School Science Lessons
2024-03-28
Brownian motion, Chemical bonds, Conservation of mass, Particles, ions
(topic11)
Contents
11.1.0 Brownian motion
11.2.0 Chemical bonds
11.3.0 Conservation of mass
11.4.0 Ionic bonding
11.5.0 Movement of ions
11.6.0 Particles, matter as particles, size of particles
11.1.0 Brownian motion
1. Brownian motion, named after Robert Brown, (1773-1858, Scotland).
He was the botanist employed by Sir Joseph Banks, (1743-1820), on the voyage of the "Investigator" to Australia in 1801.
Robert Brown published his observations in 1827.
He had noticed the jittery random movement of pollen grains thought to be caused by the pollen grains themselves.
However, he showed that similar random movement could also be seen if any solid material of similar size was observed.
A small particle suspended in a liquid executes a haphazard zigzag movement, because of the uneven buffeting from the molecules of liquid that bombard it.
This random movement of microscopic particles suspended in a liquid is caused by an imbalance of combined impact forces of solution molecules on much
larger irregularly-shaped solute particles.
In 1905, Albert Einstein was fascinated by this movement.
He wrote, "The existence of a never diminishing motion seems contrary to all experience.
This difficulty was splendidly clarified by the kinetic theory of matter".
His statistical calculations of the mass and number of molecules involved, lead to the final acceptance of the atomic theory by scientists.
2. In investigating the pollen of different plants, Robert Brown, in 1828, observed that this became dispersed in water in a great number of small particles.
They were perceived to be in uninterrupted and irregular "swarming" motion.
The phenomenon repeated itself with all possible kinds of organic substances.
He believed that he had found in these particles the "primitive molecule" of living matter.
He found later that the particles of every kind of inorganic substance presented the same phenomenon.
He drew the conclusion that all matter was built up of "primitive molecules.".
3. Albert Einstein's "simple theory of this phenomenon" was to review from consideration of diffusion and osmotic pressure,diffusion and irregular motion of molecules.
He used van 't Hoff's law, Maxwell's law of distribution of velocities and Stokes' law, (Zeit. f. Elektrochemie, 14, 1908, pp. 235-239).
Before Einstein, different scientists noted that the motion:
was caused by irregular heating, incident ligh,.
persisted unchanged for a whole year when the liquid was sealed up between two cover glasses,
is most rapid with the smallest particles,
is increased by ultraviolet light and heat rays,
has its origin in the impacts of the molecules of the liquid on the particles,
is the more lively the smaller the viscosity of the liquid,
is caused by the thermal molecular motions of the liquid and the velocity of the movement decreases with increase of size of the particles,
increases with rise of temperature
is not caused by forces exerted by the particles on one another, temperature differences, evaporation, electrical forces intensity of illumination, strong, electromagnetic fields.
4. Einstein explained that the pollen grains are buffeted by collisions with molecules of water moving randomly in all directions
Einstein's theory and Perrin's observations showed that the mean distance travelled by a pollen grain is subject to random collisions increases as the square root or time distance x time t, whereas in straight line motion, x t
5. In direct confirmation of Einstein's formula, photographs of cinnabar particles every 0.1 second showed average displacement inversely proportional to the viscosity
Jean Baptiste Perrin, 1870-1942, France, and his students verified Einstein's explanation of this phenomenon
The method was to prepare a suspension of gamboge or mastic with particles of exactly equal to be enclosed in a microscopic chamber
The distribution in height of the particles was determined, after equilibrium, by counting the particles in different layers above the bottom of the chamber
A small screen in the ocular of the microscope allowed only a few particles in the field of vision.
So a great number of observations were arranged for different sizes of particles
With the acceptance of the explanations for Brownian motion the existence of atoms and molecules became accepted
Also, Perrin calculated the value for Avogadro's number, the number of atoms in a mole
Experiments
1. Put a drop of toothpaste on a microscope slide and stir water into it until it is almost colourless.
Put a coverslip over the drop and examine it with a microscope under high power with the stage illuminated from the side.
Pay attention to one particle.
At first it appears to stay in one place.
Later you can see that it is moving with an irregular jerky movement in all directions.
The irregular movement is caused by molecules of water hitting unevenly on the sides of the particle.
This experiment does not work with the modern "gel" type of toothpaste.
2. Make a model using a tray containing many small, light beads and one large marble.
The small beads represent molecules of water and the large marble represents a particle of suspended graphite.
Shake the tray and note how the small beads hit the marble from all directions.
The forces tend to cancel so the large marble may just make very small irregular movements, but returns to the same place.
3. Observe dirty water in sunlight.
Fill a beaker with tap water and focus sunlight into the beaker with a magnifying glass.
You may see the motions of suspended particles of solid matter.
4. Observe smoke in still air.
The smoke particles suspended in the air show Brownian movement.
Use a microscope to observe the movement of particles in a smoke cell.
