the mole
DESCRIPTION
THE MOLE. Chapter 11. Writing and balancing chemical equations. Chemical equation : identities and quantities of substances involved in chemical/physical change Balance using Law of Conservation of Mass and Law of Definite Composition same # of atoms on each side - PowerPoint PPT PresentationTRANSCRIPT
• Chemical equation: identities and quantities of substances involved in chemical/physical change– Balance using Law of Conservation of
Mass and Law of Definite Composition•same # of atoms on each side •fixed ratio of elements in
compound
Writing and balancing chemical equations
1. translate statement: reactants products
2. balance atoms using stoichiometric coefficients
3. Adjust stoichiometric coefficients (if necessary)– Smallest whole # preferred
4. Check5. Specify states of matter
Writing and balancing chemical equations
Samples
• Stoichiometry: study of quantitative aspects of chemical formulas/reactions
• Mole: unit chemists use to count chemical entities by weighing them
11.1 Measuring Matter
• The Mole– SI Unit for the amount of a substance– The number of particles equal to the
number of atoms in exactly 12.0 grams of carbon-12
– Also called Avogadro’s number• 1 mol = 6.02 x 1023 particles
mole - amount of substance that contains same # of entities as atoms in 12g of carbon-12.
1 mol contains 6.022x1023 entities
Avogadro’s number (N)
The mole
1 mole H2O contains 6.022 x 1023 H2O molecules
1 mole KNO3 contains 6.022x1023 KNO3 formula units
1 mole Hg contains 6.022x1023 Hg atoms
Mole represents large quantity of microscopic particles.
mass of 1 mass of 1Substance atom (molecule) mole of atoms (molecules)CaCO3 100.09 amu 100.09 gO2 32.00 amu 32.00 gH2O 18.02 amu 18.02 gCopper 63.55 amu 63.55 g
Can weigh out grams using scale
Use mass to ‘count’ entities.
6.022 x 1023 entities
•Molar mass (M) (gmw)- mass of 1 mole of entities–M (g/mol) numerically equal to formula weight (amu)
CH4 = 1(12.10 g/mol) + 4(1.008 g/mol) = 16.04 g/mol
The mole
Relating moles to chemical formulasGlucose C6H12O6 ( M = 180.16 g/mol)
Oxygen (O)
Mass/mole of compound
6 atoms
96.00 g
Table 3.2
Carbon (C) Hydrogen (H)
Atoms/moleculeof compound
Moles of atoms/mole of compound
Atoms/mole ofcompound
Mass/moleculeof compound
6 atoms 12 atoms
6 moles of atoms
12 moles of atoms
6 moles of atoms
6(6.022 x 1023) atoms
12(6.022 x 1023) atoms
6(6.022 x 1023) atoms
6(12.01 amu) = 72.06 amu
12(1.008 amu) = 12.10 amu
6(16.00 amu) = 96.00 amu
72.06 g 12.10 g
180.16 g/mole
Interconverting Moles, Mass, and # of Chemical Entities
Mass (g) = no. of moles x gmw
1 mol
No. of entities = no. of moles x6.022x1023 entities
1 mol
Mass % of element X =
Mass Percent from Chemical Formula
moles X in formula 1 mol compound
(molar mass of X)
molar mass of compoundx 100%
i.e. Mass % of H in H2O = 2 mol H 1.008 g H 1 mol H2O 1 mol H
18.02 g H2O 1 mol H2O
x 100%
= 11.19% H by mass
Particles
• Atoms – Single elements
• Formula units– Ionically bonded compounds
• Molecules– Covalent bonded compounds
Figure 3.6
macro
micro
Avogadro’s # takes us to/from macroscopic/microscopicLaw of Conservation of Mass
(6.02 x 1023
molecules)(6.02 x 1023
molecules)(1.20 x 1024
molecules)
Particle Mole Problems
• REMEMBER– The number of particles in 1 mole of ANY
substance is ALWAYS the same (6.02 x 1023)1 mol = 6.02 x 1023 particles
– How many molecules are in 2.2 moles of water?
1 mol H2O = 6.02 x 1023 particles
Use this as your conversion factor!!!Use this as your conversion factor!!!1 mol H2O
6.02 x 1023 molecules
6.02 x 1023 molecules 1 mol H2O
OROR
Calculating amounts of reactant and product
• Balanced equation needed for stoichiometric calculations– Ratios of reactants/products to calculate amounts of
reactants/products
Particle Mole Practice Continued
How many moles of sodium carbonate contain 7.9 x 1024 formula units?
2.2 mol H2O
1
Take your given value and put it
over ONE.
