combination / separation of atoms.
J. J. Thomson (1897)
- Electrons – negatively-charged particles.
- Atoms – positively-charged sphere.
Ernest Rutherford (1911)
- Atoms – consists of a positively-charged nucleus with a cloud of electrons surrounding nucleus.
- Protons – positively-charged particles.
Niels Bohr (1913)
- Electrons – surrounding the nucleus (orbit).
James Cadwick (1932)
- Neutrons – electrically neutral subatomic particles.
- Neutrons – mass almost the same with a proton.
- Nucleus of an atom – consists of protons and neutrons.
Modern Atomic Model
- Nucleus of an atom – consists of protons and neutrons.
- Electrons – moving around the nucleus (orbits / electron shells/ quantum shells)
Atom – smallest particle of an element.
Relative atomic mass (Ar) - (an element) average mass of one atom of the element relative to 1/12 times the mass of one atom of carbon-12.
= (average mass of one atom of the element) / (1/12 x mass of one atom of C-12)
= 12 x [(average mass of one atom of the element) / (mass of one atom of C-12)]
Cations – positively-charge ions.
Example: H+, K+, NH4+ and Mg2+
Anions – negatively-charge ions.
Example: Br-, OH-, O2- and S2O32-
Molecule – a group of two or more atoms.
Relative molecular mass (Mr) – (an element or compound) average mass of one molecule of the substance relative to 1/12 times the mass of one atom of carbon-12.
= (average mass of one molecule of substance) / (1/12 x mass of one atom of C-12)
= 12 x [(average mass of one molecule of substance) / (mass of one atom of C-12)]
Proton number / Atomic number / Number of protons (Z)
- Number of protons in the nucleus of an atom.
- Number of electrons (neutral atom).
Nucleon number / Mass number / Number of nucleon (A)
- total number of protons and neutrons in the nucleus of an atom.
A = Z + N
N = number of neutrons
Isotopes (of the same element)
- atoms having the same proton number but different nucleon number.
- same number of protons, number of electrons, electronic configuration and chemical properties.
- different nucleon number, relative mass, density and rate of diffusion.
Relative isotopic mass – the ratio of the mass of one atom of the isotope relative to 1/12 times the mass of one atom of carbon-12 isotope.
= (mass of one atom of the isotope) / (1/12 x mass of one atom of C-12)
= 12 x [(mass of one atom of the isotope) / (mass of one atom of C-12)]
i. Vaporisation chamber – sample is vaporised (produce gaseous atoms or molecules).
ii. Ionisation chamber – vapour is bombarded with a stream of high-energy electrons to form positive ions. X(g) + e –> X+(g) + 2e. (produce positive ions)
iii. Acceleration chamber – positive ions are attracted towards the high negative potential plated that accelerates the positive ions to a high and constant velocity. (accelerate the positive ions).
iv. Magnetic Field – accelerated positive ions are deflected into a circular path according to the m/e ratio. (separate positive ions of different m/e ratio)
v. Ion detector – positive ions with different m/e ratios will be deflected to the ion detector that can be recorded on a moving chart. (detect the number and m/e ratio of the positive ions)
vi. Recorder – a flow of current which is amplified and recorded as peaks. (plot the mass spectrum of the sample)
- A lighter ion will deflect more than a heavier ion (the same charge)
Example: 35Cl+ will deflect more than 37Cl+
- An ion with a higher charge will deflect more than an ion with a lower charge (the same mass)
Example: 35Cl2+ will deflect more than 35Cl+
Isotopic abundance = fractional abundance = percentage abundance
One mole – the quantity of a substance that contains the same number of particles (atoms, ions or molecules) as the number of atoms in exactly 12 grams of carbon-12 isotope.
Avogadro constant, L or NA – number of particles (atoms, ions or molecules) present in a mole of substance (elements or compounds)
= 6.02 x 1023 (unit is mol-1)
Number of moles = number of atoms or molecules / Avogadro constant (mol-1)
Number of particles in a sample = number of moles x Avogadro constant (mol-1)
Mass (g) = number of moles (n) x M (Ar or Mr)
Number or moles (n) = mass (g) / molar mass (g mol-1)
Mass (g) = number of moles x molar mass (g mol-1)
Number of moles = volume of gas (dm3) / 22.4 dm3 at s.t.p. (0˚C and 1 atm or 101 kPa)
Number of moles = volume of gas (dm3) / 24 dm3 at r.t.p. (25˚C and 1 atm or 101 kPa)
Volume of gas (dm3) = number of moles x / 22.4 dm3 at s.t.p.
Volume of gas (dm3) = number of moles x / 24 dm3 at r.t.p.
Number of moles of solute = MV / 1000
(M = concentration in mol dm-3)
(V = volume in cm3)
Concentration of a solution (g dm-3) = mass of solute (g) / volume of the solution (dm3)
Concentration of a solution (mol dm-3) = number or moles of solute (mol) / volume of the solution (dm3)
MaVa / MbVb = a/b
M1V1 = M2V2