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Chap apter ter 10 Chemi mical cal Quan antities tities Se Sectio ction n 10.1 The Mol ole: : A m A meas asuremen ement t of of Mat atter ter What at is is a a Mole le? A mole (mol) of a substance is 6.02 X 10 23

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SLIDE 1

Chap apter ter 10 Chemi mical cal Quan antities tities Se Sectio ction n 10.1 The Mol

  • le:

: A m A meas asuremen ement t of

  • f Mat

atter ter

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SLIDE 2

What at is is a a Mole le?

A mole (mol) of a substance is 6.02 X 1023 representative particles of that substance. Usually atoms or molecules.

6.02 X 1023 is called Avogadro’s number

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SLIDE 3

Converting Number of Particles to Moles

Number s Avogadro' Molecules)

  • r

(Atoms Particles

  • f

Number Moles

  • f

Number = e.g. How many moles of magnesium in 1.25 X 10 23 atoms of magnesium

23 23

10 x 6.02 10 x 1.25 Moles

  • f

Number =

= 0.208 mol Mg Page e 291: #3 & 4 4

Particles Moles Av. number Molecules or atoms

Moles - Particles

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SLIDE 4

Converting Moles to Number of Particles

23

10 x 6.02 x Moles

  • f

Number Molecules)

  • r

(Atoms Particles

  • f

Number =

e.g. . How many atoms ms are in 2.12 mol of propane ane (C3H8)? )? Number of atoms = 2.12 x 6.02 x 10 23 x 11 = 1.4 x 10 25 atoms Page: e: 292 #5 & 6 6

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SLIDE 5

The mass of a Mole of an Element

The mass of a mole of an element is its mo molar ar ma mass.

How can you calculate the molar mass of a compound?

To calculate the molar mass of a compound:

  • 1. Find the number of grams of each element.
  • 2. Add the masses of the elements in the compound.

e.g. . what is the molar ar mass s of calcium lcium oxide? de? Chemical formula = CaO, Ca = 40, O = 16 Molar mass = 40 + 16 = 56g Page: e: 296, #7 & 8

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SLIDE 6

Atoms

  • r

Molecules

  • r

Particles

Number of Moles Avogadro’s Number

Moles – Atoms

Multiplies by number of atoms

If atoms

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SLIDE 7

Se Sect ction ion 10.2 .2 Mo Mole le – Ma Mass ss & Mo Mole le – Vol

  • lume

me Re Relation lationships ships

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SLIDE 8

mass Molar substance the

  • f

Mass Moles

  • f

Number =

Converting Mass ss to Moles les

e.g. . How many moles es of iron(II (III) ) oxide de are contain tained ed in n 92 92.2g g of pur ure e Fe Fe2O3? Mass = 92.2g Molar mass = 2x55.8 + 3x16 = 159.6g/mol Number of moles = 92.2/159.6 = 0.578 mol Fe2O3 Page: e: 299, # 18 & 19

Moles - Mass

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SLIDE 9

Converting Moles les to Mass ss

Mass = number of Moles x molar mass

e.g. . Wha hat t is th the ma mass s of 9. 9.45 45 mo mol of alumi uminu num m oxide? de? Chemical Formula = Al2O3 Number of moles = 9.45 mol Molar mass = 2x27 + 3x16 = 102g Al2O3 Mass = 9.45 x 102 =964g Al2O3 Page: e: 298, # 16 & 17

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SLIDE 10

The volume of a Mole of an Element

The volume of a mole of an element is called molar volume The molar volume of a gas at STP is 22.4 L Standard Temperature and Pressure To calculate the volume of a gas at STP:

Volume of gas = Number of Moles x 22.4

e.g. . Deter termine mine th the volum ume e of 0. 0.6 6 mo mol sul ulfur ur dioxide ide gas at t STP. Volume of SO2 = 0.6 x 22.4 = 13.4 L SO2 Page: e: 301, #20 & 21

Moles - Volume

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SLIDE 11

Calculating Molar Mass from Density

Molar mass = density at STP x molar volume at STP

e.g. . The he densi nsity ty of a gaseous eous compound mpound carbon bon and nd oxygen en is foun und d to be 1.964g/L g/L at STP. . What is the molar ar mass s of the compound? pound? Molar mass = 1.964 x 22.4 = 44g/mol Page: e: 302, #22 & 23

