Topic 1: Stoichiometric Relationships

Question 1

Melting

Answer 1

Solid to Liquid, Endothermic (absorbs energy)

Question 2

Vaporisation ( Boiling or Evaporation)

Answer 2

Liquid to gas, Endothermic (absorbs energy)

Question 3

Sublimation

Answer 3

Solid to Gas (No inbetween state), Endothermic (absorbs energy)

Question 4

Condensation

Answer 4

Gas to liquid, Exothermic (releases energy)

Question 5

Freezing

Answer 5

Liquid to solid, Exothermic (releases energy)

Question 6

Deposition

Answer 6

Gas to solid ( No inbetween state), Exothermic (releases energy)

Question 7

Elements:

Answer 7

Single substances composed of atoms of the same type.

Question 8

Compounds:

Answer 8

Substances composed of two or mole different elements bonded together. They may have different properties to the original.

Question 9

Molecules:

Answer 9

Two or more atoms bonded together.

Question 10

The ratio of elements in a compound

Answer 10

Find the ratio of elements in a compound by balancing the charges. Do this using the periodic table.

E.g Mg has charge +2 and Cl has charge -1, so the compound needs one Mg to two Cl to balance the charges. MgCl2

Question 11

Homogeneous mixture

Answer 11

The mixture is mixed such that the elements/compounds are distributed equally throughout.

Question 12

Heterogeneous mixture

Answer 12

The mixture is not uniform in composition

Question 13

To deduce the chemical equation:

Answer 13

1. Use the data booklet to convert the reactants and products to their corresponding symbols
2. Set up the equation with the reaction arrow and symbols.
3. Apply correct state symbols (s), (l), (g) and (aq)
4. Equate coefficients so that the same number of elements are either side.

Question 14

Relative Atomic mass (Ar)

Answer 14

The mass of an element relative to carbon-12, where Carbon-12 is defined to have a mass of 12.

Question 15

Molar mass (Mr)

Answer 15

This is calculated by summing the atomic masses of each element in a compound. The unit for molar mass is g mol-1

Question 16

What are moles?

Answer 16

Moles “n” are used to count particles because particles are so unbelievably small that a huge number of them can fit in a beaker. One mole of a something is simply 6.02 x10^23 times of it.

Question 17

Avogadro’s number:

Answer 17

The number of particles which equal one mole - 6.02 x 10^23

Question 18

To convert from moles to particles

Answer 18

multiply the number of moles by avogadro’s number, When going the other way divide instead of multiplying

Question 19

To convert from moles to grams

Answer 19

multiply the number of moles by the molar mass(Mr), When going the other way divide instead of multiplying

Question 20

Mole ratio

Answer 20

The mole ratio is the ratio between the moles of different compounds in a chemical formula.

You can find the mole ratio by looking at the coefficients in front of the compounds in the formula.
Example:
2H2+O2→2H2O
The mole ratio between the oxygen and water is 1:2, as for every one mol of oxygen reacted, 2 moles of water is produced.
It is a 1:1 mole ratio between Hydrogen and water, as they have the same coefficients.

Question 21

Empirical formula

Answer 21

This is the simplest whole-number ratio of elements in a compound. It is important that the ratio is in its simplest form. If the ratio can be simplified it must. E.g H4O2 is not an empirical formula because it can be simplified to H2O

Question 22

Molecular formula

Answer 22

This tells the number of moles of each element. It is the same as the empirical formula multiplied by some whole number.

Question 23

Finding the empirical formula given percentage composition

Answer 23

1. Assume that the compound has a mass of 100g.
2. Calculate the mass of each element by multiplying the percentage by 100g
3. Divide each mass by the atomic mass to find the moles of each element.
4. The simplest integer ratio between the moles of the element is also the ratio in the empirical formula

Question 24

Find the Molecular formula given mass and the empirical formula

Answer 24

1. With the empirical formula find the molar mass of the compound, with the simplified ratio.
2. Divide the mass given by the molar mass of the empirical formula, to get a whole number.
3. Multiply the empirical formula by this whole number to get the molecular formula.

Question 25

Limiting reactant:

Answer 25

This is the reactant that determines the theoretical yield.

Question 26

Excess reactants

Answer 26

All reactants which aren’t the limiting reactant are in excess, so they don’t affect the yield.

Question 27

Determine the limiting reactant

Answer 27

If given the chemical equation and the masses do this:

1. Convert the masses to moles
2. Divide the moles by the mole ratio coefficients
3. The reactant with the smaller number is the limiting reactant

Question 28

Experimental and theoretical yield

Answer 28

The yield gained in an experiment is often lower in an experiment due to errors with the method. The percentage yield can tell us how much of the limiting reactant actually reacted.

Question 29

Theoretical yield:

Answer 29

The yield expected theoretically, assuming all the limiting reactant is converted into products.

Question 30

Experimental yield:

Answer 30

The yield in an experiment

Question 31

Percentage yield:

Answer 31

Experimental yield divided by theoretical yield x 100

Question 32

What is an ideal gas

Answer 32

This is a gas that exhibits the five postulates of the kinetic molecular theory, as well as obeying the gas laws.
Properties of an ideal gas:
*The particles in a gas exhibit random motion
*There are no intermolecular force between particles
*All collisions are perfectly elastic
*Particles have negligible volume
*The kinetic energy is directly proportional to the temperature

Question 33

Combined gas law:

Answer 33

P1xV2 over T1 = P2xV2 over T2

Question 34

Ideal gas equation:

Answer 34

PV = nRT ( By rearranging the ideal gas equation you can find the relationship between any two components)

Question 35

Avogadro’s Law

Answer 35

States that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. This is true because a gas is mostly empty space so the the nature of the particles themselves does not affect the number of particles which are in a gas. This law is important because it means any same volume of gas also has the same number of molecules and hence the same number of moles.

Question 36

Molar volume:

Answer 36

22.7 dm3 mol-1 - This is the amount of volume one mole of idea gas takes up under STP

Question 37

Moles of gas:

Answer 37

Volume over Molar volume= V/22.7

Question 38

Concentration:

Answer 38

Moles of solute over the volume of solution (measured in mol dm-3)

Question 39

Acid-Base Titration

Answer 39

A Titration is a method of finding the concentration of an unknown solution using a solution with a known concentration.

Question 40

Titrant:

Answer 40

The solution with the known concentration in buret, can be an acid or a base

Question 41

Analyte:

Answer 41

The solution with the unknown concentration in conical flask, can be an acid or a base.