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LAW OF DEFINITE PROPORTIONS

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Title: LAW OF DEFINITE PROPORTIONS


1
LAW OF DEFINITE PROPORTIONS
  • In a compound, the ratios by mass of the
    elements in that compound are fixed independent
    of the origins or preparation of that compound.
    A compound is unique because of the specific
    arrangement and weights of the elements which
    make up that compound. That is, elements combine
    in whole numbers. Also it is not possible to
    have a compound with portion an atom.

2
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3
LAW OF DEFINITE PROPORTIONS
  • Elements combine in specific ratios to form
    compounds
  • Use the Generic equation for percent
  • ( portion / total ) 100
  • 1. What is the experimental percent of oxygen in
    CO2 if 42.0 g of carbon reacted completely with
    112.0 g of oxygen?
  • O (mass of O / mass of CO2) 100
  • O 112.0 g O / (42.0 g 112.0 g) CO2 100
    72.7 O
  • 2. What is the theoretical percent of aluminum
    in aluminum oxide?
  • Al (Atomic mass of Al / Formula mass of
    Al2O3) 100
  • Al (54 amu / 102 amu) 100 52.9
  • 3. What is the percent composition of sodium
    chloride?

Na 39.3 Cl 60.7
4
LAW OF MULTIPLE PROPORTIONS
  • When two elements form a series of compounds,
    the masses of the one element that combine with a
    fixed mass of the other element stand to one
    another in the ratio of small integers.
  • Iron oxide exists in different ratios with
    different properties
  • FeO and Fe2O3

5
LAW OF CONSERVATION OF MASS
  • In every chemical operation an equal quantity of
    matter exists before and after the operation.
    That is, the amount of matter before a reaction
    must equal the amount of matter after a reaction.
    No matter is lost.
  • The total mass of reactants total mass of
    products

6
LAW OF CONSERVATION OF MASS
  • When 0.0976 g of magnesium was heated in air,
    0.1618 g of magnesium oxide (MgO) was produced.
  • a) what is the mass of oxygen needed to produce
    0.1618 g MgO?
  • Using the LCM
  • Total mass reactants total mass products
  • mass of Mg mass O mass of MgO
  • 0.0976 g Mg mass O 0.1618 g MgO
  • mass O 0.1618 g - 0.0976 0.0642 g O

7
Law of conservation of mass Law of definite
proportions
  • When 0.0976 g of magnesium was heated in air,
    0.1618 g of magnesium oxide (MgO) was produced.
  • b) what is the percent of Mg in MgO?
  • Mg (mass Mg / Mass MgO) 100
  • (0.0976g / 0.1618 g) 100 60.3
  • c) Using only LDP, what mass of oxygen was
    needed to combine with the magnesium?
  • O 100 MgO - 60.3 Mg 39.7 O
  • O (mass O / mass MgO) 100
  • 39.7 (mass O / 0.1618 g) 100
  • mass O 0.397 ( 0.1618 g) 0.0642 g O
  • Same as using the LCM!!

8
PRACTICE PROBLEMS
  • Practicing Law of conservation of mass
  • ________1. Aluminum metal combines with oxygen to
    produce aluminum oxide. If 141.0g of aluminum
    yields 266.7 g of aluminum oxide, how many grams
    of oxygen were needed?
  • ________2. Sodium metal reacts with chlorine gas
    to produce the salt, sodium chloride. If 15.0 g
    of chlorine yields 26.5 g of salt, how much
    sodium metal is needed?
  • Practicing the law of definite proportions
  • ________3. What is the experimental percent of
    oxygen in a copper oxide if 10.0 g of copper
    reacted completely with 2.52 g of oxygen?
  • _______ 4. Based on question 1, what is the
    experimental percent composition of aluminum
    oxide?
  • _______ 5. Calculate the theoretical percent
    composition for aluminum chloride and sodium
    oxide.

9
PRACTICE PROBLEMS
  • Practicing percents
  • ________1. Pure gold is too soft a metal for many
    uses, so it is alloyed to give it more mechanical
    strength. One particular alloy is made by mixing
    29.17 g of gold, 3.81 grams of silver, and 5.91 g
    of copper. What is the percent of gold in this
    mixture?
  • ________2. If 255 g of a meat sample contains
    21.9 g of fat, what percentage of fat is present?
  • Using the LAWS
  • ________3. How many grams of CuO can be obtained
    from 1.80 g of copper (use the theoretical
    percent composition)?
  • 4. When aluminum combines with bromine gas, they
    produce the substance aluminum bromide, AlBr3.
    Write a chemical equation describing this
    reaction.
  • _______ If 56.88 g of aluminum bromide is formed
    from 5.75 g of aluminum, how many grams of
    bromine was needed?

75.0
8.6
225 g
2Al 3Br2 ? 2AlBr3
51.13 g
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