A chemical equation expresses the net change in composition associated with a chemical reaction by showing the number of moles of reactants and products. But because each component has its own molar mass, equations also implicitly define the way in which the masses of products and reactants are related. In this unit we wll concentrate on understanding and making use of these mass relations.

 

How to read and write chemical equations

Start with a balanced equation!

A chemical equation is a statement of a fact: it expresses the net change that occurs as the result of a chemical reaction. When we balance an equation, we simply make it consistent with the observed fact that individual atoms are conserved in chemical changes.

There is no set “recipe’’ for balancing ordinary chemical equations. One usually starts with one particular element or ion, readjusting coefficients (but never the formula subscripts!) in whatever way seems appropriate.

Write a balanced equation for the combustion of propane C3H8 in oxygen O2. The products are carbon dioxide CO2 and water H2O.

Answer: Begin by writing the unbalanced equation

C3H8 + O2 = CO2 + H2O

It is usually best to begin by balancing compounds containing the least abundant element, so we first balance the equation for carbon:

C3H8 + O2 = 3 CO2 + H2O

In balancing the oxygen, we see that there is no way that an even number of O2 molecules on the left can yield the uneven number of O atoms shown on the right. Don't worry about this now— just use the appropriate fractional coefficient:

C3H8 + 3 ½ O2 = 3 CO2 + H2O

Finally, we balance the hydrogens by adding more waters on the right:

C3H8 + 7/2 O2 = 3 CO2 + 4 H2O

Ah, but now the oxygens are off again— but fixing this allows us to get rid of the fraction on the left side:

C3H8 + 5 O2 = 3 CO2 + 4 H2O

It often happens, however, that we do end up with a fractional coefficient, as in this variant of the above example.

Write a balanced equation for the combustion of ethane C2H6 in oxygen O2. The products are carbon dioxide CO2 and water H2O.

Answer: Begin by writing the unbalanced equation

C2H6 + O2 = CO2 + H2O

...then balance the carbon:

C2H6 + O2 = 2 CO2 + H2O

Let's balance the hydrogen next:

C2H6 + O2 = 2 CO2 + 3 H2O

...but this leaves us with a non-integral number of dioxygen molecules on the left:

C2H6 + 7/2 O2 = 2 CO2 + 3 H2O

My preference is to simply leave it in this form; there is nothing wrong with 3 ½ moles of O2, and little to be gained by multiplying every term by two— not unless your teacher is a real stickler for doing it "by the book", in which case you had better write

2 C2H6 + 7 O2 = 3 CO2 + 4 H2O

 

Net ionic equations

Ionic compounds are usually dissociated in aqueous solution; thus if we combine solutions of silver nitrate AgNO3 and sodium chloride NaCl we are really combining four different species: the cations (positive ions) Ag+ and Na+, and the anions (negative ions) NO3– and Cl–. It happens that when the ions Ag+ and Cl– are brought together, they will combine to form an insoluble precipitate of silver chloride. The net equation for this reaction is

Ag+(aq) + Cl–(aq) → AgCl(s)

Note that

  • the ions NO3– and Cl– are not directly involved in this reaction; the equation expresses only the net change, which is the removal of the silver and chloride ions from the solution to form an insoluble solid.
  • the symbol (aq) signifies that the ions are in aqueous solution, and thus are hydrated, or attached to water molecules.
  • the symbol (s) indicates that the substance AgCl exists as a solid. When a solid is formed in a reaction that takes place in solution, it is known as a precipitate. The formation of a precipitate is often indicated by underscoring.

Predicting the outcome when dissolved salts are mixed

From the above example involving silver chloride, it is clear that a meaningful net ionic equation can be written only if two ions combine to form an insoluble compound. In order to make this determination, it helps to know the solubility rules— which all students of chemisry were at one time required to memorize, but are nowadays usually obtained from tables such as the one shown below.

