Spec point which is NOT in Triple Science
More distant galaxies have greater red-shift and therefore are receding faster because the space between all galaxies is stretching as the universe expands.
Solid
Arrangement: Particles are close together and regularly packed.
Movement: Particles vibrate around a fixed point.
Energy: Particles have less kinetic energy than both liquids and gasses.
Liquid
Arrangement: Particles are close together but irregular.
Movement: Particles are free to move.
Energy: Particles have less kinetic energy than gasses but more than solids.
Gas
Arrangement: Particles are far apart and there are no forces between them.
Movement: Particles are free to move.
Energy: Particles have more kinetic energy than liquids and solids.
Melting: When a solid is heated, the energy makes the particles vibrate fast enough so that the forces of attraction between the particles break. For example H2O(s) –> H2O(l)
Freezing: When a liquid is cooled, the particles move slow enough so that the forces of attraction between them will hold them into a solid. For example H2O(l) –> H2O(s)
Boiling: When a liquid is heated strongly, the energy makes the particles move fast enough so that all forces of attraction are broken. For example H2O(l) –> H2O(g)
Condensing: When a gas is cooled, the particles move slow enough so that the forces of attraction between them will hold them as a liquid. For example H2O(g) –> H2O(l)
Sublimation: A small number of substances have the ability to change directly from a solid to a gas when heated. For example CO2(s) –> CO2(g)
Diffusion is the spreading out of particles in a gas or liquid. There is a net movement of particles from areas of high concentration to areas of low concentration until a uniform concentration is achieved.
i) dilution of coloured solutions
Dissolving potassium manganate(VII) in water demonstrates that the diffusion in liquids is very slow because there are only small gaps between the liquid particles into which other particles diffuse.
The random motion of particles cause the purple colour to eventually be evenly spread out throughout the water.
Adding more water to the solution causes the potassium manganate(VII) particles to spread out further apart therefore the solutions becomes less purple. This is called dilution.
ii) diffusion experiments
When ammonia gas and hydrogen chloride gas mix, they react together to form a white solid called ammonium chloride.
ammonia + hydrogen chloride –> ammonium chloride
NH3(g) + HCl(g) –> NH4Cl(s)
A cotton wool pad was soaked in ammonia solution and another was soaked in hydrogen chloride solution. The two pads were then put into opposite ends of a dry glass tube at the same time.
The white ring of ammonium chloride forms closer to the hydrochloric acid end because ammonia particles are lighter than hydrogen chloride particles and therefore travel faster.
Even though these particles travel at several hundred metres per second, it takes about 5 min for the ring to form. This is because the particles move in random directions and will collide with air particles in the tube.
When a solid dissolves in a liquid:
Element: The simplest type of substances made up of only one type of atom.
Compound: A substance that contains two or more elements chemically joined together in fixed proportions.
Mixture: Different substances in the same space, but not chemically combined.
Note: elements such as oxygen (O2) are described as diatomic because they contain two atoms.
The full list of elements that are diatomic is:
Pure substances, such as an element or a compound, melt and boil at fixed temperatures.
However, mixtures melt and boil over a range of temperatures.
Example: although pure water boils at 100⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will raise the boiling point to 100.2⁰C.
Example: although pure water melts at 0⁰C, the addition of 10g of sodium chloride (NaCl) to 1000cm³ of water will lower the melting point to -0.6⁰C.
Simple distillation
This method is used to separate a liquid from a solution. For example: separating water from salt water.
The salt water is boiled. The water vapour condenses back into a liquid when passed through the condenser. The salt is left behind in the flask.
Note: cold water is passed into the bottom of the condenser and out through the top so that the condenser completely fills up with water.
Fractional distillation
This method is used to separate a mixture of different liquids that have different boiling points. For example, separating alcohol from a mixture of alcohol and water.
Water boils at 100oC and alcohol boils at 78oC. By using the thermometer to carefully control of temperature of the column, keeping it at 78oC, only the alcohol remains as vapour all the way up to the top of the column and passes into the condenser.
The alcohol vapours then condense back into a liquid.
Filtration
This method is used to separate an insoluble solid from a liquid. For example: separating sand from a mixture of sand and water.
