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Chemical Stoichiometry

Skills to Develop

Interpret the mass relationship of reactants and products.

  • How much is each of the reactants used in a chemical reaction?
  • How much will products be formed from known amounts of reactants?

Recognize unbalanced and balanced reaction equations.

  • What are the reactants and products in a reaction?
  • How to represent the reaction by an equation?
  • Is the equation properly written up and balanced?
  • What are the quantitative relations among the reactants and products?

Balance unbalanced equations.

  • What principle is applied to balance the equation?
  • What are the techniques used to balance a reaction equation?

Describe chemical reactions types and classify them.

  • How do you classify reactions?
  • How many types of chemical reactions are there?
  • Is the classification technique useful for the study of material?

Explain excess and limiting reactants.

  • Which reactant will be used up in a reaction mixture?
  • Which reactant will not be used up in a reaction mixture?
  • How does a limiting reagent affect amounts of products?

Calculate theoretical and actual yields of chemical reactions.

  • What are actual and theoretical yields?
  • For certain amounts of reactants, how much products shall be produced? How much actually is produced?


The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products. Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same. This is the basic law of nature.

From the atomic and molecular point of view, the stoichiometry in a chemical reaction is very simple. However, atoms of different elements and molecules of different substances have different weights. Thus, simple stoichiometry at the atomic level appears to be complicated when amounts (quantities) are measured in units of g, kg, L or mL. When quantities in moles are used, the relationships (or ratios) are really simple. For example, one mole of oxygen reacts with two moles of hydrogen,

\[\mathrm{2 H_2 + O_2 \rightarrow 2 H_2O}\]

or one mole of hydrogen reacts with half a mole of oxygen,

\[\mathrm{H_2 + \dfrac{1}{2} O_2 \rightarrow H_2O}\]

or one mole of carbon reacts with one mole of oxygen.

\[\mathrm{C + O_2 \rightarrow CO_2}\]

This is a major and important topic that you have to master. In order to accomplish this, you have to be able to do several things. First, you have to be able to convert amounts of substances between mass units of g (or kg) to moles and vice versa. Then, you have to understand chemical reactions (changes). In this case, you not only know what are the reactants and products, you can write a balanced equation to explain the reaction. Sometimes you may be told what the reactions are.

There are many chemical reactions, but they can be divided into a few types as a summary.

In a chemical reaction, not all reactants are necessarily consumed. One of the reactants may be in excess and the other may be limited. The reactant that is completely consumed is called the limiting reactant, whereas unreacted reactants are called excess reactants.

Amounts of substances produced are called yields. The amounts calculated according to stoichiometry are called theoretical yields whereas the actual amounts are called actual yields. The actual yields are often expressed in percentage, and they are often called percent yields.

Listed with the skills are some general questions. If you are confident in answering all these questions, you have already acquired the skills. Furthermore, we list some diagnostic problems below. Each question tests some aspect of stoichiometry. If you find a question difficult, the appropriate topic to study is suggested.

Cool Links on Stoichiometry

Typical Problems Related to Amounts of Substances

  1. What are the atomic weights of hydrogen (\(\ce{H}\)), oxygen (\(\ce{O}\)), iron (\(\ce{Fe}\)) and gold (\(\ce{Au}\))?

    Hint: Know where to find: 1.0, 16.0, 55.9, 197

  2. The element gold is a precious metal. How many moles of gold have a mass of 1.0 kg?

    Hint: 1000/97 = ?

  3. The common table salt is a compound with a formula \(\ce{NaCl}\) . What is the mass for a mole of salt? This mass is called molar mass.

    Hint: 23 + 35.5 = ?

  4. The common sugar has a chemical formula \(\ce{C12H22O11}\). You dissolved 10.0 g of sugar in 250.0 mL of solution and made it into a drink. What is the sugar concentration in the solution?

    Hint: Value depends on units: g/mL, mol/L, etc.

Chemical Reactions

Chemical reactions are changes when one or more substances convert to other substances. The best method to represent these changes is a reaction equation. Reaction equations show the molar relationship of reactants and products. For example, when we exhale via a straw into a solution containing \(\mathrm{Ca(OH)_2}\), the following reaction takes place.

\[\mathrm{Ca^{2+} + H_2O + CO_2 \rightarrow CaCO_3 + 2H^+}\]

This equation shows that a mole of \(\mathrm{Ca^{2+}}\) requires a mole of \(\mathrm{CO_2}\) to react and the products are a mole of solid \(\mathrm{CaCO_3}\) and 2 moles of \(\mathrm{H^+}\) ions. Since \(\mathrm{Ca(OH)_2}\) is basic, the hydrogen ions produced react with the \(\mathrm{OH^-}\) ions to form water. In solving stoichiometric problems, we often write the molar mass below the formula of the equation. If a sample contains 0.10 gram of \(\mathrm{Ca^{2+}}\) ions, then the amount of \(\mathrm{CO_2}\) required to react with \(\ce{Ca}\) can be evaluated in the following way:

\ce{&Ca^2+ + &&H2O + &&CO2 \rightarrow &&CaCO3 + 2H+}\\
&40.1   &&18.0  &&44.0            &&\:100.1\\
&0.10   &&      &&\:\:x


\[ \mathrm{0.10 \;\cancel{g\; Ca^{2+}} \left(\dfrac{44.0\;g\; CO_2}{40.1\; \cancel{g\; Ca^{2+}}}\right) = 0.11\;g\;CO_2}\]

The above illustrates how you may approach a stoichiometric problem. It also shows the mass relationship of chemical reactions.

