1.8 Calculations involving masses Flashcards
Relative formula mass
Calculating relative formula mass
Relative formula mass has the symbol Mr.
To calculate the Mr of a substance, add together the relative atomic masses of all the atoms shown in its formula:
Oxygen molecule - formula O2
Relative atomic mass of oxygen = 16
Relative formula mass = 2 x 16 = 32
Percentage by mass of an element in a compound given relative atomic masses
If you want to calculate the percentage by mass of an element in a compound (e.g. aluminium oxide) what you do is:
Al2O3
(2 x Al) + (3 x O)
(2 x 27) + (3 x 16)
54 + 48 = 102
If you want to find oxygen it would be 54/102
Empirical formulae
An empirical formula is the simplest whole number ratio of atoms of each element in a compound.
A 10g sample of a compound X contains 8g of carbon and 2g of hydrogen.
Write the symbol of each element
C H
Write down the mass of each element in g
8 2
Write down the Ar of each element
12 1
For each element, calculate:
8/12 = 0.667 2/1 = 2
Divide each answer by the smallest answer (0.667 here)
0.667 / 0.667 = 1 2/0.667 = 3
You may need to multiply all the numbers to remove fractions, then write out the empirical formula = CH3
Molecular formula
You can find the molecular formula of a compound from its empirical formula:
If you know its relative formula mass, Mr.
The Mr of X in the example is 30:
Calculate the Mr of the empirical formula:
Mr of CH3 = 12 + (3x1) = 15
Divide the Mr of X by answer 1:
30/15 = 2
Multiply each number in the empirical formula by answer 2:
CH3 becomes C2H6 - the molecular formula
Empirical formula experiment involving magnesium oxide
An apparatus is used for heating magnesium
Weigh some pure magnesium and heat it in a crucible to form magnesium oxide.
Weigh the mass of the magnesium oxide
Object Mass
Empty crucible and lid 19.06
Crucible, lid and Mg before heating 19.42
Crucible, lid and Mg after heating 19.66
Mass of magnesium = 19.42 - 19.06 = 0.36g
Mass of oxygen = 19.66 - 19.42 = 0.24g
Mg O
0.36 0.24
24 16
0.36/24 = 0.015 0.24/16= 0.015
MgO
Conservation of mass (closed system)
Law: the total mass of reactants and products stays constant during a chemical reaction
A closed system is when no substances can enter or leave during a reaction.
Closed systems include:
Reactions in a sealed container
Precipitation reactions = beaker
In a precipitation reaction, an insoluble product - the precipitate, is formed and the total mass in the beaker stays the same.
Non-enclosed systems
A non-enclosed system is when substances can enter or leave during a reaction.
Non-enclosed or “open” systems include:
Reactions in an open flask, where a substance in the gas state may enter or leave.
The mass of a reactive metal increases if it is heated in air because oxygen atoms combine with metal atoms to form a metal oxide.
The mass of a reactive non-metal/fuel decreases when heated because of products in the gas state escape.
This also happens to a mass of a metal carbonate but only because carbon dioxide gas is produced and this escapes from the container.
Calculating masses of reactants and products from balanced equations
Mass = moles x molar mass (of the reactants / product)
Moles = mass / molar mass
Use balancing numbers to find the moles:
2NaOH + Mg -> Mg (OH) 2 + 2NA if you had 2 moles of Mg you would form 2x2 = 4 moles of Na
Calculate the concentration of solutions in g dm-3
Concentration = mass / volume
Mass = concentration x volume
g = g/dm3 x dm3
One mole of particles of a substance
In 1 mole of a substance is the Avogadro constant: 6.02 x 10 to the power of 23 per mole.
The mass of one mole is the relative particle mass in grams.
Amount of substance = mass/relative atomic mass
The mass of 1 mol of a substance would be its atomic mass in grams. For example:
Carbon Ar = 12
Mass of 1 mol = 12g
Calculations of moles and particles
Moles = mass (g) / relative atomic mass
How many moles in 22.5 g of water?
Mol of H2O = 22.5 / 18 = 1.25 mol
What is the amount, in moles, of atoms in 22.5g of water?
3 x 1.25 = 3.75 mol
Use this answer to calculate the number of atoms
3.75 x 6.02 x 10^23 = 2.25 x 10^24
Limiting reactant and mass of product
In a reaction where there are two or more reactants, one must be used in excess to make sure that the other reactants are used up. The reactant which is used up is the limiting reactant as it limits the amount of products.
Stoichiometry
The ratio between the quantities of substances measured in moles involved in a chemical reaction.
When you balance a chemical equation, you are finding the stoichiometric equation.
Stoichiometry of a reaction
Worked example
An oxide of copper is heated with excess hydrogen, H2. In the reaction, 1.27g of copper and 0.18g of water, H2O, form. Use this information to determine the stoichiometry of the reaction. (relative atomic masses: H=1, O = 16, Cu = 63.5)
Amount of Cu = 1.27 / 63.5 = 0.02 moles
Mr of H2O = 1 + 1 + 16 = 18
Amount of H2O = 0.18 / 18 = 0.01 moles
Ratio of Cu:H2O = 0.02 : 0.01
= 2:1
right-hand side must be -> 2Cu + H2O
so the left-hand side must be Cu2O + H2
Equation is: Cu2O + H2 -> 2Cu + H2O