Cation (positive ion)
Anion (negative ion)
Soluble?
any anion alkali metal ions (Li+, Na+, K+, etc.)
yes
nitrate, NO3– any cation
yes
acetate, CH3COO– any cation except Ag+
yes
halide ions Cl–, Br–, or I– Ag+, Pb2+, Hg22+, Cu2+
no
halide ions Cl–, Br–, or I– any other cation
yes
sulfate, SO42– Ca2+, Sr2+, Ba2+, Ag+, Pb2+
no
sulfate, SO42– any other cation
yes
sulfide, S2– alkali metal ions or NH4+
yes
sulfide, S2– Be2+, Mg2+, Ca2+, Sr2+, Ba2+, Ra2+
yes
sulfide, S2– any other cation
no
hydroxide, OH– alkali metal ions or NH4+
yes
hydroxide, OH– Sr2+, Ba2+, Ra2+
slightly
hydroxide, OH– any other cation
no
phosphate, PO43–, carbonate CO32– alkali metal ions or NH4+
yes
phosphate, PO43–, carbonate CO32– any other cation
no

 

 

Write net ionic equations for what happens when aqueous solutions of the following salts are combined:

a) PbCl2 + K2SO4
b) K2CO3 + Sr(NO3)2
c) AlCl3 + CaSO4
d) Na3PO4 + CaCl2

Answer: Use the solubility rules table(above) to find the insoluble combinations:

a)Pb2+(aq) + SO42–(aq) → PbSO4(s)
b) Sr2+(aq) + CO32–(aq) → SrCO3(s)
c) no net reaction
d) 3 Ca2+(aq) + 2 PO43–(aq) → 3 Ca3(PO4)2(s)
(Note the need to balance the electric charges)

Mass relations in chemical equations

A balanced chemical equation expresses the relative number of moles of each component (product or reactant), but because each formula in the equation implies a definite mass of the substance (its molar mass), the equation also implies that certain weight relations exist between the components. For example, the equation describing the combustion of carbon monoxide to carbon dioxide

2 CO + O2 → 2 CO2

implies the following relations:

The relative masses shown in the bottom line establish the stoichiometry of the reaction, that is, the relations between the masses of the various components. Since these masses vary in direct proportion to one another, we can define what amounts to a conversion factor (sometimes referred to as a chemical factor) that relates the mass of any one component to that of any other component.

Evaluate the conversion factor that relates the mass of carbon dioxide to that of the CO consumed in the reaction.

Solution: The reacting masses based on the relative masses are (88 g CO2) / (56 g CO), or 1.57 g CO2 per g of CO.

But the same relation applies to any common set of mass or weight units, so we can simply say that the mass ratio of CO2-to-CO expressed by the above equation is 1.57, which is also just the mass ratio of these two components in the equation. This, in turn, allows us to easily handle such problems as the following:

How many tons of CO2 can be obtained from the combustion of 10 tons of CO?

Solution: (1.57 T CO2 / 1 T CO) × (10 T CO) = 15.7 T CO2

More mass-mass problems

Don't expect to pass Chemistry unless you can handle these kinds of problems!

The ore FeS2 can be converted into the important industrial chemical sulfuric acid H2SO4 by a series of processes. Assuming that the conversion is complete, how many litres of sulfuric acid (density 1.86 kg L–1) can be made from 50 kg of ore?

Solution: As with most problems, this breaks down into several simpler ones. We begin by working out the stoichiometry on the assumption that all the sulfur in the or ends up as H2SO4, allowing us to write

FeS2 → 2 H2SO4

Although this "skeleton" equation is incomplete (and thus not balanced), it is balanced in respect to the two components of interest, and this is all we need here. The molar masses of the two components are 120.0 and 98 g mol–1, respectively, so the equation can be interpreted in terms of masses as

[120 mass units] FeS2 → [2 × 98 mass units] H2SO4

Thus 50 kg of ore will yield (50 kg) × (196/120) = 81.7 kg of product.
[Check: is this answer reasonable? Yes, because the factor (196/120) is close to (200/120) = 5/3, so the mass of product should be slightly smaller than twice the mass of ore consumed.]

