1. Atoms and reactions
Revision for resit F321

2. Aims Atomic structure Isotopes and relative masses The mole
Calculations using the mole
Acids and bases
Reactions of acids and bases

3. 1 +1 1 -1 THE STRUCTURE OF ATOMS
Atoms consist of a number of fundamental particles,
the most important are ...
Mass / kg
Charge / C
Relative mass
Relative charge
PROTON
NEUTRON
ELECTRON 1 +1
1.672 x 10-27
1.602 x 10-19 1
1.675 x 10-27 1
1836 -1
9.109 x 10-31
1.602 x 10-19
Calculate the mass of a carbon-12 atom; it has 6 protons, 6 neutrons and 6 electrons
6 x x x x x x = x kg

4. MASS NUMBER AND ATOMIC NUMBER
Atomic Number (Z) Number of protons in the nucleus of an atom
Mass Number (A) Sum of the protons and neutrons in the nucleus
Mass Number (A)
PROTONS + NEUTRONS Na 23 11
Atomic Number (Z)
PROTONS

5. Relative Atomic Mass (Ar)
RELATIVE MASSES
Relative Atomic Mass (Ar)
The mass of an atom relative to the 12C isotope having a value of
Ar = average mass per atom of an element x 12
mass of one atom of carbon-12
Relative Isotopic Mass
Similar, but uses the mass of an isotope 238U
Relative Molecular Mass (Mr)
Similar, but uses the mass of a molecule CO2, N2
Relative Formula Mass
Used for any formula of a species or ion NaCl, OH¯

6. Atomic structure summary
The nucleus contains protons (positively charged) and neutrons (neutrally charged i.e. no charge).
The atomic number (proton number) is equal to the number of protons in the atom’s nucleus.
The mass number is the total number of protons and neutrons in the nucleus.
Ions do not have the same number of electrons as protons, and so have an overall charge.

7. Isotopes and relative masses
Isotopes are atoms having the same number of protons but different numbers of neutrons.
The relative atomic mass is the weighted mean mass of an atom relative to 12C, so that carbon is exactly 12 on this scale.
The average relative atomic mass is equal to the sum of each isotope’s mass for an element x its relative abundance.
The relative formula mass of a compound is equal to the sum of the individual relative atomic masses.

8. Definitions so far: Isotopes Atomic number Mass number Ion
Relative isotopic mass
Relative atomic mass
Relative molecular mass
Relative formula mass

9. ion
A positively or negatively charge atom or (covalently bonded) group of atoms (a molecular ion).
isotopes
Atoms of the same element with different numbers of neutrons and different masses.
atomic (proton) number
The number of protons in the nucleus of an atom.
mass (nucleon) number
The number of particles (protons and neutrons) in the nucleus.
relative atomic mass, Ar
The weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12.
relative formula mass
The weighted mean mass of a formula unit compared with one-twelfth of the mass of an atom of carbon-12.
relative isotopic mass
The mass of an atom of an isotope compared with one-twelfth of the mass of an atom of carbon-12.
relative molecular mass, Mr
The weighted mean mass of a molecule compared with one-twelfth of the mass of an atom of carbon-12.

10. THE MOLE – AN OVERVIEW
WHAT IS IT? The standard unit of amount of a substance -
just as the standard unit of length is a METRE
It is just a number, a very big number
It is also a way of saying a number in words
like DOZEN for 12
GROSS for 144
HOW BIG IS IT ? (approx) - THAT’S BIG !!!
It is a lot easier to write it as x 1023
And anyway it doesn’t matter what the number is
as long as everybody sticks to the same value !
WHY USE IT ? Atoms and molecules don’t weigh much so it is
easier to count large numbers of them.
In fact it is easier to weigh substances.
Using moles tells you :- how many particles you get in a certain mass
the mass of a certain number of particles

11. Empirical formula
Analysis showed that g of Mg combines with 3.995g of bromine to form a compound. Find the empirical formula: Ar Mg 24.3 Br 79.9 Mg: Br : : :2 MgBr2

12. Molecular formula
A compound has an empirical formula of CH2 and a relative molecular mass, Mr of What is its molecular formula?
Empirical formula mass of CH2: = 12 + (1x2) = 14.0
Number of CH2 units in the molecule: =4
Molecular formula: (4xCH2) = C4H8

13. The mole summary
A mole is the S.I. unit for amount of substance and has units of mol.
One mole of a substance is simply the relative formula mass for a compound, or relative atomic mass for an element in grams.
The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound.
The molecular formula is the actual number of atoms of each element in a molecule.

