Notes (this may be useful) and
(especially this)
Topic 1: Quantitative Chemistry/Stoichiometry
\topics\
Lesson 1: Basic Concepts
\basic\ \concepts\
Particle theory
Solid
Liquid
Gas
Distance between
particles
Close Together
Medium DIstance
Far Apart
Arrangement
Regular
Random
Irregular; more
random
Shape
Fixedx
Takes shape of
container
Takes shape of
container
Volume
Fixed
Fixedx
Exampandsto volume
of container
Movement Speed
Slow-vibrates
Medium
Fast
Energy
Low
Medium
High
Intermolecular Forces Strong
(forces holding
particles together)
Medium
Weak
Compressibility
Little to none
High compressibility
Changes of state
\changes\ \state\
Little to none
Cool Curves
\cool\ \curves\
The graph shows the temperature of a sample of naphthalene measured at one minute interval
as it cools from the liquid state at 90C. Explain the shape of the graph.
A
B
C
A:
●
●
●
Liquid is cooling
SLowing down
Moving closer
B:
●
●
Freezing occurs
(l) and (s)
C:
●
Solid is cooling
Elements compounds and mixtures
An element is a substance which is made of atoms. These atoms all have to have the same
number of protons: another way of saying this is that all of a particular element's atoms have the
same atomic number.
An atom is the smallest substance that cannot be broken down.
Atoms of different elements combine in fixed ratios to form compounds, which have different
properties from their component elements. Compounds contain two or more different elements
chemically bonded together. Some examples of compounds are:
●
●
●
●
●
●
●
●
H20 = water.
C6H12O6 = sugar.
NaCl = salt.
C2H6O = alcohol.
C4H10= butane.
NaHCO3 = baking soda.
N2 = nitrogen.
CH4 = methane.
Mixtures contain more than one element and-or compounds that are not chemically
bonded together and so retain their individual properties.
There are two types of mixtures: homogeneous (can’t see individual parts and
properties are the same throughout, example: air) and heterogeneous (can see
components and properties are not the same throughout, example: oil & water).
Chemical vs Physical Properties
\chemical\ \properties\ \physical\
Chemical properties dictate how something reacts in a chemical reaction
Physical properties are all the other properties of a substance such as melting point and
electrical conductivity
Chemical Equations
\equations\
\word equations\
Word Equations
Example: Hydrogen + Oxygen → Water
Substances before the arrow are called reactants. SUbstances after the arrow are called
products. The arrow means yield or makes or becomes DOES NOT MEAN equals.
\symbol equations\
Symbol Equations
Example: 2𝐻 2 𝑂2 − −> 2𝐻2 𝑂
\balancing\
Balancing Equations
●
●
●
●
Only change the coefficient
Only make one change at a time
Re-count the numbers of each atom after every change
Be Careful of Diatomic Molecules (HOFBrINCl)
EXAMPLES:
Writing chemical formulae
\formula\
Click Here for Steps (all common ions and acids are listed here and here (the ones you
gotta memorize))
Write the chemical formula of the following substances
Substance
Formula
water
H2O
Sodium phosphate
Na3+PO43−
Calcium Phosphate
CaSO4
Ammonium Nitrate
NH4NO3
Hydrochloric Acid
HCl
Nitric Acid
HNO3
Aluminum Chloride
AlCl3
Magnesium Oxide
MgO
Hydrogen Chloride
HCl
Lithium Hydrogen Carbonate
LiHCO3
Sulfuric Acid
H2SO4
Phosphoric Acid
H3PO4
Now write balanced equations from the information below:
Barium hydroxide reacts with hydrochloric acid to form barium chloride and water.
𝐵𝑎(𝑂𝐻)2 + 2𝐻𝐶𝑙 − −> 𝐵𝑎𝐶𝑙2 + 2𝐻2 𝑂
Calcium reacts with hydrochloric acid to form barium chloride and water,
3𝐶𝑎 + 2𝐻3 𝑃𝑂4 − −> 𝐶𝑎3 (𝑃𝑂4 )2 + 3𝐻2
Magnesium carbonate decomposes to form magnesium oxide and carbon dioxide.
𝑀𝑔𝐶𝑂3 − −> 𝑀𝑔𝑂 + 𝐶𝑂2
Types of Formula
\molecular\ \empirical\
Molecular Formula - The actual number of atoms of each element in a molecule
Example: 𝐻2 𝑂 (Water) or 𝐶6 𝐻6 (Benzine)
Empirical: The simplest whole number ratio of the atoms of each element in a compound in its
lowest term.
