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, know 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. When the elements are both nonmetals predicting the products is a little more difficult. 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;

Name

Formula

methane

CH4(g)

ethane

C2H6(g)

propane

C3H8(g)

butane

C4H10(g)

pentane

C5H12(l)

hexane

C6H14(l)

heptane

C7H16(l)

octane

C8H18(l)

nonane

C9H20(l)

decane

C10H22(l)

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

Double Displacement/Replacement Reactions

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).

What happens when an ionic compound dissolves in water? Below is a figure of a small crystal of sodium chloride surrounded by water molecules. We all know when sodium chloride is added to water the crystals of sodium chloride disappear, that is, they dissolve.

What happens when the sodium chloride dissolves? Below is a picture showing a chloride ion (larger sphere) and a sodium ion (smaller sphere) dissolved in water.

Notice how water surrounds (hydrates) each ion. For the negatively charged chloride ion the hydrogen's of each water molecule orient themselves towards the chloride ion. For the positively charged sodium ion the oxygen's of each water molecule orient themselves towards the sodium ion.

The specific example of what happens when sodium chloride is added to water can be used for the general case of any ionic compound.

Remember we talked about the conductivity of solutions containing dissolved ionic compounds. You will be testing the conductivity of soluble ionic compounds in laboratory later this semester.

To begin we looked at the conductivity of a sample of deionized water (water with little or no ions) which could be viewed as pure water. Looking at the image on the left the light bulb is not glowing. This can be explained in terms of the nature of the species in the sample of water. That the light does not glow suggests there are no ions in the sample. This is an important measurement to establish subsequent measurements in deionized water. Any observed conductivity will due to the presence of the solute.

Next we measured the conductivity of an aqueous solution of sodium chloride. Here the light bulb glows brightly indicating the presence of ions in the solution. The ability of an aqueous solution of sodium chloride to conduct electricity is characteristic of a strong electrolyte. How do we express this observation using a chemical equation?

NaCl(s) ---H2O--> Na+(aq) + Cl-(aq)

Soluble ionic compounds are also strong electrolytes, i.e., behave the same way as sodium chloride...forming ions when dissolved in water. How do we know whether an ionic compound dissolves in water? That is determined by experiment. A Solubility Table summarizes the experimental observations of the solubility behavior of a large group of ionic compounds. We'll discuss a Solubility Table shortly.

When we tested the conductivity of an aqueous solution of sucrose the light bulb did not glow. This type of behavior is characteristic of a nonelectrolyte. No ions in this solution, yet the sucrose dissolved. How do we write a chemical equation in this case?

C12H22O11(s) ---H2O--> C12H22O11(aq)

In general soluble covalent compounds do not dissociate into ions when dissolved in water.

In laboratory you will investigated the solubility of a large collection of ionic compounds. For those ionic compounds that dissolved in water you were expected to write the formula of ions formed. In the post-laboratory you were to write a chemical equation describing what happens when a soluble ionic compound dissolves in water.

For example, sodium chloride dissolves in water. The chemical equation that can be written to describe the solubility of sodium chloride is,

NaCl(s) ---H2O--> Na+(aq) + Cl-(aq)

We can write the equation to describe what happens when copper(II) chloride dissolves in water,

CuCl2(s) ---H2O--> Cu2+(aq) + 2Cl-(aq)

You must be able to write the ions formed and the chemical equation which describe the behavior of soluble ionic compounds.

The goal of Experiment #2 is to organize all the experimental observations into something meaningful. This data can be organized into a Solubility Table. The Solubility Table summarizes the solubility behavior of ionic compounds in terms of anions and cations.

A Solubility Table summarizes the solubility behavior of a large group of ionic substances.

How to interpret a Solubility Table?

We can use the Solubility Table to determine whether an ionic compound exist as ions in aqueous solution (soluble) or as a solid (insoluble). Once we know the compound we use the Solubility Table to determine its solubility.

For example, consider the following compounds; NaCl, BaSO4, NaC2H3O2, and CaS. Determine the solubility in water for these ionic substances.

The solubility of each of these compounds can be determined by using the Solubility Table.

NaCl (all chlorides are soluble except...) SOLUBLE;

BaSO4 (all sulfates are soluble except...) INSOLUBLE;

NaC2H3O2 (all sodium compounds are soluble) SOLUBLE;

CaS (all sulfides are insoluble...) INSOLUBLE

For those compounds that are soluble write a chemical equation which describes what happens when the solid is added to water.