The mixture is poured into the filter paper. The sand does not pass through and is left behind (residue) but the water passes through the filter paper and is collected in the conical flask (filtrate).
Crystallisation
This method is used to obtain a salt which contains water of crystallisation from a salt solution. For example: hydrated copper sulfate crystals (CuSO4.5H2O(s)) from copper sulfate solution (CuSO4(aq)).
Gently heat the solution in an evaporating basin to evaporate some of the waterIf instead the solution is heated until all the water evaporates, you would produce a powder of anhydrous copper sulfate (CuSO4(s)).
Paper chromatography
This method can be used to separate the parts of a mixture into their components. For example, the different dyes in ink can all be separated and identified.
The coloured mixture to be separated (e.g. a food dye) is dissolved in a solvent like water or ethanol and carefully spotted onto the chromatography paper on the baseline, which is drawn in pencil so it doesn’t ‘run or smudge’.
The paper is carefully dipped into the solvent and suspended so the baseline is above the liquid solvent, otherwise all the spots would dissolve in the solvent. The solvent is absorbed into the paper and rises up it as it soaks into the paper. The choice of solvent depends on the solubility of the dye. If the dye does not dissolve in water then normally an organic solvent (e.g. ethanol) is used.
As the solvent rises up the paper it will carry the dyes with it. Each different dye will move up the paper at different rates depending on how strongly they stick to the paper and how soluble they are in the solvent.
Paper chromatography can be used to investigate the composition of a mixture.
A baseline is drawn on the paper. The mixture is spotted onto the baseline alongside known or standard reference materials. The end of the paper is then put into a solvent which runs up the paper and through the spots, taking some or all of the dyes with it.
Different dyes will travel different heights up the paper.
The resulting pattern of dyes is called a chromatogram.
In the example shown, the mixture is shown to contain the red, blue and yellow dyes. This can be seen because these dots which resulted from the mixture have travelled the same distance up the paper as have the red, blue and yellow standard reference materials.
When analysing a chromatogram, the mixture being analysed is compared to standard reference materials by measuring how far the various dyes have travelled up the paper from the baseline where they started.
For each dye, the Rf value is calculated. To do this, 2 distances are measured:
The Rf value is calculated as follows:
If the Rf value of one of the components of the mixture equals the Rf value of one of the standard reference materials then that component is know to be that reference material.
Note that because the solvent always travels at least as far as the highest dye, the Rf value is always between 0 and 1.
Dyes which are more soluble will have higher Rf values than less soluble dyes. In other words, more soluble dyes move further up the paper. The extreme case of this is for insoluble dyes which don’t move at all (Rf value = 0). The other aspect affecting how far a dye travels is the affinity that dye has for the paper (how well it ‘sticks’ to the paper).
Common solvents are water or ethanol. The choice of solvent depends on whether most of the dyes are soluble in that solvent.
Atom: An atom is the smallest part of an element.
Atom: An atom is the smallest part of an element.
Molecule: A molecule is made of a fixed number of atoms covalently bonded together.
An atom consists of a central nucleus, composed of protons and neutrons.
This is surrounded by electrons, orbiting in shells (energy levels).
Atoms are neutral because the numbers of electrons and protons are equal.
| Mass | Charge | |
|---|---|---|
| Proton | 1 | +1 |
| Neutron | 1 | 0 |
| Electron | negligible (1/1836) | -1 |
Atomic number: The number of protons in an atom.
Mass number: The number of protons and neutrons in an atom.
Relative atomic mass (Ar): The average mass of an atom compared to 1/12th the mass of carbon-12.
Atomic number: The number of protons in an atom.
Mass number: The number of protons and neutrons in an atom.
Isotopes: Atoms of the same element (same number of protons) but with a different number of neutrons.
Relative atomic mass (Ar): The average mass of an atom compared to 1/12th the mass of carbon-12.
