Chemistry Notes Chapter 11

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Chemistry Notes Chapter 11

• THE MOLE

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Discovery Lab (A Look At The Mole)

• Chemists use the SI unit of the mole when
  counting large numbers.
• In order to see how large a mole is open your
  book to page 309 and do the discovery lab.
• Procedure
• Measure the length of a paper clip to the nearest
  0.1 cm.
• If a mole is 6.02 x 1023 items, how far will a
  mole of paper clips placed end to end,
  lengthwise, reach into space?
• Analysis
• How many light years (ly) would the paper clips
  extend into space? (1 light year 9.46 x 1015m)
• How does the distance you calculated compare with
  the following astronomical distances nearest
  star other than the sun 4.3 ly, center of our
  galaxy 30,000 ly, nearest galaxy 2 x 106 ly?

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Expected Results

• The Length of a 1 paper clip is 3.2 cm.
• One mole of paper clips reaches 1.9 x 1025 cm.
• Analysis
• 6.02 x 1023 paper clips x 3.2 cm/1paperclip x
  1m/100cm x 1ly/9.46 x 1015m 2.0 million ly.
• This compares to the distance to the nearest
  galaxy.
• So as you can see a mole of something is quite a
  lot

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What is a mole?

• The mole, commonly abbreviated mol, is the SI
  base unit used to measure the amount of a
  substance.
• One mole of anything contains 6.0221367 x 1023
  representative particles.
• This number is called Avogadros number, in honor
  of Amedeo Avogadro, who first determined the
  volume of one mole of a gas.
• Avogadros Number is usually rounded to 6.02 x
  1023
• When written out completely Avogadros number
  would be 602,000,000,000,000,000,000,000

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Converting Moles to Particles and Particles to
Moles

• Converting the number of moles to the number of
  particles is pretty simple.
• All you have to do is use conversion factors.
• For example If you want to find the number of
  particles there are in 3 moles of copper (II)
  chloride. All you would have to do take the
  number of moles and multiply it by the conversion
  factor of 6.02 x 1023
  representative particles/mole and you would get
• 3 moles CuCl2 x 6.02 x 1023 molecules CuCl2
  1 mole CuCl2
• And end up with 1.81 x 1023 molecules CuCl2

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Practice Problems

• Determine the number of atoms in 2.50 mole of Zn.
• 2.50 mol Zn x 6.02 x 1023 molecules of Zn
  1.505 x 1024 molecules of Zn
• 1 mol Zn
• Given 3.25 mol AgNO3, determine the number of
  formula units
• 3.25 mol AgNO3 x 6.02 x 1023 molecules of AgNO3
  1.96 x 1024 formula units 1 mole
  AgNO3
• Calculate the number of molecules in 11.5 mol
  H2O.
• 11.5 mole x 6.02 x 1023 molecules of H2O
  6.92 x 1024 molecules H20 1 mol
  H2O

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The Mass Of The Mole

• The Mass in grams of one mole of any pure
  substance is called its molar mass.
• The molar mass of any element is numerically
  equal to its atomic mass and has the units g/mol.
• An atom of manganese has an atomic mass of 54.94
  amu.
• So the molar mass of manganese is 54.94 g/mol.
• When you measure out 54.94 g of manganese on a
  balance you indirectly count 6.02 x 1023 atoms of
  manganese

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Using Molar Mass

• In order to calculate the mass of a certain
  amount of moles all you have to do is take the
  amount of moles you are seeking and multiply that
  number by the number of grams per mole of that
  substance.
• So Number of moles x Number of grams mass
• 1 mole

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Using Molar Mass

• In order to go from mass of substance to moles of
  substance you would take the mass of the
  substance and divide it by the molar mass to get
  the number of moles.
• Mass x 1 mole number of moles
• number of grams

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Practice Problems

• Lets work problems 11 and 12 on page 316 together.

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Conversions from mass to atoms

• In order to find the number of atoms of a given
  mass of a substance you need to find the number
  of moles and multiply that number by Avogadro's
  number of 6.02 x 1023.
• Mass x 1 mole number of moles
• number of grams
• of moles x 6.02 x 1023 of particles

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Conversions from atoms to mass

• In order to get the calculate the mass of a set
  number of atoms all you have to do is take the
  number of particles and divide it by the number
  of moles then multiply it by the number of grams
  per mole.
• of particles/ 6.02 x 1023 of moles
• of moles x (number of grams/mole) mass

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Chemical Formulas and the Mole

• A chemical formula tells us the types of atoms
  and the number of each contained in one unit of a
  compound.
• In order to convert from moles of a compound to
  moles of individual atoms in the compound or from
  moles of an individual compound to moles of the
  compound you need a conversion factor.