Example: 𝐶6 𝐻6(Molecular) → CH (Empirical)
C: 6 , H: 6
→ C: 1 , H: 1
NOTE: because of their structure, ionic and giant covalent compounds do not form molecules
so empirical formula is the only one relevant. More on this in the bonding topic (ie. ionic which
are a metal and a nonmetal can’t have molecular only empirical)
You will meet other types in the organic chemistry unit including structural displayed and
skeletal
Examples:
Substance
Molecular Formula
Empirical Formula
𝑂2
𝑂
Water
𝐻2 𝑂
𝐻2 𝑂
Sodium Chloride
𝑁𝑎𝐶𝑙
𝑁𝑎𝐶𝑙
Ethane
𝐶2 𝐻6
𝐶𝐻3
Glucose
𝐶6 𝐻12 𝑂6
𝐶𝐻2 𝑂
𝐶𝑢𝑆𝑂4
𝐶𝑢𝑆𝑂4
𝐶𝑙2
𝐶𝑙
𝐶3 𝐻8
𝐶3 𝐻8
𝑂3
𝑂
𝐻2 𝑆𝑂4
𝐻2 𝑆𝑂4
𝐶2 𝐻5 𝑂𝐻
𝐶2 𝐻5 𝑂𝐻
𝑁𝑎𝑂𝐻
𝑁𝑎𝑂𝐻
Oxygen
Copper (II) Sulphate
Chlorine Gas
Propane
Ozone
Sulfuric acid
Ethanol
Sodium Hydroxide
A bit about the Mole:
\mole\ \mol\
● A mole is the mass of substance containing the same number of grams as the amount of
atoms in 12.000 g of 12C
● A mole = 6.02 ∗ 1023 atoms. This number is avogadro's number referred to from now as
an avocado (�).
● Therefore 1 mol of carbon 12 is 12 grams and has an � # of atoms
● Used for calculating atoms/molecules/particles/
● Think of an � as the ratio of atoms to moles (it is constant no matter which element)
● Variables:
○ Mole
■ Symbol: 𝑛
■ Found by: Equation (below)
■ Equation: 𝑔𝑟𝑎𝑚𝑠 / 𝑔𝑚𝑜𝑙 −1 OR 𝑎𝑡𝑜𝑚𝑠 𝑜𝑟 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒𝑠 𝑜𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 / 🥑
■ Meaning: The amount of amu’s of an element inside the amount of grams
of a sample (it’s a unit of measurement)
○
Molecular Weight (amu)
■ Symbol: 𝑔𝑚𝑜𝑙 −1
■ Found by: Periodic Table
■ Equation: 𝑔𝑟𝑎𝑚𝑠 / 𝑚𝑜𝑙𝑒𝑠
■ Meaning: Complicated process to find by scientists. It is it’s own unit of
measurement to measure a single element’s weight.
○
Grams
■ Symbol: 𝑔
■ Found by: Measuring
■ Equation: 𝑚𝑜𝑙 ∗ 𝑔𝑚𝑜𝑙 −1
■ Meaning: The amount of grams of a sample (unit of measurement)
○
Atoms/Particles/Molecules
■ Symbol: n/a
■ Found by: Equation (below)
■ Equation: 𝑚𝑜𝑙 ∗ 🥑
■ Meaning: The number of particles in a sample
Molar Ratios
\molar ratio\ \mole ratios\ \ratio\ \ratios\ \grams from chemical equation\ \moles from chemical
equation\ \n(wanted)\ \n(given)\ \wantedS\ \givenS\
If we have the reaction of sodium with oxygen:
4 𝑁𝑎(𝑠) + 𝑂2 (𝑔) − −> 2 𝑁𝑎2 𝑂
And we are given that the mass of oxygen is 3.20g, we can calculate everything else using
molar ratios.