75% of chlorine atoms are the type 35Cl (have a mass number of 35)
25% of chlorine atoms are of the type 37Cl (have a mass number of 37)
In order to calculate the relative atomic mass (Ar) of chlorine, the following steps are used:
![\[ A_r = \frac{( (35 \times 75) + (37 \times 25) )}{100} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-0113f8af79de4c65c49213ff5f0f1314_l3.png "Rendered by QuickLaTeX.com")
![\[ A_r = 35.5 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-01950bf9b03b8605a6a2508f3e0a27ea_l3.png "Rendered by QuickLaTeX.com")
Example question:
A sample of bromine contained the two isotopes in the following proportions: bromine-79 = 50.7% and bromine-81 = 49.3%.
Calculate the relative atomic mass (Ar) of bromine.
![\[ A_r = \frac{( (79 \times 50.7) + (81 \times 49.3) )}{100} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-070b5d40d7a43c6544b0e3bdc25ecfcd_l3.png "Rendered by QuickLaTeX.com")
![\[ A_r = 79.99 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-9d467cc4bda5e7b370a0c2f489469759_l3.png "Rendered by QuickLaTeX.com")
The elements in the Periodic Table are arranged in order of increasing atomic number.
Columns are called Groups. They indicate the number of electrons in the outer shell of an atom.
Rows are called Periods. They indicate the number of shells (energy levels) in an atom.
Electrons are found in a series of shells (or energy levels) around the nucleus of an atom.
Each energy level can only hold a certain number of electrons. Low energy levels are always filled up first.
Rules for working out the arrangement (configuration) of electrons:
Example – chlorine (Cl)
1) Use the periodic table to look up the atomic number. Chlorine’s atomic number (number of protons) is 17.
2) Remember the number of protons = number of electrons. Therefore chlorine has 17 electrons.
3) Arrange the electrons in levels (shells):
Therefore the electron arrangement for chlorine (17 electrons in total) will be written as 2,8,7
4) Check to make sure that the electrons add up to the right number
The electron arrangement can also be draw in a diagram.
Electron arrangement for the first 20 elements:
Metals
Non – Metals
Metals
Non – Metals
Metals on the left of the Periodic Table.
Non-Metals on the top-right, plus Hydrogen.
Elements in the same group have the same number of electrons in their outer shell.
This is why elements from the same group have similar properties.
Elements in the same group of the periodic table have the same number of electrons in their outer shells, which means they have similar chemical properties.
The noble gases are inert (unreactive) because they have a full outer shell of electrons.
Example:
Sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).
Word equation:
sodium + water –> sodium hydroxide + hydrogen
Writing the chemical equation
A chemical equation represents what happens in terms of atoms in a chemical reaction.
Step 1: To write a chemical equation we need to know the chemical formulae of the substances.
Na + H2O –> NaOH + H2
Step 2: The next step is to balance the equation: write a large number before each compound so the number of atoms of each element on the left hand side (reactants) matches the number on the right (products). This large number is the amount of each compound or element.
During this balancing stage the actual formulas for each compound must not be changed. Only the number of each compound changes.
2Na + 2H2O –> 2NaOH + H2
If asked for an equation, the chemical equation must be given.
State symbols are used to show what physical state the reactants and products are in.
| State symbols | Physical state |
|---|---|
| (s) | Solid |
| (l) | Liquid |
| (g) | Gas |
| (aq) | Aqueous solution (dissolved in water) |
Example:
A solid piece of sodium (Na) reacts with water (H2O) to produce a solution of sodium hydroxide (NaOH) and hydrogen gas (H2).
2Na(s) + 2H2O(l) –> 2NaOH(aq) + H2(g)
Relative formula mass (Mr) is mass of a molecule or compound (on a scale compared to carbon-12).
It is calculated by adding up the relative atomic masses (Ar) of all the atoms present in the formula.
Example:
The relative formula mass (Mr) for water (H2O) is 18.
Water = H2O
Atoms present = (2 x H) + (1 x O)
Mr = (2 x 1) + (1 x 16) = 18
In Chemistry, the mole is a measure of amount of substance (n).
The abbreviation for mole is mol.
The mass of 1 mole of a substance is the relative formula mass (Mr) of the substance in grams.
Example:
The Mr of water is 18.
Therefore the mass of 1 mol of water equals 18 g.
