Godly Calculations of Chemistry
1. Relative atomic mass (Ar)
Relative atomic mass is the average mass of naturally occurring atoms
of an isotopic element on a scale where the carbon-12 atom has a
mass of exactly 12 units.
To find relative atomic mass from percentage abundances of isotopes,
multiply each isotope’s mass number with their abundance, add them
all and divide by 100.’
E.g. Mg-24 (78.6%), Mg-25 (10.1%), Mg-26 (11.3%)
[(24 x 78.6) + (25 x 10.1) + (26 x 11.3)] / 100 = 24.3
Relative molecular Mass (Mr)
Relative molecular mass is the sum of the relative atomic masses of all
the atoms shown in the formula.
We can put ‘amu’ as the unit when we find relative masses.
2. Molar Mass
Molar mass is the mass of 1 mole of a substance (molar mass) is the Ar
or Mr in grams. The unit is g/mol.
3. Mole Concept
Calculating Number of Moles of Single Substance
Moles = Mass / Molar Mass
Molar Volume
Moles = Volume/ Molar Volume
Molar Volume = 24 dm3 or 24,000 cm3 at rtp
The unit will be mol for both.
Calculating the Concentration of a substance
Concentration(mol/dm3) = Mole(mol)/Volume(dm3)
Another unit = g/dm3
From mol/dm3 to g/dm3  multiply with Mr
From g/dm3 to mol/dm3  divide by Mr
Calculating the Percentage Yield
Percentage yield(purity) = actual yield/theoretical yield * 100
Percentage by mass
Mass percentage = mass component/total mass x 100%
4. Empirical vs Molecular Formula
Empirical formula (E.F) shows simplest whole number ratio of atoms
present in a compound.
The molecular formula (M.F) shows the actual number of atoms of
each element present in a molecule (covalent compound) or formula
unit (ionic compound) of a compound.
We divide the given mass/g of each element with its molar mass. Then,
we divide each number with the smallest number to get the mole ratio.
(Round it up if decimal does not end in 5, multiply again so every mole
ratio has a whole number if decimal ends in 5)
To find the molecular formula, we divide the compound’s given molar
mass by the empirical mass. Then, we multiply the empirical formula
with that number.
E.g. 16.66% of C and 3.49% of H, compound weighs 58 g.
Mass
Mole
Mole Ratio
E.F
Carbon
16.66
16.66/12
1.39
1.39/1.39
1
2
C2H5
Hydrogen
3.49
3.49/1
3.49
3.49/1.39
2.5
5
N = molar mass/empirical mass
= 180/ 2 x (C) + 5 x (H)
= 58/ 2 x 12 + 5 x 1 = 58/29 = 2
M.F = (C2H5)2 = C4H10
Finding the ratio of water from the water of crystallization (Unknown
H2O amount questions)
Find the mass of the anhydrous compound. Find the mass of the
water. Find the empirical formula using the ratio of the compound and
the water. The mole ratio of H2O is the ratio of water.
Finding the amount in moles of a substance formed
Use the unitary method. Use the mole ratios given in the equation.
Then, find how much mole would be formed using the given mass
amount(after changing it into mole). Use the limiting reagent when
calculating with the unitary method for ratios.
E.g. 2Fe + 3Br2  2FeBr3
10g of iron reacts with bromine.
Amount in mole of iron that reacted
10/56 = 0.18 mol
Amount in moles of bromine reacted
If 2 mol of Fe  3 mol of Br2
Then, 0.18 mol of Fe  ? = 0.27 mol of Br2
Amount of Fe2Br3 formed
If 2 mol of Fe  2 mol of FeBr3
0.18 mol of Fe  ? = 0.18 mol of FeBr3
Mass of Fe2Br3 formed
Mole x Mr = 0.18 * 296 = 53.28g
Calculating which substance is in excess
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
3 mol of Mg reacts with 5 mol of HCl.
If 1 mol of Mg  2 mol of HCl – (equation)
3 mol of Mg  6 mol of HCl. – (should be amount)
But in given amounts, 3 mol of Mg  5 mol of HCl. (given)
There is less HCl than there should be so it is limited. Mg would be in
excess.
Balancing Equations
Balance the numbers of atoms of an element on the LHS and RHS.
Numbers can only be written in front of a chemical formula.
For groups of atoms (SO4, NO3), we need to count individual atoms.
Generally, balance the metal first, then the other non-metals.
Leave hydrogen and oxygen atoms last as they are always included in
all chemical equations.
For organic equations (contains carbon and hydrogen compounds), we
do not multiply it. Instead, we can use fractions to balance the two
sides.
C7H17 + O2  CO2 + H2O | 4 C7H17 + 45 O2  28 CO2 + 34 H2O
General Notes
When the question asks ‘mass of 1 mole’, use g/mol as the unit.
Make sure to show volume’s units as ‘…. at rtp’.
Make sure the units are the same and appropriate when calculating.
E.g. when calculating the concentration, the volume unit must be in
dm3.
If the concentration of something is given, use the concentration
formula to find volume instead of the molar volume formula.