Moles - Density

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SLIDE 12

Mass

Number of Moles Molar Mass

Moles – Mass

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SLIDE 13

Volume

Number of Moles

22.4

Moles – Volume

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SLIDE 14

Molar mass

Density

22.4

Moles – Density

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SLIDE 15

Se Sect ction ion 10.3 .3 Per ercent cent Co Composit position ion and d Ch Chemical emical Fo Formulas mulas

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SLIDE 16

The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100

100 compound

  • f

mass element

  • f

mass element

  • f

mass %  =

e.g. When a 13.6 g sample of a compound containing only magnesium and oxygen. 5.4 g of oxygen is obtained. What is the percent composition of this compound?

The e percent cent comp mpos

  • sitio

ition is the relative amounts of the elements in a compound. Mass of compound = 13.6 g Mass of oxygen = 5.4 g Mass of magnesium = 13.6 – 5.4 = 8.2 g

% 7 . 39 100 6 . 13 4 . 5 100 compound

  • f

mass O

  • f

mass O % =  =  = % 3 . 60 100 6 . 13 2 . 8 100 compound

  • f

mass Mg

  • f

mass Mg % =  =  =

Page e 30 306, 6, # 32 # 32 & 33 33

Calculating Percent Composition from Mass Data

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SLIDE 17

Calculating Percent Composition from the Chemical Formula

100 compound

  • f

mass molar compound mole 1 in element

  • f

mass element

  • f

mass %  =

e.g. Propane (C3H8) is one of the compounds obtained from petroleum. Calculate the percent composition of propane. % 7 . 81 100 08 . 44 . 36 100 propane

  • f

mass C

  • f

mass C % =  =  = % 3 . 18 100 08 . 44 08 . 8 100 propane

  • f

mass H

  • f

mass H % =  =  = Mass of C in C3H8 = 36.0 g Mass of H in C3H8 = 8.08 g Molar mass of C3H8 = 44.08 g/mol

Page: e: 307, # 34 & 35

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SLIDE 18

Empirical Formula

The empirical formula of a compound shows the smallest whole-number ratio of the atoms in the compound. It shows the kinds and lowest relative count of atoms or moles of atoms in molecules or a compound. An empirical formula may or may not be the same as a molecular

  • formula. For example, the lowest ratio of hydrogen to oxygen in hydrogen

peroxide is 1:1 So the empirical formula of hydrogen peroxide is HO The actual molecular formula of hydrogen peroxide has twice the number

  • f atoms as the empirical formula.

The molecular formula is (HO) x 2

  • r

H2O2 Notice ice that the ratio of hydrogen to oxygen is still the same. The molecular formula tells the actual number of each kind of atom present in a molecule of the compound.

For carbon dioxide, the empirical and molecular formulas are the same – CO2

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SLIDE 19

e.g. A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? 25.9% N 74.1% O Step 1. Assume that you have 100 g of the compound, 25.9 g 74.1 g Step 2. Divide each by its atomic mass. 25.9 74.1 14 16 Step 3. Divide each by the smallest number of moles. The empirical formula is N2O5 Page: e: 310, # 36 & 3 37

1.85 mol 4.63 mol

1.85 1.85 4.63 1.85

1 2.5

  • Multiply all by 2
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SLIDE 20

Molecular ecular Fo Formu mula la

is the same as the empirical formula or it is a simple whole-number multiple of its empirical Once you determined the empirical formula of a compound, you can determine its molecular formula, but you must know the compound’s molar mass and empirical formula mass. Ste tep 1. Calculate the empirical formula mass e.g. Calculate the molecular formula of a compound whose molar mass is 60 g/mol and empirical formula is CH4N Empirical formula = CH4N Molar mass = 60 g/mol Molecular formula = C?H?N? Empirical formula mass = 1x12 + 4x1.01 + 1x14 = 30.04 g/mol Step 2. Divide the molar mass by the empirical formula mass 60 = 2 30.04 Molecular formula = 2 x CH4N = C2H8N2 Step 3. Multiply the formula subscripts by this value (factor) Page: e: 312, # 38 & 3 39