1.6.1. Chemical Reactions and Equations. In subsection 1.4.7, we learned about the separation of a mixture of sulfur and iron powders using a magnet (Figure 1-19). This separation is a physical phenomenon that is based on the difference in magnetic properties of iron and sulfur.

Let us now see what happens if a mixture of iron and sulfur is heated. As carrying out this experiment at home is hardly appropriate, just watch the demonstrations shown in Videos 1-13, 1-14, and 1-15 below. The chemical reaction that occurs between Fe and S is spectacular. In this transformation, the two reagents (also called reactants), Fe and S, are consumed and a new substance, iron sulfide (FeS), is formed.

The reaction product, iron sulfide (FeS), has physical and chemical properties that are totally different from those of iron and sulfur, the substances it was made from. Iron (Fe) is attracted by magnets, whereas iron sulfide (FeS) is not. Sulfur (S) is yellow in color, whereas iron sulfide (FeS) produced in the reaction is greyish-brown in color. The melting point of FeS (1,194 °C) is different from those of iron (1,538 °C) and sulfur (115 °C).

The chemical equation for this reaction is as follows.

Fe + S = FeS

This equation shows that one atom of iron (Fe) reacts with (+) one atom of sulfur (S) to give (=) one molecule of iron sulfide (FeS). The equals sign in chemical equations is often replaced with an arrow, as shown below.

Fe + S → FeS

The arrow indicates the direction in which the reaction proceeds. The equals and arrow signs are often used interchangeably in chemical equations.

1.6.2. Signs of Chemical Reactions. There are many signs suggesting that a chemical reaction might be occurring. These include effervescence (evolution of a gas), the formation of a precipitate, the outbreak of fire as a result of spontaneous ignition, a flare of light, a change in color, odor, or temperature, etc. Of all of these effects, perhaps fire is the only unambiguous indication of a chemical reaction. No other sign is an absolute guarantee that a chemical reaction is taking place. For example, bubbles of steam produced by boiling water are obviously not due to a chemical reaction. Similarly, precipitation of a solid from its saturated solution on cooling has nothing to do with a chemical transformation. We also have to keep in mind that some chemical reactions occur without any visible signs.

1.6.3. Simple Chemical Calculations. Stoichiometry. The Law of Conservation of Mass (Matter) and the Law of Definite (Constant) Proportion. Iron sulfide, FeS, the product of the reaction between iron (Fe) and sulfur (S), is sometimes used for groundwater and soil remediation. Suppose we need to make some quantity of FeS from Fe and S for this application. For that, we have to figure out in what ratio iron and sulfur should be used for the preparation or, in other words, the stoichiometry of the reaction. In a broader sense, stoichiometry is the relationship between the amounts of substances that react together in a particular chemical reaction and the amounts of products that are formed. To understand this relationship (the stoichiometry) for our reaction between Fe and S, we should learn two fundamental laws of chemistry, the law of conservation of mass and the law of definite (constant) proportion.

- The law of conservation of mass (matter) states that in chemical reactions atoms are neither created nor destroyed. As the number of atoms remains the same, the overall mass remains the same, too. This law was deduced from the work of two great scientists, Antoine-Laurent Lavoisier (France) and Mikhail Lomonosov (Russia), whose portraits are on display in Figure 1-20.

- The law of definite (constant) proportion, also known as the law of definite composition, states that regardless of its origin and method of preparation, a chemical compound always contains its component elements in a fixed ratio. This law was discovered by the great French chemist Joseph Proust (Figure 1-20).

Figure 1-20. Left to right: Antoine-Laurent Lavoisier (1743-1794) (source), Mikhail Lomonosov (1711-1765) (source), and Joseph Proust (1754-1826) (source).


Now, let us see how these two laws can be applied to solve our problem. Suppose we need to make 88 g of FeS. From the law of conservation of mass we conclude that the total mass of our mixture of iron and sulfur for the preparation of 88 g of FeS should also be 88 g.