NaCl(s) ---H2O--> Na+(aq) + Cl-(aq)

NaC2H3O2(s) ---H2O--> Na+(aq) + C2H3O2-(aq)

 

We'll also use the information in a Solubility Table to help identify the phase of ionic substance in a chemical equation. The chemical reaction types where the Solubility Table is important are;

 

Lets consider a double displacement reaction problem;

Write the formula and identify the phase for the product(s) and balance the following reaction.

Na2SO4(aq) + CaCl2(aq) --->

Since this is a double replacement reaction we can write the formulas of the products by exchanging the cations and anions. When you do this remember to keep track of the individual charges on the cations and anions.

Na2SO4(aq) + CaCl2(aq) ---> CaSO4(?) + 2NaCl(?)

Now we'll use the Solubility Table to predict the phases of the products. According to the table CaSO4 is INSOLUBLE and NaCl is SOLUBLE.

Na2SO4(aq) + CaCl2(aq) ---> CaSO4(s) + 2NaCl(aq)

Let's consider another example.

Write the formula and identify the phase for the product(s) and balance the following reaction.

AgNO3(aq) + Na2CO3(aq) --->

Since this is a double replacement reaction we can write the formulas of the products by exchanging the cations and anions. When you do this remember to keep track of the individual charges on the cations and anions.

2AgNO3(aq) + Na2CO3(aq) --->Ag2CO3(?) + 2NaNO3(?)

Now we'll use the Solubility Table to predict the phases of the products. According to the table Ag2CO3 is INSOLUBLE and NaNO3 is SOLUBLE.

2AgNO3(aq) + Na2CO3(aq) ---> Ag2CO3(s) + 2NaNO3(aq)

So lets write the ionic and net ionic equations for the two equations above. The first equation we'll convert is;

Na2SO4(aq) + CaCl2(aq) ---> CaSO4(s) + 2NaCl(aq)

To change the above equation to an ionic equation the aqueous ionic substances must be written as ions and any solid, liquid or gas remains in its molecular form.

2Na+(aq) + SO42-(aq) + Ca2+(aq) + 2Cl-(aq) ---> CaSO4(s) + 2Na+(aq) + 2Cl-(aq)

Now we'll cancel all species common to both sides of the equation;

2Na+(aq) + SO42-(aq) + Ca2+(aq) + 2Cl-(aq) ---> CaSO4(s) + 2Na+(aq) + 2Cl-(aq)

The final net ionic equation is;

SO42-(aq) + Ca2+(aq) ---> CaSO4(s)


The second equation we'll convert is;

2AgNO3(aq) + Na2CO3(aq) ---> Ag2CO3(s) + 2NaNO3(aq)

To change the above equation to an ionic equation the aqueous ionic substances must be written as ions;

2Ag+(aq) + 2NO3-(aq) + 2Na+(aq) + CO32-(aq) ---> Ag2CO3(s) + 2Na+(aq) + 2NO3-(aq)

The net ionic equation is;

2Ag+(aq) + CO32-(aq) ---> Ag2CO3(s)

Neutralization Reactions

To begin with we need to be able to list some acids and bases that were are familiar with...here are a few I think everyone should know...already knows from our discussion in Chapter 6 on nomenclature.

Important Acids

Name

Formula

Sulfuric acid

H2SO4

Sulfurous acid

H2SO3

Nitric acid

HNO3

Nitrous acid

HNO2

Phosphoric acid

H3PO4

Phosphorus acid

H3PO3

Carbonic acid

H2CO3

Perchloric acid

HClO4

Acetic acid

HC2H3O2

 

Formula

Name

HF(aq)

Hydrofluoric acid

HCl(aq)

Hydrochloric acid

HBr(aq)

Hydrobromic acid

HI(aq)

Hydroiodic acid

H2S(aq)

Hydrosulfuric acid

HCN(aq)

Hydrocyanic acid

 

Important Bases (All of the Group IA and IIA hydroxides)

Name of Base

Formula of Base

Sodium hydroxide

NaOH

Potassium hydroxide

KOH

Barium hydroxide

Ba(OH)2

Ammonia

NH3

Calcium hydroxide

Ca(OH)2

Aluminum hydroxide

Al(OH)3

Let's consider a few neutralization reactions and how we write the equations.

Consider the reaction between hydrochloric acid and sodium hydroxide;

HCl(aq) + NaOH(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water.

