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Stoichiometry
στοιχεῖον (stoicheion, meaning element]) & μέτρον (metron, meaning measure)
Balancing Chemical Equations
• 2 Mg + O2 -> 2 MgO
Balancing Chemical Equations
• 2 Mg + O2 -> 2 MgO
1.2 x 1024 + 6 x 1023 form 1.2 x 1024 formula units
atoms O2 molecules MgO
Mg
Balancing Chemical Equations
• 2 Mg + O2 -> 2 MgO
2 moles + 1 mole form 2 moles
Mg O2 MgO
Mole Ratio
Balancing Chemical Equations
• 2 Mg + O2 -> 2 MgO
48.6 g + 32 g form 80.6 grams
Mg O2 MgO
Mass
Balance
The total mass of the product(s) must equal the total mass of the reactants
Yield
• Building Bicycles
Yield• Building Bicycles1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels
Finding Product Quantities
• How much O2 gas is formed when
49.89 g KClO3 decomposes?
KClO3 -> KCl + O2
Balance the Equation
• How much O2 gas is formed when
49.89 g KClO3 decomposes?
2 KClO3 -> 2 KCl + 3 O2
Balance the Equation
• How much O2 gas is formed when
49.89 g KClO3 decomposes?
2 KClO3 -> 2 KCl + 3 O2
1 mole49.89 g KClO3 x 122.45 g = 0.41 moles KClO3
3 moles O2
2 moles KClO3 x 0.41 moles KClO3
= 0.615 moles O2
Balance the Equation
• How much O2 gas is formed when
49.89 g KClO3 decomposes?
2 KClO3 -> 2 KCl + 3 O2
49.89 g KClO3 ?
0.615 moles O2 x 32 g/molee = 19.68 grams O2
Standard Temperature & Pressure
• STP
Temperature = 0 °C (273.15 K; 32.00 °F)
Pressure = 101.325 kPa (1 atm; 760 Torr)
For Gases: Volume = 22.4 Liters/mole
Balance the Equation
• How much O2 gas is formed when
49.89 g KClO3 decomposes?
2 KClO3 -> 2 KCl + 3 O2
49.89 g KClO3 ?
0.615 moles O2 x 22.4 L/mole = 13.776 Liters O2
Yield• Building Bicycles1 Frame + 1 Seat + 1 Handlebar + 2 Pedals + 2 Wheels
= 1 bicycle
Yield
• Theoretical Yield
• How much O2 gas is formed when
49.89 g KClO3 decomposes?
2 KClO3 -> 2 KCl + 3 O2
49.89 g KClO3 6.12 Liters O2
What if only 5.5 Liters of O2 are produced?
Yield
• Theoretical Yield
2 KClO3 -> 2 KCl + 3 O2
49.89 g KClO3 6.12 Liters O2
What if only 5.5 Liters of O2 are produced?
5.5 Liters
6.12 Liters x 100% = 89.9% actual yield
Gas Yield
• Nitroglycerin decomposes to
produce N2, O2, CO2 and H2O
Gas Yield
• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O
227 g how much gas results?
Gas Yield
• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O
1 mole
227 g x 227 g = 1 mole nitroglycerin
Gas Yield
• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O
1.5 moles N2
1 mole nitroglycerin -> 0.25 mole O2
3 moles CO2
0.25 mole H2O
Gas Yield
• 4 C3H5N3O9 -> 6 N2 + O2 + 12 CO2 + H2O
1.5 moles N2 x 22.4 L/ mole = 33.6 L
0.25 mole O2 x 22.4 L/mole = 5.6 L
3 moles CO2 x 22.4 L/mole = 67.2 L
0.25 mole H2O x 22.4 L/mole = 5.6 L
112 Liters !
Making Aspirin
• White willow tree bark = pain reliever
Making Aspirin
• Salicylic acid acetylsalicylic acid
+ acetic anhydride -> and acetic acid
C7H6O3 + C2H6O3 -> C9H8O4 + C2H4O2
each aspirin tablet 325 mg
How much salicylic acid would be needed if we assume 90% yield?
Making Aspirin
C7H6O3 + C2H6O3 -> C9H8O4 + C2H4O2
0.325 g x 1 mole/180 g = 0.0018 moles of aspirin
90 % of 0.0018 = 0.9 x 0.0018 = 0.00162 moles
0.00162 moles x 138 g/mole = 0.224 grams of salicylic acid
Rocket Fuel
N2H4(l) + N2O4(l) -> N2(g) + H2O(g)
Is the reaction balanced?