1. The coefficients in front of a compound shows the ratio. In this case it is
4𝑁𝑎 ∶ 2𝑂 =
2 𝑁𝑎 ∶ 1 𝑂
2. Calculate the moles of the compound
𝑔
𝑔 𝑚𝑜𝑙 −1
3.20
𝑚𝑜𝑙 =
31.998
𝑚𝑜𝑙(𝑜𝑥𝑦𝑔𝑒𝑛) = 0.100 𝑚𝑜𝑙
𝑚𝑜𝑙 =
3. Multiply by the ratio to find the moles of the other compound
0.100 ∗ 2
𝑚𝑜𝑙(𝑠𝑜𝑑𝑖𝑢𝑚) = 0.200 𝑚𝑜𝑙
4. Now calculate the grams of whatever you need
𝑔 = 𝑚𝑜𝑙(𝑔𝑚𝑜𝑙 −1 )
𝑔 = 0.200(22.99)
𝑔(𝑠𝑜𝑑𝑖𝑢𝑚) = 2.299𝑔
5. You can also find the total mass of the products by adding the two masses
2.29 + 3.2
𝑚𝑎𝑠𝑠(𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠) = 5.49𝑔
There is also an algebraic way to calculate #2 (and sort of skip it without having to think)
EXAMPLE: What quantity in moles of H3PO4 is required to just neutralize 2.70 mol NaOH in:
3NaOH
+
H3PO4
→
Na3PO4
+
H2O
This is pretty obvious it would just be 2.70/3 = 0.9 but you can do it algebraically by using the
equation:
𝑤𝑎𝑛𝑡𝑒𝑑𝑆
𝑛(𝑤𝑎𝑛𝑡𝑒𝑑) = 𝑛(𝑔𝑖𝑣𝑒𝑛) ∗
𝑔𝑖𝑣𝑒𝑛𝑆
where:
● n(wanted) is the moles you want (in this case n(H3PO4))
● n(given) is the moles you already know (in this case n(NaOH) = 2.70 mol)
● wantedS is the coefficient of what you want (in this case it is 1)
● givenS is the coefficient of what you have (in this case it is 3)
Plug and chug:
𝑤𝑎𝑛𝑡𝑒𝑑𝑆
𝑛(𝑤𝑎𝑛𝑡𝑒𝑑) = 𝑛(𝑔𝑖𝑣𝑒𝑛) ∗
𝑔𝑖𝑣𝑒𝑛𝑆
1
𝑛(𝐻3𝑃𝑂4) = 𝑛(𝑁𝑎𝑂𝐻) ∗
3
1
𝑛(𝐻3𝑃𝑂4) = 2.70 ∗
3
𝑛(𝐻3𝑃𝑂4) = 0.9 *notice how it’s the same as when we used common sense to do it
Relative Atomic Mass, Ar
\relative\ \ar\
The relative mass (Ar) of an element is the average mass of all the isotopes of an element in a
naturally occurring sample of an element relative to a 12th of the mass of 12 𝐶 (carbon 12).
Example: Ar of
12
𝐶 is just 12 (look at the periodic table for Ar)
Relative Molecular Mass, Mr
\mr\
The relative molecular mass (Mr) of a compound is the mass of a molecule of that compound
relative to a 12th of the mass of 12 𝐶.
Example: Mr of 𝐶𝑂2 is:
Element and Quantity
Ar (Relative Atomic Mass)
Product
𝐶-1
12.01
1 * 12.01 = 12.01
𝑂-2
16
2 * 16 = 32
Then we take the sum of the products (= 44.01)
ஃ the Relative Molecular Mass of 𝐶𝑂2 is 44.01
Calculating Relative Molecular Mass
It is the sum of the relative atomic masses of the individual atoms making up a molecule.
Formula
Mass
Formula
𝐻𝐶𝑙
1.01 + 35.46 = 36.46
𝐶2 𝐻4
28.06
𝑀𝑔(𝑂𝐻)2
58
𝐵𝑟2
159.8
(𝑁𝐻4 )2 𝑆𝑂4
132
𝐶6 𝐻12 𝑂6
180
𝐶3 𝐻8
3,611.08
𝐻2 𝑆𝑂4
Mass
98.09
Percentage composition by mass
\percentage composition\ \percentage\ \composition\ \%\
How much percent an atom takes out of the total compound
% by mass of an element =
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ∗ 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑎𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠
𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
∗ 100
Calculating percentage composition by mass
Calculate the percentage by mass of each element in ethanol, 𝐶2 𝐻5 𝑂𝐻
𝐶2 : 12*2 = 24 → 24/46*100 = 52%
𝐻6 : 1*6 = 6 → 6/46*100 = 13%
𝑂: 1*16 = 16 → 16/46*100 = 35%
TOTAL: 24+6+16 = 46
Calculate the percentage by mass of each element in ethanol, 𝐶𝑢𝑆𝑂4 ⋅ 5𝐻2 𝑂
NOTE: The dot means that it is mixed with water. This means there is a veeeery weak
bond with ethanol and water that can be broken by heating (which is why the formula is
different than that above it). (ie. Substance's water of crystallization/hydrated. Anhydrous
version would be 𝑪𝒖𝑺𝒐𝟒 )
𝐶𝑢: 63 * 1 = 63 → 63 / 249 * 100 = 25%
𝑆: 32 * 1 = 32 → 32 / 249 * 100 = 13%
𝑂9 : 16 * 9 = 144 → 144 / 249 * 100 = 58%
𝐻10: 1 * 10 = 10 → 10 / 249 * 100 = 4%
TOTAL: 63+32+144+10 = 249
WORD PROBLEMS
Example 1:
Calculate the mass of oxygen present in 2.20g of carbon dioxide.