5. Use a microscope to observe the movement of particles in a smoke cell.
11.2.0 Chemical bonds
11.2.1 Chemical bonds
11.2.2 Bond energy (bond strength)
11.2.3 Construct molecular models
11.2.4 Covalent bonds
11.2.5 Hydrogen bonds
11.2.6 Intermolecular bonds
11.2.7 Ionic bonds, electrovalent bonds, e.g. sodium chloride
11.2.8 Liquids with different viscosity, hydrogen bonds
11.2.9 Metallic bonds
11.2.10 Network solids
11.2.11 van de Waals forces (van de Waals bonds)
11.3.0 Conservation of mass
11.1.01 Chemical equations and ionic equations, conservation of mass
Burn
Experiments
11.1.5 Burn steel wool and burn iron filings
11.1.2 Burn steel wool, change in weight
11.1.3 Burn sulfur in oxygen
11.1.4 Copper cycle reactions
11.1.5 Effervescent tablets, health salts, sodium bicarbonate, (baking soda)
11.4.0 Ionic bonding
Ionic bonding occurs when positive and negative ions are held together in a crystal lattice by electrostatic forces.
When ionic or covalent chemical bonds form between different elements, a chemical compound is obtained, which can be represented by a chemical formula.
11.5.0H Movement of ions
11.5.1 Movement of copper ions and chromate ions
11.5.2 Movement of copper ions in ammonium nitrate solution
11.5.3 Movement of ions between microscope slides
11.5.4 Movement of ions, potassium permanganate solution
11.5.5 Movement of ions, sodium sulfate solution
11.5.6 Movement of suspended aluminium powder
11.6.0 Particles
11.3.1 Matter
11.3.2 Clay soil suspension
11.3.3 Colour change of diluted potassium permanganate
11.3.4 Particles of matter and dilution
11.3.5 Size of carbon atom in stearic acid molecule
11.3.6 Size of colloidal particles
11.3.7 Size of oil molecule
11.3.8 Size of stearic acid molecule
11.1.01 Chemical equations and ionic equations, conservation of mass
Chemical reactions are influenced by the conditions under which they take place and, being reversible, may reach a state of equilibrium.
Chemical reactions occur at different rates and changing the nature of the reactants, temperature, or concentration, or introducing a catalyst, may alter these.
Chemical reactions may be reversible.
Reversible chemical reactions may reach a state of dynamic balance known as equilibrium, which, when disturbed, will be re-established.
Specific criteria can be used to classify chemical reactions.
Redox reactions involve a transfer of electrons and a change in oxidation number.
Precipitation reactions result in the appearance of a solid from reactants in aqueous solution.
Acid-base reactions involve transfer of protons from donors to acceptors.
Polymerization reactions produce large molecules with repeating units.
Chemical reactions involve energy changes.
All chemical reactions involve energy transformations.
The spontaneous directions of chemical reactions are towards lower energy and greater randomness.
Chemical reactions, but not nuclear reactions, obey the law of conservation of mass.
The mass of the products is equal to the mass of the reactants.
In a balanced chemical equation, the number of atoms of each element in the reactants is equal to the number of atoms in the product.
Ionic equations show soluble ionic compounds in solution as separate ions and the sum of charges on the product side is equal to the sum of charges on the reactant side.
Chemical equation: HCl (aq) + NaOH (aq) --> NaCl (aq) + H2O (l)
Ionic equation: H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) --> H2O (l) + Cl- (aq) + Na+ (aq)
The Cl- and Na+ ions are called spectator ions, because they are on both sides of the equation and do not react.
Cancel the spectator ions to leave the net ionic equation: H+ + OH- --> H2O
11.1.1 Burn steel wool and burn iron filings
1. Collect oxygen in test-tubes with stoppers.
Store test-tubes in a test-tube rack and remove the stoppers just before inserting the burning element.
Fasten steel wool to wire.
Heat the steel wool in a burner flame.
Put it into a test-tube of oxygen.
The steel wool burns with bright sparkles to form black-grey iron oxide.
4Fe (s) + 3O2 (g) --> 2 Fe2O3 (s)
iron + oxygen --> iron oxide
2. Repeat the experiment by placing iron filings in a sieve and shaking it over a Bunsen burner flame.
A shower of sparks occurs as in some fireworks.
3. Hold an iron nail against a grinding wheel.
Friction breaks the iron into little pieces and heats them until they ignite and become white hot, like sparks from a fire.
11.1.2 Burn steel wool, change in weight
Roll some steel wool into a ball in your hands and weigh it.
Hold it with tongs over a sheet of paper.
Heat the steel wool until red hot; remove the flame, and blow gently on the red hot steel wool until it stops burning.
When cold, weigh the steel wool and any fragments fallen onto the sheet of paper, on the balance.
When iron burns, the product formed, iron oxide, is heavier than the iron.
4 Fe (s) + 3 O2 (g) --> 2 Fe2O3 (s)
11.1.3 Burn sulfur in oxygen
Dip a wire loop into sulfur powder. Ignite the sulfur in a burner flame and then put it into a test-tube of oxygen.
The sulfur burns with a bright blue flame to form the colourless gas sulfur dioxide.
S (s) + O2 (g) --> SO2
sulfur + oxygen --> sulfur dioxide
Some sulfur trioxide may also form in this reaction.
11.1.4 Copper cycle reactions
See diagram 1.13a: Simple fume hood
Step 1. Convert copper metal to copper nitrate
1. Weigh 1.000 g of copper wire.
It must be clean, bright and shiny.
Twist the wire into a flat spiral and put it in a beaker in a fume hood.
Slowly add 4.0 mL of 16 M nitric acid. Be careful!
Note the brown fumes of nitric oxide (NO) nitrogen dioxide (NO2)
and dinitrogen tetroxide (N2O4), i.e. NOx.
