Calculations involving masses Flashcards
How do we calculate the Mr of a compound
Add up the Ar of all the elements then multiply it by the stoichiometric coefficient BUT FOR CALCULATING MOLES YOU USE THE AR NOT MR SO YOU DONT MULTIPLY
Calculate the formulae of simple compounds from reacting masses or percentage composition if given their masses and understand that these are empirical formulae
- an empirical formula gives the simplest whole number ratio of atoms of each element in the compound
- to find empirical formula from reacting masses first separate each of the elements, then do the mass/Ar to find the moles. Then find which element has the lowest number of moles. Then divide each of the number of moles by the smallest number of moles. That gives the ratio of the compound. Ie: if the ratio is 1:2 then the first element in the compound has just one of it but the 2nd element has 2 of it.
How do you deduce the molecular formula out of the empirical formula given the Mr of the product
Find the Mr of the empirical formula
Divide the Mr of the product by Mr of the empirical formula
Times that number by all the numbers of how many of each element there are in the empirical formula (ie: if the result is 2 and the empirical formula is C2H3O then the molecular formula is C4H6O2)
Deduce and experiment to find out the empirical formula of Magnesium Oxide
● weigh some pure magnesium
● Heat magnesium to burning in a crucible to form magnesium oxide, as the magnesium will react with the oxygen in the air
● weigh the mass of the magnesium oxide
● Known quantities: mass of magnesium used & mass of magnesium oxide produced
● Required calculations:
○ mass oxygen = mass magnesium oxide - mass magnesium
○ moles magnesium = mass magnesium ÷ molar mass magnesium
○ moles oxygen = mass oxygen ÷ molar mass oxygen
○ calculate ratio of moles of magnesium to moles of oxygen
○ use ratio to form empirical formula
Explain the law of conservation of mass applied to:
a closed system including a precipitation reaction in a closed flask
a non-enclosed system including a reaction in an open flask that takes in or gives out a gas
● Law of conservation of mass: no atoms are lost or made during a chemical
reaction so the mass of the products = mass of the reactants
● With a precipitation reaction – precipitate that forms is insoluble and is a solid,
as all the reactants and products remain in the sealed reaction container then it
is easy to show that the total mass is unchanged
● Does not hold for a reaction in an open flask that takes in or gives out a gas,
since mass will change from what it was at the start of the reaction as some
mass is lost when the gas is given off
mass (__) = ___(mol) * ____
g
volume/amount
Ar/Mr (g/mol)
one mole of particles of a substance is defined as: the _____ ______ (_______ atoms, molecules, formulae or ions): number of particles of that substance and a mass of ‘relative particle mass’
Avogrado Constant
6.02*10^23
Calculate masses of reactants and products from balanced equations, given the mass of one substance
find moles of that one substance: moles = mass / molar
mass
● Use balancing numbers to find the moles of desired
reactant or product (e.g. if you had the equation:
2NaOH + Mg -> Mg(OH) 2 + 2Na, if you had 2 moles of
Mg, you would form 2x2=4 moles of Na)
● Mass = moles x molar mass(of the reactant/product) to
find mass
Recall that one mole of particles of a substance is defined
as:
the Avogadro constant number of particles (6.02 x 10 23 atoms,
molecules, formulae or ions) of that substance
and a mass of ‘relative particle mass’ g
Explain why, in a reaction, the mass of product formed is
controlled by the mass of the reactant which is not in excess
● In a chemical reaction with 2 or more reactants you will often use one in excess
to ensure that all of the other reactant is used
o The reactant that is used up / not in excess is called the limiting reactant
since it limits the amount of products
● if a limiting reagent is used, the amount reactant in excess that actually reacts is
limited to the exact amount that reacts with the amount of limiting reagent you
have, so you need to use the moles/mass of the limiting reagent for any
calculations
Deduce the stoichiometry of a reaction from the masses of
the reactants and products
● Stoichiometry refers to the balancing numbers in front of compounds/elements
in reaction equations
● Balancing numbers in a symbol equation can be calculated from the masses of
reactants and products:
○ convert the masses in grams to amounts in moles (moles = mass/Mr)
○ convert the numbers of moles to simple whole number ratios
● e.g. for the reaction: Cu + O 2 -> CuO (not balanced), 127 g Cu react, 32g of
oxygen react and 159g of CuO are formed. Work out the balanced equation
using the masses given:
○ moles: (moles = mass/Mr)
Cu: moles = 127 / 63.5 = 2
O 2 : moles= 32 / (16 x 2) = 32/32 = 1
CuO moles = 159 / (16 + 63.5) = 2
● therefore you have a ratio of 2:1:2 for Cu:O 2 :CuO, making the overall
balanced equation 2Cu + O 2 -> 2CuO
Calculate the percentage yield of a reaction from the actual yield and
the theoretical yield
Percentage yield = Amount of product produced x 100//Maximum amount of product possible
● Amount of product obtained is known as yield
Describe that the actual yield of a reaction is usually ____ than the
theoretical yield and that the causes of this include:
● causes of actual yield being less than theoretical yield:
○ incomplete reactions (not all of the reactants have reacted)
○ practical losses during the experiment (some product has been left in the
weighing boat etc)
○ side reactions (some of the products react to form other products than
those you wanted)
Atom economy definition and formula
atom economy- a measure of the amount of starting materials that end up as
useful products
atom economy = (Mr of desired product from reaction / sum of Mr of all
reactants) x 100
Explain why a particular reaction pathway is chosen to
produce a specified product, given appropriate data
● look for a high atom economy, high yield, fast rate, equilibrium position to the
right (towards products) and useful by-products – be prepared to look for these
within given information for the question and present them as an answer
The volume occupied by one mole of any gas is
the same
so the equation for the volume of gas in a space at room temperature is:
Volume (dm 3 ) of gas at RTP = Mol. x 24
Volume (cm 3 ) of gas at RTP = mol x 24,000
Use the molar volume and balanced equations in
calculations involving the masses of solids and volumes of gases
ie: ● if you are given a balanced equation, the mass/volume of a reactant and are
asked to calculate the mass/volume of a product:
○ calculate moles of the reactant
if given a mass: moles=mass ÷ molar mass
if given a volume: moles = volume ÷ 24
● work out the mole ratio and so work out how many moles of the product
you have
● calculate mass/volume using moles
for calculating mass, mass=moles x molar mass
for calculating volume, volume=moles x 24
Use Avogadro’s law to calculate volumes of gases involved
in a gaseous reaction, given the relevant equation
● avogadro’s law: one mole of a substance contains 6.02 x 10^23 particles (this is
why we use moles in calculations- the number of particles is massive)
● this means if you had 10 moles of a substance, you would have 10 x 6.02 x 10^23
particles = 6.02 x 10^24