Multiply by the conversion factor that allows you to cancel out the
top unit. Leaving you with the unit you WANTED.6.02 x 1023 molecules 1 mol H2OX = 1.3244 x 1024
ANSWER: 1.3 x 1024 molecules of H2O
11.2 Mass and the Mole
• Atomic mass – The mass of an atom relative to the mass assigned
to carbon-12
• Molar mass– The mass in grams of one mole of any pure
substance– Use the average atomic mass off the periodic table
• Molar mass of an element– Atomic mass in grams per mole (g/mol)
• Molar mass of oxygen= 16.00 g/moL
• Molar mass of helium= 4.00 g/moL
• Calculate moles of O2 consumed when 10 mol of H2O are produced (using balanced equation from Table 3.5)?
– C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
• Calculate mass of CO2 produced burning 1.00 g butane (C4H10).
– 2C4H10(l) + 13O2(g) 8CO2(g) + 10H2O(l)
• Practice, practice, practice!!!
Calculating amounts of reactant and product
11.3 Moles of Compounds
• Formula mass – the sum of the atomic masses of all the atoms in a compound
H2O
H: 2 x 1.01amu = 2.02 amuO: 1 x 16.00amu = 16.00 amu
18.02 amu
• Limiting reagent- reactant that forms fewer moles of product– Limits amount of product formed (chair
analogy)– To decide which reagent is limiting
reagent:1. If information given in g, convert to
moles2. Use ratio from balanced equation to
find moles of final product possibly produced3. Reagent that produces least possible
moles of product is limiting reagent
Chemical rxts. that involve limiting reagents
What mass of NH3 is produced from the rxt. of 1.00 g H2(g) w/ 1.00 g N2(g)?
(Info on both reactants given)
3H2(g) + N2(g) 2NH3(g)
• Is H2 or O2 the limiting reagent?– H2
Chemical rxts. that involve limiting reagents
Microscopic Picture
Moles of Compounds Continued
• Molar mass – The mass in grams of 1 mole of a
substance
– Formula mass of H2O = 18.02 amu
– Molar mass of H2O = 18.02g/mol
1 mol of H2O = 18.02 g = 6.02 x 1023 molecules
Mass Mole Problems
• Changing the mass to moles or vice versa using the molar mass
• How many moles are in 11.2g of NaCl?1. Determine the molar mass.
- Na: 1 x 23.00 = 23.00g- Cl: 1 x 35.45 = 35.45g
= 58.45g/mol
Therefore, 1 mol NaCl = 58.45g
Mass Mole Problems
2. Convert between the molar mass and the moles.
1 mol of NaCl = 58.45g1 mol of NaCl58.45g NaCl
58.45g NaCl 1 mol of NaCl
OROR
11.2g NaCl1
Take your given value and put it
over ONE.
Multiply by the conversion factor that allows you to cancel out the
top unit. Leaving you with the unit you WANTED.1 mol NaCl
58.45g NaClX =0.1916167665 mol NaCl
ANSWER: 0.192 mol NaCl
More Practice
• What is the mass of 2.50 mol of NaCl?– Find the molar mass…
1 mol of NaCl = 58.45g
1 mol of NaCl58.45g NaCl
58.45g NaCl 1 mol of NaCl
OROR
2.50 mol NaCl1
Take your given value and put it
over ONE.
Multiply by the conversion factor that allows you to cancel out the
top unit. Leaving you with the unit you WANTED.58.45g NaCl
1 mol NaClX = 146.125g NaCl
ANSWER: 146.13g NaCl
Multi-Step Conversions
• Mass-Particleg mol particles
• Particle-MassParticles mol g
• What is the mass of 8.2 x 1022 atoms of calcium?
1 mol Ca = 6.02 x 1023 atoms Ca1 mol Ca = 40.08g Ca
Mole Volume Problems
• Equal volumes of gases at the same temperature and pressure contain the same number of particles.
• Molar volume– The volume of 1 mol of gas at standard
conditions (STP)– STP
• standard temperature and pressure: 0oC and 1 atm
– 1 mol = 22.4 liters
Mole Volume Practice Problem
• What is the volume of 0.35 moles of helium gas at STP?
0.35 mol He1
Take your given value and put it
over ONE.
Multiply by the conversion factor that allows you to cancel out the
top unit. Leaving you with the unit you WANTED.22.4 L
1 mol HeX = 7.84L He
ANSWER: 7.84L He
Percent Composition
• Percent Composition– The percent by mass of each element
in a compound
mass of elementmass of compoundX 100= % composition