Using the simple example of the reaction when aqueous solutions of sodium chloride and silver nitrate are mixed. The chemical equation you would usually write is as follows: (Note: Arrows may not show up properly on your monitor. If you see something odd (like ¾ R), it’s supposed to be a reaction arrow.)
Molecular equation: NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)
(or Overall equation)
Ionic equation: Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) ¾® AgCl (s) + Na+ (aq) + NO3- (aq)
(or total ionic equation)
Net ionic equation: Cl-(aq) + Ag+ (aq) ¾® AgCl (s)
which is usually re-written as Ag+ (aq) + Cl-(aq) ¾® AgCl (s)
NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)
It is actually rather odd to call this equation “molecular” since there are no molecules in the reaction (other than the solvent, water). All of the reactants and products are ionic compounds. None of them are molecules.
How do you know whether the reactants and products are ionic or molecular? If you have not learned this yet, or perhaps forgotten this, you should certainly learn it immediately. To tell whether a compound is molecular or ionic, the general rule is that an ionic compound is made of metal cations and nonmetal anions. A molecular compound is made of nonmetals only. You would only need to check the first element listed in the formula. If it is a metal, it is most likely to be ionic. Of course, this is only a general rule, and there are exceptions. One of the important exceptions is the ammonium ion (NH4+), which is a cation, but does not contain a metal.
Ionic compounds: Metal cations + Nonmetal anions
Molecular substances: All nonmetals
Important exception: Compounds with NH4+ are ionic even though they have no metals.
Why then is “NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)” called a “molecular equation” when there are no molecules? The reason is because the formulas are shown with no charges even though the compounds contain ions with charges. For example, “NaCl (aq)” represents Na+ and Cl- ions, each surrounded by water molecules. Technically speaking we should really write Na+Cl- (aq) and Ag+NO3- (aq). But, by convention, we don’t. These compounds are written as if they were molecules, with no charges showing. These equations are therefore called “molecular equations”. They are useful in that they show the correct stoichiometry and physical states without the cumbersome charges. The formulas are, of course, written so that the positive charge from the cations equal to the negative charge from the anions. In other words, the formulas of ionic compounds must have a net charge of zero. At this point, you should have already learned how to write these formulas: Na+ and SO42- make Na2SO4 and not NaSO4- and not NaSO4. (If you need practice writing formulas correctly, click here.)
Ionic equations are written to help you get to the net ionic equations, which are equations that show the essence of the reaction. When you become experienced at this, you would not have to write ionic equations in order to determine what the net ionic equations are. But, for now, you should learn to do this stepwise.
Going back to our original molecular equation:
NaCl (aq) + AgNO3 (aq) ¾® AgCl (s) + NaNO3 (aq)
The essence of this reaction is the Ag+ cations combining with the Cl- ions to form the insoluble AgCl precipitate.
Ag+ (aq) + Cl- (aq) ¾® AgCl (s)
This is called a net ionic equation.
The Na+ and NO3- ions are of no concern. They are called spectator ions. The cation (Na+) can be replaced by other cations such as K+, NH4+, Mg2+, etc., and the anion (NO3-) can be replaced by other anions such as ClO3-, C2H3O2-, and the essence of the reaction would still be the same: Ag+ (aq) + Cl- (aq) ¾® AgCl (s).
To get a full discussion of this, you must refer to your textbook. I can only give you a summary:
In writing molecular equations:
1. Predict the formula of the products solely according to the charges of the ions.
· For metal cations that have variable charges, you can deduce their charges by looking at the charge of the anion.
For example, in FeCl2, since you know the chloride ion has a charge of 1– each, the iron must have a charge of 2+.
In some cases it may not be possible to predict
what the charge is on the cation. In that
case you may refer to the activity series to see which charge to use.
For example, in the reaction (Fe + HCl ¾® ?), you cannot deduce the charge of Fe after the reaction. By examining the Activity Series, you see that Fe is likely to form 2+, so you would expect the product to be FeCl2.
· Most of the charges of the anions you should already know (such as sulfate having a charge of 2–, bromide having a charge of 1-, etc.)
· Occasionally you will run into anions that you had not encountered before. In such cases you can deduce the charge of the anion by looking at the charge of the caiton.
For example, in Na2S2O3, since you know each sodium ion is 1+, the anion must have a charge of 2–. Incidentally this is the thiosulfate ion, S2O32–.