2∗16
% by mass = 44.01 ∗ 100=72.71% ← percent of oxygen in carbon dioxide
Mass of oxygen = 72.71% * 2.2 = 1.6g
Example 2:
3*16 +1*1+14*1 = 63
What mass of nitric acid contains 2.00g of oxygen?
% by mass =
3∗16
=76.19%
63
Mass of nitric acid * 76.19 = 2
Mass of nitric acid = 2/76.19
Mass of nitric acid = 2.625g
Going back to empirical/molecular formula, we are now going to calculate them from
percent composition by mass (How much percent an atom takes out of the total
compound)
\empirical\ \molecular\ \percent\ \composition\ \percent composition\ \%\
Follow the steps below (we are using the example 𝐶 − 0.681𝑔, 𝐻 − 0.137𝑔, 𝑂 − 0.181𝑔):
1. Write the symbols and the mass or % composition below them
Element
Carbon
Hydrogen
Oxygen
Mass (g)
0.681
00.137
0.181
2. Divide each mass or % composition by the Ar (relative atomic mass) of the element to
get the number of moles
Element
Carbon
Hydrogen
Oxygen
m / Ar (Equation)
0.681 / 12.01
0.137 / 1.01
0.181 / 16
Equation Answer
(= # of moles)
0.0567 mol
0.16 mol
0.0113 mol
3. Divide all the moles by the smallest mole (in this case it would be 0.0113 mo)
Element
Carbon
Hydrogen
Oxygen
mole / smallest
mole (Equation)
0.0567 / 0.0113 =
0.136 / 0.0113 =
0.0113 / 0.0123 =
Equation Answer
5.02
12.04
1
4. Multiply to remove fractions not always needed (in this case it’s not because they’re
all pretty much whole numbers)
5. Now these 3 numbers are your numbers for each element in your compound. (Element
is C5H12O)
6. To \convert from empirical to molecular\ Divide Mr by formula mass or opposite
depending which one helps you more tbh and multiply the empirical formula by this
number
To get Mr (Ar from Ptable and Quantity from empirical formula):
Element
Carbon
Hydrogen
Oxygen
Quantity * Ar
5 * 12 =
12 * 1 =
1 * 16 =
Answer
60
Total
60
12
+
16
12
+
16
=
88
To get formula mass just look at the question (“𝐶 − 0.681𝑔, 𝐻 − 0.137𝑔, 𝑂 − 0.181𝑔”). In this
case it gives us the separate masses by grams so just add them all up.
(0.681+0.137+0.181=0.999)
𝑀𝑟/𝐹𝑜𝑟𝑚𝑢𝑙𝑎 𝑀𝑎𝑠𝑠
= 88/0.999
= 88.09
Now multiply the empirical formula by this number to get the molecular:
𝐶5𝐻12𝑂 ∗ 88 =
𝐶440𝐻1056𝑂88 ← and that's the molecular formula
*NOTE: Example questions in notebook page “09/08/19”
Going back to empirical/molecular formula, we are now going to calculate them from
Combustion data
\empirical\ \molecular\ \combustion\
From the combustion masses we need to find the empirical formula of Menthol.
In this example question, Menthol contains carbon, hydrogen, and oxygen. A 0.2010 g of
Menthol is combusted, producing 0.5658g of carbon dioxide and 0.2318g of water (h2o) are
formed. Calculate the empirical formula.
𝐶𝑥 𝐻𝑦 𝑂𝑧 − −> 𝐶𝑂2 + 𝐻2 𝑂
1. Write the elements produced from the combustion formula and calculate the # of
moles of each compound. (Mass of compound is the mass of each compound in
grams and Mr is relative molecular mass.)