When all the copper is dissolved, add 100 mL of deionized water.
Cu (s) + 8HNO3 (aq) --> 3Cu(NO3)2 (aq)+ 4H2O (l) + 2NO (g) [dilute nitric acid]
Cu (s) + 4HNO3 (aq) --> Cu(NO3)2 (aq) + 2H2O (l) + 2NO2 (g) [concentrated nitric acid]
N2O4 < --> 2NO2 [in equilibrium]
Step 2 Convert copper nitrate to copper hydroxide
Add 30.0 mL of 3.0 M sodium hydroxide to the solution while stirring.
If red litmus paper does not turn blue in the solution, add more sodium hydroxide.
Note the precipitate of copper hydroxide, an ionic solid.
Cu(NO3)2 (aq) + 2NaOH (aq) --> Cu(OH)2 (s) + 2NaNO3 (aq)
Step 3 Convert copper hydroxide to copper oxide
Heat the solution on a hot plate while continually stirring to prevent bumping caused by steam bubbles.
Note the precipitate changing to a black solid, copper oxide, CuO.
Carefully decant the liquid, add deionized water and decant again.
Heat the precipitate until it becomes a firm mass.
Cu(OH)2 (s) + heat --> CuO (s) + H2O (l)
Step 4 Convert copper oxide to copper sulfate
Reaction 4: Converting copper oxide to copper sulfate
Add 15 mL of 6.0 M sulfuric acid to the copper oxide while swirling, not stirring, the copper oxide to help it dissolve.
CuO (s) + H2SO4 (aq) --> CuSO4 (aq) + H2O (l)
Step 5 Convert copper sulfate to copper metal
CuSO4 (aq) + Zn (s) --> ZnSO4 (aq) + Cu (s)
In the fume hood add 2.0 g of zinc metal and keep stirring until the solution becomes colourless.
The zinc is oxidized as it reduces the copper.
Add one drop of the solution to 1 mL of concentrated ammonia solution in a test-tube.
If the ammonia turns deep blue, some unreduced copper is till in the solution so the reaction is not finished.
When all the copper is reduced, decant the liquid and add 20.0 mL of 6M hydrochloric acid to dissolve excess zinc.
Note the bubbles of hydrogen gas until the reaction is complete.
Cool the beaker and observe the metallic copper settling on the bottom.
Carefully decant the solution and wash the copper metal with deionized water and decant again.
Transfer the copper to an evaporating dish using a small amount of deionized water.
Decant excess water evaporating dish then wash the precipitate with methanol.
Decant the liquid the gently heat the precipitate on a hot plate.
If you heat the copper precipitate too strongly, it will oxidize to copper oxide.
Transfer the dried copper metal to a preweighed beaker and calculate the mass of recovered copper.
Percentage recovery = mass of recovered copper / initial mass of copper × 100.
11.1.5 Effervescent tablets, health salts, sodium bicarbonate, (baking soda)
See diagram 12.1.1a: Conservation of matter |
When a chemical reaction occurs, matter is neither created nor destroyed.
The mass of the reactants = the mass of the products.
1. Effervescent tablets or "fruit salts" contain sodium hydrogen carbonate, and dry citric acid or tartaric acid.
Put the tablet or fruit salts in water in a test-tube.
Carbon dioxide forms as bubbles and any other substance in the tablet or fruit salts dissolves easily.
Tests for carbon dioxide by holding a test-tube containing limewater at an angle near the mouth of the test-tube containing the effervescent tablet.
The carbon dioxide given off by effervescence is heavier than air and will roll into the limewater test-tube where the limewater will turn milky, because of the presence of carbon dioxide.
Weigh an effervescent tablet or fruit salts and put into the bottom corner of a small plastic bag.
Twist the bag above the corner and tie around the twist with thin string or wire.
Weigh the plastic bag + string + effervescent substance.
Pour a known amount of water into the open bag then tie string tightly to close the bag so that no liquid or gas can escape.
The weight of the water + bag + string + effervescent substance is now known.
Undo the string around the twisted part of the bag and untwist the bag.
The acid and sodium hydrogen carbonate dissolves in the water and react to produce a salt and carbon dioxide.
Weigh the bag and products of the reaction.
The weight is the same.
2. Place a small amount of water in a plastic cup and place on the pan of the electronic scale.
Cover the top with a piece of paper or metal foil.
Place two Alka-Seltzer tablets (ant-acid tablets), on top of the cover.
Record the initial mass.
Tilt the cover so that the tablets drop into the water and immediately replace the cover so water droplets will not escape.
Record the mass reading until it is constant.
Repeat this experiment with the cup enclosed in a sealed container.
3. Place a small amount of water in a large 2 litre plastic drink bottle.
Break two Alka-Seltzer tablets in pieces that will fit in the bottle.
Weigh the bottle, bottle cap, and Alka-Seltzer tablets together.
Drop the Alka-Seltzer tablet pieces into the bottle and quickly replace the bottle cap tightly and place back on the scale.
Record the mass reading until it is constant.
The results differ from the experiment done in an open container.
Remove the bottle from the scale, loosen the bottle cap, and measure again the weight of the bottle and its cap.
