SAT Chemistry Chemical Formulas - Laws Of Definite Composition and Multiple Proportions

SAT Chemistry Chemical Formulas - Laws Of Definite Composition and Multiple Proportions

In the problems involving percent composition, we have depended on two things: each unit of an element has the same atomic mass, and every time the particular compound forms, it forms in the same percent composition. That this latter statement is true no matter the source of the compound is the Law of Definite Composition. There are some compounds formed by the same two elements in which the mass of one element is constant, but the mass of the other varies. In every case, however, the mass of the other element is present in a small-whole- number ratio to the weight of the first element. This is called the Law of Multiple Proportions. An example is H2O and H2O2.
In H2O the proportion of H: O = 2:16 or 1:8
In H2O2 the proportion of H: O = 2:32 or 1:16
The ratio of the mass of oxygen in each is 8:16 or 1:2 (a small-whole-number ratio).

An equation is a simplified way of recording a chemical change. Instead of words, chemical symbols and formulas are used to represent the reactants and the products. Here is an example of how this can be done. The following is the word equation of the reaction of burning hydrogen with oxygen:
Hydrogen + oxygen yields water.
Replacing the words with the chemical formulas, we have
H2 + O2 —> H2O

We replaced hydrogen and oxygen with the formulas for their diatomic molecular states and wrote the appropriate formula for water based on the respective oxidation (valence) numbers for hydrogen and oxygen. Note that the word yields was replaced with the arrow.
Although the chemical statement tells what happened, it is not an equation because the two sides are not equal. While the left side has two atoms of oxygen, the right side has only one. Knowing that the Law of Conservation of Matter dictates that matter cannot easily be created or destroyed, we must get the number of atoms of each element represented on the left side to equal the number on the right. To do this, we can only use numbers, called coef¬ficients, in front of the formulas. It is important to note that in attempting to balance equations THE SUBSCRIPTS IN THE FORMULAS MAY NOT BE CHANGED.
Looking again at the skeleton equation, we notice that if 2 is placed in front of H2O the numbers of oxygen atoms represented on the two sides of the equation are equal. However, there arc now four hydrogens on the right side with only two on the left. This can be corrected by using a coefficient of 2 in front of H2. Now we have a balanced equation:
2H2 + O2 —> 2H2O

This equation tells us more than merely that hydrogen reacts with oxygen to form water. It has quantitative meaning as well. It tells us that two molecular masses of hydrogen react with one molecular mass of oxygen to form two molecular masses of water. Because molecular masses are indirectly related to grams, we may also relate the masses of reactants and products in grams.
This aspect will be important in solving problems related to the masses of substances in a chemical equation.
Here is another, more difficult example: Write the balanced equation for the burning of butane (C4H10) in oxygen. First, we write the skeleton equation:
C4H10 + O2 yields CO2 + H2O.

Looking at the oxygens, we see that there are an even number on the left but an odd number on the right. This is a good place to start. If we use a coefficient of 2 for H2O, that will even out the oxygens but introduce four hydrogens on the right while there are ten on the left. A coefficient of 5 will give us the right number of hydrogens but introduces an odd number of oxygens. Therefore, we have to go to the next even multiple of 5, which is 10. Ten gives us 20 hydrogen atoms on the right. By placing another coefficient of 2 in front of C4H10, we also have 20 hydrogen atoms on the left. Now the carbons need to be balanced. By placing an 8 in front of CO2, we have eight carbons on both sides. The remaining step is to balance the oxygens. We have 26 on the right side, so we need a coefficient of 13 in front of the O2 on the left to give us 26 oxygens on both sides. Our balanced equation is:
2C4H10 + 13O2 -> 8CO2 + 10H2O

Once an equation is balanced, you may choose to give additional information in the equation. This can be done by indicating the phases of substances, telling whether each substance is in the liquid phase (l), the gaseous phase (g), or the solid phase (s). Since many solids will not react to any appreciable extent unless they are dissolved in water, the notation (aq) is used to indicate that the substance exists in a water (aqueous) solution. Information concerning phase is given in parentheses following the formula for each substance. Several illustrations of this notation are given below:
An example of phase notation in an equation:
2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)
In words, this says that a water solution of hydrogen chloride (called hydrochloric acid) reacts with solid zinc to produce zinc chloride dissolved in water plus hydrogen gas.

At times, chemists choose to show only the substances that react in the chemical action. These equations are called ionic equations because they stress the reaction and production of ions. If we look at the preceding equation, we see the complete cast of “actors”:

Notice that nothing happened to the chloride ion. It appears the same on both sides of the equation. It is referred to as a spectator ion. In writing the net ionic equation, spectator ions are omitted, so the net ionic equation is:
2H+(aq) + Zn(s) → Zn2+(aq) + H2(g)


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