# Mass and balance

Classified in Chemistry

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## The Mole

### Given the equation above, we can tell the number of moles of reactants and products. A mole simply represents Avogadro's number (6.022 x 10^{23}) of molecules. A mole is similar to a term like a dozen. If you have a dozen carrots, you have twelve of them. Similarly, if you have a mole of carrots, you have 6.022 x 10^{23} carrots. In the equation above there are no numbers in front of the terms, so each coefficient is assumed to be one (1). Thus, you have the same number of moles of AgNO_{3}, NaCl, AgCl, NaNO_{3}.

Converting between moles and grams of a substance is often important. This conversion can be easily done when the atomic and/or molecular mass of the substance(s) are known. Given the atomic or molecular mass of a substance, that mass in grams makes a **mole** of the substance. For example, calcium has an atomic mass of 40 atomic mass units. So, 40 grams of calcium makes one mole, 80 grams makes two moles, etc.

## Balancing Chemical Equations

### Sometimes, however, we have to do some work before using the coefficients of the terms="http:> to represent the relative number of molecules of each compound. This is the case when the equations are not properly balanced. We will consider the following equation:

_{3}O

_{4}---> Al

_{2}O

_{3}

+ Fe

Since no coefficients are in front of any of the terms, it is easy to assume that one (1) mole of Al and one (1) mole of Fe_{3}O_{4} react to form one (1) mole of Al_{2}O_{3}. If this were the case, the reaction would be quite spectacular: an aluminum **atom** would appear out of nowhere, and two (2) iron atoms and one (1) **oxygen** atom would magically disappear. We know from the **Law of Conservation of Mass **(which states that matter can neither be created nor destroyed) that this simply cannot occur. We have to make sure that the **number** of atoms of each particular element in the reactants equals the number of atoms of that same element in the products. To do this we have to figure out the relative number of molecules of each **term** expressed by the term's coefficient.

Balancing a simple chemical **equation** is essentially done by trial and error. There are many different ways and systems of doing this, but for all methods, it is important to know how to count the number of atoms in an equation. For example we will look at the following term.

_{3}O

_{4}

This term expresses two (2) molecules of Fe_{3}O_{4}. In each **molecule** of this substance there are three (3) Fe atoms. Therefore in two (2) molecules of the substance there must be six (6) Fe atoms. Similarly there are four (4) oxygen atoms in one (1) molecule of the substance so there must be eight (8) oxygen atoms in two (2) molecules.

Now let's try balancing the equation mentioned earlier:

_{3}O

_{4}---> Al

_{2}O

_{3}+ Fe

Developing a strategy can be difficult, but here is one way of approaching a problem like this.

- Count the number of each atom on the reactant and on the product side.
- Determine a term to balance first. When looking at this problem, it appears that the oxygen will be the most difficult to balance so we'll try to balance the oxygen first. The simplest way to balance the oxygen terms is:Al + 3 Fe
_{3}O_{4}---> 4 Al_{2}O_{3}+ FeBe sure to notice that the subscript times the coefficient will give the number of atoms of that element. On the reactant side, we have a coefficient of three (3) multiplied by a subscript of four (4), giving 12 oxygen atoms. On the product side, we have a coefficient of four (4) multiplied by a subscript of three (3), giving 12 oxygen atoms. Now, the oxygens are balanced.

- Choose another term to balance. We'll choose iron, Fe. Since there are nine (9) iron atoms in the term in which the oxygen is balanced we add a nine (9) coefficient in front of the Fe. We now have:Al +3 Fe
_{3}O_{4}---> 4Al_{2}O_{3}+ 9Fe - Balance the last term. In this case, since we had eight (8) aluminum atoms on the product side we need to have eight (8) on the reactant side so we add an eight (8) in front of the Al term on the reactant side.

Now, we're done, and the balanced equation is:

_{3}O

_{4}---> 4Al

_{2}O

_{3}+ 9 Fe