Moles
Important
Ar = relative atomic mass
Mr = relative molecular mass
RELATIVE = NO UNITS!
Calculations from periodic table (eg Ar, Mr) = 1 dp
All other calculations = 3 sf
Percentage composition of a compound
% mass of element in a sample = No. of atoms of the element x Ar of element x 100%
Mr of the compound
Mass of element in a sample = No. of atoms of the element x Ar of element x Mass of sample
Mr of the compound
Mass of element in a sample = % of the element x Mass of sample
100%
Eg: Calculate the percentage mass of sodium in sodium carbonate
Relative formula mass of Na2CO3 = (23.0 x 2) + 12.0 + (16.0 x 3)
= 106.0 (1 dp)
% Mass of Na = (46.0 / 106.0) x 100%
= 43.4% (3 sf)
Mole
Mole = Avogadro’s number = 6.02 x 1023
Eg:
* 1 mole of carbon = 6.02 x 1023 carbon atoms
* 1 mole of oxygen gas = 6.02 x 1023 oxygen molecules
* 1 mole of sodium chloride = 6.02 x 1023 sodium ions and 6.02 x 1023 chloride ions
* 2 moles of ammonia gas = 1.204 x 1024 ammonia molecules
* 2 moles of ammonia gas = 2 moles of nitrogen atoms, 6 moles of hydrogen atoms
* 0.5 moles of calcium chloride = 0.5 moles of calcium ions, 1 mole of chloride ions
* 0.5 moles of calcium chloride = 3.01 x 1023 calcium ions and 1.204 x 1024 chloride ions
Molar mass
Mass of 1 mole of any substance
g/mol
Read off periodic table (1 dp)
Calculation of mass (g) (3sf)
Mass of chemical = no. of moles x molar mass of chemical
Eg:
*
*
*
*
*
Molar mass of carbon = 12 g/mol
Molar mass of oxygen (O2) = 32 g/mol
Molar mass of sodium chloride = 58.5 g/mol
Calculate the mass of 3 moles of helium gas
o Molar mass of helium gas = 4.0 g/mol
o Mass of 3 moles = 4.0 x 3 = 12.0 g
Calculate the number of moles of sulphur dioxide present in 32g of sulphur dioxide
o Molar mass of sulphur dioxide = 32.1 + 32.0 = 64.1 g/mol
o No. of moles present = 32g / 64.1 = 0.50mol (3 sf)
Molar volume
Volume of 1 mole of any gas at a fixed temperature and pressure
Room temperature and pressure (rtp) = 24dm3/mol
Standard temperature and pressure (stp) = 22.4dm3/mol
Increase temp, increase vol. Increase pressure, decrease vol.
dm3/mol
1 dm3 = 100 cm3
Volume = molar volume x no. of moles
Eg:
* Calculate the volume, at rtp, of
* 0.4 mol of oxygen
o 0.4 x 24 = 9.6dm3
* 1.9g of fluorine
o Amount of fluorine = 1.9 / 38.0 = 0.05mol
o Volume of fluorine = 0.05mol x 24dm3 = 1.2dm3
Density = Mass / volume
Empirical formula
Simplest whole number ratio of different atoms present
Must know:
* Types of elements present
* Mass of each element present
Can conclude name from empirical formula for ionic compound, but not covalent
Eg: Must draw table!
Element
Iron (Fe)
Mass of each element / g
14
Molar mass / g/mol
55.8
No. of moles present / mol
0.251
Molar ratio
1
Simplest whole number ratio
2
The empirical formula of the substance is Fe2O3, Iron (III) Oxide.
Oxygen (O)
6
16.0
0.375
1.5
3
Molecular formula
To find molecular formula from empirical formula, we must know = Relative molecular mass
Eg:
An organic compound (Mr = 180) has the empirical formula of CH2O. Find the molecular formula of this compound.
* Let its molecular formula be (CH2O)n
* Relative mass of (CH2O)n = (12.0 + 18.0)n = 30.0n
* Therefore 30.0n = 180
* n = 180 / 30.0 = 6
* Molecular formula is (CH2O)6 = C2H12O6
balanced chemical equations
Relationship between amounts (mol) of reactants and products involved in a chemical reaction = stoichiometry
Eg: The products from the complete combustion of methane are carbon dioxide and steam. For the complete
combustion of 4.0g methane, calculate the mass of steam produced.
Write down balanced chemical equation for reaction
Decide which substances are used
No. of moles of each
Calculate no. of moles of methane
Determine no. of moles of steam produced
Convert no. of moles of steam into mass in g
1 mole of methane > 2 moles of steam
Amt of methane = 4.0 / 16.0 = 0.25mol
Amt of steam = 0.25mol x 2 = 0.5mol
Mass of steam = 0.5 x 18.0 = 9.0g
Limiting reactants
Chemical completely consumed = LIMITING REACTANT (the one you don’t have enough of)
Chemical left unreacted = IN EXCESS
Eg:
What volume of carbon monoxide is produced by the reaction between 12g of carbon and 10dm of carbon dioxide?
Amount of carbon = 12 / 12.0 = 1mol
Amount of CO2 = 10 / 24 = 0.4167mol
Since no. of moles of C:CO2 from equation = 1:1
Limiting reactant is CO2
Amount of CO2 = 0.4167 x 2 = 0.8334mol
Volume of CO2 = 0.8334 x 24 = 20.0dm3
Concentration
@ Amt of solute dissolved in 1dm3 of solution (1dm3 = 1000cm3)
@ Mol/ dm3 (molarity) or g/ dm3
Percentage yield
@ Experimental yield/theoretical yield x100%
@ Assume that it is pure
Percentage purity
@ Mass of pure substance/mass of sample used x100%
@ Assume theoretical yield (all reacted)