Do not try to balance the equation yet!!
2. Decide whether the reaction will take place.
For precipitation reactions,
use solubility rules to decide whether a ppt (precipitate) will form.
Generally, all neutralization reactions will occur.
For redox (single replacement-type), use the activity series to decide.
3. Write in physical states of each product according to solubility rules, and physical states of pure substances. If you have not learned the formulas and physical states of pure substances (element not bonded to another element) you should study Unit II of the Nomenclature Tutorial. You will find it invaluable! Click here to get to the Nomenclature Tutorial.
4. Balance the equation by adding only coefficients, not by changing subscripts.
Example: Predict the products to this reaction:
Na2CO3 (aq) + Cu (s) ¾®
Since copper has a variable charge, you should look up in the activity series, the charge for the copper ion (which turns out to be Cu2+). Thus the product would be CuCO3 (s) + Na (s). (You should know that the carbonate has a charge of 2–) and Na should not be written as Na2.
Na2CO3 (aq) + Cu (s) ¾® CuCO3 (s) + Na (s)
To balance the number of Na, write “2” in front of Na.
Na2CO3 (aq) + Cu (s)
(s) + 2 Na (s)
Will the reaction occur? Examine the Activity Series. Cu is lower than Na, so Cu cannot replace Na.
No, the reaction will not occur. Be careful you do not say that the reaction will occur “because a precipitate has formed!” This is not a precipitation-type reaction, so we are not concerned as to whether a solid is formed.
The ionic equation, or total ionic equation, is the intermediary step between molecular equation and net ionic equation. You take the molecular equation and convert all the strong electrolytes into their ionic form.
What are strong electrolytes? They are defined as compounds that completely dissociate into ions when placed in water, and thus conduct electricity well. They consist of mainly two types of compounds. Learn this general rule well!!
1) Ionic compounds that are soluble and dissolved in water.
2) Strong acids dissolved in water.
Note that both are dissolved in water, and they are easily recognized in a molecular equation in that they MUST have the physical state (aq). In addition to having (aq) designated, you must check that they are ionic and not molecular, OR they must be a strong acid.
What is a strong acid? How do you know whether it is a strong acid?
The answer is simple. First of all you can easily recognize what is an acid. Their formulas usually start with H.
For example: HF, HCl, HNO2, H2SO4, H3PO4
How do you know whether they are strong acids? The answer is also very simple: You memorize the names and formulas of the 6 common strong acids. All other acids can be assumed to be weak.
6 common strong acids are: HNO3, H2SO4, HClO4,
Take the time to memorize this now! You will find it MOST helpful to remember this list throughout the semester and the next (in Gen Chem II and organic chemistry).
HERE IS THE IMPORTANT RULE FOR DECIDING WHETHER TO SEPARATE A FORMULA
The scientific term for separating a formula into ions is "dissociate" which means to "split" into ions:
In writing total ionic equations (your textbook may refer to them as just ionic equations):
Convert into separate ions, only the following:
1. Ionic compounds that are dissolved in water (indicated by (aq)).
2. Strong acids dissolved in water (also indicated as (aq)).
All others should be left intact!
In the following exercise write down what you think the answers should be, then scroll down to check your answers.)
e.g. K3PO4 (aq) becomes ________________
3Na2CO3 (aq) becomes ________________
CH3OH (aq) becomes ________________
Ba(CH3CO2)2 (aq) becomes ________________
2(NH4)2SO4 (aq) becomes ________________
NaCl (s) becomes ________________
H2O (l) becomes ________________
H2S (g) becomes ________________
HNO2 (aq) becomes ________________
C12H22O11 (aq) (This is table sugar which you well know can dissolve in water.)
Zn (s) becomes _________________
H2 (g) becomes _________________
Answers to above exercise:
K3PO4 (aq) becomes 3K+ (aq) + PO43- (aq)
3Na2CO3 (aq) becomes 6Na+ (aq) + 3CO32- (aq)
CH3OH (aq) becomes CH3OH (aq) (unchanged because it is not ionic)
Note that this is not an acid.
