HTML University
Science Dept.
Chemistry CHE 100 Notes
K. Stoichiometry
- Stoichiometry
- Definition:
- Types of Problems
- Mass to Mass
- Mass to Volume or Volume to Mass
- Volume to Volume
- The Road Map
- Mass to Moles Using the Periodic Table of the Elements
- Moles to Moles Using the Balanced Equation
- Moles to Mass Using the Periodic Table of the Elements
- Periodic Table of the Elements
- Atomic Mass or Weight
- Mole Concept
- Atomic Number
- The Balanced Equation
- See Guidelines in Chemical Equations
- Example: The complete oxidation of ethane
- _____ C2H6(g)
+ _____ O2(g) ----->
_____ CO2(g) + _____
H2O(g) + E
- Let's Solve a Problem: Calculate the mass
of oxygen needed for the complete oxidation of
72 grams of ethane
- (72g ethane) x (mole ethane/mass ethane) = (moles of ethane)
- The relationship needed is between mass of ethane
and moles of ethane
- From the Periodic Table:
- C = 12 amu
- H = 1.0 amu
- C2H6 =
30 amu
- One mole of
C2H6 =
30 grams
- (72g ethane) x (one mole ethane/30g ethane) =
(2.4 moles of ethane)
- (2.4 moles ethane) x (moles O2/moles ethane)
= (moles of oxygen)
- The relationship needed is between moles of ethane
and moles oxygen
- From the Balanced Equation:
- There are 2 moles of ethane to 7 moles oxygen gas
- (2.4 moles ethane) x (7 moles O2/2 moles ethane)
= (8.4 moles of O2)
- (8.4 moles O2) x
(mass of O2/moles O2)
= (mass of O2)
- The relationship needed is between one mole of O2
and the mass of O2
- From the Periodic Table:
- O = 16 amu
- O2 = 32 amu
- one mole of O2 = 32 grams
- (8.4 moles O2) x
(32g of O2/one mole O2)
= (268.8g of O2)
- Mass to Volume
- What is the volume of 268.8 g (or 8.4 moles)
of O2?
- (8.4 moles O2) x (22.4l/mole of gas)
= 188.2 liters of O2 at STP