Basic-laws of Chemcombi

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    In the study of matter, we observe certain

    phenomena which can be reproduced at will.These observed phenomena are called

    scientific facts. Often, a large number of

    scientific facts can be summarized into broad

    or sweeping statements called laws.

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    LAW OF CONSERVATION OF MASS

    -which states that when an ordinary chemicalreaction occurs, there is no detectable change in

    the masses of the substances involved before and

    after the reaction.

    masses of the reactants = masses of the products

    before the reaction after the reaction

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    LAW OF CONSERVATION OF MASS

    However, Albert Einstein, a Nobel laureate

    physicist and one of the most creative intellects in

    human history, recognized the quantitative

    interconversion of mass and energy in reactions,as shown by his famous equation.

    E = mc

    2

    Where: E = energy

    m = mass

    c = velocity of light

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    LAW OF CONSERVATION OF MASS

    The interconvertibility of mass and energy holdstrue in energetic processes like radioactivity and

    atomic explosions. Strictly speaking, the

    interconversion of energy and mass cannot

    possibly allow a constancy of one or the other.

    However, the amount of energy and mass change

    are too small to be detected in the laboratory.

    Therefore, we can still count on the law ofconservation of mass for practical use.

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    LAW OF DEFINITE PROPORTION

    -states that a pure compound always contains the

    same elements combined in the same proportions

    by mass.

    -when two elements combine to form a given

    compound, they always do so in a fixedproportion.

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    LAW OF DEFINITE PROPORTION

    To illustrate the law of definite composition, refer

    to the table

    Trial Mass of C

    (g)

    Mass of O2(g)

    Mass of CO2(g)

    1

    2

    3

    2.00

    15.00

    5.00

    5.34

    40.05

    13.36

    7.34

    55.05

    18.36

    These are the mass data for the amounts of C,

    O2, and CO2 in the reaction.

    C + O2 CO2

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    LAW OF DEFINITE PROPORTION

    Trial Mass of C

    (g)

    Mass of O2

    (g)

    Mass of CO2

    (g)

    1

    2

    3

    2.00

    15.00

    5.00

    5.34

    40.05

    13.36

    7.34

    55.05

    18.36

    Find the percentages of carbon and oxygen.

    % of C =

    x 100

    % of O =

    x 100

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    LAW OF DEFINITE PROPORTION

    Trial Mass of C

    (g)

    Mass of O2(g)

    Mass of CO2(g)

    1

    2

    3

    2.00

    15.00

    5.00

    5.34

    40.05

    13.36

    7.34

    55.05

    18.36

    Trial 1 Trial 2 Trial 3

    C =2.0

    7.34 x 100

    = 27.2%

    O =5.34

    7.34 x 100

    = 72.8%

    C =15.0

    55.05 x 100

    = 27.2%

    O =40.05

    55.05 x 100

    = 72.8%

    C =5.0

    18.36 x 100

    = 27.2%

    O =13.36

    18.36 x 100

    = 72.8%

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    LAW OF DEFINITE PROPORTION

    Trial 1 Trial 2 Trial 3

    C =2.0

    7.34 x 100

    = 27.2%

    O =

    5.34

    7.34 x 100= 72.8%

    C =15.0

    55.05 x 100

    = 27.2%

    O =

    40.05

    55.05 x 100= 72.8%

    C =5.0

    18.36 x 100

    = 27.2%

    O =

    13.36

    18.36 x 100= 72.8%

    These show that the percentage of carbon in CO2

    which is 27.2% and the percentage of oxygen inCO2 which is 72.8% are definite.

    The experimental data in the example also verify

    and confirm the law of conservation of mass.

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    LAW OF MULTIPLE PROPORTIONS

    -this states that when two elements combine toform more than one compound, the masses of

    one element which combine with a fixed mass of

    the other element are in a ratio of small whole

    numbers such as 2:1, 1:1, 2:3, etc.

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    LAW OF MULTIPLE PROPORTIONS

    Sample Exercise:Two different compounds elements C and D

    were found to have the following compositions:

    C D

    First compound 2.276 g 0.792 gSecond compound 1.422 g 0.948 g

    Find the formulas of the two compound.

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    Solution: Mass may be fixed at C.

    C D

    First compound

    2.276 g/2.276g = 1 0.792 g/2.276 g = 0.348

    Second compound

    1.422 g/1.422 g = 1 0.948 g/1.422 g = 0.667

    Therefor, the formulas of the two compounds are:Ratio C D Formula

    First 1 0.348/0.348 =1 CD

    compound

    Second 1 0.667/0.348 = 2 CD2compound

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    Practice exercises

    Find the formula of the compounds having the

    following composition:X Y

    First compound 50 g 56 g

    Second compound 35 g 82 g