But, how much of this total mass of 88 g should be sulfur and how much iron? To answer this question, we apply Proust's law of constant proportion. As follows from this law, no matter what chemical transformation is selected for making iron sulfide, no matter what scale the reaction is run on and under what conditions, and regardless of the ratio of reagents used, the ratio of Fe atoms to S atoms in the iron sulfide product will always be 1:1. Even if we use a huge pile of iron and a pinch of sulfur (or vice versa) for the reaction, each sulfur atom will combine with only one iron atom and the iron sulfide produced will have the same composition and, consequently, formula, FeS. What will happen to the sulfur or iron atoms that turn out to be in excess? Nothing. They will not react. How do we know that? From many thorough experimental studies conducted by many different scientists. It is precisely from those studies that we know the composition of iron sulfide and the chemical equation for its formation.

Fe + S = FeS

As already explained, this equation indicates that one atom of iron (Fe) reacts with one atom of sulfur (S) to give one molecule of iron sulfide (FeS). The atomic weights (or masses) of these two elements can be found in the periodic table. The mass of one sulfur atom is 32.066 ≈ 32 and that of an iron atom is 55.845 ≈ 56 atomic mass units (a.m.u.), also known as Dalton (Da). [One atomic unit equals the mass of one-twelfth of a carbon atom and is approximately 1.66 × 10⁻²⁴ g. We will talk about the a.m.u. in more detail later; for the time being, all we need to know are just the numbers.] Knowing the formula (FeS) and the atomic masses of Fe (56) and S (32), the molecular mass of FeS is easily calculated: 56 + 32 = 88 a.m.u. Therefore, to make 88 mass units of FeS (grams, ounces, pounds, kilos, tons, etc.) of FeS, we should use 56 mass units (grams, ounces, pounds, kilos, tons, etc.) of iron and 32 mass units (grams, ounces, pounds, kilos, tons, etc.) of sulfur for the reaction. To make 88 g of FeS, we need 56 g of Fe and 32 g of S. If we needed to make 88 pounds of FeS, we would use 56 pounds of Fe and 32 pounds of S. To make 88 kilos of FeS, 56 kilos of iron and 32 kilos of sulfur would be needed, etc.

What if we wanted to make twice as much FeS, 176 g? How much sulfur and iron would we use? Of course, twice the amounts calculated above for the preparation of 88 g FeS, which is 56 x 2 = 112 g of Fe and 32 x 2 = 64 g of S.

Let us now consider a slightly less straightforward case. Suppose we need to make 100 g of FeS. How much Fe and S should we use? To calculate the quantities needed for the preparation, we will use simple proportions. From the atomic and molecular masses, we know that to make 88 g of FeS we need 56 g of Fe and 32 g of S. Think logically, as follows.

To make 88 g of FeS, we need 56 g of Fe. In a shortened form:

88 g of FeS ------ 56 g of Fe;

To make make 100 g of FeS, we need x g of Fe. In a shortened form:

100 g of FeS ------ x g of Fe

Our proportion therefore is:

88 g of FeS ------ 56 g of Fe
100 g of FeS ------ x g of Fe

Solving such proportions is easy. The number on the same line as the unknown (100 in our case) is multiplied by the number above (or below) the unknown (above in our case, 56), and divided by the number diagonal to the unknown (88 in our case):

x = (56 x 100)/88 = 63.6 g.

This is the amount of Fe needed to make 100 g of FeS.

Since now we know that we need 63.6 g of Fe to make 100 g of FeS, the amount of S for the reaction can be calculated by simply subtracting 63.6 from 100, 100 - 63.6 = 36.4 g. Alternatively, we can determine the amount of sulfur by solving a proportion. Think the same way:

To make 88 g of FeS, we need 32 g of S. In a shortened form:

88 g of FeS ------ 32 g of S

To make 100 g of FeS, we need y g of S. In a shortened form:

100 g FeS ------ y g S

Our proportion therefore is:

88 g of FeS ------ 32 g of S
100 g of FeS ------ y g of S

To solve this proportion for y, we multiply 100 by 32 and divide the result by 88.

y = (32 x 100)/88 = 36.4 g

Unsurprisingly, the same number (36.4 g) is obtained, regardless of what calculation method is used. Thus, to make 100 g of FeS, we need 36.4 g of sulfur (S) and 63.6 g of iron (Fe).

Do you find it hard to understand the calculations above? No problem! In many ways, chemistry is like cooking. In both, the ingredients should be used in certain proportions. Suppose you want to make an apple pie. The recipe says that to make one pie, you need 2 apples and one cup of flour (among other ingredients). You have 5 apples and want to make a bigger pie using all of them. How much flour do you need to use? The amount of flour (as well as all other ingredients) in the recipe should be increased proportionally, as follows.

