Stoichiometry
Part 2: Relative masses of atoms and molecules
Part 3: The mole and Avogadro’s constant
1
Learning
objectives
2
• Different atoms have different masses.
NB
definition!
Introduction
These values tell you that:
1. A magnesium atom has twice the mass of a carbon atom, and 24 times more mass
than a hydrogen atom.
2. Hydrogen atoms have 12 times less mass than a carbon atom.
3
• Relative molecular mass (Mr) is defined as →
The sum of the relative atomic masses within a
molecule.
• Relative formula mass (M ) is defined as
Relative
→ The sum of the relative atomic masses within ionic
molecular mass
compounds.
Mr
r
To calculate these values, we add the relative atomic masses of the
atoms in the compound.
4
Calculate the relative molecular mass of the following
molecules:
Practice
1.
2.
3.
4.
Hydrogen (H2)
Water
Ammonia
Carbon dioxide
Calculate the relative formula mass of the following ionic
compounds:
1.
2.
3.
4.
Potassium iodide
Potassium carbonate
Calcium hydroxide
Ammonium sulfate
5
6
Mr Practice
7
Mole Calculations
8
THE MOLE
pair
2
dozen
12
score
20
gross
144
mole
602204500000000000000000
6.02 x 1023
9
A mole is a measure of the amount of matter.
1 mole of a substance = 6.02 x 1023 particles
Example:
Moles
1 mole of
= 6.02 x 1023 frappuccinos
2 moles of
= 2 x 6.02 x 1023 donuts
10
1 mole = 6.02 x 10
6.02 x 10
23
23
particles
= Avogadro's Number (NA)
•1 mole of sodium (Na) contains 6.02 x 1023 atoms of sodium
•1 mole of hydrogen (H2) contains 6.02 x 1023 molecules of hydrogen
•1 mole of sodium chloride (NaCl) contains 6.02 x 1023 ions of Na and Cl
11
N = n x NA
Where
Calculations
using NA
N = number of particles
NA = 6.02 x 1023
n = moles
12
1. Calculate the number of water molecules in 3.5 mol
water.
Practice Calculations
using NA
2. Calculate:
a) The number of SO2 molecules in 3.0 mol of SO2.
b) The number of atoms in 3 mol of SO2
3. Calculate the number of sodium ions in 2.0 mol Na2O.
13
The concentration of a solution is
the amount of solute dissolved in
a solvent to make 1 dm3 (one
cubic decimetre) of solution.
Concentration
The greater the amount of
solute in a given volume,
the greater the concentration
Units for concentration
= moles per decimetre cubed
= mol.dm-3 or mol/dm3
14
• The solvent is usually water
• 1000 cm3 = 1dm3 = 1L
To convert from cm3 to dm3
→ divide by 1000
Calculations
using
concentration
Look out for this:
Concentration in g.dm-3 = concentration in mol.dm-3 x Mr
15
1. Calculate the volume of water required to
make a 2.5 mol.dm-3 solution using 0.5 mol
NaOH.
Calculations
using
concentration
2. Calculate the concentration in mol.dm–3 of
sodium hydroxide, NaOH, if 250 cm3 of a
solution contains 0.05 mol of sodium
hydroxide.
16
Solutions and
concentration
s
practice
17
At room temperature and pressure (@ RTP T= 25℃ and P = 1 atm)
One mole of any gas occupies a volume of 24dm3
Vgas = n x 24
Volume of gas
Where
V = volume of gas (dm3)
n = moles (mol)
Remember: volume must be in units of dm3
cm3
mL
Divide by 1000
dm3
L
18
1. Calculate the volume of 2.5 mol of chlorine (Cl2) gas at
rtp.
Calculations
using volume
of gas
2. How many moles of gas occupy a volume of 24cm3 at
rtp?
3. Calculate the volume of gas occupied by 1.24 x 1024
molecules of CO2 at rtp.
4. Calculate the number of SO2 molecules that occupy a
volume of 250cm3 at rtp.
19
Mass (g)
Moles (mol)
Molar mass (g.mol-1)
Calculations
using mass
1. Calculate the mass of 0.25 moles of potassium chloride.
2. How many moles are in 340g of magnesium chloride?
3. How many moles are in 4.7kg of carbon dioxide?
20
Find this on Google Classroom!