11.2.1 Chemical bonds
See diagram 3.01: Water molecule 1
See diagram 3.02: Water molecule 2
1. Chemical bonds are forces of attraction between the atoms together in a molecule or a crystal.
All chemical bonds are caused by attractive force between positive and negative particles.
2. The three main types of bonding are ionic, covalent and metallic bonding.
3. Chemical bonding is about atoms and electrons arranging into lower energy states.
11.2.2 Bond energy (bond strength)
The bond energy needed to break both of the two O-H bonds in water, H-O-H, is 459 kJ mol-1.
Relative bond strengths
The van der Waals bonds strengths × 10 = hydrogen bonds strengths × 10 = non-polar covalent bonds strengths, then covalent bonds
strengths, then ionic bonds strengths, then metallic bonds strengths.
11.2.3 Construct molecular models
The physical and chemical properties of chemical substances are largely influenced by their shape, e.g. the polarity of a molecule and subsequent intermolecular forces.
This can be demonstrated by the construction of chemical models requiring the use of geometric skills.
Use models to show the different kinds of bonding between atoms to form molecules and ions.
1. Use polystyrene spheres to represent ions in the sodium chloride crystal lattice.
Cut holes in a plate to take the bottom layer of spheres then construct layers of spheres.
Join the spheres with small springs to show how the ions may vibrate about mean positions.
2. Use corks to represent atoms or ions and use hooks or pins to represent bonds.
2.1 Construct ionic crystals: halite NaCl, calcite CaCO3.
2.2 Construct covalent crystals: diamond C, quartz SiO2, ice H2O, sulfur S8.
2.3 Construct alkanes: methane CH4, ethane C2H6, propane C3H8, butane C4H10, 2-methylpropane C4H10, CH3.CH.CH3.CH3
2.4 Construct alkanoic acids: methanoic acid HCOOH, ethanoic acid CH3COOH, propanoic acid C2H5COOH, butanoic acid, (butyric acid), C3H7COOH, CH3CH2CH2.COOH, 2-methylpropanoic acid, (isobutyric acid) C4H8O2, (CH3)2-CH-COOH
2.5 Construct organic molecules: PVC (polyvinyl chloride, poly-chloroethane) (CH2Cl.CH3)n, polypropylene (-CH3.CH.CH2-)2, polystyrene (-C6H5.CH.CH2-)n,
nylon-66, -NH2(CH2)6NH2CO(CH2)CO-}n
2.6 Construct biochemical models: (+) glucose C6H12O6, sucrose C12H22O11 starch / cellulose (C6H11O5)n, fat C57H110O6, tripalmitin) polypeptide (-CH[CH3]CO.NH-)n, DNA (deoxyribonucleic acid).
3. Molecular models set, e.g. "Molymod", stick and ball, organic set, inorganic set (toy product)
3.1 The plastic straw, represents a single covalent bond or a "C=O" double bond.
3.2 The flexible straw, 2 flexible straws represent a "C=C" double bond.
3.3 The 4-pronged tetrahedral carbon centre (black), represents C in "-C-".
3.4 The 3-pronged trigonal planar carbon centre (black), represents C in ">C=O".
3.5 The one-pronged hydrogen centre (white), represents "H".
3.6 The one-pronged oxygen centre (red), represents O in ">C=O".
3.7 The 2-pronged V-shaped oxygen centre (red), represents O in "-O-H".
11.2.4 Covalent bonds
1. Covalent bonds form by sharing of electrons.
Pairs of atoms may be bound together by the sharing of electrons between them in a covalent bond.
Two or more atoms bound together by one or more covalent bonds form a molecule, with definite size, shape and arrangement of bonds.
An atom or group of atoms covalently bound together may gain or lose one or more electrons to form ions.
2. Non-polar covalent bonds exist between two or more, usually identical, atoms, e.g. H: H.
Each atom donates one electron to form a pair of electrons that are equally shared by the two atoms.
Non-polar covalent bonds (co-ordinate covalent bonds), in ammonium ion NH4+, have electrons shared between the atoms and are not soluble in polar solvents, e.g. water, but are soluble in non-polar solvents, e.g. hexane.
3. Polar covalent bonds form when atoms of two different elements each donate one electron to form a shared pair of electrons, but one atom has a greater share than the other atom.
Some people say that the shared electrons spend more time around one atom than the other atom, so this kind of covalent bond has a sort of ionic character.
Polar covalent bonds exist as a dipole, e.g. the H-Cl bond in hydrogen chloride.
The H end has a small +ve charge and the Cl end has a small -ve charge.
Water, H2O, is a polar molecule, a dipole with a definite positively charged end and a definite negatively charged end, because each H has a small +ve charge, H+, and the O has a small -ve charge, O2-.
Dipole-dipole bonds occur between polar molecules where the slightly positive end of one molecule is attracted to the slightly negative end of another molecule.
Dipole-dipole forces are intermolecular forces between molecule dipoles at δ + and δ - areas.
Solids with covalent bonds within the molecules with van der Waals forces or weak dipole to dipole intermolecular forces have low melting points and no electrical conductivity.
5. Molecules may be saturated, i.e. have only single bonds, --, or unsaturated, i.e. contain multiple bonds, e.g. double bond =, or triple bond, e.g. carbon monoxide, CO, C≡O.
11.2.5 Hydrogen bonds
1. Hydrogen bonds are weak intermolecular forces of attraction between polar molecules that contain hydrogen, i.e. between the H atom of one molecule
and the negative charged atoms in another molecule.