For 2 apples ------ 1 cup of flour is needed
For 5 apples ------ x cups of flour are needed

x = (5 x 1)/2 = 2.5 cups of flour.

1.6.4. Limiting Reagent. Let us now consider a slightly different apple pie problem. Suppose we have 6 apples and 4 cups of flour. We want to make as big a pie as possible using the ingredients we have. Since the recipe says 2 apples per each cup of flour, there are obviously three possibilities (though only one is correct).

1. We have more flour than needed for 6 apples. If this is the case, the size of our pie to be made is limited by the number of apples we have. We could say that the apples are our limiting ingredient. We use all 6 apples and calculate how many cups of flour are needed to make a pie with 6 apples.

2. We have less flour than needed for 6 apples. If this is the case, the size of our pie to be made is limited by the amount of flour we have. Our limiting ingredient is flour then. We use all of our flour and calculate how many apples are needed to make a pie with that much flour.

3. The amount of flour we have is exactly how much is needed for 6 apples by the recipe.

To find out which of the three possibilities is true in our case, we calculate how many cups of flour are needed to make a pie using 6 apples.

For 2 apples ------ 1 cup of flour is needed
For 6 apples ------ x cups of flour are needed

x = (6 x 1)/2 = 3 cups of flour.

So, to use all 6 apples following the recipe, 3 cups of flour are needed. We have 4 cups of flour, more than necessary to make a pie with 6 apples. Therefore, it is the apples that are the limiting ingredient in our case, which is possibility number 1 above.

Switching back to chemistry, suppose that we have 64 g of sulfur and 120 g of iron to make FeS. How much of each of these two chemicals should we use to make the largest possible amount of FeS? Following the same simple apple-flour logic, we need to find out if it is the sulfur or the iron that is the limiting ingredient. In chemistry, a limiting ingredient is called a limiting reagent or limiting reactant.

The atomic masses of sulfur (32) and iron (56) indicate that to fully react with 32 g of sulfur, 56 g of iron is needed. We have 64 g of S, not 32 g. How many grams of Fe (x) do we need for the reaction with 64 g of S?

32 g S ------ 56 g Fe
64 g S ------ x g Fe

x = (64 x 56)/32 = 112 g

So, for the reaction of the entire amount of sulfur we have at our disposal (64 g), 112 g of iron is needed. As this is less than the amount of iron we have (120 g), our limiting reagent is sulfur. We therefore use for the reaction all 64 g of the sulfur and 112 g of iron out of the 120 g we have. The amount of FeS made from these quantities of S and Fe will be 64 g + 112 g = 176 g.

What would happen if we used all of our iron and sulfur for the reaction? Of the 120 g of Fe, 112 g would react with the 64 g of S. The rest of the iron, 120 g - 112 g = 8 g would remain unchanged or, as chemists say, unreacted. After the reaction, the product would consist of 176 g of FeS produced and 8 g of the unreacted iron metal.

1.6.5. Exercises.

1. Just like iron, zinc (Zn) reacts with sulfur. The reaction produces zinc sulfide, ZnS. By the way, ZnS is an important compound that has a broad variety of applications in materials for electronics and optical devices, as a pigment, and also as a catalyst. A catalyst is a compound that can accelerate certain chemical reactions. We will talk about catalysts and the phenomenon of catalysis later. But in the meantime,

(A) Knowing the formula of zinc sulfide (ZnS), write a chemical equation for its formation from zinc (Zn) and sulfur (S);

(B) Using the equation and the atomic masses of S and Zn from the periodic table, calculate the amount of Zn in grams to make ZnS by the reaction with 16 g of sulfur. How much ZnS will be obtained? Round the atomic masses to one a.m.u. and quantities in grams to 0.1 g; and

(C) You have 30 g of sulfur and 30 g of zinc. How much ZnS can you make using these chemicals? Which of the two is the limiting reagent?

Answer

2. As we found out at the end of the previous subsection (1.6.4), if we used all 120 g of iron for the reaction with 64 g of sulfur, only 112 g of the iron would react. The rest of the iron, 120 g - 112 g = 8 g would remain unreacted, and the reaction product would consist of 176 g of FeS and 8 g of Fe. Suggest a simple method to separate the unreacted iron metal from the FeS product? Answer