Types of Chemical Reactions

Two reactants undergo a combination reaction to form a new compound, for example:

\[\mathrm{H_2 + Cl_2 \rightarrow 2 HCl}\]

The formation of \(\ce{AgCl}\) when \(\ce{NaCl}\) and \(\mathrm{AgNO_3}\) solutions are mixed gives rise to an exchange reaction:

\[\mathrm{NaCl + AgNO_3 \rightarrow AgCl + NaNO_3}\]

Reaction with oxygen or air is often called combustion,

\[\mathrm{C_2H_5OH + O_2 \rightarrow 2 CO_2 + 3 H_2O}\]

A combustion reaction is often an oxidation and reduction reaction. Since the oxidation of one substance involves the reduction of another, this type of reaction is often called redox reaction. In the following reaction,

\[\mathrm{2 Al + Fe_2O_3 \rightarrow 2 Fe + Al_2O_3}\]

The element \(\ce{Al}\) is oxidized, but \(\ce{Fe}\) is reduced. This reaction is also called a displacement reaction because \(\ce{Al}\) displaces \(\ce{Fe}\) in the oxide.

Limiting and Excess Reactants

When one or more reactants (or reagents) react, the one that is exhausted first is called the limiting reactant or limiting reagent. When one reactant is exhausted, no more reaction can take place. The reactants left are called excess reactants or excess reagents. For example, when equal masses of \(\ce{CO}\) and \(\mathrm{O_2}\) react, \(\ce{CO}\) will be used up first, because the equation is

\[\mathrm{2 CO + O_2 \rightarrow 2 CO_2}\]

Since equal masses are used, we assume 28 g each of \(\ce{CO}\) and \(\mathrm{O_2}\) are used. The amount of \(\mathrm{O_2}\) required can be calculated to be:

\[\mathrm{( 28 \;g\; CO) \left( \dfrac{1\; mole \; CO}{28\;g\;CO} \right) \left( \dfrac{1\; mol\; O_2}{2\; mol\; CO} \right) \left(\dfrac{32\; g\; O_2}{1 \; mole\; O_2} \right) = 16\; g\; O_2\; required}\]

The amount of \(\mathrm{O_2}\) left will be \(\mathrm{(28 - 16)\: g = 8\: g\: O_2}\). Thus, \(\ce{CO}\) is the limiting reactant and \(\mathrm{O_2}\) is the excess reactant.

Theoretical and Actual Yields

For known amounts of reactants, theoretical amounts of products can be calculated in a chemical reaction or process. Calculated amounts of products are called theoretical yield. In these calculations, the limiting reactant is the limiting factor for the theoretical yields of all products. However, in a reaction to prepare a compound, you may get less than the theoretical yield, because of incomplete reactions or loss. The amount recovered divided by the theoretical yield gives a percent yield (% yield) or actual yield.

Confidence Building Questions

  1. What is oxidized in this REDOX reaction?

    \[\mathrm{Zn + 2 HCl \rightarrow ZnCl_2 + H_2}\]

    If your answer is...I'm lost!
    Well, the oxidation state for \(\ce{Zn}\) is zero, but for \(\ce{ZnCl2}\) the oxidation state for \(\ce{Zn}\) is 2.

    If your answer is...\(Zn\)
    The oxidation state increased for \(\ce{Zn}\).

  2. At the same temperature and pressure, equal volumes of \(\ce{CO}\) and \(\mathrm{O_2}\) are mixed for a reaction. What is the limiting reactant?

    If your answer is...I'm lost!
    At the same temperature and pressure, equal volumes of gases contain the same number of moles.

    If your answer is...\(O_2\)
    The reaction equation is \(\mathrm{2 CO + O_2 \rightarrow 2 CO_2}\).

    If your answer is...\(CO\)
    The required volume for \(\ce{CO}\) is twice the volume for \(\ce{O2}\).

  3. How much \(\ce{Fe}\) should be produced when the limiting reactant is exhausted, if 10 g each of \(\ce{Al}\) and \(\mathrm{Fe_2O_3}\) are used for the thermite reaction?

    \[\mathrm{2 Al + Fe_2O_3 \rightarrow 2 Fe + Al_2O_3}\]

    Since 10 g \(\mathrm{Fe_2O_3}\) requires 3.4 g \(\ce{Al}\), we assume \(\mathrm{Fe_2O_3}\) completely reduced.

    \[ \mathrm{(10\; g\; Fe_2O_3) \left (\dfrac{1\;mol}{159.6\; g\; Fe_2O_3}  \right ) \left (\dfrac{2\; mol\; Al}{1 \;mol\; Fe_2O_3}  \right ) \left (\dfrac{27.0\; g \;Al}{1\; mol\; Al}  \right ) = 3.4\; g\; Al}\]

    Amount \(\ce{Fe}\) produced is calculated this way:

    \[ \mathrm{(10\; g\; Fe_2O_3) \left (\dfrac{1 \;mol}{159.6\; g \;Fe_2O_3}  \right ) \left (\dfrac{1\; mol\; Fe}{1\; mol\; Fe_2O_3}  \right ) \left (\dfrac{55.8\; g\; Fe}{1\; mol\; Al}  \right ) =\; ?\; g\; Fe}\]