14. THE MOLE molar mass g mol-1 or kg mol-1 MOLES = MASS MOLAR MASS
CALCULATING THE NUMBER OF MOLES OF A SINGLE SUBSTANCE
moles = mass / molar mass
mass = moles x molar mass
molar mass = mass / moles
UNITS
mass g or kg
molar mass g mol-1 or kg mol-1
MOLES = MASS
MOLAR MASS
MASS
MOLES x MOLAR
COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED

15. MOLES OF A SINGLE SUBSTANCE
1. Calculate the number of moles of oxygen molecules in 4g
oxygen molecules have the formula O2
relative mass will be 2 x 16 = 32 ; molar mass will be 32g mol-1
moles = mass = g = mol
molar mass g mol -1
What is the mass of 0.25 mol of Na2CO3 ?
Relative Molecular Mass of Na2CO3 = (2x23) (3x16) = 106
Molar mass of Na2CO = 106g mol-1
mass = moles x molar mass = x = g

16. REACTING MASS CALCULATIONS
CaCO HCl ———> CaCl CO H2O
1. What is the relative formula mass of CaCO3? (3 x 16) = 100
2. What is the mass of 1 mole of CaCO g
3. How many moles of HCl react with 1 mole of CaCO3? 2 moles
4. What is the relative formula mass of HCl? =
5. What is the mass of 1 mole of HCl? g
6. What mass of HCl will react with 1 mole of CaCO3 ? 2 x 36.5g = 73g
7. What mass of CO2 is produced ? moles of CO2 = moles of CaCO3
moles of CO2 = moles
mass of CO2 = x 44 = g

17. CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
UNITS concentration mol dm-3
volume dm3
BUT IF... concentration mol dm-3
volume cm3
MOLES = CONCENTRATION x VOLUME
MOLES
CONC x VOLUME
COVER UP THE VALUE YOU WANT AND THE METHOD OF CALCULATION IS REVEALED
MOLES = CONCENTRATION (mol dm-3) x VOLUME (dm3)
MOLES = CONCENTRATION (mol dm-3) x VOLUME (cm3)
1000

18. CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
THE MOLE
CALCULATING THE NUMBER OF MOLES OF SOLUTE IN A SOLUTION
concentration of solution in the graduated flask = mol dm-3
volume pipetted out into the conical flask = cm3
25cm3
250cm3
250cm3
The original solution has a concentration of mol dm-3
This means that there are mols of solute in every 1 dm3 (1000 cm3) of solution
Take out cm3 and you will take a fraction 25/1000 or 1/40 of the number of moles
moles in 1dm3 (1000cm3) =
moles in 1cm3 = /1000
moles in 25cm3 = x / = x 10-3 mol

19. MOLE OF SOLUTE IN A SOLUTION MOLES = CONCENTRATION x VOLUME
1 Calculate the moles of sodium hydroxide in 25cm3 of 2M NaOH
moles = conc x volume in cm3
1000
= 2 mol dm-3 x 25cm3 = moles
2 What volume of 0.1M H2SO4 contains moles ?
volume = x moles (re-arrangement of above)
(in cm3) conc
= x = 20 cm3
0.1 mol dm-3

20. STANDARD SOLUTION
‘ONE WHOSE CONCENTRATION IS KNOWN ACCURATELY’
4.240g of Na2CO3 was placed in a clean The solution was transferred
beaker and dissolved in de-ionised water quantitatively to a 250 cm3
graduated flask and made up
to the mark with de-ionised
(or distilled) water.
What is the concentration of the solution in mol dm-3 ?
mass of Na2CO3 in a 250cm3 solution = g
molar mass of Na2CO3 = 106g mol -1
no. of moles in a 250cm3 solution = g / 106g mol = mol
Concentration is normally expressed as moles per dm3 of solution
Therefore, as it is in 250cm3, the value is scaled up by a factor of 4
no. of moles in 1000cm3 (1dm3) = 4 x = mol ANS mol dm-3

