5.3: Moles and Chemical Equations
5.3: Rule for Using Chemical Equations
Mole concept is used to interpret chemical
equations.
2 H2 (g) + O2 (g) → 2 H2O (l)
1) Ask yourself “Am I in moles”
if not,
2 molecules Hydrogen gas reacts with 1 molecule
of Oxygen gas to form 2 molecules of water
OR
2 moles hydrogen gas reacts with 1 mole of
oxygen gas to form 2 moles of water
NOT 2 g H2 reacts with 1 g O2
CONVERT TO MOLES !
!!!
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
Example:
Calculate the number of moles of magnesium oxide
which can be obtained by burning 1 mole of magnesium
metal in air.
CALCULATIONS - STEP 1
Calculate the number of moles of magnesium oxide
which can be obtained by burning 1 mole of magnesium
metal in air.
Write and Balance the equation
2 Mg (s) + O2 (g) → 2 MgO (s)
Interpretation:
2 moles of Mg atoms react with 1 mole of oxygen
molecules to give 2 moles of MgO solid
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
Calculate the number of moles of magnesium oxide
which can be obtained by burning 1 mole of magnesium
metal in air.
CALCULATIONS - STEP 2
Calculate the number of moles of magnesium oxide
which can be obtained by burning 1 mole of magnesium
metal in air.
CALCULATIONS - STEP 3
Set out data and identify what is “known” and
“unknown” from the information given in the question
Use the ratio from the BALANCED equation to
calculate the moles of unknown (MgO):
Unknown
Known: Moles of Mg = 1 mole
2 Mg (s) + O2 (g) → 2 MgO (s)
2 moles MgO
× 1 mole Mg
2 moles Mg
Unknown: Moles of MgO = ?
Known
5
= 1 mole of MgO
Amount (in moles!)
of known
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5.3: Stoichiometry Problems
Via Ratio from
Balanced Equation
Moles of Mg
Example:
Moles of MgO
Calculate the number of moles of aluminium chloride
which can be obtained by burning aluminium metal in
0.80 moles chlorine gas.
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
Calculate the number of moles of aluminium chloride
which can be obtained by burning aluminium metal in
0.80 moles chlorine gas.
Calculate the number of moles of aluminium chloride
which can be obtained by burning aluminium metal in
0.80 moles chlorine gas.
CALCULATIONS - STEP 1
CALCULATIONS - STEP 2
Write and Balance the equation
Set out data and identify what is “known” and
“unknown” from the information given in the question
2 Al (s) + 3Cl2 (g) → 2 AlCl3 (s)
Known: Moles of Cl2 = 0.80 moles
Interpretation:
2 moles of Al atoms react with 3 moles of chlorine
molecules to give 2 moles of AlCl3 solid
Unknown: Moles of AlCl3 = ?
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5.3: Stoichiometry Problems
Via Ratio from
Balanced Equation
Calculate the number of moles of aluminium chloride
which can be obtained by burning aluminium metal in
0.8 moles chlorine gas.
Moles of Cl2
Moles of AlCl3
CALCULATIONS - STEP 3
Use the ratio from the BALANCED equation to
calculate the moles of unknown (MgO):
Unknown
Known
2 Al (s) + 3 Cl2 (g) → 2 AlCl3 (s)
2 moles AlCl3
× 0.80 moles Cl2
3 moles Cl2
Amount (in moles!)
= 0.53 moles of AlCl3
of known
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
Example:
Calculate the mass of magnesium oxide which can be
obtained by burning 20.0 g of magnesium metal in air.
CALCULATIONS - STEP 1
Calculate the mass of magnesium oxide which can be
obtained by burning 20.0 g of magnesium metal in air.
Balance the equation (if not already given):
2 Mg (s) + O2 (g) → 2 MgO (s)
All problems of this sort can be solved
using FIVE simple steps
Interpretation:
2 moles of Mg atoms react with 1 mole of oxygen
molecules to give 2 moles of MgO solid
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
Calculate the mass of magnesium oxide which can be
obtained by burning 20.0 g of magnesium metal in air.
Calculate the mass of magnesium oxide which can be
obtained by burning 20.0 g of magnesium metal in air.
CALCULATIONS - STEP 2
CALCULATIONS - STEP 3
Set out data and identify what is “known” and
“unknown” from the information given in the question
Calculate the number of moles of the “known”
Moles of Mg = mass of Mg (g) / Molar mass of Mg (g.mol-1)
Known: Mass of Mg = 20.0 g
= 20.0 g / 24.31 g.mol-1
Unknown: Mass of MgO = ?
= 0.8227 moles of Mg
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
Calculate the mass of magnesium oxide which can be
obtained by burning 20.0 g of magnesium metal in air.
Calculate the mass of magnesium oxide which can be
obtained by burning 20.0 g of magnesium metal in air.
CALCULATIONS - STEP 4
CALCULATIONS - STEP 5
Use the ratio from the BALANCED equation to
calculate the moles of unknown (MgO):
Calculate quantity (usually mass) of unknown:
2 Mg (s) + O2 (g) → 2 MgO (s)
Mass of MgO = moles of MgO × Molar mass of MgO (g.mol-1)
= 0.8227 mol × 40.31 g.mol-1
2 moles MgO
× 0.8227 moles Mg
2 moles Mg
= 33.2 g of MgO
= 0.8227 moles of MgO
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5.3: Stoichiometry Problems
Via Ratio from
Balanced Equation
Moles of Mg
How much CO2 is produced per km by the average small car?