Mass of Compound:
CO2: 0.5658g
H2O: 0.2318g
Mr (relative molecular mass):
CO2: C : 12*1 + O : 16*2 = 44.01
H2O: H : 1*2 + O : 1*16 = 18.02
Element
Carbon Dioxide
Water
Equation (Mass of
compound / Mr)
0.5658 / 44.01
0.2318 / 18.02
Moles (n)
n(CO2) = 0.0128 mol
n(H2O) = 0.1286 mol
2. Find the moles of all the elements in the compound (only find the carbon and
hydrogen. Oxygen if present will be calculated in the next step).
1 mole of carbon dioxide has one mole of carbon (find this out by looking at formula). Therefore,
0.0128 moles of carbon dioxide is 0.0128 moles of carbon.(0.0128*1)
1 mole of water has two moles of hydrogen (find this out by looking at formula). Therefore,
0.01286 moles of water is 0.02572 moles of hydrogen. (0.01286*2)
3. Now deduce the mass and the moles of oxygen (skip this if no oxygen in the
compound)
*In this case notice how oxygen is produced in the compound and is in the reactants therefore
we do not skip this step.
To get the moles of oxygen, we first need the mass. To get this we will convert the carbon
moles and hydrogen moles (the 2 other elements) to grams and subtract them from the total
grams.
Element
Carbon
Hydrogen
Mol * Atomic Mass
(Equation)
0.0128*12.01
0.0257*1.01
Equation Answer (Grams)
0.15 g
0.0259 g
Now, subtract total by those 2 0.2010 − 0.15 − 0.0259
= 0.0205 𝑔 of oxygen
Now we can find the moles of oxygen using 𝑔𝑟𝑎𝑚𝑠/𝑔𝑚𝑜𝑙 −1
=0.0205/15.999
=0.00128 moles of oxygen
Now that we know the moles of everything we continue from the other methods’s steps:
4. Divide all the moles by the smallest mole (in this case it would be 0.00128 mol)
Element
Carbon
Hydrogen
Oxygen
mole / smallest
mole (Equation)
0.0128 / 0.00128 =
0.0257/0.00128
0.00128/0.00128
Equation Answer
9.98
19.97
1
5. Multiply to remove fractions (not always needed) in this case it isn’t needed since
they’re all pretty much whole numbers.
6. Now these 3 numbers are your numbers for each element in your compound. (Element
is C10H20O)
NOTE: CTRL+F \.convert from empirical to molecular.\ OR click here for converting from
empirical to molecular
NOTE: Example questions in notebook page “09/08/19” ones that are marked with blue
explosion
Solutions & Concentration
\solution\ \solute\ \solvent\ \concentration\
A Solute is a substance that is dissolved in a solvent. These are often solids but don’t have to
be (ex. Hydrochloric acid is a solute when dissolved in water)
A Solvent is what a solute is dissolved in. It is always a liquid or gas.
A Solution is formed when a solute is dissolved in a solvent
Concentration is the strength of a solution (more solute in solvent = more concentration)
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 =
𝑚𝑜𝑙𝑒𝑠 (𝑛)
,
𝑣𝑜𝑙𝑢𝑚𝑒 (𝑑𝑚3 )
the correct notation is 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑚𝑜𝑙 𝑑𝑚−3
NOTE: The correct unit for volume is 𝑑𝑚3 . This is equal to one litre. To convert between 𝑐𝑚3
and 𝑑𝑚3 :
←⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ *1,000⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯
𝑐𝑚
3
𝑑𝑚3(which equals to 1 litre)
⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ÷1,000⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ →
Formulas (c = concentration (mol dm^-3), n = moles (mol), v = volume (dm^3))
𝑐 =
𝑛
𝑣
𝑛 = 𝑐𝑣
𝑣=
𝑛
𝑐
𝑐1 𝑣1
𝑛1
=
𝑐2 𝑣2
𝑛2
*for the third equation click here
The 2 last formulas can be used to skip the steps below. However, they won’t get you full
method marks
Example Questions:
25.0 cm^3 of a solution of hydrochloric acid contains 0.100 mol HCl. What is it’s concentration?
1. Write down all the information you know (concentration, moles, and volume are the 3
things)
HCl
V = 25cm^3
= 25/1000
V = 0.025 dm^3
n = 0.100
C=?
2. Calculate anything you can
𝑚𝑜𝑙
𝑣𝑜𝑙𝑢𝑚𝑒
𝑛
0.1
𝑐 =
=
𝑣
0.025
𝑐 =
= 4 mol dm^-3
Water is added to 4.00g NaOH to produce a 2.00 mol dm^-3 solution. What volume should the
solution be in cm^3?
1. Write down all the info you know
NaOH
m = 4g
c = 2mol dm^-3
n=?
v =?
