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The Mole Mrs. Coyle Chemistry

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Part I The Mole Molar Mass Moles Particles

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-How do we measure the mass of chemical quantities? -amu’s -grams -moles

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Atomic Mass Unit (amu): = 1/12 of the mass of a 12 C atom

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Relationship of amu to gram 1 amu = 1.66054 x 10 -24 g

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Atomic Mass: - The weighted average of the masses of the naturally occurring isotopes of an element. -The atomic mass of an element is listed on the periodic table.

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Why the Mole? Are the quantities scientists use on a lab scale at the level of the atom or at the level of grams?

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The Mole = 6.02 x 10 23 One mole of atoms (6.02x10 23 atoms) of an element is contained in an amount equal to its atomic mass measured in grams. This mass is called molar mass.

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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

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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: H 2 O H 2 SO 4 Mg(NO 3 ) 2 CuSO 4 ∙ 5 H 2 O

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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 12 C.

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How many atoms are there in 12 g of C-12 ? 6.022145 x 10 23 atoms 1 mole of atoms (mol) Avogadro’s Number

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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 6.022x10 23 books, just like a dozen books is 12 books.

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“Mole of Atoms of Hydrogen” 1 atomAtomic Mass:1.008 amu 1 mole of atoms: 6.022 x 10 23 atoms Molar Mass: 1.008 g

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“Mole of Molecules of Hydrogen” 1 molecule= 2 atoms Molecular Mass: 2.016 amu 1 mole of molecules= 6.022 x 10 23 molecules= 2x6.022 x10 23 atoms Molar Mass: 1.008 g

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Problem Solving Methods Using Mathematical Formulas Factor-Label Method

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Mathematical Formulas #mol= Mass of Sample (g) Molar Mass (g/mol) # particles=#mol x 6.022 x10 23 particles/mol particles can be atoms, molecules, formula units, etc.

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Equalities Used in Factor- Label Method 1 mol = Molar Mass g 1mol= 6.022 x10 23 particles (atoms, molecules, formula units, etc)

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Problem # 1- Moles to Atoms How many atoms are there in 6 moles of Fe?

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Solution using Factor-Label Method 6 mol of Fe x 6.02 x 10 23 atoms Fe = 1 mol Fe =3.6 x 10 24 atoms

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Solution using Math Formula #atoms=#mol x 6.022 x10 23 atoms/mol= =6 mol of Fe x 6.02 x 10 23 atoms Fe/mol= =3.6 x 10 24 atoms

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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

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Problem #3- Atoms to Moles How many moles of H 2 O are there in 18.06 x 10 23 molecules of H 2 O? A: 3 moles of H 2 O

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Part II Moles Mass Problem

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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: a)#mol Fe= mass Fe(g) = 20g Fe molar mass Fe(g/mol) 55.85g/mol #mol Fe= 0.358 mol Fe atoms

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Problem # 4 Cont’d b) #atoms Fe= #mol x (6.022 x 10 23 atoms/mol)= = (0.358 mol Fe)(6.022 x 10 23 atoms/mol) #atoms Fe= 2.156x10 23 atoms Fe

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Problem # 5 For a sample of 150.0 g of H 2 SO 4, calculate: a)The number of moles of H 2 SO 4 present. b)The number of oxygen atoms present. Solution: MM H 2 SO 4 = H: (2 atoms)(1.008g/mol) S: (1 atom)(32.07g/mol) O: (4 atoms)(16.00g/mol) 98.09g/mol

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Problem # 5 # mol H 2 SO 4 = Mass (g) =150.0g H 2 SO 4 MM (g/mol) 98.09g/mol # mol H 2 SO 4 = 1.529 mol H 2 SO 4

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Problem # 5 Cont’d b) #atoms of O = (#mol O)(6.022x10 23 atoms)= = (4 x 1.529 mol O) (6.022x10 23 atoms) = = 3.683 x10 24 atoms O

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Part III Molar Volume of an Ideal Gas at STP

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Molar Volume of an Ideal Gas At STP conditions, Standard Temp of 0 deg C and Standard Pressure of 1 atm, the molar volume of an ideal gas is 22.4 L

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For an ideal gas at STP: #mol= Volume (L) 22.4 (L/mol) or 1mol = 22.4 L

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Problem # 6-Volume to Atoms How many atoms of oxygen are there in 44.8L of oxygen gas at STP? Solution: #mol O 2 = Volume (L) = 44.8 L = 2.00 mol O 2 22.4 (L/mol) 22.4 L/mol #atoms O =#mol x 6.022 x10 23 atoms/mol = 2 x (2.00mol O)(6.022 x10 23 atoms/mol) #atoms O= 2.41 x 10 24 atoms O

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Problem #7- Molecules to Volume How many L at STP are contained in a sample of 8.6 x10 24 molecules of CO 2 gas? A: 320L

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Part IV Density of a Gas at STP from Molar Mass.

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Density D= Mass = Molar Mass Volume Molar Volume At STP, Molar Volume of an Ideal Gas = 22.4 L

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Problem #8- Density of a Gas from MM What is the density of NO 2 gas at STP? A: 2.05g/L

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