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The Mole. Mrs. Coyle Chemistry. Part I. The Mole Molar Mass Moles <->Particles. -How do we measure the mass of chemical quantities? -amu’s -grams -moles. Atomic Mass Unit (amu):. = 1/12 of the mass of a 12 C atom. Relationship of amu to gram. 1 amu = 1.66054 x 10 -24 g.
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The Mole Mrs. Coyle Chemistry
Part I • The Mole • Molar Mass • Moles <->Particles
-How do we measure the mass of chemical quantities? -amu’s -grams -moles
Atomic Mass Unit (amu): = 1/12 of the mass of a 12C atom
Relationship of amu to gram 1 amu = 1.66054 x 10 -24 g
Atomic Mass: -The weighted averageof the masses of the naturally occurring isotopes of an element. -The atomic mass of an element is listed on the periodic table.
Why the Mole? Are the quantities scientists use on a lab scale at the level of the atom or at the level of grams?
The Mole = 6.02 x 1023 One mole of atoms (6.02x1023 atoms) of an elementis contained in an amount equal to its atomic mass measured in grams. This mass is called molar mass.
Molar Mass of an Element: an amount equal to the atomic mass of the element expressed in grams. • Example: for Neon, Atomic Mass = 20.18 amu Molar Mass = 20.18 g
Molar Mass of a Compound:an amount equal to the molecular mass expressed in grams. Also known as, Gram-Formula Mass, Molecular Weight.Examples: • What is the molar mass of: • H2O • H2SO4 • Mg(NO3)2 • CuSO4∙ 5 H2O
The Definition of The Mole The mass of an element or compound that contains as many elementary (representative) particles (atoms, molecules, ions, etc.) as there are atoms in exactly 12 grams of 12C.
How many atoms are there in 12 g of C-12 ? 6.022145 x 1023 atoms1 mole of atoms (mol)Avogadro’s Number
Avogadro’s Number • Can be used the same way the word “dozen” is used to mean 12 for any object. • One mole of books is 6.022x1023 books, just like a dozen books is 12 books.
Problem Solving Methods • Using Mathematical Formulas • Factor-Label Method
Mathematical Formulas #mol=Mass of Sample (g) Molar Mass (g/mol) #particles=#mol x 6.022 x10 23 particles/molparticles can be atoms, molecules, formula units, etc.
Equalities Used in Factor- Label Method 1 mol = Molar Mass g 1mol= 6.022 x10 23 particles (atoms, molecules, formula units, etc)
Problem # 1- Moles to Atoms • How many atoms are there in 6 moles of Fe?
Solution using Factor-Label Method • 6 mol of Fe x 6.02 x 1023 atoms Fe = 1 mol Fe=3.6 x 10 24 atoms
Solution using Math Formula • #atoms=#mol x 6.022 x10 23 atoms/mol= =6 mol of Fe x 6.02 x 1023 atoms Fe/mol= =3.6 x 10 24 atoms
Problem #2 –Moles to Atoms • How many iron atoms are there in 6 moles of iron (III) oxide?A: 7.2 x 10 24 atoms
Problem #3- Atoms to Moles • How many moles of H2O are there in 18.06 x 10 23 molecules of H2O? • A: 3 moles of H2O
Part II • MolesMass Problem
Problem # 4 –Mass to Moles, Mass to Atoms For a sample of 20.0g of iron, Fe, calculate: a) the number of moles of atoms b) the number of atoms • Solution using Math Formula Method: • #mol Fe= mass Fe(g) = 20g Fe molar mass Fe(g/mol) 55.85g/mol #mol Fe = 0.358 mol Fe atoms
Problem # 4 Cont’d b) #atoms Fe= #mol x (6.022 x 1023 atoms/mol)= = (0.358 mol Fe)(6.022 x 10 23 atoms/mol) #atoms Fe= 2.156x1023 atoms Fe
Problem # 5 For a sample of 150.0 g of H2SO4, calculate: • The number of moles of H2SO4 present. • The number of oxygen atoms present.Solution: MM H2SO4 = H: (2 atoms)(1.008g/mol) S: (1 atom)(32.07g/mol) O: (4 atoms)(16.00g/mol) 98.09g/mol
Problem # 5 • # mol H2SO4 = Mass (g) =150.0g H2SO4 MM (g/mol) 98.09g/mol # mol H2SO4= 1.529 mol H2SO4
Problem # 5 Cont’d b) #atoms of O = (#mol O)(6.022x1023 atoms)= = (4 x 1.529 mol O) (6.022x1023 atoms) = = 3.683 x1024 atoms O
Part III • Molar Volume of an Ideal Gas at STP
Molar Volume of an Ideal Gas • At STP conditions, Standard Temp of 0 deg Cand Standard Pressure of 1 atm, the molar volume of an ideal gas is 22.4 L
For an ideal gas at STP: #mol= Volume (L) 22.4 (L/mol)or 1mol = 22.4 L
Problem # 6-Volume to Atoms • How many atoms of oxygen are there in 44.8L of oxygen gas at STP? • Solution: #mol O2= Volume (L) = 44.8 L = 2.00 mol O2 22.4 (L/mol) 22.4 L/mol #atoms O =#mol x 6.022 x10 23atoms/mol = 2 x (2.00mol O)(6.022 x10 23atoms/mol) #atoms O= 2.41 x 1024 atoms O
Problem #7- Molecules to Volume • How many L at STP are contained in a sample of 8.6 x10 24 molecules of CO2 gas?A: 320L
Part IV • Density of a Gas at STP from Molar Mass.
Density • D= Mass = Molar Mass Volume Molar Volume At STP,Molar Volume of an Ideal Gas = 22.4 L
Problem #8- Density of a Gas from MM • What is the density of NO2 gas at STP? • A: 2.05g/L