The following formula allows for the interconversion between a mass in grams and a number of moles for a given substance:
![\[Amount(mol) = \frac{mass(g)}{M_r} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-b5d70ea918d5537fea1b3b37d62bf10a_l3.png "Rendered by QuickLaTeX.com")
Example 1:
Calculate the amount, in moles, of 8.8 g of carbon dioxide (CO2).
Step 1: Calculate the relative formula mass (Mr) of carbon dioxide (CO2).
![\[Carbon\,dioxide = CO_2 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-50233b71a0f9bf05b486880c58c17ef8_l3.png "Rendered by QuickLaTeX.com")
![\[Atoms\,present = (1 \times C) + (2 \times O) \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-5f04e4c9e47460508edd35b75ea7eeb6_l3.png "Rendered by QuickLaTeX.com")
![\[M_r = (1 \times 12) + (2 \times 16) = 44 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-8192e5b101c415aa00f3d4692be7b767_l3.png "Rendered by QuickLaTeX.com")
Step 2: Use the formula to calculate the amount in moles.
![\[Amount = \frac{mass(g)}{M_r} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-b920b64cfcdaf260756cbfb31ea117d3_l3.png "Rendered by QuickLaTeX.com")
![\[Amount = \frac{8.8}{44} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-486a6a6252e55bfe10ad925c0704b373_l3.png "Rendered by QuickLaTeX.com")
![\[Amount = 0.2 mol \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-d5557f6fc0e5727846b988bb54cfd62e_l3.png "Rendered by QuickLaTeX.com")
Example 2:
Calculate the mass of 2 mol of copper(II) sulfate (CuSO4).
Step 1: Calculate the relative formula mass (Mr) of copper(II) sulfate (CuSO4).
![\[copper(II)\,sulfate = CuSO_4 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-bec21c05b5bd9349031f68949418b2e9_l3.png "Rendered by QuickLaTeX.com")
![\[Atoms\,present = (1 \times Cu) + (1 \times S) + (4 \times O) \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-7e191cd9a58c6db9dd71df477db822c3_l3.png "Rendered by QuickLaTeX.com")
![\[M_r = (1 \times 63.5) + (1 \times 32) + (4 \times 16) = 159.5 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-9f5f48d491f801b7f180a30c45c907ee_l3.png "Rendered by QuickLaTeX.com")
Step 2: Rearrange the formula to calculate the mass.
![\[mass = amount \times M_r \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-01418823888b5b435a64ba72def6e2be_l3.png "Rendered by QuickLaTeX.com")
![\[mass = 2 \times 159.5 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-6b7cdee8f8a062b97cc0a15aa0e86475_l3.png "Rendered by QuickLaTeX.com")
![\[mass = 319g\]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-f13fbfa9d626fca93d125951d7fd8a09_l3.png "Rendered by QuickLaTeX.com")
Example: When calcium carbonate (CaCO3) is heated calcium oxide is produced. You can use reacting mass calculations to calculate the mass of calcium oxide produced when heating 25 g of calcium carbonate.
CaCO3 –> CaO + CO2
Step 1: Calculate the amount, in moles, of 25 g of calcium carbonate (CaCO3)
![\[M_r(CaCO_3)=(1 \times 40) + (1\times12) + (3 \times 16) \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-ce03f22311561a1ef86664fdf89ddcca_l3.png "Rendered by QuickLaTeX.com")
![\[M_r(CaCO_3)= 100\]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-6727c4b663acbdd68c6005c2bf76b06d_l3.png "Rendered by QuickLaTeX.com")
![\[Amount = \frac{25}{100} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-f6972a61a76a0fa4983ac76308cea61d_l3.png "Rendered by QuickLaTeX.com")
![\[Amount = 0.25 mol \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-264bbe4930deee22dfe9afc7786c937d_l3.png "Rendered by QuickLaTeX.com")
Step 2: Deduce the amount, in moles, of CaO produced from 0.25 mol of CaCO3.
This step involves using the ratio of CaCO3 to CaO from the chemical equation.
CaCO3 –> CaO + CO2
From the equation you can see that the ratio of CaCO3 to CaO is 1:1.
Therefore if you have 0.25 mol of CaCO3 this will produced 0.25 mol of CaO.