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Chemical Formulas and the Mole

• Going from moles of a compound to moles of
  individual atoms
• If you have 3 moles of H2O and you need to find
  out how many moles of hydrogen you have all you
  have to do is use the conversion factor 2 mol H
  / 1 mol H2O
• So 3mol H2O x 2 mol H 6 mol H
• 1mol H2O

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Chemical Formulas and the Mole

• Going from moles of individual atoms to moles of
  a compound
• If you have 12 moles of hydrogen present in an
  over abundance of oxygen how many moles of water
  can be produced?
• The conversion factor would be 1mol
  H2O/2mol H
• So 12 mol H x 1mol H2O 6 mol H2O
• 2 mol H

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The Molar Mass of Compounds

• The mass of a mole of a substance is equal to the
  sum of the mass of every particle in it.
• What is the molar mass of rust (Fe2O3)?
• 2 mol Fe x 55.85 g Fe / 1mol Fe 111.16 g
• 3 mol O x 16 g O / 1 mol O 48 g
• 111.16 48 159.16 g Fe2O3
• Do Practice Problems 25 and 26 on pg 322

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Converting Moles Of a Compound to Mass

• When converting from moles of a compound to mass
  of compound all you have to do is multiply the
  number of moles you have of each part of the
  compound by the molar mass of that part and add
  all of these together.

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Example problem

• The molar mass of (C3H5)2S is calculated like
• 1mol S x 32.07 g S 32.07 g S
• 1 mol S
• 6 mol C x 12.01 g C 72.06 g C
• 1 mol C
• 10 mol H x 1.008 g H 10.08 g H
• Molar mass of (C3H5)2S 114.21 g/mol (C3H5)2S
• Do practice problems 27-29 on page 323

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Converting the Mass of a Compound to Moles

• In order to convert from mass of a compound to
  moles all you have to do is take the mass of the
  compound and divide it by the molar mass of the
  compound.
• Ca(OH)2 has a molar mass of 74.1 g/mol
• If we are given 650 g of Ca(OH)2 how many moles
  of it do we have?
• Solution 650 g Ca(OH)2 8.77 mol Ca(OH)2
• 74.1 g Ca(OH)2
• Do Practice Problem 30 on page 324

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Converting Mass of a Compound to Number of
Particles

• There is no direct conversion possible between
  mass and number of particles.
• You must first convert the given mass to moles
  by multiplying the inverse of the molar mass.
  Then you can convert moles to number of
  representative particles by multiplying by
  Avogadro's number.

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Example Problem

• How many particles are there in 35.6 g AlCl3?
• First you must determine the molar mass of AlCl3.
  
• 1mol Al x 26.98 g Al/mol 26.98 Al
• 3mol Cl x 35.45 g Cl/mol 106.35 g Cl
• So the molar mass of AlCl3 is 133.33 g/mol AlCl3.
• Second you must multiply by the inverse of the
  molar mass as a conversion factor to convert the
  mass of AlCl3 to moles.
• 35.6 g AlCl3 x 1mol AlCl3 0.267 mol AlCl3
  
• 133.33g AlCl3

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Quick reminder for all the formulas

• This Diagram is a good tool to use when doing
  calculations. You may copy it down on the back of
  your periodic table.

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Percent Composition

• The percent by mass of any element in a compound
  can be found by dividing the mass of the element
  by the mass of the compound and multiplying by
  100.
• Mass of element x 100 by mass
• mass of compound
• Because percent means parts per 100, the percents
  by mass of all the elements of a compound must
  always add up to 100.
• The percent by mass of each element in compound
  is called percent composition of a compound.

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Percent Composition

• The general equation of calculating the percent
  by mass of any element in a compound is
• Mass element in 1 mol of compound x 100 by
  mass element
• Molar mass of compound
• Do Practice problems 4245 on page 331

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Empirical Formula

• The empirical formula for a compound is the
  formula with the smallest whole number mole ratio
  of the elements.
• The empirical formula may or may not be the same
  as the actual molecular formula.
• If the two formulas are different , the molecular
  formula will always be a simple multiple of the
  empirical formula.
• The subscripts in an empirical formula are in a
  ratio of the smallest whole numbers of moles of
  the elements in the compound.

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Calculating Empirical Formula (from
Composition) convert of each element to grams
based on 100 grams of the compound. multiply
grams of each element by 1/molar mass that
elementcompare ratio of moles of each element
and divide each by the smallest number that you
calculated.V. Calculating Empirical Formula
(from experimentally determined masses) multiply
the mass of each element (in grams) by 1/molar
mass of that elementcontinue with steps 3 4
from IV above.
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Molecular Formula

• The molecular formula for a compound can be
  determined by finding the integer by which the
  mass of the empirical formula differs from the
  molar mass of the compound
• A formula for determining molecular formula would
  be
• Experimentally determined molar mass of compound
• Mass of empirical formula
• This will give you the integer by which mass
  differs from the empirical formula and then all
  you have to do is multiply all subscripts and
  coefficients by that number.
• So molecular formula (empirical formula)n Where
  n stands for the integer that you must multiply
  by.

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Naming Hydrates

• A hydrate is a compound that has a specific
  number of water molecules bound to its atoms.
• The formula for a hydrate consists of the formula
  for the ionic compound and the number of water
  molecules associated with on formula unit.
• The Name of a hydrate consists of the compound
  name followed by the word hydrate with a prefix
  indicating the number of water molecules
  associated with one mole of compound.
• Anhydrous compounds are formed when hydrates are
  heated and the water that is in them is driven
  off.