In water, there is weak attraction between the hydrogen in a water molecule with the oxygen (in the OH) of another water molecule.
In water, each hydrogen nucleus is bound to the central oxygen atom by a pair of electrons shared between them, a covalent chemical bond.
Two of the six outer shell electrons of oxygen are used, leaving four electrons as two non-bonding pairs.
The four electron pairs around the oxygen are as far apart as possible to reduce repulsion between their negative charges.
However, the two non-bonding pairs stay closer to the oxygen atom and exert a stronger repulsion against the two covalent bonding pairs, and so pushing the two hydrogen atoms closer together.
These forces cause the H-O-H angle, 104.5o.
The positive and negative charges are not distributed uniformly.
The negative charge is concentrated at the oxygen end of the molecule, because of the non-bonding electrons and oxygen's high nuclear positive charge.
This charge displacement causes an electric dipole.
The partially-positive hydrogen atom on one water molecule is electrostatically attracted to the partial negative oxygen on a neighbouring molecule by a process is called hydrogen bonding.
That hydrogen bond is weak.
In his 1931 book, "The Nature of the Chemical Bond", Linus Pauling described the hydrogen bond as an electrostatic attraction between hydrogen atoms and areas of high electron density in certain other elements, such as oxygen or nitrogen.
2. Intermolecular force between strongly electronegative atom, e.g. F, O, N, and a H atom covalently bonded to an electronegative atom.
The hydrogen ion is the positive ion, H+, a proton, formed when a hydrogen atom loses its electron.
The hydrogen ion may become solvated to form the hydrated ion H3O+, the hydronium ion.
Hydrogen bonds are weak bonds caused by the electrostatic attraction between strongly electronegative atoms and a hydrogen atom covalently linked to another electronegative atom.
The polarized water molecule is a dipole with each H+ linked to the one O2- by a polarized covalent bond.
The hydrogen bonds are inter-molecular bonds between polar molecules containing hydrogen.
So, in water, the H+ of one water molecule is attracted to the O2- of another water molecule.
Hydrogen bonds between water molecules only, cause the force of cohesion and surface tension.
Hydrogen bonds between water molecules and other molecules, e.g. the molecules in glass, cause the force of adhesion.
Electronegativity refers to an atom attracting to itself electrons in covalent bonds.
11.2.6 Intermolecular bonds
Intermolecular bonds exist between molecules and intramolecular bonds exist between the atoms of Intermolecular bonds hold together solids and liquids,
but they are stronger in solids and cause the close-packed ordered arrangement of particles.
Strong intramolecular bonds in liquids cause high boiling points.
Smaller molecules usually have lower boiling points.
11.2.7 Ionic bonds, electrovalent bonds, e.g. sodium chloride.
Ionic bonds, electrovalent bonds, form by transfer of electrons
Ionic bonding occurs when positive and negative ions are held together in a crystal lattice by electrostatic forces.
Ionic bonds are electrical links between atoms caused by the distribution of electrons around the nuclei of the bonded atoms.
Ionic solids are usually transparent and have high melting points and boiling points.
When an ionic solid, e.g. sodium chloride, NaCl, dissolves in water, the cations and anions separate, e.g. Na+ and Cl-.
Solids with ionic bonds have high melting points and low electrical conductivity.
However, when molten or in solution, their electrical conductivity is high.
11.2.8 Liquids with different viscosity, hydrogen bonds
The viscosity's are different mainly, because of difference in hydrogen bonding.
Pour 100 mL of the different liquids into separate conical flasks, e.g. | glycerol (glycerine | glycol | water | methylated spirit |
benzene | concentrated sulfuric acid 18 M.
BE CAREFUL!
Swirl each flask equally and note the relative time until the disappearance of each vortex.
Use this method to show that a mixture of two organic liquids is more viscous than the viscosity of either pure substance.
This is caused by hydrogen bond formation.
11.2.9 Metallic bonds
In solid metals or alloys, the atoms exist in a lattice of positive ions with electrons moving freely between them.
Metals (metallic bonds) have high melting points and are good electrical conductors, because the outer electrons are held loosely.
Metals are ductile and malleable, because the metallic bonds are weak.
However, alloys with other metals strengthen the lattice and so improve mechanical strength.
11.2.10 Network solids
Network solids have atoms in a lattice of strong covalent bonds, e.g. C (diamond) Si, P (red phosphorus) B, and Ge.
11.2.11 van de Waals forces (van de Waals bonds)
The van der Waals forces (Johannes Diderik van der Waals 1837-1923, Netherlands) are the weak forces between molecules,
including H bonding forces in H2O and HF, dipole-dipole forces between HCl molecules and dispersion forces between Cl and Cl.
The van der Waals equation: (P + a / V2)(V-b) = RT, where a = mutual attraction constant between molecules, b = space occupied by molecules,
and V = volume of the gas.
The van der Waals attraction and repulsion forces between atoms and between molecules, are caused by fluctuating polarization of atomic particles and are quite different from the forces of covalent and ionic bonding.
Although the water molecule is electrically neutral, the distribution of charge is not symmetrical, causing a net attraction between molecules from momentary dipoles,(cohesion of water molecules), that affects viscosity and surface tension.
Even non-polar molecules have some van der Waals bonding, although nonpolar liquids have low surface tension and low boiling points.