2. Calculate anything you can
𝑛 =
𝑚
4
=
−1
𝑔𝑚𝑜𝑙
40
n = 0.1 mol
𝑣 =
𝑛
0.1
=
𝑐
2
v = 0.05 dm^3
PAY ATTENTION TO UNITS (GOTTA CONVERT FROM DM TO CM)
v = 0.05 dm^3 * 1000
v = 50cm^3
It is found by titration that 25cm^3 of an unknown solution of sulfuric acid is neutralised by
adding 11.3 cm^3 of 1.00 mol dm^-3 sodium hydroxide. What is the concentration of sulfuric
acid in the sample?
H2SO4 + 2 NaOH → Na2SO4 + 2 H2O
1. Write down what you know
H2SO4
+
2 NaOH
v = 25 cm^3
v = 11.3 cm^3
v = 0.025 dm^3
v = 0.0113 dm^3
c=?
c = 1 mol dm^-3
n=?
n=?
2. Calculate what you can
H2SO4
+
2 NaOH
c = not calculable
n = cv = 0.0113*1
n = not calculable
n = 0.0113 mol
3. If there’s something missing and you can’t calculate something it probably means you
should probably use molar ratios. (if not be smart and do something else and if worst
comes to worst use the plain algebra method)
H2SO4
+
2 NaOH
* Since there are 2 NaOH’s and only 1 H2SO4 you can divide the moles of NaOH by 2 to get
H2SO4
+
2 Na OH
n = n(2 NaOH)/2
n = 0.0113 / 2
n = 0.00565
Now we have enough information to calculate concentration:
𝑛 0.00565
𝑐 = =
𝑣
0.025
c = 0.226 mol dm^-3
4. Plain algebra method if worst comes to worst or if you just want to solve quickly. As I
stated before using these.
H2SO4
2 NaOH
𝑐1 𝑣1 𝑐2 𝑣2
=
𝑛1
𝑛2
𝑐1 (0.025)
𝑛
=
1∗0.0113
*notice
2𝑛
how we have to use molar ratios here too
𝑐(0.025) 1 ∗ 0.0113
=
̶𝑛̶
2 ̶𝑛̶
1 ∗ 0.0113
𝑐 =
/0.025
2
c = 0.226 mol dm^-3
Reaaaaaaaaaaaaaaaly Small Concentrations
\concentration\ \ppm\ \parts per million\ \small concentration\ \small concentrations\
When dealing with very small concentrations we use parts per million (ppm)
Example: 1 gram of a solute is present in 1 million grams of a solution
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑥 106
𝐶𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑝𝑝𝑚 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
1 𝑔𝑟𝑎𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑒 ∗ 106
=
1 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 𝑔𝑟𝑎𝑚𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
106
= 6
10
= 1 ppm
This is pretty obvious because 1 gram of a solute in 1 million grams of a solution should be 1
part per million
Example 2: A sample of 252.10g of water is found to contain 2.03mg of cyanide, what is the
cyanide concentration in ppm?
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑦𝑎𝑛𝑖𝑑𝑒 = 2.03 ∗ 10−3 𝑔
2.03 ∗ 10−3 ∗ 106
𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑐𝑦𝑎𝑛𝑖𝑑𝑒 𝑖𝑛 𝑝𝑝𝑚 =
252.10
2.03 ∗ 103
=
252.10
2030
=
252.10
= 8.05ppm
*NOTE: This notation is commonly used to discuss the concentrations of various pollutant
gasses in the air. In this case it would be: 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑝𝑝𝑚 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑔𝑎𝑠 ∗ 106
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑖𝑟
Example of Working out the concentration of ions:
Calculate the number of moles of chloride ions present in 50.0cm^3 of a 0.0500 mol dm^-3
solution of iron (III) chloride (FeCl3) and the total concentration of all the ions present
1. First, calculate the number of moles
𝑛 = 𝑐𝑣
𝑛(𝐹𝑒𝐶𝑙3) = 0.0500 ∗ 0.05 ← * don’t be confused that they’re the same number the first
one just represents the concentration and the second is the volume in dm^3 (50
cm^3/1000 = 0.05 dm^3)
n(FeCl3) = 2.50*10^-3
2. Find the concentration of all the ions
a. Find the molar mass of: Cl3
Since it is 3 Cl’s with a molar mass of 2.50*10^-3 we just multiply the 2 numbers
to get the molar mass of Cl3
𝑛(𝐶𝑙 − (𝑎𝑞)) = 3 ∗ 2.50 ∗ 10−3
= 0.0075
b. Now get the total concentration from the amount of ions multiplied by the
concentration of the total compound (4 ions in FeCl3)
𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑖𝑜𝑛𝑠 𝑝𝑟𝑒𝑠𝑒𝑛𝑡 = 4 ∗ 0.0500
total concentration of ions present = 0.200 mol dm^-3
Molar Ratios in Conjunction with Limiting Reactant/Reagent
\limiting\ \limiting reactant\ \limiting reagent\ \reagent\ \excess\ \theoretical yield\ \yield\ \%yield\
\percent yield\ \percentage yield\
Using molar ratios is easy enough. If you click on the link, you only need to do up to step
number 2 in order to work with limiting reactants (also known as limiting reagents.)