Practice
21
Reacting masses
What a chemical equation means (molar ratios)
22
N2 + 3 H 2
2 NH3
1 molecule
2 molecules
3 molecules
12
36
24
1 dozen
3 dozen
2 dozen
6 x 1023
18 x 1023
12 x 1023
600000000000000000000000
1800000000000000000000000
1200000000000000000000000
1 mole
3 moles
2 moles
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© www.chemsheets.co.uk
AS 1028
25-Jun-2016
Molar ratios
practice
24
1. Calculate the number of moles of the substance
whose mass you have (n = m/Mr)
2. Use mole ratios to determine moles of products
Steps for
mass-mass
calculations
3. Calculate mass of product (m = n x Mr)
Example:
Calculate the mass of ammonia that is produced
when 56g of Nitrogen reacts with excess hydrogen.
N2 + H2 → NH3
NB: You must ALWAYS
balance the equation first
25
1. Calculate the mass of water that is produced
when 70.0g of oxygen reacts with excess
hydrogen.
H 2 + O 2 → H 2O
Practice:
mass-mass
calculations
2. Calculate the volume of ammonia that is
produced when 100.0g of Nitrogen reacts with
excess hydrogen.
N2 + H2 → NH3
26
REACTING MASS CALCULATIONS (Example 1)
1)
What mass of oxygen reacts with 12 g of magnesium?
✔
?
2Mg + O2 → 2MgO
Moles Mg = mass
Mr
= 12
24
= 0.5
Moles O2 = 0.25
Mass O2 = Mr x moles = 32 x 0.25
= 8g
27
© www.chemsheets.co.uk
GCSE 1116
24-Dec-2016
REACTING MASS CALCULATIONS (Example 2)
2)
What mass of calcium hydroxide is made from 14 kg of calcium oxide?
?
✔
CaO + H2O → Ca(OH)2
Moles CaO = mass
Mr
= 14000
56
= 250
Moles Ca(OH)2 = 250
Mass Ca(OH)2 = Mr x moles = 74 x 250
= 18500 g
28
© www.chemsheets.co.uk
GCSE 1116
24-Dec-2016
REACTING MASS CALCULATIONS (Example 3)
3)
What mass of aluminium is needed to react with 640 g of iron oxide?
✔
?
Fe2O3 + 2 Al → 2 Fe + Al2O3
Moles Fe2O3 = mass = 640
Mr
160
= 4
Moles Al = 8
Mass Al = Mr x moles
= 27 x 8 = 216 g
29
© www.chemsheets.co.uk
GCSE 1116
24-Dec-2016
REACTING MASS CALCULATIONS (Example 4)
4)
What mass of titanium chloride reacts with 460 g of sodium?
?
✔
TiCl4 + 4 Na → Ti + 4 NaCl
Moles Na = mass
Mr
= 460
23
= 20
Moles TiCl4 = 5
Mass TiCl4 = Mr x moles
= 190 x 5
= 950 g
30
© www.chemsheets.co.uk
GCSE 1116
24-Dec-2016
You do not need to convert to moles first.
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
When working
with ALL
gases
1 mol
2 mol
1 mol
2 mol
24 dm3
48 dm3
24 dm3
48 dcm3
What volume of ammonia (NH3) will be produced if
10cm3 of nitrogen reacts with 20cm3 of hydrogen?
N2 (g) + 3H2 (g) → 2NH3 (g)
10cm3
30cm3
20cm3
31
Given to you
Calculate using
mass-mass
calculations
Percentage
yield
Example:
Calculate the percentage yield when 0.2g of NH3 is
formed from 10g of N2.
N2 (g) + 3H2 (g) → 2NH3 (g)
1.
2.
3.
Calculate theoretical yield (moles)
Convert theoretical moles to theoretical mass
Plug into equation
32
1. Calculate the % yield when 24g of hydrogen produces
2g of water.
2H2 + O2 → 2H2O
Practice Percentage
yield
2. A reaction has a percentage yield of 72%. What mass of
nitrogen is required to form 3.0g of NH3?
N2 + 3H2 → 2NH3
33
3. Calculate the percentage yield when 2g of sodium
oxide is formed from 50g of sodium
4Na + O2 → 2Na2O
Practice Percentage
yield
4. Calculate the percentage yield when 48g of magnesium
reacts to form 4g of magnesium oxide
2Mg + O2 → 2MgO
34
5. Calculate the percentage yield when 20g of SO2 oxide
reacts with oxygen to form 5g of SO3
2SO2 + O2 → 2SO3
Practice Percentage
yield
35
Percentage
yield
calculations
Quiz: https://chemquiz.net/per/
36
Empirical formula:
The smallest whole number ratio of elements in a
compound
NB: The final answer must be a whole number
Empirical vs
molecular
formula
Molecular formula:
The actual number of each type of atom in a compound
37
1. Calculate the molar mass of the empirical formula
2. Divide the relative formula mass given to you by
your answer
To find
molecular
formula
3. Multiply each element in the EF by the answer
Example:
If EF = CH and relative formula mass is 78g.mol-1
1. Mr (CH) = 13
2. 78/13 = 6
3. MF = C6H6
38
A compound was found to contain 75.46% carbon, 4.43% hydrogen,
20.10% oxygen. What is the empirical formula of the compound?
