Moles and Empirical Formula CGP pages 26-32
Mass of an Atom
Mass number – number of protons + neutrons
Atomic number – number of protons
Isotopes are atoms of the same element with a
different number of neutrons but same number of
protons.
!
Relative atomic mass (Ar) the weighted average mass of all the isotopes of that element compared to !"
the mass of carbon-12
Relative abundance is the probability of the existence of an isotope
∈(%&'()*+& *-.)./*0 1(-- × (3456(50&)
Ar =
!88
e.g.
("9×:;)<(!!×"=)<(!8×">)
Magnesium isotopes !88
= 24.31 g/mol
79% of Mg24
11% of Mg25
10% of Mg26
Relative formula mass à Relative atomic masses can be used to find the relative formula mass of a
compound. To find the relative formula mass (Mr) of a compound, you add together the relative atomic
mass values (Ar values) for all the atoms in its formula.
NaCl
Mr = the relative atomic mass added so…… à 23+35.5 = 58.5
CaSO4
40+32+(16×4) = 136
, Moles
A mole is just a number: 6.022×1023 𝒎𝒂𝒔𝒔
6.022×1023 atoms of carbon has a mass of 12g which is its RAM Moles =
𝐀𝐫 𝐨𝐫 𝐌𝐫
So, one mole of atoms will have a mass in grams equal to the Ar or Mr
Examples
1) 24g of carbon atoms 5) 9g aluminium atoms 9) 1 mole of nitrogen gas
24 ÷ 12 = 2mol 9 ÷ 27 = 0.33 mol 14 × 2 = 28g
2) 36g of water molecules 6) 1 mole of nickel atoms 10) 0.5 moles of water (H2O)
36 ÷ (16×2) = 2 mol 59g à the atomic mass 2 + 16 = 18 ÷ 2 = 9g
3) 120g of calcium atoms 7) 2 moles of sodium atoms 11) 1.5 moles of ammonia (NH)
120÷ 40 = 3 mol 2 × 23 = 46g 14 + 1 = 15 × 1.5 = 22.5
4) 48g of magnesium atoms 8) 0.5 moles of sulphur atoms 12) 0.75 moles of methane (CH4)
48 ÷ 24 = 2 mol 32 ÷ 2 = 16g 12 + 4 = 16 × 0.75 = 12g
How to do the empirical formula
1. Start with the number of grams of each element, given in the problem.
2. Convert the mass of each element to moles using the molar mass from the periodic table.
3. Divide each mole value by the smallest number of moles calculated.
4. Round to the nearest whole number. This is the mole ratio of the elements and is. M = 230
r
C2H6O = 24 + 6 + 16 = 46
Example questions "G8
= 5.34 = 5
0.6g of carbon, 0.15g of hydrogen, 0.4g of oxygen with a Mr of MF being 230g/mol 9>
C12 H1 O16
Mass or % 0.6 0.15 0.4
Moles 0.6 ÷ 12 = 0.05 0.15 ÷ 1 = 0.15 0.4 ÷ 16 = 0.025
Ratio 0.05 ÷ 0.025 = 2 0.15 ÷ 0.025 = 6 0.025 ÷ 0.025 = 1
Multiple 2 6 1
C 2H 60
The empirical formula is a formula that gives the simplest whole number ratio Mr = 710
of the different atoms in a compound P2O5 = 62 + 80 = 142
:!8
!9"
=5
Calculate the empirical formula of a compound that contains 43.7% P and 56.3% O by mass, with an Mr of
MF being 710g/mol
P31 O16
Mass or % 43.7 56.3
Moles 43.7 ÷ 31 = 1.409 56.3 ÷ 16 = 3.51
Ratio 1.409 ÷ 1.409 = 1 3.51 ÷ 1.409 = 2.5
Multiple 2 5
P 2 O5
, You use Avogadro’s constant when you want to convert from moles to atoms
Mass of an Atom
Mass number – number of protons + neutrons
Atomic number – number of protons
Isotopes are atoms of the same element with a
different number of neutrons but same number of
protons.
!
Relative atomic mass (Ar) the weighted average mass of all the isotopes of that element compared to !"
the mass of carbon-12
Relative abundance is the probability of the existence of an isotope
∈(%&'()*+& *-.)./*0 1(-- × (3456(50&)
Ar =
!88
e.g.
("9×:;)<(!!×"=)<(!8×">)
Magnesium isotopes !88
= 24.31 g/mol
79% of Mg24
11% of Mg25
10% of Mg26
Relative formula mass à Relative atomic masses can be used to find the relative formula mass of a
compound. To find the relative formula mass (Mr) of a compound, you add together the relative atomic
mass values (Ar values) for all the atoms in its formula.
NaCl
Mr = the relative atomic mass added so…… à 23+35.5 = 58.5
CaSO4
40+32+(16×4) = 136
, Moles
A mole is just a number: 6.022×1023 𝒎𝒂𝒔𝒔
6.022×1023 atoms of carbon has a mass of 12g which is its RAM Moles =
𝐀𝐫 𝐨𝐫 𝐌𝐫
So, one mole of atoms will have a mass in grams equal to the Ar or Mr
Examples
1) 24g of carbon atoms 5) 9g aluminium atoms 9) 1 mole of nitrogen gas
24 ÷ 12 = 2mol 9 ÷ 27 = 0.33 mol 14 × 2 = 28g
2) 36g of water molecules 6) 1 mole of nickel atoms 10) 0.5 moles of water (H2O)
36 ÷ (16×2) = 2 mol 59g à the atomic mass 2 + 16 = 18 ÷ 2 = 9g
3) 120g of calcium atoms 7) 2 moles of sodium atoms 11) 1.5 moles of ammonia (NH)
120÷ 40 = 3 mol 2 × 23 = 46g 14 + 1 = 15 × 1.5 = 22.5
4) 48g of magnesium atoms 8) 0.5 moles of sulphur atoms 12) 0.75 moles of methane (CH4)
48 ÷ 24 = 2 mol 32 ÷ 2 = 16g 12 + 4 = 16 × 0.75 = 12g
How to do the empirical formula
1. Start with the number of grams of each element, given in the problem.
2. Convert the mass of each element to moles using the molar mass from the periodic table.
3. Divide each mole value by the smallest number of moles calculated.
4. Round to the nearest whole number. This is the mole ratio of the elements and is. M = 230
r
C2H6O = 24 + 6 + 16 = 46
Example questions "G8
= 5.34 = 5
0.6g of carbon, 0.15g of hydrogen, 0.4g of oxygen with a Mr of MF being 230g/mol 9>
C12 H1 O16
Mass or % 0.6 0.15 0.4
Moles 0.6 ÷ 12 = 0.05 0.15 ÷ 1 = 0.15 0.4 ÷ 16 = 0.025
Ratio 0.05 ÷ 0.025 = 2 0.15 ÷ 0.025 = 6 0.025 ÷ 0.025 = 1
Multiple 2 6 1
C 2H 60
The empirical formula is a formula that gives the simplest whole number ratio Mr = 710
of the different atoms in a compound P2O5 = 62 + 80 = 142
:!8
!9"
=5
Calculate the empirical formula of a compound that contains 43.7% P and 56.3% O by mass, with an Mr of
MF being 710g/mol
P31 O16
Mass or % 43.7 56.3
Moles 43.7 ÷ 31 = 1.409 56.3 ÷ 16 = 3.51
Ratio 1.409 ÷ 1.409 = 1 3.51 ÷ 1.409 = 2.5
Multiple 2 5
P 2 O5
, You use Avogadro’s constant when you want to convert from moles to atoms
