stoichiometry Flashcards
what does emperical formula tell us
The empirical formula tells you the simplest whole-number ratio of atoms in a compound
what does the molecular formula tell us
The molecular formula tells you the actual number of atoms of each element in one molecule of the compound or element e.g. H2 has 2 hydrogen atoms, HCl has 1 hydrogen atom and 1 chlorine atom
what are the valencys of group 1- 8
I ) 1 II ) 2 III ) 3 IV ) 4 V) 3 VI ) 2 VII) 1 VIII) 0
what is the formula of aluminium sulfide
Write out the symbols of each element and write their combining powers underneath:
Al S
3 2
The formula is then calculated by cross multiplying each atom with the number opposite, hence the formula for aluminium sulfide is Al2S3
what is the formula of aluminium sulfate (ionic compounds )
Write out the formulae of each ion, including their charges
Al3+ SO42-
Balance the charges by multiplying them out:
Al3+ x 2 = +6 and SO42- x 3 = -6; so +6 – 6 = 0
So the formula is Al2(SO4)3
for naming compounds how do you name them If one is a metal and the other a nonmetal
the name of the metal atom comes first and the ending of the second atom is replaced by adding –ide
Eg. NaCl which contains sodium and chlorine thus becomes sodium chloride
for naming compounds how do you name them if both atoms are nonmetals and one of those is hydrogen
hydrogen comes first
Eg. hydrogen and chlorine combined is called hydrogen chloride
for naming compounds how do you name them if its a combaniation of non metals
nonmetals as a general rule, the element that has a lower Group number comes first in the name Eg. carbon and oxygen combine to form CO2 which is carbon dioxide since carbon is in Group 4 and oxygen in Group 6
balance the following equation: Al + CuO → Al2O3 + Cu
ALUMINIUM: There is 1 aluminium atom on the left and 2 on the right so if you end up with 2, you must start with 2. To achieve this, it must be 2Al
2Al + CuO → Al2O3 + Cu
OXYGEN: There is 1 oxygen atom on the left and 3 on the right so if you end up with 3, you must start with 3. To achieve this, it must be 3CuO
2Al + 3CuO → Al2O3 + Cu
COPPER: There is 3 copper atoms on the left and 1 on the right. The only way of achieving 3 on the right is to have 3Cu
2Al + 3CuO → Al2O3 + 3Cu
what is the symbol for relative atomic mass
The symbol for the relative atomic mass is Ar
what is the symbol for relative molecuular mass
The symbol for the relative molecular mass is Mr
how do you calculate the relative formula molecure mass
To calculate the Mr of a substance, you have to add up the Relative Atomic Masses of all the atoms present in the formula
what is one mole of water
For a compound we add up the relative atomic masses
So one mole of water would have a mass of 2 x 1 + 16 = 18g
what is one mole of helium
one mole of helium atoms would have a mass of 4g
what is molar mass
This is the volume that one mole of any gas
Calculate the percentage of oxygen in CO2
Step 1 – Calculate the molar mass of the compound
Molar mass CO2 = (2 x 16) + 12 = 44
Step 2 – Add the atomic masses of the element required as in the question (oxygen)
16 + 16 = 32
Step 3 – Calculate the percentage
% of oxygen in CO2 = 32/44 x 100 = 72.7%
25.0 cm3 of 0.050 mol / dm3 sodium carbonate was completely neutralised by 20.00 cm3 of dilute hydrochloric acid. Calculate the concentration in mol / dm3 of the hydrochloric acid.
