transcript
- Slide 1
- Slide 2
- Stoichiometry: The Mole
- Slide 3
- Stoichiometry The study of the quantitative aspects of chemical
reactions.
- Slide 4
- The Mole A counting unit 1 mole = 602 billion trillion:
602,000,000,000,000,000,000,000 6.02 X 10 23 (in scientific
notation) 1 dozen eggs = 12 1 mole of eggs = 6.02 x 10 23 eggs 1
dozen cars = 12 cars 1 mole of cars = 6.02 x 10 23 cars Note that
the NUMBER is always the same, but the MASS is very different! Mole
is abbreviated mol
- Slide 5
- = 6.02 x 10 23 C atoms = 6.02 x 10 23 H 2 O molecules A Mole of
Particles A Mole of Particles Contains 6.02 x 10 23 particles 1
mole C 1 mole H 2 O Note that a particle could be an atom OR a
molecule!
- Slide 6
- 6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles
Avogadros Number as a Conversion Factor
- Slide 7
- 1. Number of atoms in 0.500 mole of Al a) 500 Al atoms b) 6.02
x 10 23 Al atoms c) 3.01 x 10 23 Al atoms 2.Number of moles of S in
1.8 x 10 24 S atoms a) 1.0 mole S atoms b) 3.0 mole S atoms c) 1.1
x 10 48 mole S atoms Learning Check
- Slide 8
- The Mass of 1 mole (in grams) Equal to the numerical value of
the average atomic mass (get from periodic table) 1 mole of C
atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of Cu atoms =63.5 g
Molar Mass
- Slide 9
- Find the molar mass: (usually we round to the tenths place)
Learning Check! A.1 mole of Br atoms B.1 mole of Sn atoms =79.9
g/mole = 118.7 g/mole
- Slide 10
- Chemical Formulas of Compounds Formulas give the relative
numbers of atoms or moles of each element in a formula unit -
always a whole number ratio (the law of definite
proportions).Formulas give the relative numbers of atoms or moles
of each element in a formula unit - always a whole number ratio
(the law of definite proportions). NO 2 2 atoms of O for every 1
atom of N NO 2 2 atoms of O for every 1 atom of N 1 mole of NO 2 :
2 moles of O atoms to every 1 mole of N atoms If we know or can
determine the relative number of moles of each element in a
compound, we can determine a formula for the compound.If we know or
can determine the relative number of moles of each element in a
compound, we can determine a formula for the compound.
- Slide 11
- Mass in grams of 1 mole sum of the atomic masses 1 mole of CaCl
2 = 111.1 g/mol 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol =
111.1 g/mol CaCl 2 1 mole of N 2 O 4 = 92.0 g/mol Molar Mass of
Molecules and Compounds
- Slide 12
- A.Molar Mass of K 2 O = ? Grams/mole B. Molar Mass of antacid
Al(OH) 3 = ? Grams/mole Learning Check!
- Slide 13
- The artificial sweetener aspartame (Nutra-Sweet) formula C 14 H
18 N 2 O 5 is used to sweeten diet foods, coffee and soft drinks.
How many moles of aspartame are present in 225 g of aspartame?
Learning Check!
- Slide 14
- molar mass Avogadros number Grams Moles particles You can
convert Grams to Particles, but you always must go through moles
first. Calculations
- Slide 15
- Flowchart Atoms or Molecules Moles Mass (grams) Divide by 6.02
X 10 23 Multiply by 6.02 X 10 23 Multiply by atomic/molar mass from
periodic table Divide by atomic/molar mass from periodic table
- Slide 16
- Atoms/Molecules and Grams How many atoms of Cu are present in
35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X 10 23 atoms Cu 63.5 g Cu 1
mol Cu = 3.4 X 10 23 atoms Cu
- Slide 17
- Learning Check! How many atoms of K are present in 78.4 g of
K?
- Slide 18
- Learning Check! What is the mass (in grams) of 1.20 x 10 24
molecules of glucose (C 6 H 12 O 6 )?
- Slide 19
- Using Equations Chemical equations also show mole ratios. In
the equation: 2LiOH + CO 2 Li 2 CO 3 + H 2 O For every 2 moles of
LiOH and 1 mole of CO 2, 1 mole of Li 2 CO 3 and 1 mole of H 2 O
are produced.
- Slide 20
- Mole-Mass Conversions Most of the time in chemistry, the
amounts are given in grams instead of moles We still go through
moles and use the mole ratio, but now we also use molar mass to get
to grams Example: How many grams of chlorine are required to react
completely with 5.00 moles of sodium to produce sodium chloride? 2
Na + Cl 2 2 NaCl 5.00 moles Na 1 mol Cl 2 70.90g Cl 2 2 mol Na 1
mol Cl 2 = 177g Cl 2
- Slide 21
- Learning Check For each of the following equations, calculate
the moles of product that would be produced by reacting 4.0 moles
of oxygen, O 2. Show your work. 4Fe + 3O 2 Fe 2 O 3 2Mg + O 2 2MgO
2Cu + O 2 2CuO
- Slide 22
- Learning Check Given the following reaction: N 2 + O 2 2NO
a)How many moles of NO can form when 40.0 g of N 2 react with
excess oxygen? b)How many grams of NO can form, using the amounts
in part A?