Step 3: Calculate the mass of 0.25 mol of CaO.
![\[M_r(CaO)=(1 \times 40) + (1 \times 16) \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-9ef704378ea8798987d3f4b3ad9c5afa_l3.png "Rendered by QuickLaTeX.com")
![\[M_r(CaO)= 56\]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-d9b8e678774deea58320e542933473ac_l3.png "Rendered by QuickLaTeX.com")
![\[Mass(CaO)=amount \times M_r \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-352d09e6ef6633f0116cd84cab4b56e2_l3.png "Rendered by QuickLaTeX.com")
![\[Mass(CaO)= 0.25 \times 56\]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-380e95f0cd76abdb353183566643391b_l3.png "Rendered by QuickLaTeX.com")
![\[Mass(CaO)= 14g\]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-ef2a433a285005ae9c0f1c317b9e7817_l3.png "Rendered by QuickLaTeX.com")
A simple format for laying out this method can be used.
Example: What mass of ammonia (NH3) is formed when 7 g of nitrogen (N2) is combined with hydrogen (H2).
Yield is how much product you get from a chemical reaction.
The theoretical yield is the amount of product that you would expect to get. This is calculated using reacting mass calculations.
In most chemical reactions, however, you rarely achieved your theoretical yield.
For example, in the following reaction:
CaCO3 –> CaO + CO2
You might expect to achieve a theoretical yield of 56 g of CaO from 100 g of CaCO3.
However, what if the actual yield is only 48 g of CaO.
By using the following formula, the % yield can be calculated:
![\[yield= \frac{actual\,amount\,of\,product}{theoretical\,amount\,of\,product} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-1a5e41f62300bb75bdfbdb51418ae77b_l3.png "Rendered by QuickLaTeX.com")
![\[yield= \frac{48}{56} \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-e573a8a95cc6bbec0ebbe4ad0e8991e0_l3.png "Rendered by QuickLaTeX.com")
![\[yield=0.86 \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-02f808752b18acc81c2a4d2a1663f6f3_l3.png "Rendered by QuickLaTeX.com")
![\[\% yield=86\% \]](https://www.tutormyself.com/wp-content/ql-cache/quicklatex.com-a90a5367437b5cac4a36d935fa1726ec_l3.png "Rendered by QuickLaTeX.com")
Finding the formula of a metal oxide experimentally
The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.
Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)
Finding the formula of a metal oxide experimentally
The formulae of metal oxides can be found experimentally by reacting a metal with oxygen and recording the mass changes.
Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
Method:
• Weigh a crucible and lid
• Place the magnesium ribbon in the crucible, replace the lid, and reweigh
• Calculate the mass of magnesium
(mass of crucible + lid + Magnesium – mass of crucible + lid)
• Heat the crucible with lid on until the magnesium burns
(lid prevents magnesium oxide escaping therefore ensuring accurate results)
• Lift the lid from time to time (this allows air to enter)
• Stop heating when there is no sign of further reaction
(this ensures all Mg has reacted)
• Allow to cool and reweigh
• Repeat the heating , cooling and reweigh until two consecutive masses are the same
(this ensures all Mg has reacted and therefore the results will be accurate)
• Calculate the mass of magnesium oxide formed (mass of crucible + lid + Magnesium oxide – mass of crucible + lid)
Finding the formula of a salt containing water of crystallisation
When some substances crystallise from solution, water becomes chemically bound up with the salt. This is called water of crystallisation and the salt is said to be hydrated. For example, hydrated copper sulfate has the formula CuSO4.5H2O which formula indicates that for every CuSO4 in a crystal there are five water (H2O) molecules.
When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. If the hydrated copper sulfate (CuSO4.5H2O) are strongly heated in a crucible then they will break down and the water lost, leaving behind anhydrous copper sulfate (CuSO4). The method followed is similar to that for metal oxides, as shown above. The difference of mass before and after heating is the mass of the water lost. These mass numbers can be used to obtain the formula of the salt.
The empirical formula shows the simplest whole-number ratio between atoms/ions in a compound.
The molecular formula shows the actual number of atoms of each type of element in a molecule.