Molecular equation: HCl(aq) + NaOH(aq) ---> NaCl(aq) + H2O(l)

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case HCl(aq), NaOH(aq), and NaCl(aq) must be written as aqueous ions and H2O(l) remains in its molecualr form.

ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) ---> Na+(aq) + Cl-(aq) + H2O(l)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation

ionic equation: H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) ---> Na+(aq) + Cl-(aq) + H2O(l)

net ionic equation: H+(aq) + OH-(aq) ---> H2O(l)

 

Consider the reaction between nitric acid and calcium hydroxide;

HNO3(aq) + Ca(OH)2(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water. Be sure to balance this equation.

Molecular equation: 2HNO3(aq) + Ca(OH)2(aq) ---> Ca(NO3)2(aq) + 2H2O(l)

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case HNO3(aq), Ca(OH)2(aq), and Ca(NO3)2(aq) must be written as aqueous ions and H2O(l) remains in its molecualr form.

ionic equation: 2H+(aq) + 2NO3-(aq) + Ca2+(aq) + 2OH-(aq) ---> Ca2+(aq) + 2NO3-(aq) + 2H2O(l)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation

ionic equation: 2H+(aq) + 2NO3-(aq) + Ca2+(aq) + 2OH-(aq) ---> Ca2+(aq) + 2NO3-(aq) + 2H2O(l)

net ionic equation: 2H+(aq) + 2OH-(aq) ---> 2H2O(l)

or

net ionic equation: H+(aq) + OH-(aq) ---> H2O(l)

 

Consider the reaction between hydrobromic acid and ammonia;

HBr(aq) + NH3(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water. Be sure to balance this equation.

Molecular equation: HBr(aq) + NH3(aq) ---> NH4Br(aq)

Notice in this particular neutralization equation no water is formed. Since there is no hydroxide ions we can not write water as a product. So when ammonia is one of the reactants we do not include water as a product. Water is present since the reaction occurs in aqueous solution, we just do not write it as a product.

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case HBr(aq) and NH4Br(aq) must be written as aqueous ions and NH3(aq) remains in its molecualr form.

ionic equation: H+(aq) + Br-(aq) + NH3(aq) ---> NH4+(aq) + Br-(aq)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation

ionic equation: H+(aq) + Br-(aq) + NH3(aq) ---> NH4+(aq) + Br-(aq)

net ionic equation: H+(aq) + NH3(aq) ---> NH4+(aq)

 

Consider ONE more reaction, between sulfuric acid and barium hydroxide;

H2SO4(aq) + Ba(OH)2(aq) --->

To write the products we combine the anion of the acid with the cation of the base and write the correct formula following the principle of electroneutrality. The other product is water. Be sure to balance this equation.

Molecular equation: H2SO4(aq) + Ba(OH)2(aq) ---> BaSO4(s) + 2H2O(l)

So the molecular form of the equation is shown above. To write the ionic equation we must separate all aqueous species into their ions and leave any solid, liquid or gaseous substance in its molecular form. So in this case H2SO4(aq) and Ba(OH)2(aq) must be written as aqueous ions and BaSO4(s) and 2H2O(l) remains in their molecualr form.

ionic equation: 2H+(aq) + SO42-(aq) + Ba2+(aq) + 2OH-(aq) ---> BaSO4(s) + 2H2O(l)

Finally to write the net ionic equation we must cancel all species common to both sides of the equation. But there are no species common to both sides of the equation!

net ionic equation: 2H+(aq) + SO42-(aq) + Ba2+(aq) + 2OH-(aq) ---> BaSO4(s) + 2H2O(l)

So the net ionic equation and the ionic equaton are the same.

Looking at the list of acids and bases at the top of the page you can imagine ALL the possibilities. So practice a few on your own until you get comfortable with writing neutralization equations.

Single replacement Reactions

We looked at the reactions of lithium with water, sodium with water, potassium with water and barium with water. All of these are examples of single replacement reactions and following the equation as show about.

We also looked at the reaction of magnesium with hydrochloric acid. In general only the most reactive metal (Group IA) react with water. Group IIA element are not as reactive with water. To get magnesium to react with water requires the water to be hot. Calcium and barium both react with water, but the reactions are not as impressive at the Group IA elements. To get the Group IIA elements to liberate hydrogen we need to react the metals with an acid like hydrochloric acid. Zinc will also react with hydrochloric acid like magnesium.

Decomposition Reactions

The only example of a decomposition reaction I showed was that of potassium chlorate decomposing when heated.

2KClO3(s) ---> 2KCl(s) + 3O2(g)