From the density information we find that the volume of liquid H2SO4 is

(81.7 kg) ÷ (1.86 kg L–1) = 43.9 L
[Check: is this answer reasonable? Yes, because density tells us that the number of litres of acid will be slightly greater than half of its weight.]

 

Barium chloride forms a crystalline hydrate, BaCl2.xH2O, in which x molecules of water are incorporated into the crystal lattice for every unit of BaCl2. This water can be driven off by heat; if 1.10 g of the hydrated salt is heated and reweighed several times until no further loss of weight (i.e., loss of water) occurs, the final weight of the sample is 0.937 g. What is the value of x in the formula of the hydrate?

Solution: The first step is to find the number of moles of BaCl2 (molecular weight 208.2) from the mass of the dehydrated sample.

(0.937 g) / (208.2 g mol-1) = 0.00450 mol

Now find the moles of H2O lost when the sample was dried:

(1.10 – .937)g / (18 g mol-1) = .00905 mol

Allowing for a reasonable amount of measurement error, it is apparent that the mole ratio of BaCl2:H2O = 1:2. The formula of the hydrate is BaCl2.2H2O.

 

 

Limiting reactants

Most chemical reactions that take place in the real world begin with more or less arbitrary amounts of the various reactants; we usually have to make a special effort if we want to ensure that stoichiometric amounts of the reactants are combined. This means that one or more reactants will usually be present in excess; there will be more present than can react, and some will remain after the reaction is over. At the same time, one reactant will be completely used up; we call this the limiting reactant because the amount of this substance present will control, or limit, the quantities of the other reactants that are consumed as well as the amounts of products produced.

Limiting reactant problems are handled in the same way as ordinary stoichiometry problems with one additional preliminary step: you must first determine which of the reactants is limiting— that is, which one will be completely used up. To start you off, consider the following very simple example:

For the hypothetical reaction 3 A + 4 B → [products], determine which reactant will be completely consumed when we combine
a) equimolar quantities of A and B;
b) 0.57 mol of A and 0.68 mol of B.

Solution:

a) Simple inspection of the equation shows clearly that more moles of B are required, so this component will be consumed (and is thus the limiting reactant), leaving behind ¾ as many moles of A.

b) How many moles of B will react with .57 mol of A? The answer will be (4/3 × 0.57 mol). If this comes to less than 0.68 mol, then B will be the limiting reactant, and you must continue the problem on the basis of the amount of B. If the limiting reactant is A, then all 0.57 mol of A will react, leaving some of the B in excess. Work it out!

The concept of limiting reactants has immense practical importance.

Sulfur and copper, when heated together, react to form copper(I) sulfide, Cu2S. How many grams of Cu2S can be made from 10 g of sulfur and 15 g of copper?

Solution: From the atomic weights of Cu (63.55) and S (32.06) we can interpret the he reaction 2 Cu + S → Cu2S as

[2 × 63.55 = 127.1 mass units] Cu + [32.06 mass units] S → [159.2 mass units] Cu2S

10 g of S will require (10 g S) × (127.1 g Cu)/(32.06 g S) = 39.6 g Cu
...which is a lot more than what is available, so copper is the limiting reactant here.
[Check: is this answer reasonable? Yes, because the chemical factor (127/32) works out to about 4, indicating that sulfur reacts with about four times its weight of copper.]

The mass of copper sulfide formed will be determined by the mass of copper available:

(15 g Cu) × (159.2 g Cu2S) / (127.1 g Cu) = 18.8 g Cu2S
[Check: is this answer reasonable? Yes, because the chemical factor (159.2/127.1) is just a bit greater than unity, indicating that the mass of the product will slightly exceed that of the copper consumed.]

What you need to know

Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.

  • Given the formulas of reactants and products, write a balanced chemical equation for the reaction.
  • Given the relative solubilities, write a balanced net ionic equation for a reaction between aqueous solutions of two ionic compounds.
  • Calculate the masses of all components of a chemical reaction when the mass of any single component is specified in any system of units.
  • When the masses of two or more reactants are specified, identify the limiting reactant and find the masses of all components present when the reaction is complete.

 

Concept Map

chemical formula calculations concept map

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