For example, for ethane:
Molecular formula = C2H6
Empirical formula = CH3
Here are some more examples:
| Name | Molecular formula | Empirical Formula |
|---|---|---|
| pentene | C5H10 | CH2 |
| butene | C4H8 | CH2 |
| glucose | C6H12O6 | CH2O |
| hydrogen peroxide | H2O2 | HO |
| propane | C3H8 | C3H8 |
Notice from the table that several different molecules can have the same empirical formula, which means that it is not possible to deduce the molecular formula from the empirical formula without some additional information. Also notice that sometimes it is not possible to simplify a molecular formula into simpler whole-number ratio, in which case the empirical formula is equal to the molecular formula.
Calculating the Empirical Formula
Example: What is the empirical formula of magnesium chloride if 0.96 g of magnesium combines with 2.84 g of chlorine?
Step 1: Put the symbols for each element involved at the top of the page.
Step 2: Underneath, write down the masses of each element combining.
Step 3: Divide by their relative atomic mass (Ar).
Step 4: Divide all the numbers by the smallest of these numbers to give a whole number ratio.
Step 5: Use this to give the empirical formula.
(If your ratio is 1:1.5 then multiple each number by 2
If your ratio is 1:1.33 then x3. If your ratio is 1:1.25 x4)
Calculating the Molecular Formula
If you know the empirical formula and the relative formula mass you can work out the molecular formula of a compound.
Example: A compound has the empirical formula CH2, and its relative formula mass is 56. Calculate the molecular formula.
Step 1: Calculate the relative mass of the empirical formula.
Step 2: Find out the number of times the relative mass of the empirical formula goes into the Mr of the compound.
Step 3: This tells how many times bigger the molecule formula is compared to the empirical formula.
Empirical formula calculations involving water of crystallisation
When you heat a salt that contains water of crystallisation, the water is driven off leaving the anhydrous (without water) salt behind. For example, barium chloride crystals contain water of crystallisation, and therefore would have the formula BaCl2.nH2O where the symbol ‘n’ indicates the number of molecules of water of crystallisation. This value can be calculated using the following method.
Example: If you heated hydrated barium chloride (BaCl2.nH2O) in a crucible you might end up with the following results.
Step 1: Calculate the mass of the anhydrous barium chloride (BaCl2) and the water (H2O) driven off.
Step 2: Use the empirical formula method to find the value of n in the formula.
To determine the formula of a metal oxide by combustion.
Example: When magnesium is burned in air, it reacts with oxygen (O2) to form magnesium oxide (MgO).
2Mg + O2 –> 2MgO
Method:
It is then possible to use this data to calculate the empirical formula of the metal oxide.
To determine the formula of a metal oxide by reduction
Example: When copper (II) oxide is heated in a stream of methane, the oxygen is removed from the copper (II) oxide, producing copper, carbon dioxide and water:
4CuO + CH4 –> 4Cu + CO2 + 2H2O
By comparing the mass of copper produced with the mass of copper oxide used it is possible to determine the formula of the copper oxide.
As an alternative, hydrogen gas could be used instead of methane:
CuO + H2 –> Cu + H2O
It is then possible to use this data to calculate the empirical formula of the metal oxide.
There is an important safety point to note with both versions of this experiment. Both methane and hydrogen are explosive if ignited with oxygen. It is important that a good supply of the the gas is allowed to fill the tube before the gas it lit. This flushes out any oxygen from the tube, so the gas will only burn when it exits the tube and comes into contact with oxygen in the air.
Ions are electrically charged particles formed when atoms lose or gain electrons.
They have the same electronic structures as noble gases.
Metal atoms form positive ions (cations).
Non-metal atoms form negative ions (anions).
