Topic 1: Formulae, Equations and Amounts of Substance

Question 1

Key terms: Atom, Element, Ion, Molecule, Compound, Empirical formula and Molecular formula

Answer 1

-atom: is the smallest part of an element that can exist, all substances are made up of atoms
-element: a substance made up of only one type of atom, all with the same number of protons
-ion: is what’s formed when an atom either gains or loses electrons, this gives it an overall charge + a positive if it has lost at east one electron and a negative charge if it has gained at least one electron
-molecule: consists of two or more atoms that have been bonded together chemically
-compound: is a substance that combines two or more different elements through the formation of chemical bonds
-empirical formula: smallest whole number ratio of atoms of each element in a compound
-molecular formula: actual number of atoms of each element in a compound

Question 2

Mole and the Avogrado Constant:

Answer 2

-> the mole is a unit of measurement for substances, it always contains the same number of particles
L = 6.02 x 10^23 particles
-this number is the Avogrado Constant (L) and allows the number of particles present in a sample of a substance with known mass to be found:
Number of particles = n (number of moles) x L (Avogrado Constant)
-the mole is a very important unit of measurement in many calculations:
Moles = mass/Mr = concentration (dm^3) x volume (cm^3) /1000

Question 3

Balanced Equations:

Answer 3

-full or ionic chemical equations must be balanced before they can be used in calculations
-this is because the reaction ratios must be correct
-for a chemical equation to be balanced, it must have the same number and type of each atom present on both sides of the equation
-it can be useful to also include state symbols so it is clear what might be observed during the reaction, for example, bubble of gas, a precipitate forming, or a colour change that may infer a displacement reaction
-there are four state symbols:
-(s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water)
-these balanced equations can then be used to calculate reacting masses, percentage yield and atom economy

Question 4

Relative Atomic Mass:

Answer 4

-> the mean mass of an atom of an element, divided by 1/12 of the mass of an atom of the carbon-12 isotope

Question 5

Relative Molecular Mass and Relative Formula Mass:

Answer 5

-> the mean mass of a molecule of a compound, divided by 1/12 of the mass of an atom of the carbon-12 isotope
-the molecular mass can be calculated by adding together the atomic masses (Ar) of all the atoms in that compound
-example: To calculate the Mr of the compound C2H5OH the Ar’s must be used:
C = 12, O = 16, H = 1
C2H5OH = (2x12) + (6x1) + (16x1) Mr = 46
-relative formula mass refers to compounds that have a giant structure

Question 6

Empirical and Molecular formula:

Answer 6

True Mr = Mr of empirical formula x multiplier
-example: The empirical formula of a molecule containing 5 atoms of oxygen for every 2 atoms of phosphorus has an Mr of 284. What is its molecular formula?
Empirical formula = P2O5
Mr of empirical formula = (31x2) + (16x5) = 142
Multiplier = 284 divided by 142 = 2
Molecular formula = 2(P2O5) = P4O10

Question 7

Molar Mass:

Answer 7

-> the molar mass of a substance is its mass in grams per mole and has the units g mol^-1. It can be calculated using the following equation:
Molas mass = mass/number of moles

Question 8

Parts Per Million (ppm):

Answer 8

-concentration can be given in parts per million (ppm)
-this gives the units of mass of that particular species within 1,000,000 total units of mass
-it is most commonly used to represent the concentrations of gases

Question 9

Concentration calculations:

Answer 9

-the concentration of a solution can be measured in mol dm^-3 and g dm^-3 which can be calculated using the following equations:
-Concentration mol dm^-3 = Number of moles (mol) / volume (dm^3)
-Concentration (g dm^3) = Mass (g) / volume (dm^3)

Question 10

Experimental Data:

Answer 10

-can be used to work out empirical and molecular formulas and reaction stoichiometries
-these calculations require the use of equations given in this section, along with some others, these include:
-Mol = concentration x volume
-True Mr = Mr of empirical formula x multiplier
-Volume of gas (dm^3) = 24 x number of moles (at room temperature and pressure)
-Number of particles = n x L
-Mass = Mr x mol

Question 11

Volume of gases: Molar volume of gases

Answer 11

-one mole of any gas at room temperature and pressure will take up the same volume, regardless of its composition
-this volume is 24,000 cm^3, or 24 dm^3, and is known as the molar volume of gases
-this relationship gives the following equation that can be used to work out the volume of a gas if its amount (number of moles) is known and vice versa
Volume of gas (dm^3) = 24 x number of moles (at room temperature and pressure)

Question 12

The Ideal Gas Law:

Answer 12

-when under standard conditions, gases and volatile liquids follow certain trends:
-pressure is proportional to temperature
-volume is proportional to temperature
-pressure and volume are inversely proportional
-these relationships can be combined to give the ideal gas equation:
pV = nRT = mRT/Mr
-in order to use this equation, the variables must be in the correct standard units:
p = pressure in pascals
V = volume in m^3
T = temperature in Kelvin
n = moles
m = mass in grams
R is the ideal gas constant, equal to 8.31 JK^-1 mol^-1

Question 13

Percentage Yield:

Answer 13

-> indicates how much of the maximum amount of product you obtained during an experiment
-a low percentage yield could indicate an incomplete reaction, or the loss of product during purification
% yield = experimental mass/theoretical mass x 100

Question 14

Atom Economy:

Answer 14

-> is a measure of efficiency since it measures the proportion of reactant atoms which are converted into the desired product
% atom economy = Mr of desired product/total Mr of all products x 100
-in industrial chemical processes, it is desirable to have a high atom economy for a reaction
-this means there is little or no waste product, only the desired product
-therefore it means the process is more economically viable for industrial-scale manufacture

Question 15

Displacement reactions:

Answer 15

Br2 (aq) + 2Kl (aq) -> I2 (aq) + 2KBr (aq)
-in this reaction, the more reactive bromine displaces the less reactive iodide in potassium iodide
-this can also be seen in the ionic equation for the reaction
Br2 (aq) + 2I^- (aq) -> I2 (aq) + 2Br^-(aq)

Question 16

Neutralisation reactions:

Answer 16

-these can be identified by the presence of reactant acids and bases as well as the formation of a neutral salt solution and water (and sometimes other compounds such as carbon dioxide)
HCl (aq) + NaOH3 (aq) -> NaCl (aq) + H2O (l)
Na2CO3(aq) + 2HNO3 (aq) -> 2NaOH3 (aq) + H2O (l) + CO2 (g)
-the ionic equations can more clearly demonstrate the neutralisation of an acid and a base:
H^+ (aq) + OH^- (aq) -> H2O (l)
2H+ (aq) + CO3 ^2- (aq) -> H2O (l) + CO2 (g)

Question 17

Precipitation Reactions:

Answer 17

-there are shown by the reaction of two aqueous solutions to form products which include one solid
BaCl2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaCl (aq)
-the ionic equation shows the precipitation reactions more clearly as there are no other products considered
Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s)