11.3.1 Matter
Matter, states of matter, solid (s), liquid (l), gas (g), aqueous solution, (dissolved in water), (aq)
Matter is composed of atoms, which, in turn, contain protons and neutrons in a nucleus, and electrons outside the nucleus.
The number of positively charged protons is equal to the number of negatively charged electrons in a neutral atom, and determines all the chemical properties of an atom.
Materials may be elements, compounds or mixtures.
11.3.2 Clay soil suspension
Shake clay soil with water and leave it to settle.
Note the humus layer at the surface and particles of rock and mineral at the bottom.
Filter the liquid.
The filtrate is still cloudy, because clay particles have passed through the filter paper.
Suspension particles may take days to settle, but they will settle eventually.
Colloid particle size is about 10-7 to 10-5 cm.
Divide the filtrate into two parts in test-tubes.
Keep one part as a control.
To the other test-tube add drops of barium chloride solution or an aluminium salt solution.
Note whether the filtrate remains cloudy.
The same effect may occur when a clay suspension in a river meets the salts contained in sea water.
11.3.3 Colour change of diluted potassium permanganate
Put one crystal of potassium permanganate in a test-tube.
Add 1 mL of water.
Dissolve the crystal completely by shaking vigorously with the thumb over the end of the test-tube.
Then add water to a total volume of 10 mL.
This is a "X 10" dilution.
Pour this 10 mL of purple solution into a 100 mL beaker and then fill up the beaker with water.
This is now a "X 100" dilution.
Fill the 10 mL test-tube with this solution and discard the rest.
Dilute this again in the beaker to 100 mL.
This is now a "X 1000" dilution.
Continue diluting the solution.
Some purple colour is still visible.
This shows that if matter consists of particles, the particles must be very small.
11.5.1 Movement of copper ions and chromate ions
See diagram 3.2.89: Movement of copper ions and chromate ions.
Copper chromate forms two coloured ions, the blue-green positive copper ion and the orange negative chromate ion.
Prepare copper chromate by adding 100 mL of 1 M copper (II) sulfate solution to 100 mL of 1 M potassium chromate solution.
Use a Buchner funnel fitted with a filter pump to separate the copper chromate precipitate.
Wash the precipitate with demineralized water then transfer it to a beaker.
Dissolve the precipitate in the minimum volume of dilute hydrochloric acid.
Dissolve excess urea in the copper chromate solution.
Add dilute hydrochloric acid to a U-tube until it is one third full.
Position the jet of a pipette full of copper chromate solution at the bottom of the U-tube.
Slowly deliver the copper chromate solution so that it pushes the hydrochloric acid up and forms a separate layer below.
Take out the pipette carefully to avoid mixing.
The carbon electrodes must be in contact with the hydrochloric acid and also connected to a 20 V DC supply.
After some minutes, note the blue-green colour of the copper on the negative side, and the orange chromate colour on the positive side.
The boundaries of these coloured ions will move very slowly towards the electrodes.
11.5.2 Movement of copper ions in ammonium nitrate solution
1. Dissolve 10 g of ammonium nitrate in 125 mL of water.
Add 10 mL of concentrated aqueous ammonia solution.
Dip a piece of filter paper in this solution of ammonium nitrate and then twine the filter paper round a test-tube by using two pieces of copper wire.
Leave no air bubbles between the filter paper and the wall of the test-tube.
Connect the copper wires to a source of direct current and apply a voltage of 12 volts.
Note a greenish blue colour moving from the positive terminal to the negative terminal.
2. Dissolve 10 g of ammonium nitrate in 125 mL water.
Add 10 mL concentrated aqueous ammonia solution.
Pour the solution into a Petri dish or beaker.
Use a 12- volt battery.
Use crocodile clips to attach the positive pole to a piece of copper gauze (mesh) and attach the negative pole to a strip of copper.
Put the copper electrode 5 cm apart in the solution.
Observe the blue streaks of copper ions moving from the +ve electrode towards the -ve electrode.
11.5.3 Movement of ions between microscope slides
See diagram 3.2.90: Movement of ions between microscope slides.
Show the movement of positive coloured ions towards a negative electrode.
The electrolyte is held in a strip of filter paper sandwiched between two microscope slides.
Use carbon rod electrodes to lead the current through the filter paper.
With a 10 to 20 V DC. supply, use the width of the slide, not as in the diagram, because greater voltage is needed if the length of the slide is used.
Cut a strip of dry filter paper 1 cm wide.
Make a pencil mark across the centre of the paper.
Moisten the paper with tap water so that it is damp, but not wet.
Use a fine capillary tube to apply the coloured ion solution, e.g. Cu2+ or CO2+ ions, along the pencil mark.
Fix the strip of filter paper between the two slides and fold the ends around the carbon rods.
Use a paper clip to the slides together.
Connect the carbon rods to the 20 V DC supply.
Wait for some minutes and observe the coloured ions moving towards the negative electrode.
By contrast, repeat the experiment with potassium manganate (VII) and observe the coloured permanganate ion moving towards the positive electrode.
If you want to make a dispenser for the solution of coloured ions, fold a strip of filter paper 1 cm wide around a thin piece of plastic material to form a firm wick.
Wedge the wick between a split cork and put it in a test-tube containing the solution of coloured ions.