Assume the fuel is pure octane (C8H18) and 52.35 g is
consumed per kilometre.
{Based on 7.5 L/100 km and density of octane 0.698 g.mL-1}
Moles of MgO
Write and Balance the equation
Via Molar Mass
(g.mol-1)
Via Molar Mass
(g.mol-1)
Mass of Mg
2 C8H18 (g) + 25 O2 (g) →
16
8 CO2 (g) + 18
9 H2O (g)
Mass of MgO
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
How much CO2 is produced per km by the average small car?
Assume the fuel is pure octane (C8H18) and 52.35 g is
consumed per kilometre.
{Based on 7.5 L/100 km and density of octane 0.698 g.mL-1}
How much CO2 is produced per km by the average small car?
Assume the fuel is pure octane (C8H18) and 52.35 g is
consumed per kilometre.
{Based on 7.5 L/100 km and density of octane 0.698 g.mL-1}
Calculate the number of moles of the “known”
Determine known and unknown
2 C8H18 (g) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
Known:
Unknown:
Moles of octane = mass of octane (g)
/ Molar mass of octane (g.mol-1)
mass of octane = 52.35 g
mass of CO2 = x g
= 52.35 g / 114.26 g.mol-1
= 0.4582 moles of octane
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5.3: Stoichiometry Problems
5.3: Stoichiometry Problems
How much CO2 is produced per km by the average small car?
Assume the fuel is pure octane (C8H18) and 52.35 g is
consumed per kilometre.
{Based on 7.5 L/100 km and density of octane 0.698 g.mL-1}
How much CO2 is produced per km by the average small car?
Assume the fuel is pure octane (C8H18) and 52.35 g is
consumed per kilometre.
{Based on 7.5 L/100 km and density of octane 0.698 g.mL-1}
Use the ratio from the BALANCED equation to
calculate the moles of unknown (CO2):
Calculate quantity of unknown (carbon dioxide):
Mass of CO2 = moles of CO2 × Molar mass of CO2 (g.mol-1)
2 C8H18 (g) + 25 O2 (g) → 16 CO2 (g) + 18 H2O (g)
= 3.6656 mol × 44.01 g.mol-1
16 moles CO2
× 0.4582 moles octane
2 moles octane
= 161.3 g of CO2
= 3.6656 moles of CO2
per km!
20000 km / year produces 3.2 tonnes CO2 !
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Rule for Using Chemical Equations
Via Ratio from
Balanced Equation
Moles of Octane
REMINDER
Moles of CO2
REMINDER
REMINDER
REMINDER
When using ratios from a chemical equation,
Ask yourself “Am I in moles”
Via Molar Mass
(g.mol-1)
Via Molar Mass
(g.mol-1)
if not,
CONVERT TO MOLES !
Mass of Octane
Mass of CO2
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5.3: Stoichiometry Problems
Via Ratio from
Balanced Equation
How much water is produced in the combustion
of 10 g of glucose C6H12O6?
Moles of Glucose
Moles of H2O
CALCULATIONS - STEP 1
Write and Balance the equation
Via Molar Mass
(g.mol-1)
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Via Molar Mass
(g.mol-1)
Interpretation:
1 mole of glucose produces 6 moles of water
Mass of Glucose
Mass of H2O
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5.3: Stoichiometry Problems
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5.3: Stoichiometry Problems
How much water is produced in the combustion
of 10 g of glucose C6H12O6?
How much water is produced in the combustion
of 10 g of glucose C6H12O6?
CALCULATIONS - STEP 2
CALCULATIONS - STEP 3
Determine known and unknown
Calculate the number of moles of the “known”
Known:
Unknown:
Moles of glucose = mass of glucose (g)
/ Molar mass of glucose (g.mol-1)
mass of glucose = 10 g
mass of water = ?
= 10 g / 180.18 g.mol-1
= 0.05550 moles of glucose
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5.3: Stoichiometry Problems
Via Ratio from
Balanced Equation
Moles of Glucose
How much water is produced in the combustion
of 10 g of glucose C6H12O6?
Moles of H2O
CALCULATIONS - STEP 4
Via Molar Mass
(g.mol-1)
Via Molar Mass
(g.mol-1)
Use the ratio from the BALANCED equation to
calculate the moles of unknown (water):
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
6 mole H2O
× 0.05550 moles glucose
1 mole glucose
Mass of Glucose
Mass of H2O
= 0.3330 moles of H2O
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5.3: Stoichiometry Problems
Via Ratio from
Balanced Equation
Moles of Glucose
How much water is produced in the combustion
of 10 g of glucose C6H12O6?
Moles of H2O
CALCULATIONS - STEP 5
Calculate quantity of unknown (water):
Via Molar Mass
(g.mol-1)
Via Molar Mass
(g.mol-1)
Mass of H2O = moles of H2O × Molar mass of H2O (g.mol-1)
= 0.334 mol × 18.02 g.mol-1
Mass of Glucose
Mass of H2O
= 6 g of H2O
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(1 sig. fig.)
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Rule for Using Chemical Equations
REMINDER
REMINDER
REMINDER
REMINDER
When using ratios from a chemical equation,
Ask yourself “Am I in moles”
if not,
CONVERT TO MOLES !
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