Click here to find out how to do it algebraically
In a reaction we can describe the reactants as being ‘limiting’ or ’excess’
Limiting - This is the reactant that runs out (also called reagent)
Excess - The reactant that will not run out
More definitions that you should know:
● Theoretical Yield - the maximum amount of product you would make if the limiting
reagent was fully converted to product
● The actual amount of product collected in a reaction (this is usually less due to
substances sticking on the side of the beaker, evaporation, etc.)
●
𝑎𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
%yield = 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑 ∗ 100
In this example, instead of reactants we will use tires, headlights, and cars (to symbolise the
reactants):
4
+ 2
→ �
If we have 20 tires and 14 headlights we can only make 5 cars.
We have a limiting reagent of tires.
We have an excess of headlights.
We find this by dividing 20 / 4 = 5
14 / 2 = 7
The smallest number is 5 so the limiting reagent is tires.
For reactants we do the EXACT same thing. Except the actual number of components= moles.
the coefficient is still the coefficient
Therefore, the equation for finding the limiting reagent is:
moles / coefficient
Do this for each reactant and the one that is the smallest is the limiting reagent.
EXAMPLES:
2𝐻2 + 𝑂2 − −> 2𝐻2 𝑂
If you have 2.0 mol of H2 and 2.0 mol of O2
- H2: 2.0 mol / 2 = 1
- O2: 2.0 mol / 1 = 2
Therefore
- H2 is the limiting reagent
- O2 is excess
*NOTE: Another common example of excess in oxygen in combustion because there is a looot
more oxygen in the air than there is whatever reactant is combusting.
Below is another example that asks for the amount of product produced in moles and steps to
go through with it. (note you can’t just use molar ratios because one of the reactants is in
excess so if you use molar ratios with the excess one it would be pretending as if there wasn’t a
limiting reagent)
EXAMPLE:
What quantity in moles, of MgCl2 can be produced by reacting 10.5g magnesium with 100cm^3
of 2.50 mol dm^-3 hydrochloric acid solution?
1) Write the equation and balance it
Mg + 2HCl → MgCl2 + H2
2) Calculate whatever moles you can for the reactants (don’t worry about the products for now)
Mg
+
m = 10.5g
Mr = 24.3
n = 10.5/24.3 = 0.432 mol
2HCl
→
MgCl2
+
H2
v = 0.10dm^3
c = 2.5 mol dm^-3
n = cv = 0.25 mol
3) Divide the number of moles by the coefficient to figure out the limiting reagent
Mg: 0.431 / 1 = 0.432
2HCl: 0.25 / 2 = 0.125
Since HCl is smaller, therefore HCl is the limiting reagent
4) Now use the molar ratios using the LIMITING REAGENT ONlY, DO NOT use the one in
excess (that’s why I said don’t worry about the products because we can’t calculate the
products without knowing which is limiting first)
n(MgCl2) = n(HCl) / 2
n(MgCl2) = 0.25 / 2
= 0.125 moles of MgCl2
Extra Note: for questions that ask how much of which reactant would remain?
Find the moles of the excess (going back to the above one):
n(Mg) = 0.432 / 2
= 0.216 moles of Mg
This number isn’t the actual number of moles (0.432 still is) but is the number if it wasn’t in
excess.
So subtract 0.432 - 0.216
= 0.216 moles of Mg leftover
^^^^^ This number is the number of moles of Mg leftover (since it is in excess)
NOTE: another little trick is that if you have an acid and a metal for the reactants then the
product that is the compound of the metal/acid will be the same volume as the acid. If
that doesn’t make sense than for example:
2HCl + Zn —> ZnCl2 + H2
The volume of ZnCl2 is the same as the volume of 2HCl
Molar Ratios for Gases
\gasses\ \gas moles\ \gas mole\ \gas mol\ \moles for gasses\ \mol for gas\ \mol gas\ \gas mol\
\mole gas\ \gas mole\ \gas mol\ \STP\ \stp\ \RTP\ \rtp\ \gas pressure\ \gas temperature\ \22.7\
\pressure\
Avagrado's law states that equal volumes of all gasses at the same temperature and pressure
have the same number of moles.