First touch the wick on absorbent to remove excess solution and then lightly touch it on the pencil mark.
11.5.4 Movement of ions, potassium permanganate solution
See diagram 3.4.1: Potassium permanganate ions
1. Bind a strip of damp filter paper to a test-tube with two pieces of copper wire.
Do not leave any air bubbles between the filter paper and the wall of the test-tube.
Fix the test-tube horizontally on an iron stand.
Place a length of cotton thread previously dipped into potassium permanganate solution round the middle of the filter paper.
Connect the two pieces of copper wire to a 12 volts or more electric source of direct current.
A colour of purplish red can be seen to leave the cotton thread, moving gradually towards the positive terminal.
2. Cut a 1 cm wide strip of dry filter paper.
Draw a pencil mark across the centre of the paper.
Moisten with tap water so that it is damp, but not very wet.
Make a potassium permanganate solution.
Use a fine capillary tube to put the coloured ion solution along the pencil mark.
Hold the strip of filter paper between two microscope slides.
Attach carbon electrodes across the slides to lead the current through the filter paper.
Use 12 volts or more direct current.
11.5.5 Movement of ions, sodium sulfate solution
Dissolve sodium sulfate in water and add drops of Universal Indicator.
The solution should be green showing that the solution is neutral.
Dissolve 1 g of powder agar or agar gel in 100 mL of hot water.
Mix the two solutions then pour into an U-tube so that the arms are half full.
When the gel has set, pour dilute sulfuric acid into one arm and dilute sodium hydroxide into the other arm.
Put platinum or carbon electrodes into the solutions.
Connect the terminal in the sulfuric acid to the positive terminal of a 12-volt battery.
Connect the terminal in the sodium hydroxide to the negative terminal.
Turn on the electric current and note the colour changes at each arm.
The violet colour in the gel below the sodium hydroxide solution is due to movement of hydroxide ions into the gel.
The red colour in the gel below the sulfuric acid solution occurs, because of the movement of hydrogen ions into the gel.
During electrolysis, there is a flow of ions in opposite directions caused by the electric field.
11.5.6 Movement of suspended aluminium powder
Add a small amount of aluminium powder to a beaker of the treated water.
Add a few drops of detergent to the water to clean the very small pieces of aluminium and prevent them from sticking together.
The pieces probably had oils or grease on them.
Stir the mixture a little to make the aluminium powder mix freely with the water.
Darken the room as much as possible and shine the light of a film projector through the liquid.
Watch the pieces of suspended aluminium powder from the side.
As the eyes become accustomed to the darkness, the larger suspended aluminium particles appear not to change, but the smallest ones will twinkle like tiny stars.
11.3.4 Particles of matter and dilution
Put a crystal of potassium manganate (VII) in a test-tube.
Add 1 mL of water and fix a stopper.
Shake the test-tube to dissolve the crystal.
Add water to a total volume of 10 mL.
This is a "10 times" dilution.
Pour this 10 mL of purple solution into a 100 mL container and then fill the container with water.
This is now "100 times" dilution.
Fill the 10 mL test-tube with this solution and throw the rest away.
Dilute this again in the container to 100 mL.
It is now a "1 000 times" dilution.
Note how often the solution can be diluted by a factor of 10 before the colour is so pale that it is only just visible.
The final dilution factor shows that if matter is particulate, the size of the particles must be very small.
11.3.5 Size of carbon atom in stearic acid molecule
Fill a 14 cm diameter watch glass with water and measure the diameter of the water surface.
Add drops of 0.10 g /litre stearic acid solution in hexane to the water until one drop sits on the water, i.e. the drop extra to the monolayer of stearic acid in the water.
Calculate the mass of pure stearic acid in the total number of drops of 0.10 g /litre stearic acid solution in hexane in the monolayer after evaporation of the hexane.
The density of solid stearic acid is 0.85 g / mL.
The area of the monolayer = π × radius of water surface2.
The thickness of the monolayer, L = volume of stearic acid / area of water surface = length of the stearic acid molecule.
If the stearic acid molecule contains 18 closely packed carbon atoms in vertical stacks, the diameter of a carbon atom = L / 18, and the volume of a carbon atom = (L / 18)2.
Compare the experimental result with the accurately measured diameter of a carbon atom = 1.54 × 10-10 m.
11.3.6 Size of colloidal particles
1. Use three 50 mL, beakers.
Pour 20 mL copper (II) sulfate solution into Beaker 1.
Add 20 mL skim milk solution to Beaker 2. and Beaker 3.
Add 5 mL 3 M acetic acid to Beaker 2. to form the precipitate casein.
Filter the contents of each beaker into three large test-tubes 1, 2 and 3.
Use two 20 cm lengths of dialysis tubing, soak for ten minutes in distilled water to soften and tie up one end of each length.
Pour the copper (II) sulfate solution from Beaker 1 into one set of dialysis tubing.
Pour the skim milk from Beaker 2 into the other set of dialysis tubing.
Tie the ends of both sets of tubing, rinse with deionized water and put into two separate beakers of water.
Observe what happens after some time.
2. Repeat the last experiment using a mixture of starch and glucose solution inside the dialysis tubing.
After 30 minutes test the water outside the dialysis tubing for water for glucose and starch.
The next day, repeat the test for glucose and starch.
Filter the starch and glucose solutions and observe whether the filter papers contain any residues.
