Ch 10 chemical quantities
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Ch. 10 Chemical Quantities. 3 Methods of Measuring. Counting Mass Volume. Example 1. If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?. Example 1. Count: 1 dozen apples = 12 apples Mass: 1 dozen apples = 2.0 kg apples

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Ch. 10 Chemical Quantities

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Ch. 10 Chemical Quantities


3 Methods of Measuring

  • Counting

  • Mass

  • Volume


Example 1

  • If 0.20 bushel is 1 dozen apples, and a dozen apples has a mass of 2.0 kg, what is the mass of .050 bushel of apples?


Example 1

  • Count: 1 dozen apples = 12 apples

  • Mass: 1 dozen apples = 2.0 kg apples

  • Volume: 1 dozen apples = 0.20 bushels

    apples

    Conversion Factors:

  • 1 dozen2.0 k.g1 dozen

    12 apples 1 dozen 0.20 bushels


Example 1

  • 0.50 bushel x 1 dozen x 2.0 kg =

    0.20 bushel 1 dozen

    = 5.0 kg


Avogadro’s Number

  • Named after the Italian scientist Amedo Avogadro di Quaregna

  • 6.02 x 10 23


Mole (mol)

  • 1 mol = 6.02 x 10 23 representative particles

  • Representative particles: atoms, molecules ions, or formula units (ionic compound)


Mole (mol)

  • Moles= representative x 1 mol

    particles 6.02 x 10 23


Example 2 (atoms  mol)

  • How many moles is 2.80 x 10 24 atoms of silicon?


Example 2

  • 2.80 x 10 24 atoms Si x 1 mol Si

    6.02 x 10 23 atoms Si

    = 4.65 mol Si


Example 3 (mol  molecule)

  • How many molecules of water is 0.360 moles?


Example 3

  • 0.360 mol H2O x 6.02 x 10 23 molecules H2O1 mol H2O

    =2.17 molecules H2O


The Mass of a Mole of an Element

  • The atomic mass of an element expressed in grams = 1 mol of that element = molar mass

Molar mass S

Molar mass Hg

Molar mass C

Molar mass Fe


6.02 x 10 23 atoms S

6.02 x 10 23 atoms Hg

6.02 x 10 23 atoms C

6.02 x 10 23 atoms Fe


Example 4 (mol  gram)

  • If you have 4.5 mols of sodium, how much does it weigh?


Example 4

  • .45 mol Na x 23 g Na = 10.35 g Na = 1.0 x 10 2 g Na

    1 mol Na


Example 5 (grams  atoms)

  • If you have 34.3 g of Iron, how many atoms are present?


Example 5

  • 34.3 g Fe x 1 mol Fe x 6.02 x 10 23 atoms

    55.8 g Fe 1 mol Fe

    =3.70 x 10 23 atoms Fe


The Mass of a Mole of a Compound

  • To find the mass of a mole of a compound you must know the formula of the compound

  • H2O  H= 1 g x 2

    O= 16 g

    18 g = 1 mole = 6.02 x 10 23

    molecules


Example 6 (gram  mol)

  • What is the mass of 1 mole of sodium hydrogen carbonate?


Example 6

  • Sodium Hydrogen Carbonate = NaHCO3

  • Na=23 g

  • H=1 g

  • C=12 g

  • O=16 g x3

  • 84 g NaHCO3 = 1 mol NaHCO3


Mole-Volume Relationship

  • Unlike liquids and solids the volumes of moles of gases at the same temperature and pressure will be identical


Avogadro’s Hypothesis

  • States that equal volumes of gases at the same temperature and pressure contain the same number of particles

  • Even though the particles of different gases are not the same size, since the gas particles are spread out so far the size difference is negligible


Standard Temperature and Pressure (STP)

  • Volume of a gas changes depending on temperature and pressure

  • STP= 0oC (273 K)

    101.3 kPa (1 atm)


Standard Temperature and Pressure (STP)

  • At STP, 1 mol = 6.02 X 1023 particles = 22.4 L of ANY gas= molar volume


Conversion Factors

  • AT STP

  • 1 mol gas22.4 L gas

    22.4 L gas 1 mol gas


Example 7

  • At STP, what volume does 1.25 mol He occupy?


Example 7

  • 1.25 mol He x 22.4 L He = 28.0 L He

    1 mol He


Example 8

  • If a tank contains 100. L of O2 gas, how many moles are present?


Example 8

  • 100. L O2 X 1 mol O2 = 4.46 mol O2

    22.4 L O2


Calculating Molar Mass from Density

  • The density of a gas at STP is measured in g/L

  • This value can be sued to determine the molar mass of gas present


Example 9

  • A gaseous compound of sulfur and oxygen has a density of 3.58 g/L at STP. Calculate the molar mass.


Example 9

  • 1 mol gas x 22.4 L gas X 3.58 g gas =

    1 mol gas 1 L gas

    Molar Mass= 80.2 g


Percent Composition

  • The relative amounts of the elements in a compound

  • These percentages must equal 100


Percent Composition

  • %element = mass of element x 100

    mass of compound


Example 10

  • Find the percentage of each element present in Al2 (CO3)3


Example 10

  • Al2(CO3)3

  • Al= 27 g x 2 = 54 g / 234 g x 100=23%

  • C= 12 g x 3 = 36 g/ 234 g x 100= 15%

  • O = 16 g x 9 = 144 g / 234 g x 100=62%

    234 gAl2(CO3)3


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