## SAT Subject Test Chemistry

__PART 2__

__REVIEW OF MAJOR TOPICS__

__CHAPTER 4__

__Chemical Formulas__

__WRITING AND BALANCING SIMPLE EQUATIONS__

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

H_{2} + O_{2} → H_{2}O

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 **coefficients**, 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.

**TIP **

You cannot change subscripts of formulas to attempt to balance an equation!

Looking again at the skeleton equation, we notice that if 2 is placed in front of H_{2}O the numbers of oxygen atoms represented on the two sides of the equation are equal. However, there are 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 H_{2}. Now we have a balanced equation:

2H_{2} + O_{2} → 2H_{2}O

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 (C_{4}H_{10}) in oxygen. First, we write the skeleton equation:

C_{4}H_{10} + O_{2} yields CO_{2} + H_{2}O

**TIP **

First deal with multiatomic reactants.

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 H_{2}O, 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 C_{4}H_{10}, we also have 20 hydrogen atoms on the left. Now the carbons need to be balanced. By placing an 8 in front of CO_{2}, 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 O_{2 }on the left to give us 26 oxygens on both sides. Our balanced equation is:

2C_{2}H_{4} + 13O_{2} → 8CO_{2} + 10H_{2}O

For more practice in balancing equations, see Chapter 12.

**TIP **

Be sure you have included all sources of a particular element since it may occur in two or more compounds.