SCH 3U – Unit 3 – Calculations in Chemistry Review Note – Part 1
6.3 The Mole
•
The mole is a unit of measurement, like a dozen, that represents 6.02 x 1023 particles of
atoms, molecules, electrons, etc. – very small things!
6.5 Calculating number of entities
We can calculate the number of particles in a given number of moles using Avogadro’s
constant: 6.02 x 1023
• Formula: N=nNA
N: number of particles
n: number of moles
NA: Avogadro’s constant: 6.02 x 1023
•
Determine the number of entities
• 4 moles of sodium
• 0.0025 moles of potassium
• 2.76 moles chlorine
• 4.3 x 10-6 moles of iodine
Determine the number of moles
• 7.18 x 1023 atoms of hydrogen
• 2.2 x 1010 molecules of methane
• 3.5 x 1025 crackers
6.4 Molar Mass
•
•
•
Molar mass is the mass, in grams, of one mole of substance – Units: g/mol
We get the molar mass of the elements from the periodic table
The molar mass of a compound is the combined molar masses of the elements in the
compound
Determine the molar mass of the following compounds:
• Na3PO4
• H2SO4
• MgCl2
• Al2O3
Calculating moles using molar mass
• We can convert mass to moles using this formula: n = m/M
m: mass in grams
n: moles
M: molar mass
• We can rearrange this formula to solve for mass or molar mass
m = nM
M = m/n
Determine the number of moles
• 45 g of aluminum
• 25.362 g of hydrogen
• 186.997 g of potassium oxide
Determine the mass
• 17 moles of sodium
• 6 moles of carbon dioxide
• 3.67 x 10-4 moles of calcium phosphate
6.6 Percent Composition
•
Percent Composition: The % by mass of each element in a compound (NOT the ratio)
% 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑖𝑛 1 𝑚𝑜𝑙
× 100
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑 𝑖𝑛 1 𝑚𝑜𝑙
1. Calculate molar mass for elements and compound
2. Plug the numbers into the formula
Example: Calculate the % composition of potassium permaganate (KMnO4)
1. Calculate the molar mass
K
=
1(39.1) =
Mn = 1(54.9) =
54.9
O
=
6(16.0) =
MM =
39.1
64.0
158 g/mol
2. Use % formula
%𝐾 =
% 𝑀𝑛 =
%𝑂 =
39.1 𝑔/𝑚𝑜𝑙
× 100 = 24.7 %
158 𝑔/𝑚𝑜𝑙
54.9 𝑔/𝑚𝑜𝑙
× 100 = 34.7 %
158 𝑔/𝑚𝑜𝑙
64.0 𝑔/𝑚𝑜𝑙
× 100 = 40.5 %
158 𝑔/𝑚𝑜𝑙
Try these


Determine the % composition of ethanol (C2H5OH).
Determine the % composition of sodium oxalate (Na2C2O4).
6.7 Empirical Formula
•
The simplest form of a molecular formula
An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen.
Calculate the empirical formula.
1. Determine the number of grams of each element
Al: 4.151g
and
O: 3.692g
2. Convert mass to moles
nAl =
nO =
m/MM
=4.151/26.98
=0.1539 mol Al
m/MM
=3.692/16.00
= 0.2308 mol O
3. Find ratio by dividing each element by the smallest amount
0.1539 𝑚𝑜𝑙 𝐴𝑙
= 1𝑚𝑜𝑙 𝐴𝑙
0.1539 𝑚𝑜𝑙 𝐴𝑙
0.2308 𝑚𝑜𝑙 𝑂
= 1.5 𝑚𝑜𝑙 𝑂
0.1539 𝑚𝑜𝑙 𝐴𝑙
4. Multiply by a factor to get whole numbers
O = 1.5 x 2
Al = 1 x 2
Therefore: Al2O3
Try these:
•
•
2.000 g of iron metal is heated in air. It reacts with oxygen to achieve a final mass of 2.573
g. Determine the empirical formula.
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80%
hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.
6.8 Molecular Formula
•
The molar mass of the molecular formula will always be a whole number multiple of the
molar mass of the molecular formula.
1. Find your whole number factor using division
2. Multiple subscripts by that number
•
A white powder is found to have an empirical formula of P2O5. The compound has a molar
mass of 283.88g. What is the compound’s molecular formula?
1. Find molar mass of the empirical formula
P= 2(30.97) =
O = 5(16.00) =
2. Divide MM by empirical MM
61.94
80.00
MM =
141.94
283.88/141.94 = 2
3. Multiply subscripts
P2O5  P4O10
Try this:
• A compound has an experimental molar mass of 78.12 g/mol. Its empirical formula is CH.
What is its molecular formula?