A chemical equation is a symbolic representation of all of the substances involved in a chemical reaction. We use the chemical formulas of substance to represent each chemical specie involved in the reaction. We also use the notation (g), (l), (s), or (aq) following the chemical formula to identify the phase of the substances in the equation.

As an example of a chemical reaction we watched the reaction between iron, Fe(s), and sulfur, S8(s), on video. There are some additional reactions on our Web site that you can review. We noted the physical properties of both iron and sulfur before the reaction and the physical property of the product of the reaction. The magnetic property of iron, present when iron was in its elemental form, was absent in the product. The yellow color of the sulfur, present in the elemental form as a reactant was absent in the product. Yet both iron and sulfur are in the product. Since the properties changed the product is a new substance.

How do we tell if a chemical reaction occurs? The best evidence is a difference in the physical and chemical properties of the reactants and the products. This can be obvious as in the case of the reaction between iron and sulfur. Other characteristics to watch for when mixing two or more substances;

The form of a chemical equation involves writing the formulas of the reactants (the substances that are mixed together) on the left, using '+' when more than one substance is involved and the formula(s) of the product(s) on the right. The reactants and the products are separated by an arrow '--->'. Sometimes additional information about the reaction is placed above or below the arrow which separates the reactants and products. Such information include;

Well, how about writing the equation for the reaction between iron and sulfur? To do that we have to know the formula for both elements and their phases;

Fe(s) + S8(s) --heat-->

(Note: I can not use a delta symbol on the WEB easily). Now we have to know what the products are. This is a little more difficult but in this case is it easy. We are reacting a metal with a nonmetal so we know the product is an ionic compound. Since there are only two elements reacting the product has to be binary. So we need the formula for a binary ionic compound. You know how to do that. Metals form cations and the nonmetal forms an anion and we need to use the principle of electroneutrality to balance the charges. Our only other problem is that iron is a transition metal and as such it can have several different charges. For our purposes we limit the possibilities to 2+ or 3+. If we use 2+ for iron and we know sulfur is always 2-, the formula of the product would have to be, FeS(s). We know it is a solid because we saw on the video it was solid, but also ALL ionic compounds are solids!

So the reaction is,

Fe(s) + S8(s) --heat--> FeS(s)

Another reaction (side reaction) which occurred as the above reaction occurred was between sulfur and oxygen in the air. This side reaction was evident because of the bluish flame that appeared as the sulfur melted. At one point in the video as we looked the crucible, containing the reaction, from the side the blue flame was erupting from the crucible. The reaction was surpressed when the hand placed the crucible lid onto the crucible. But what was the reaction?

S8(s) + O2(g) --heat-->

This is a harder question because we are combining two nonmetals. The product of this reaction we know will be a binary covalent compound. But we do not know what the formula of the product might be because we have not discussed any rules for determining formulas of covalent compounds. So what do we do? Well, unfortunately, we have to know some formulas of binary compounds containing sulfur and oxygen. You know sulfate, SO42-, and sulfite, SO32- are either of these likely? No! And the reason is the charged nature of those species. We will not form a charged substance in this particular case. What else you ask? How about SO? Is that possible? Well, no it is not. If we try to find such a compound in the Merck Index or the CRC we will not. So how about SO2? In this case you would be right on! That is the correct formula. Sulfur dioxide is an evil smelling, colorless, gas. So the reaction is;

S8(s) + O2(g) --heat--> SO2(g)

These two reactions we have discussed are called formation reactions. (Our textbook also refers to these reactions as synthesis or combination reactions.)

Formation reactions are characterized by the fact that the reactant are elements in their standard state and the product is a compound containing the reactant elements combined in some way. We have seen two formation reactions so far.

Formation reactions involving metals and nonmetals are the more straightforward. Writing the formula of the product is guided by the knowledge that the metal becomes positively charged and the nonmetal becomes negatively charged. We use the principal of electroneutrality (balance the charges) to write the formula of the product.

All ionic compounds are solids in their standard state (25 degrees Celsius and 1 atmosphere). Ionic compounds are composed of ions; elements or molecules which are positively charged are called cations and elements or molecules which are negatively charged are called anions. A simple model to be used to imagine what ions (cations or anions) look like at the atomic level, is a sphere. For simple monoatomic ions and most polyatomic ions we can think of as charged spheres of different sizes.

In the state the spheres (ions) are arranged so that any particular ion is surrounded by the oppositely charged ion. What ever pattern which arises depends on several factors which we will not be discussing at this point in our class. Below is a picture of a model of a simple ionic compound. Notice the spheres (ions) of different sizes and how each sphere (ion) is surrounded by the other spheres (ion).

When the elements are both nonmetals predicting the products is a little more difficult. The product is predicted based on our 'chemical intuition'. Our chemical intuition is based on what compounds we have been exposed to during the semester.