21. How to work out how much to weigh out
STANDARD SOLUTION
How to work out how much to weigh out
A chemist needs to make up a 250cm3 standard solution of 0.100M sodium carbonate from anhydrous sodium carbonate. How much will they need to weigh out?
What concentration is the solution to be? = mol dm-3
How many moles will be in 1 dm3 ? = mol
How many moles will be in 250cm3 ? = /4 = mol
What is the formula of anhydrous sodium carbonate? = Na2CO3
What is the relative formula mass? = 106
What is the molar mass? = 106g mol -1
What mass of Na2CO3 is in moles = x = g
of Na2CO3 ? (mass = moles x molar mass)
ANS. The chemist will have to weigh out 2.650g, dissolve it in water
and then make the solution up to 250cm3 in a graduated flask.

22. VOLUMETRIC CALCULATIONS
Care must be taken when dealing with reactions that do not have a 1:1 molar ratio. If you don’t understand what an equation tells you, it is easy to make a mistake.
2NaOH H2SO4 ——> Na2SO H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
i.e moles of NaOH = x moles of H2SO4
or moles of H2SO4 = moles of NaOH 2
REMEMBER... IT IS NOT A MATHEMATICAL EQUATION
2 moles of NaOH DO NOT EQUAL 1 mole of H2SO4
More examples follow

23. VOLUMETRIC CALCULATIONS
Calculate the volume of sodium hydroxide (concentration mol dm-3)
required to neutralise 20cm3 of sulphuric acid of concentration mol dm-3.
2NaOH H2SO4 ——> Na2SO H2O
you need 2 moles of NaOH to react with every 1 mole of H2SO4
therefore moles of NaOH = 2 x moles of H2SO4
moles of H2SO4 = x 20/ (i)
moles of NaOH = x V/ (ii) where V is the volume of alkali in cm3
substitute numbers moles of NaOH = x moles of H2SO4
x V/ = x x 20/1000
cancel the 1000’s x V = x x 20
re-arrange Volume of NaOH (V) = x x = cm3
0.100

24. MOLAR VOLUME stp = standard temperature and pressure (298K and 105 Pa)
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at stp
1. Calculate the volume occupied by 0.25 mols of carbon dioxide at stp
1 mol of carbon dioxide will occupy a volume of 24 dm3 at stp
0.25 mol of carbon dioxide will occupy a volume of 24 x 0.25 dm3 at stp
0.25 mol of carbon dioxide will occupy a volume of 6 dm3 at stp
2. Calculate the volume occupied by 0.08g of methane (CH4) at stp
Relative Molecular Mass of CH4 = (4x1) = 16
Molar Mass of CH4 = 16g mol-1
Moles = mass/molar mass 0.08g / 16g mol = mols
1 mol of methane will occupy a volume of 24 dm3 at stp
0.005 mol of carbon dioxide will occupy a volume of 24 x dm3 at stp
0.005 mol of carbon dioxide will occupy a volume of dm3 at stp
stp = standard temperature and pressure (298K and 105 Pa)
ONE MOLE OF ANY GAS OR VAPOUR OCCUPIES 24dm3 at room temperature and pressure

25. Calculations using the mole summary
Mass calculations: calculate the amount of substance and then use the chemical equation to deduce the moles of required substance.
Gas calculations: 1 mol of any gas occupies 24  000 cm3 or 24 dm3 at room temperature.
Solution calculations: the amount of substance dissolved is equal to the concentration x the volume of solution (in dm3).
A dilute solution consists of a small amount of dissolved solute. A concentrated solution consists of a large amount of solute

26. Some more definitions Amount of substance Avogadro constant The mole
Molar mass
Empirical formula
Molecule
Molecular formula
Molar volume
Concentration
Standard solution
Stoichiometry