State your conclusions the sizes of colloids compared with the true solutions in these experiments.
11.3.7 Size of oil molecule
See diagram 3.2.57: Oil layer and powder on surface of water
1. Oil floats on the surface of water as a one molecule thick layer, if allowed to spread out freely.
When the water has a large enough surface area, the oil spreads out in a layer one molecule thick, because it does not form "hills" of molecules.
Knowing the volume of the oil and the area of the surface, calculate the thickness of the monomolecular layer.
2. Use a thin petroleum distillate or a pure vegetable oil and a flat tray containing water.
Lightly sprinkle the surface of the water with talcum powder.
Pour oil into a burette and measure the volume of fifty drops of oil.
Let another drop of oil to fall on a piece of plastic.
Pick up oil from the oil drop with the point of a glass rod and transfer it to the water surface by touching the surface lightly.
This oil spreads out pushing out the talcum powder.
Estimate the area of the oil by comparing it with a piece of graph paper.
Estimate how much oil is removed from the oil drop by using the glass point to pick up equal amounts of oil from the oil drop until it is all removed.
Knowing the volume of the oil on the water, calculate the thickness of the oil layer.
The volume of the oil drop = area of oil picked up by point × number of "pick ups" × depth of floating oil.
The depth of floating oil = diameter of the water molecule.
The diameter of the water molecule is in the range 10-6 to 10-7mm.
3. Oil floats on the surface of water as a one molecule thick layer, if allowed to spread out freely.
Use a tray with a glass bottom.
Put graph paper under the tray.
Put water in the tray.
Select a suitable oil to pour on the water, e.g. 1% oleic acid / methylated spirit solution, petroleum distillate.
Use a burette to find the volume of 100 drops of oil, then calculate the volume of 1 drop of oil.
Sprinkle the surface of the water with a very fine light powder.
Let 1 drop of oil fall from the burette onto the water.
The oil spreads out over the surface of the water, but must not touch the sides of the tray.
The oil on the water pushes the powder aside so you can easily see the area covered by the oil.
Look down on the graph paper to measure the approximate area over which it spreads.
The volume of oil put on the water = area of the oil on the water X thickness of the oil layer.
The approximate dimension of a single molecule of oil is 1 X 10-7 m.
4. Another way to calculate the volume of a drop of oil is to let the drop fall on a piece of flat plastic.
Touch the oil with the point of a glass rod then touch the water surface.
Oil leaves the glass rod and spreads over the water.
Use graph paper to measure the approximate area over which the oil spreads.
To calculate the volume of oil placed on the water, keep using the glass
rod to remove successive fractions from the flat plastic until no oil remains.
11.3.8 Size of stearic acid molecule
Stearic acid has the molecular formula, C18H36O2.
Make a 0.3700 g / L to 0.3750 g / L solution of stearic acid in benzene (benzol).
BE CAREFUL! Benzene may be carcinogenic..
Use safety glasses and nitrile chemical-resistant gloves.
Make only enough solution for immediate use.
Add the solution drop by drop to the surface of water in a Petri dish.
The dropper must have an outlet small enough so that there are more than 50 drops to make a total of 1 mL of the benzene solution.
The benzene evaporates and leaves the stearic acid to spread over the entire water surface.
The hydroxyl group of the organic acid molecule sticks into the water, because of its hydrophilic property.
The hydrocarbon portion of the molecule resists entering the water surface, because of its hydrophobic property.
This behaviour leads to a monomolecular film in which all the stearic acid molecules are compactly and fully arranged on the water surface.
The effective section area A cm2 of each stearic acid molecule can be calculated by using the following equation: A = M × S × V / m × NA × Vd × (d-1),
where M = molar weight of stearic acid, CH3(CH2)16COOH (284 g / mol) | S = total area of monomolecular film, | m / V = concentration of stearic acid in benzene solution (0.3700-0.3750 g / L)
| NA = Avogadro's Constant (L) (6.022 × 1023 mol-1) | Vd = volume of a drop of stearic acid solution | d = number of drops of stearic acid solution.
Before each determination, it is necessary to wash the Petri dish by using sodium carbonate solution and then tap water, finally distilled water 2-3 times to clear off stearic acid and base.
Repeated determination should use the same Petri dish.
Control dropping evenly to make the volume of every drop be the same.
Determine S:
Measure the inside diameter of the Petri dish to be used (~200 mm) in three directions with inside calipers (or a straight edge).
Take the average value and calculate the area (π × r2).
Determine Vd:
Use a dropper to add benzene (benzol), to a dry, small graduated cylinder until the volume reaches 1.0 mL, and count the drops.
Calculate the Vd.
Determine (d-1):
Use another dropper to add the stearic acid in benzene (benzol), solution vertically and gently at a point above the Petri dish and 1-2 cm from the water surface.
Wait for a fallen drop to diffuse until oil beads cannot be seen, and then give another drop.
Diffusion of the stearic acid solution gradually slows.
When a fallen drop cannot diffuse any more in two minutes, a monomolecular film has been formed, looking like a lens.
Record the number of drops, d, from which subtract the final drop to obtain the actual drop number needed to form the monomolecular film of stearic acid, i.e. d-1.
Repeat the determination and get the average value.
Calculate A according to the equation mentioned above.
(A = 2.2 × 10-15 cm2)