Ba(CH3CO2)2 (aq) becomes Ba2+ (aq) + 2 CH3CO2- (aq)
2(NH4)2SO4 (aq) becomes 4NH4+ (aq) + 2SO42- (aq)
NaCl (s) becomes NaCl (s) (unchanged because it is not dissolved in water)
H2O (l) becomes H2O (l) (unchanged because H2O is not on our list of strong acids)
H2S (g) becomes H2S (g) (unchanged because it is not dissolved in water, also it is not on our list of strong acid)
HNO2 (aq) becomes HNO2 (aq) (unchanged because it is not on our list of strong acids)
C12H22O11 (aq) (sugar dissolved in water)
becomes C12H22O11 (aq) (unchanged because it is not ionic)
Zn (s) becomes Zn (s) (unchanged because Zn is not an ionic compound)
It does NOT become Zn2+!
H2 (g) becomes H2 (g) (unchanged because is not ionic)
It does NOT become H+!
For example, for the molecular equation
Sr (s) + 2H2O (l) ¾® Sr(OH)2 (aq) + H2 (g)
the ionic equation is
Sr (s) + 2H2O (l) ¾® Sr2+ (aq) + 2OH– (aq) + H2 (g).
It is not...
Sr2+ (s) + 2H2O (l) ¾® Sr2+ (aq) + (OH)22– (aq) + 2H+ (g).
Sr (s) is not ionic! It is made of neutral atoms.
Neither is H2, which is a molecule. It is not made up of ions.
that we do not write (OH)22– because you should recognize
(OH) as the hydroxide ion
When you write (OH)22– you are implying that the two OH’s are bonded together: (HOOH)2–.
Instead we are talking about two hydroxide ions:
2OH– which stands for 2 individual ions: OH– and OH–.
example of HNO2 (aq).
First you should recognize it as an acid. Second you should recognize it as being a
WEAK acid (because it is not on our list of strong acids).
It should therefore not be dissociated and should be written intact as HNO2 (aq).
The hardest part is writing the ionic equation. Once you have the correct ionic equation, the net ionic equation is easy to arrive at. All you have to do is cancel out the "spectator ions", which are ions that appear on both sides of the equation. What remains is the net ionic equation. Be sure to check that the coefficients are reduced to the lowest ratio.
Example: NaCl (s) + Ag+ (aq) + NO3- (aq) ¾® AgCl (s) + Na+ (aq) + NO3- (aq)
becomes NaCl (s) + Ag+ (aq) ¾® AgCl (s) + Na+ (aq)
Note that this is different from the reaction described at the top of the page. Here, the reactant NaCl is provided as the solid sodium chloride rather than the aqueous solution.
molecular equation: H2SO4 (aq) + 2KOH (aq) ¾® 2H2O (l) + K2SO4 (aq)
ionic equation: 2H+ (aq) + SO42- (aq) + 2K+ (aq) + 2OH- (aq) ¾® 2H2O (l) + 2K+ (aq) + SO42- (aq)
net ionic equation: 2 H+(aq) + 2OH- (aq) ¾® 2H2O (l)
which needs to be reduced to H+ (aq) + OH-(aq) ¾® H2O (l)
Spectator ions are SO42- and K+.
The charges in the net ionic equation are conserved:
The net charge on the left side of the equation must equal to the net charge on the right side.
This gives you a quick check on whether you made any careless mistakes.
NaCl (s) + Ag+ (aq) ¾® AgCl (s) + Na+ (aq)
the left side has a net charge of +1, and so does the right side.
In H+ (aq) + OH-(aq) ¾® H2O (l), the net charge on the left side is zero, and so is the net charge on the right.
no new rule that applies to weak
acids. Weak acids do not fit the
criterion for being dissociated. As
pointed out in an earlier example, HNO2 (aq) is such a
case. It does dissolve in water, but as
a weak acid, it exists primarily as the molecule HNO2 and does not
split up into ions.
HNO2 (aq) + NaOH (aq) ¾® H2O (l) + NaNO2 (aq)
becomes HNO2 (aq) + Na+ (aq) + OH- (aq) ¾® H2O (l) + Na+ (aq) + NO2- (aq)
Net ion eqn: HNO2 (aq) + OH-(aq) ¾® H2O (l) + NO2- (aq)
The spectator ions are Na+.
Incidentally, did you ever wonder why H2O is not considered an acid (with H written in front like the other acids)?
Well, it actually is an acid, just a very, very weak acid. You will work with H2O as a weak acid in Gen Chem II.
What do you with weak bases just as NH3 (aq)? The answer is, no different from any formulas that do not fit the criteria for dissociation.