27. amount of substance
The quantity whose unit of the mole. Chemists use ‘amount of substance’ as a means of counting atoms.
Avogadro constant, NA
The number of atoms per mole of the carbon-12 isotope (6.02 × 1023 mol–1).
mole
The amount of any substance containing as many particles as there are carbon atoms in exactly 12 g of the carbon-12 isotope.
molar mass, M
The mass mole of a substance. The units of molar mass are g mol–1.
molar volume
The volume per mole of a gas. The units of molar volume are dm3 mol–1. At room temperature and pressure the molar volume is approximately 24.0 dm3 mol–1.
empirical formula
The simplest whole-number ratio of atoms of each element present in a compound.
molecular formula
The number of atoms of each element in a molecule.
molecule
A small group of atoms held together by covalent bonds.

28. concentration
The amount of solute, in mol, per 1 dm3 (1000 cm3) of solution.
standard solution
A solution of known concentration. Standard solutions are normally used in titrations to determine unknown information about another substance.
stoichiometry
The molar relationship between the relative quantities of substances taking part in a reaction.

29. Acids and Bases 3 common acids: Sulfuric acid H2SO4
Hydrochloric acid HCl
Nitric acid HNO3
Acids are proton donors
Weaker acids:
Ethanoic acid CH3COOH
Methanoic acid HCOOH
Citric acid C6H8O7
All acids contain H+ ions this is the active ingredient in acids and it is responsible for acid reactions. They have a pH of less than 7

30. Bases and Alkalis H+(aq) + OH-(aq)  H2O(l)
A base is a proton acceptor. An alkali is a base that dissolves is water forming OH- ions
Common Alkalis
Sodium hydroxide NaOH
Potassium hydroxide KOH
Ammonia NH3
Common bases:
Metal oxides: MgO, CuO
Metal hydroxides: NaOH, Mg(OH)2
Ammonia: NH3
H+(aq) + OH-(aq)  H2O(l)

31. Acid base summary
An acid is a hydrogen ion (H+) or proton donor in solution, whereas a base is a hydrogen ion or proton acceptor in solution.
Hydrochloric acid (HCl), sulfuric acid (H2SO4) and nitric acid (HNO3) are common acids.
Bases include metal oxides (e.g. MgO), metal hydroxides (e.g. NaOH) and ammonia (NH3).
Alkalis are soluble bases and form hydroxide ions, OH-, in solution.

32. Acid base reactions Acid + carbonate  salt + CO2 + H2O
Acid + base  salt + water
Acid + alkali  salt + water
Metal + acid  salt + hydrogen
You must be able to write balanced equations for all of these reactions.

33. Water of crystallisation
Or how much water is in a compound How would you calculate the water of crystallisation? Mass of hydrated salt Mass of anhydrous salt Mass of water that was in the Hydrated salt

34. Determine the formula of hydrated magnesium sulfate
Mass of hydrated salt: 4.312g Mass of anhydrous salt: 2.107g So mass of H2O in MgSO4.xH2O = Calculate the amount in mol of anhydrous MgSO4 MgSO4 = (16.0 x 4) = 120.4gmol-1 n(MgSO4) = = mol Calculate the amount in mol of water n(H2O) = = MgSO4 : H2O : (divide by the smallest number) 1: 7 MgSO4. 7 H2O

35. Acid base reactions summary
Salts are formed when a hydrogen ion from the acid is replaced by a metal ion, or an ammonium ion.
Acids react with bases to form a salt and water only; they react with metal carbonates to form a salt, water and carbon dioxide gas.
Metals react with acids to form a salt and hydrogen gas.
Salts may chemically combine with water as water of crystallisation in hydrated salts. (Without water in anhydrous salts.)

36. Final definitions
Salt
Hydrated
Anhydrous
Water of crystallisation

37. Salt
A chemical compound formed from an acid, when a H+ ion from the acid has been replaced by a metal ion or another positive ion, such as the ammonium ion, NH4+.
hydrated
Crystalline and containing water molecules.
anhydrous
A substance that contains no water molecules.
water of crystallisation
Water molecules that form an essential part of the crystalline structure of a compound.