In other words, at a given temperature and pressure, the volume occupied by one mole of gas is
the same as any other given gas.
This is true-ish (we'll find out why "ish" in the next lesson).
To generalize pressure and temperature of a gas we have STP (standard temperature and
pressure) and we also have RTP (room temperature and pressure):
STP = 273 Kelvin and 1.01*10^5 Pascals. At STP the volume of gas is 22.7dm^3 mol^-1
RTP = 298 Kelvin and 1.01*10^5 Pascals. At RTP the volume of gas is 24.5dm^3 mol^-1
This above information makes it easy to convert between volume and moles of gasses given the
dm^3 per moles (listed above for STP and RTP), most questions will use STP though.
Equations:
𝑣
22.7
𝑣 = 𝑛(22.7)
22.7 = 𝑣/𝑛 *NOTE: This 22.7 can be exchanged with any dm^3 per mole; _TP for questions
answering finding the dm^3 per mole
𝑛 =
EXAMPLE: At STP, 30cm^3 ethane reacts with 60cm^3 oxygen. What volume CO2 is
produced?
1. First, write out the BALANCED equation with everything you know and calculate the moles
using the above equations to find the limiting reagent (I'm quickly going over this step but if you
want more in-depth click here except there it calculates using grams and concentration while we
are using STP (same concept, different equations basically))
2C2H6
+
7O2
-->
6H2O
+
4CO2
v = 0.03 dm^3
v = 0.06 dm^3
n = v/22.7 = 0.0013 moles
n = v/22.7 = 0.0026 mol
÷2
÷7
= 6.60*10^-4
= 3.77*10^-4
Therefore Oxygen is limiting
2. Now just find CO2's moles
n(wanted) = n(given) * wanteds/givens
n(CO2) = n(O2) * 4/7
n(CO2) = 0.0026 * 4/7
n(CO2) = 0.00151
3. Now, you can calculate the volume CO2 using the above equations
v = n(22.7)
v = 0.00151(22.7)
v = 0.034 dm^3 <-- final answer
Ideal Gas Calculations
\ideal\ \ideal gas\ \nRT\ \pv\ \pv = nRT\ \pv=nRT\ \pv=nrt\ \pv = nrt\
This is the same as above but just different equation. Pretty simple really…
You HAVE to use this one if you have enough information to use it, if not use the one above
EQUATIONS:
𝑃𝑉 = 𝑛𝑅𝑇 OR
●
●
●
●
●
𝑃1 𝑉1
𝑇1
=
𝑃2 𝑉2
𝑇2
, where:
P = pressure in Pa
V = volume in m^3 (NOT dm^3)
n = moles of a gas
R = gas constant = 8.31 JK^-1 mol^-1 (joule per kelvin per mole)
T = temperature in kelvin
These are assumptions this equation uses (just memorize them):
● Particles occupy no volume (are less and less valid at high pressure)
● Particles have no intermolecular forces (are less and less valid and low temperatures)
EXAMPLE 1 (FOR FIRST EQUATION):
* NOTE: This equation is used for gases where nothing changes
1.048g of an unknown gas A, occupies 846cm^3 at 500K and standard pressure. What is its
molecular mass?
pv = nRT
p = 100KPa = 100,000 Pa
v = 846cm^3 = 846*10^-6 m^3
r = 8.31 JK^-1 mol^-1
T = 323K
N = 45.4 mol
PV = Nrt
P = Nrt/V
P = 45.4(8.31)(323) / 3000*10^-6
P = 40,619,834 Pa
EXAMPLE 2 (FOR SECOND EQUATION):
* NOTE: This equation is used for gases where stuff changes
The volume of an ideal gas at 54 kelvin is increased from 3.00m^3 to 6.00m^3. At what
temperature in kelvin will the gas have original pressure?
P1 = ? P2 = ?
V1 = 0.003 V2 = 0.006
T1 = 327 T2 = ?
P1V1/T1 = P2V2/T2
P1(3)/54 = P2(6)/T2
* P1 = P2 because same room therefore:
3/54 = 6/T2
T2 = 108 Kelvin