Step 1 – Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing by 1000 to convert cm3 to dm3
Amount of Na2CO3 = (25.0 x 0.050) ÷ 1000 = 0.00125 mol
Step 2 – Calculate the amount, in moles, of hydrochloric acid reacted
1 mol of Na2CO3 reacts with 2 mol of HCl, so the Molar Ratio is 1 : 2
Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl
Step 3 – Calculate the concentration, in mol / dm3 of the Hydrochloric Acid
1 dm3 = 1000 cm3
Volume of HCl = 20 ÷ 1000 = 0.0200 dm3
Concentration HCl (mol / dm3) = 0.00250 ÷ 0.0200 = 0.125
Concentration of Hydrochloric Acid = 0.125 mol / dm3
what is the limiting reactant
The limiting reactant is the reactant which is not present in excess in a reaction
It is always the first reactant to be used up which then causes the reaction to stop
9.2g of sodium is reacted with 8.0g of sulfur to produce sodium sulfide, NaS. Which reactant is in excess and which is the limiting reactant?
Step 1 – Calculate the moles of each reactant
Moles = Mass ÷ Ar
Moles Na = 9.2/23 = 0.40
Moles S = 8.0/32 = 0.25
Step 2 – Write the balanced equation and determine the molar ratio
2Na + S → Na2S so the molar ratios is 2 : 1
Step 3 – Compare the moles. So to react completely 0.40 moles of Na require 0.20 moles of S and since there are 0.25 moles of S, then S is in excess. Na is therefore the limiting reactant.
Calculate the Mass of Magnesium Oxide that can be made by completely burning 6 g of Magnesium in Oxygen
Magnesium (s) + Oxygen (g) → Magnesium Oxide (s)
Symbol Equation:
2Mg + O2 → 2MgO
Relative Formula Mass: Magnesium : 24 Magnesium Oxide : 40
Step 1 – Calculate the moles of Magnesium Used in reaction
Moles = Mass ÷ Mr Moles = 6 ÷ 24 = 0.25
Step 2 – Find the Ratio of Magnesium to Magnesium Oxide using the balanced Chemical Equation
extended_Calculating reacting masses_magnesium table, IGCSE & GCSE Chemistry revision notes
Step 3 – Find the Mass of Magnesium Oxide
Moles of Magnesium Oxide = 0.25
Mass = Moles x Mr Mass = 0.25 x 40 = 10 g
Mass of Magnesium Oxide Produced = 10 g
A compound that contains 10 g of Hydrogen and 80 g of Oxygen has an Empirical Formula of H2O. This can be shown by the following calculations:
Amount of Hydrogen Atoms = Mass in grams ÷ Ar of Hydrogen = (10 ÷ 1) = 10 moles
Amount of Oxygen Atoms = Mass in grams ÷ Ar of Oxygen = (80 ÷ 16) = 5 moles
The ratio of moles of hydrogen atoms to moles of oxygen atoms:Since equal numbers of moles of atoms contain the same number of atoms, the ratio of hydrogen atoms to oxygen atoms is 2:1
Hence the empirical formula is H2O
find the percentage yield of Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)
Step 1: Calculate the Amount, in Moles of Zinc Reacted
Moles of Zinc = 6.5 ÷ 65 = 0.10 moles
Step 2: Calculate the Maximum Amount of Copper that could be formed from the
Molar ratio
Maximum Moles of Copper = 0.10 moles (Molar ratio is 1:1)
Step 3: Calculate the Maximum Mass of Copper that could be Formed
Maximum Mass of Copper = ( 0.10 x 64 ) = 6.4 g
Step 4: Calculate the Percentage of Yield of Copper
Percentage Yield = ( 4.8 ÷ 6.4 ) x 100 = 75%
Percentage Yield of Copper = 75%
find the percentage purity of CaCO3 (s) → CaO(s) + CO2 (g)
Step 1: Calculate the relative formula masses
1 mole CaCO3 → 1 mole CO2
40+12+(3×16) 12+(2×16)
100 → 44
Step 2: Calculate the theoretical mass of calcium carbonate used if pure
From 2.5g CO2 we would expect 2.5/44 x 100 = 5.68g
Step 3: Calculate the percentage purity
(Mass of pure substance / mass of impure substance) x 100
= 5.68/7.0 x 100
= 81.1%