Besides the one example of this type above, others we have discussed include;

Several of these formation reaction we have discussed are not balanced. So let's take a moment and talk about balancing chemical equations.

How about balancing these reactions? Balancing equations is the process of equating the elements on both sides of the reaction arrow. A balanced equation has equal numbers of each element on both sides of the equation. Balancing consists of introducing coefficients which proceed the formula in the equation. Subscripts are NEVER changed when balancing equations.

For many reactions a method of trail and error is used. By practicing you're skills will sharpen. As you balance more and more equations certain 'rules' begin to emerge which can be useful. One rule which is very useful is to balance elemental forms last when balancing equations. This rule derives from the fact that changing the coefficient preceding an element only effects that element. Changing a coefficient preceding a compound changes all the elements in the compound and may effect another substance in the equation.

So shall we balance a few of the equations we've covered so far?

Let's begin with;

Fe(s) + S8(s) --heat--> FeS(s)

In this equation if we begin with the element iron, we see there one Fe atom on the left and one Fe atom on the right. So iron is balanced. Sulfur has eight atoms on the left and only one on the right. Sulfur is not balanced. To balance the sulfur atoms we need to replace the coefficient preceeding FeS with the number 8.

Fe(s) + S8(s) --heat--> 8FeS(s)

Now the sulfur atoms are balanced on both sides, but by adding the coefficient we've unbalanced the iron atoms. To fix that we need to place a coefficient of 8 before the elemetn Fe on the left side of the equation.

8Fe(s) + S8(s) --heat--> 8FeS(s)

Now the equation is balanced.

Balance the following equations;

a) S8(s) + O2(g) --heat--> SO2(g) Answer

b) Fe3O4(s) + H2(g) ----> Fe(s) + H2O(l) Answer

We've used the reaction between iron and sulfur and between sulfur and oxygen as examples of formation reactions. There was another reaction which occurred when the iron and sulfur mixture was heated. That reaction occurred when we lit the Bunsen burner to heat the mixture of iron and sulfur. The reaction that was used to heat the mixture was a combustion reaction between methane (CH4) and oxygen (O2).

CH4(g) + O2(g) --heat-->

Methane is an example of a hydrocarbon. It is not the only example of a hydrocarbon. But more about that shortly. The products of this combustion reaction are carbon dioxide gas and water vapor (liquid).

CH4(g) + O2(g) --heat--> CO2(g) + H2O(g)

Other hydrocarbons you are familiar with include;























You should know the name and formula for the first ten hydrocarbons (memorize).

Methane is a colorless, odorless gas which is used as a fuel in most gas stoves to heat/cook food. Gas companies add a compound which has an odor to help detect gas leaks should they arise. The compound I mentioned in class that was added to natural gas was CH3SH. This is an error, the compound that is added to natural gas is call t-butyl mercaptan and has the formula C4H9SH. So lets write the combustion reaction of this compound.

C4H9SH(g) + O2(g) --heat--> CO2(g) + H2O(g) + SO2(g)

Hey we get more practice balancing equations!

Balance the following equations;

a) CH4(g) + O2(g) --heat--> CO2(g) + H2O(g) Answer

b) C4H9SH(g) + O2(g) --heat--> CO2(g) + H2O(g) + SO2(g) Answer

c) C4H10(g) + O2(g) ----> CO2(g) + H2O(l) Answer

We'll consider some other chemical reaction types in the next Chapter. The formation and combustion reaction are sufficient for us to consider at this time.

When we look at the periodic table each element has two numbers associated with it. The atomic number (the number of protons in the elements) and the atomic number (the relative weighted average atomic mass). Since the atomic number is a number it has no units. The average atomic mass has units of atomic mass units (u). The average atomic mass is the mass in u's of a single atom of the element. Previously we had defined an atomic mass unit as one twelveth the mass of one atom of the isotope of 12C.

1 amu = 1.66054 x 10-24 grams

So for the element sodium it average atomic mass is 22.99 u. It is possible to calculate the mass of one atom of carbon in grams. We can do this by converting from u's to grams using the conversion factor that relates u's to grams.

So the mass of one atom of carbon is determined by the following conversion;

22.99 u * (1.66057 x 10-24 grams/1 u) = 3.818 x 10-23 grams

This is the mass of one atom of sodium in grams.

What is the mass of 100 atoms of sodium?

100 atoms * (3.818 x 10-23 grams/1 atom) = 3.818 x 10-21 grams

What is the mass of 10,000 atoms of sodium?

10,000 atoms * (3.818 x 10-23 grams/1 atom) = 3.818 x 10-19 grams

What is the mass of 6.022 x 1023 atoms of sodium ?

6.022 x 1023 sodium atoms * (3.818 x 10-23 grams/1 atom) = 22.99 grams

hmmm...now isn't that interesting!