Rocket Fuel
2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g)
What is the mole ratio
of N2H4 to N2?
Rocket Fuel
2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g)
What is the mole ratio
of N2H4 to N2?
3 moles N2
2 moles N2H4
Rocket Fuel
2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g)
How many Liters of N2 will be
produced from 16000 g of N2H4?
16000 g x 1 mole = 500 moles N2H4
32g
3 moles N2
500 moles N2H4 x 2 moles N2H4
= 750.0 moles N2
= 16800 Liters of N2
Rocket Fuel
2 N2H4(l) + N2O4(l) -> 3 N2(g) + 4 H2O(g) ENERGY
How many Liters of N2 will be
produced from 16000 g of N2H4?
16000 g x 1 mole = 500 moles N2H4
32g
3 moles N2
500 moles N2H4 x 2 moles N2H4
= 750.0 moles N2
= 16800 Liters of N2
Chemical Change
• Energy is absorbed or released
Exothermic vs. Endothermic
Enthalpy ΔH
• Total energy of a system:Internal energy + (pressure) x (Volume)
Enthalpy ΔH
• Only way to measure ΔH is when heat is added or removed:
Molar Enthalpy Change: ΔH = C ΔT
(Recall Specific Heat Capacity: Q = m C ΔT)
Molar Enthalpy Change
• ΔH = CΔT
What is ΔH if an aluminum can is cooled from 25° C to 4° C?
Temperature Scales
Molar Enthalpy Change
• ΔH = CΔT
What is ΔH if an aluminum can is cooled from 25° C to 4° C? (C for Al = 24.2 J/K mole)
25° C = 298 K
4° C = 277 K
ΔH = CΔT = 24.2 J/K mole (21 K)
= 508.2 J/mole
Enthalpy Change
• Negative = Exothermic
C(s) + 1/2 O2(g) -> CO(g) ∆H = - 110.5 kJ
• Positive = Endothermic
C2H6(g) -> 2C(s) + 3H2(g) ∆H = + 4.83 kJ
Enthalpy Change (Endothermic)
A + B ----> C + D
Enthalpy Change (Exothermic)
W + X --> Y + Z
Enthalpy in Chemical Changes
Enthalpy in Chemical Changes
Enthalpy in Chemical Changes
Stoichiometric Enthalpy
• ∆H = ? When 4.8 g C reacts with O2?
• C + O2 -> CO2 ∆H = -393.5 kJ/mole
4.8 g x 1mole/12g = 0.4 moles
0.4 moles x (-393.5 kJ/mole)
= 157.4 kJ
Hess’ Law
• Total ∆H = Sum of ∆H for all steps
• Example:
H2S(g) + 2O2(g) ->SO3(g) + H2O(l)
∆H = ???????
Hess’ Law
H2S(g) + 2O2(g) -> H2SO4(l) ∆H= -628KJ
H2SO4(l) -> SO3(g) + H2O(g) ∆H= +164KJ
H2O(g) -> H2O(l) ∆H= -88KJ
H2S(g) + 2O2(g) ->SO3(g) + H2O(l) ∆H = ?
Hess’ Law
Germain Henri Hess
• Enthalpy change (∆H) depends only on the initial and final states of the reaction
(not on the intermediate stages)
Hess’ cycle
Hess’ Law
• CH4 + 2 O2 ----> CO2 + 2 H2O ∆H = ??
CH4 + 2 O2 -----> CO + 2 H2O + ½ O2
∆H = -607 kJ
CO + 2 H2O + ½ O2 ------> CO2 + 2 H2O ∆H = -283 kJ
Standard Enthalpy of Formation
• Formation from the composing elements:
• Glucose:
• 6 C + 6 H2 + 3 O2 --> C6H12O6
∆H = –1273.3 kJ
Hess’ Law
• CH4 + 2 O2 ----> CO2 + 2 H2O ∆H = -890 kJ
CH4 + 2 O2 -----> CO + 2 H2O + ½ O2
∆H = -607 kJ
CO + 2 H2O + ½ O2 ------> CO2 + 2 H2O ∆H = -283 kJ
Hess’ Cycle
Unit 6 Recap
• Stoichiometry Mole Ratios
Yield = actual/theoretical
Limiting and extra reactants
1 mole = 22.4 Liters at STP
• Enthalpy Change ∆Hendothermic vs. exothermic
∆H = C∆THess’ Law