Step 1: change the % to grams.
75.46%→ 75.48g of C
4.43%→4.43g of H
20.10%→ 20.10g of O
Calculations to find
empirical
formula
Step 2: Find number of moles (n = m/Mr)
75.46 / 12 = 6.288 of C
4.43/1 = 4.43 of H
20.10/ 16 = 1.256 of O
Step 3: Divide each of the moles by the smallest number of moles e.g.
divide by 1.256
6.2866 / 1.256 = 5
4.43 / 1.256 = 3.5
1.256 / 1.256 = 1
Step 4: If you do not end up with whole numbers, times all values by the
same whole number until all fractions are eliminated
5 x 2 = 10
C10 H7 O 2
3.5 x 2 = 7
1x2=2
39
The compound on the previous slide has a formula mass of
318.31 g/mol. Find the molecular formula.
Practice Calculations to find
molecular
formula
Answer:
1. Mr (C10H7O2) = 159
2. 318.31/159 = 2
3. MF = C40H14O4
40
Practice Calculations to find
empirical /
molecular
formula
1. A compound was found to contain 74.1% carbon,
8.6% hydrogen, 17.3% oxygen. The molas mass is
160 g/mol.
a. What is the empirical formula of the compound?
b. Find the molecular formula of the compound.
2. A compound was found to contain 59.0 % carbon, 7.1%
hydrogen, 26.2 % oxygen. The molas mass is
180 g/mol.
(a) What is the empirical formula of the compound?
(b) Find the molecular formula of the compound.
41
Calculating
empirical &
molecular
formulae
42
1 to 3
available
Reacting
masses
practice
43
We can use the formula of a compound and relative atomic masses to
calculate percentage by mass of a particular element in a compound.
Percentage
composition
by mass
Example
Calculate the percentage by mass of iron in iron(III) oxide, Fe2O3.
(Ar values: Fe = 55.8, O = 16.0)
Practice
Calculate the percentage by mass of carbon in ethanol, C2H5OH.
(Ar values: C = 12.0, H = 1.0, O = 16.0)
44
Moles of pure sample
X 100
Moles of total sample
Example
Percentage
purity
When 24.0 g of an impure sample of aluminium was reacted with iron
oxide, 45.7 g of iron was formed. Calculate the percentage purity of the
sample of impure aluminium.
Step 1: Calculate the number of moles of Fe - this is the pure, known
chemical.
Step 2: Use the equation to work out how many moles of Al would have
been needed and convert this to a mass.
Step 3: Percentage purity = mass pure/mass impure x 100
45
The limiting reagent limits the amount of product formed.
Excess = more than what is needed
Limiting
reagents
Example
You are provided with 100g of Cl2 and 100g of Na. Calculate how many
moles of NaCl will form.
2Na + Cl2 → 2NaCl
1. Calculate moles of Na: n=m/Mr
= 100/23 =4.35 mol
2. Calculate moles of Cl2: n=m/Mr
=100/71 = 1.41
3. Use molar ratios to identify how many moles of product will form
using the moles from step 1 and step 2
4.35 mol Na : 4.35 mol NaCl
1.41 mol Na : 2.82 mol NaCl → Limiting reagent
4. Calculate mass of NaCl using moles of limiting reagent
46
Practice Limiting
reagents
47
The mole
This is the mass of a substance containing the same number of
atoms/ions/molecules as there are atoms in exactly 12.00 g of 12C
Relative atomic mass
Definitions
The average mass of the isotopes of an element compared to 1/12th of
the mass of an atom of 12C”.
Covalent bond
Covalent compounds are formed when electrons
are shared between atoms that are non-metal elements.
48
Video:
The mole: https://youtu.be/wI56mHUDJgQ
Revision quiz:
Mass Ratios, Percent Composition & Empirical Formulas
https://chemquiz.net/mas/
Revision
videos
Concentrations and solutions
https://chemquiz.net/sol/
Significant figures
https://chemquiz.net/sig/
49
When 5.00 g of crystals of hydrated tin (II) chloride, SnCl2.xH2O, are
heated, 4.20 g of anhydrous tin (II) chloride are formed.
Calculate the number of molecules of water of crystallisation are in
SnCl2.xH2O (i.e. the value of x).
Waters of
crystallisation
50