When given this information of the following ions, it is possible to work out the formulae of ionic compounds which include these ions.
| Name of Ion | Formula | Charge |
|---|---|---|
| Sulfate | SO42- | -2 |
| Carbonate | CO32- | -2 |
| Nitrate | NO3- | -1 |
| Hydroxide | OH- | -1 |
| Ammonium | NH4+ | +1 |
| Silver ion | Ag+ | +1 |
| Zinc ion | Zn2+ | +2 |
| Hydrogen ion | H+ | +1 |
| Copper (II) ion | Cu2+ | +2 |
| Iron (II) ion | Fe2+ | +2 |
| Iron (III) ion | Fe3+ | +3 |
| Lead (II) ion | Pb2+ | +2 |
Ion charges on the periodic table
| Name of Ion | Formula | Charge |
|---|---|---|
| Sulfate | SO42- | -2 |
| Carbonate | CO32- | -2 |
| Nitrate | NO3- | -1 |
| Hydroxide | OH- | -1 |
| Ammonium | NH4+ | +1 |
| Silver ion | Ag+ | +1 |
| Zinc ion | Zn2+ | +2 |
| Hydrogen ion | H+ | +1 |
| Copper (II) ion | Cu2+ | +2 |
| Iron (II) ion | Fe2+ | +2 |
| Iron (III) ion | Fe3+ | +3 |
| Lead (II) ion | Pb2+ | +2 |
Ion charges on the periodic table
Sodium chloride, NaCl
Calcium oxide, CaO
Ionic bonding: a strong electrostatic attraction between oppositely charged ions.
Ionic compounds have high melting and boiling points because they have a giant structure with strong electrostatic forces between oppositely charged ions that require a lot of energy to break.
Giant 3D lattice of sodium chloride (NaCl)
Ionic compounds do not conduct electricity when solid.
However, ionic compounds do conduct electricity if molten or in solution.
A covalent bond is formed between two non-metal atoms by sharing a pair of electrons in order to fill the outer shell.
Covalent bonding: a strong attraction between a shared pair of electrons and two nuclei.
Carbon dioxide (CO2) has a simple molecular structure. This just means that it is made up of molecules.
Within each molecule are atoms bonded to each other covalently. These covalent bonds inside the molecules are strong.
However, between the molecules are weak forces of attraction that require little energy to break. These forces are not covalent bonds. This is why simple molecular substances such as carbon dioxide have a low boiling point.
So when carbon dioxide changes from a solid to a gas, for example, that process can be represented as:
CO₂ (s) → CO₂ (g)
Notice that even though there has been a dramatic change of state from solid to gas, the substance before and after the change is always made up of carbon dioxide molecules. During the change of the state the covalent bonds within each molecule remain unbroken. Instead it is the weak forces of attraction between the molecules which have been overcome.
Larger molecules tend to have higher boiling points.
This is because larger molecules (molecules with more mass) have more forces of attraction between them. These forces, although weak, must be overcome if the substance is to boil, and larger molecules have more attractions which must be overcome.
Diamond has a high melting point because it is a giant covalent structure with many strong covalent bonds that require a lot of energy to break.
Allotropes are different forms of the same element. Three different allotropes of carbon are shown here as examples: diamond, graphite and C60 fullerene.
Diamond is made up of only carbon atoms, in a giant 3D lattice, where each of those atoms has a strong covalent bonds to 4 other carbon. Every one of carbon’s 4 outer electrons is involved in one of these strong covalent bonds.
Diamond is extremely hard because it is a giant covalent structure with many strong covalent bonds.
Because it is hard, diamond is used in high speed cutting tools, eg diamond-tipped saws.
Graphite is also made of only carbon atoms, and is also a giant structure, but it is formed of layers where each carbon atom has a strong covalent bond to 3 other carbons. This means each carbon atom has one electron not involved in a covalent bond, and these electrons form a sea of delocalised electrons between the layers.
Even though it is a non-metal, graphite can conduct electricity because the delocalised electrons are free to move.
Each layer is a giant structure, with weak forces of attraction between the layers. These layers can easily slide over each other.
Graphite is soft and slippery because it has weak forces of attraction between layers. It is used as a lubricant and in pencils because it is soft and slippery.
C60 fullerene which is a simple molecular structure (also known as a buckyball) is also made of only carbon atoms, but it forms molecules of 60 carbon atoms. The molecule has weak intermolecular forces of attraction between them which take little energy to overcome. Hence C60 fullerene has a low melting point, and it is soft.
C60 fullerene cannot conduct electricity. Although in each molecule every carbon is only covalently bonded to 3 others and the other electrons are delocalised, these electrons cannot jump between different molecules.
