Edexcel GCSE 9-1 Chemistry: SC9 - Calculations Involving Masses

Paper 1 + 2 Specification: https://qualifications.pearson.com/content/dam/pdf/GCSE/Science/2016/Specification/GCSE_Chemistry_Spec.pdf

C1.43 Calculate relative formula mass given relative atomic masses.

The relative formula mass of a substance is the sum of the relative atomic masses of all the atoms or ions in its formula. (Mr)
e.g. CO2
Mr of CO2 = 12 + (2 x 16)
= 44

C1.44 Calculate the formulae of simple compounds from reacting masses and understand that these are empirical formulae.

C : H : O (symbol for each element)
2.0g : 0.33g : 2.67g (mass in g)
2/12 : 2/0.33 : 2.67/16 (divide mass by rel. mass)
0.16 : 0.33 : 0.148125
0.16/0.148125 : 0.33/0.148125 : 0.148125/0.148125
1 : 2: 1 (divide each answer by the smallest answer)
COH2

Empirical Formula

The simplest whole number ratio of elements present in a substance

C1.45 Deduce:
a the empirical formula of a compound from the formula of its molecule

a) Molecular formula of propane: C3H6
Empirical formula of propane: CH2
- the simplest

C1.45 Deduce:
b the molecular formula of a compound from its empirical formula and its relative molecular mass.

b) Empirical formula for glucose is CH2O and its relative formula mass is 180.
- Find the empirical formula mass
12 + (2 x 1) + 16 = 30
-divide Mr by empirical formula.
180/30 = 6
The molecular mass is six times the empirical formula, so the molecular formula is C6H12O6

C1.46 Describe an experiment to determine the empirical formula of a simple compound such as magnesium oxide.

Magnesium oxide can be made by heating magnesium ribbon in a limited oxygen supply. If the reactant and product are 'weighed', the empirical formula for magnesium oxide can be calculated.

Law of conservation of mass

Law of conservation of mass - the total mass of reactants and products stays constant during a chemical reaction

C1.47 Explain the law of conservation of mass applied to:
a a closed system including a precipitation reaction in a closed flask

A closed system is a situation in which no substances can enter or leave during a reaction - sealed container (flask with bung) / precipitation reaction.
In a precipitation reaction, two soluble reactants form an insoluble product - precipitate. Total mass of the closed system stays constant during the formation of precipitate

C1.47 Explain the law of conservation of mass applied to:
b a non-enclosed system including a reaction in an open flask that takes in or gives out a gas.

A non-enclosed system is a system in which substances can enter or leave during a reaction.
- reactions in an open flask, where substances in the gas state may enter or leave.
-Mass is conserved but, the mass of reactive metal increases if heated in air - oxygen atoms combine with metals, form metal oxides.
- the mass of a reactive non-metal or a fuel decreases if it is heated in air - products in gas state will escape
-Mass of metal carbonate decrease if it is heated - carbon dioxide will escape.

C1.48 Calculate masses of reactants and products from balanced equations, given the mass of one substance

1 - write balanced equation
2 - calculate relative formula masses of the substances needed.
3 - calculate ratio of masses (multiply Mr values by the balancing numbers shown in equation)
4 - Work out mass for 1 g of reactant or product.
5 - Scale up or down (from 1 g to the mass you are given)

C1.49 Calculate the concentration of solutions in g dm-3.

concentration =
mass of solute in g/ volume of solution dm^3

C1.50 H Recall that one mole of particles of a substance is defined as:
a the Avogadro constant number of particles (6.02 × 10^23 atoms, molecules, formulae or ions) of that substance

one mole of particles of a substance is defined as the Avogadro constant number of particles
(6.02 × 10^23 atoms, molecules, formulae or ions) of that substance

C1.50 H Recall that one mole of particles of a substance is defined as:
b a mass of 'relative particle mass' g.

one mole of particles of a substance is defined as a mass of 'relative particle mass' g.

C1.51 H Calculate the number of:
a moles of particles of a substance in a given mass of that substance and vice versa
b particles of a substance in a given number of moles of that substance and vice versa
c particles of a substance in a given mass of that substance and vice versa.

mass g = moles / Ar or Mr
moles = mass g / Ar or Mr
Ar or Mr = mass g / moles

C1.52 H Explain why, in a reaction, the mass of product formed is controlled by the mass of the reactant which is not in excess.

The mass of a product formed is controlled by the mass of the reactant that is not in excess. The reaction continues until all the particles of the limiting reactant have been used up .
The mass of product is directly proportional to the mass of the limiting reactant.

stoichiometry

the ratio of reactants to products

C1.53 H Deduce the stoichiometry of a reaction from the masses of the reactants and products.

1 calculate number of moles (mass/Ar or Mr)
2 divide by the smaller answer
3 simplest whole number ratio

The Avogadro constant

is the number of particles/atoms/ions/molecules in 1 mole.
6.02x10^23