What is the mass of 1 mol of sodium atoms?

The answer to this question is found in the definition of a mol,

a mol and is defined as the number of 12C atoms in 12 g of 12C. This number has been experimentally determined as 6.022 x 1023 atoms. A mol of any substance is defined as the amount of the substance which contains the same number of units as are in 12 g of 12C or 6.023 x 1023 units of the substance.

Remember a unit is an atom, a molecule or a formula unit.

So if a mol of sodium atoms is 6.022 x 1023 atoms than the mass of one mol of sodium atoms is,

6.022 x 1023 sodium atoms * (3.818 x 10-23 grams/1 atom) = 22.99 grams

hmmm...now isn't that interesting!

The mass in grams of a mol of an element is equal to the atomic mass. In the case of sodium, its atomic mass is 22.99 u (this is the relative weighted average atomic mass for sodium in atomic mass units). The mass of 1 mol of sodium atoms is the atomic mass expressed in grams, or 22.99 g.


Consider the balanced equation

We can balance the equation and we recognize it is an example of a formation reaction. How do we read this equation?

This equation reads 1 atom of magnesium combines with one molecule of chlorine to form one formula unit of the ionic compound magnesium chloride. If we wanted to do this reaction to prepare some magnesium chloride it would be difficult to find a single atom of magnesium and a single molecule of chlorine, even for the most experienced chemist. In the chemistry laboratory we must measure amounts of substances by mass. What is the relationship between the amount of substance and the number of atoms or molecules? How are we able to perform quantitative chemical reactions? There is an answer!

So if I have 1 mol of 12C, how many atoms of 12C do I have? Answer

If I have 1 mol of Mg atoms, how many Mg atoms do I have? Answer

If I have 1 mol of Cl2 molecules, how many molecules of Cl2 do I have? Answer

How many atoms of chlorine do I have in 1 mol of Cl2 molecules? Answer

How many formula units are in 1 mol of MgCl2? Answer

According to our definition of a mol 1 mol of 12C weighs 12 g. Interestingly the relative atomic mass of 12C is 12.000 u. 12.000 u is the relative atomic mass of one 12C atom. 12.000 g is the mass of 1 mol of 12C atom. And 1 mol of 12C has 6.02 x 1023 12C atoms. The mass of one 12C atom in grams is;

12.000 g / 6.022 x 1023 12C atoms = 1.99 x 10-23 g per atom

A mol of 12C atoms weighs the same in grams as the atomic mass of 12C, 12 g. The weight of any substance which contains 6.023 x 1023 units is given by its atomic or formula mass expressed in grams and is called the molar mass of the substance.

How much does 1 mol of Mg atoms weigh? Answer

How much does 1 mol of Cl2 weigh? Answer

How much does 1 mol of MgCl2 weight? Answer

How much does 1 mol of C6Hl2O6 weigh? Answer

A molar mass is the mass in grams of 1 mol of a substance. So for the compounds listed above,


Molar Mass (g)









It is now possible to 'count' atoms, molecules or formula units when weighing chemicals in the laboratory. For the reaction we discussed earlier,

weighing 24.3 grams of magnesium atoms and 71 g of chlorine means we have 6.022 x 1023 atoms of magnesium and 6.022 x 1023 molecules of chlorine. According to the balanced chemical equation magnesium and chlorine combine in equal numbers. The reaction will produce 6.022 x 1023 formula units of MgCl2 which weighs 95.3 g.

So the equation can be read as 1 mol of magnesium reacts with 1 mol of chlorine to produce 1 mol of magnesium chloride.

The relationship between a mol, Avogadro's number and the atomic or formula weight expressed in grams are important unit factors.

A sample problem will demonstrate the use of these unit factor.

Here are few additional calculations.

Calculate how many mol are in 10.0 grams of glucose (C6Hl2O6).

To answer this question we need to convert grams to mol. Since 1 mol of has a mass of 180 g we can set up the conversion as;

How many molecules of glucose are in 10.0 g of C6Hl2O6?

We can now convert from mol of C6Hl2O6 to moelcules using Avogadro's number.

How many oxygen atoms are there in 10.0 grams of C6Hl2O6?



Sample Problem:

Determine the number of moles of N2O5 in 45.6 g of N2O5 gas and the number of nitrogen atoms.


The molar mass of N2O5 is 108 g­mol-1. The sample is 45.6 g so we know there is less than a mol since a mol of N2O5 weighs 108 g. To calculate the number of moles we setup the problem in the following way;

To determine the number of nitrogen atoms in the sample we must first determine how many N2O5 molecules.

Now we can determine the number of nitrogen atoms;


Calculate the number of magnesium atoms in 10.0 g of magnesium.



Here is one of my favorite problems;