Basic Chemical's Laws (Law of Definite Proportions)

            In 1799, Joseoh Proust found that the elements constructed calcium carbonate (CaCO3), both natural and synthetically made in laboratory have a constant number in proportion and mass. How about other compounds? For example is CuO compound. The comparison of the number of elements in CuO is Cu and O, It always has equal proportions (1:1). Reaction between copper and oxigen gas can be written us follows.

2Cu + O2 ----> 2CuO

          The  proportion of relative atomic mass of elements in CuO is shown in the table below.

            Mean while, based on the experiment from reacting copper and oxygen (by heating procces), can be result as show below.

         The mass of copper oxide is equal to the mass of copper before the heating considering that all copper perfectly react. Whereas the mass of oxygen in copper oxide is the difference between the mass of copper oxide and copper.

        Based on the table, we caan conclude that mass proportion of copper and oxygen is constat, about 3,9 : 1. That proportion equals to the comparison or relative atomic mass of copper and oxygen in cupric oxides.

          From hi sobservation on mass proportion, Proust declared that the mass of each element which formed a compound has a definite proportion. It is called the law of definite proportion. By using that law, we can predict the mass of element reacted, the mass of element left, and the mass of the product.

Example

The mass proportion of carbon and oxygen in carbon dioxides (CO2) is 3:8

a. How many gram of carbon can be reacted with 24 gram of oxygen?

b. If 6 gram of carbon is rected with 12 gram of oxygen, is there any element left? How many gram of carbon dioxides formed?

c. How many gram of carbon and oxygen that must be reacted to form 33 gram of carbon dioxide compound?

Solution

Equal of reaction is : C + O2 -----> CO2

C : O = 3:8, means taht each 3 gram of C can react with 8 gram of oxygen to produce 11 gram of carbon dioxide.

a. To react 24 gram of O2, needs C as:

    3/8 x 24 gram = 9 gram

b. To determine recatant mass taht has been reacted completely and left, determine that element mass that has been reacted completely by dividing each gram of the element with its standard number of comparison. Element which has smallest division in result, means it has been reacted completely.

For carbon = 6/3 = 2

For oxygen = 12/8 = 1,5 (the smallest division in result, means oxygen has been reacted completely)

Remember!

The division result is only to determine which one is empty (reacted completely) or have remains, and is not included in next calculation.

Oxygen reacted = 12 gram

Carbon reacted = 3/8 x 12 gram = 4,5 gram

Rest of carbon = 6 gram - 4,5 gram = 1,5 gram

Carbon dioxides product = 12 gram + 4,5 gram = 16,5 gram

c. To form 33 gram of carbon dioxides need:

Carbon (C) as 3/11 x 33 gram = 9 gram

Oxygen (O2) as 8/11 x 33 gram = 24 gram

Exercise

The proportion of iron (Fe) mass and sulfur (S) mass in ferrous sulfate (FeS) is 7:4

a. How many gram of iron can be reacted with 16 gram of sulfur?

b. If 35 gram of iron is rected with 25 gram of sulfur, how many gram of sulfur taht has not reacted?

c. How many gram of iron and sulfur have to be reacted to form 44 gram of ferrous sulfide compound?