Equations and Mass Relationships

Submitted by ChemPRIME Staff on Wed, 12/08/2010 - 23:46

A balanced chemical equationA representation of a chemical reaction in which chemical symbols represent reactants on the left side and products on the right side. such as

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)      (1)

not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula. that is involved. Equation (1) says that 4 NH3 molecules can react with 5 O2 molecules to give 4 NO molecules and 6 H2O molecules. It also says that 4 mol NH3 would react with 5 mol O2 yielding 4 mol NO and 6 mol H2O.

The balanced equationA representation of a chemical reaction that has values of the stoichiometric coefficients of reactants and products such that the number of atoms of each element is the same before and after the reaction. does more than this, though. It also tells us that 2 × 4 = 8 mol NH3 will react with 2 × 5 = 10 mol O2, and that ½ × 4 = 2 mol NH3 requires only ½ × 5 = 2.5 mol O2. In other words, the equation indicates that exactly 5 mol O2 must react for every 4 mol NH3 consumed. For the purpose of calculating how much O2 is required to react with a certain amount of NH3 therefore, the significant information contained in Eq. (1) is the ratio

\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}}

We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (1),

\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{NH}_{\text{3}}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}}\text{                                                                      (2}\text{)}

The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compoundA substance made up of two or more elements and having those elements present in definite proportions; a compound can be decomposed into two or more different substances.) against another.

EXAMPLE 1 Derive all possible stoichiometric ratios from Eq. (1)

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Any ratio of amounts of substance given by coefficients in the equation may be used:

\text{S}\left( \frac{\text{NH}_{3}}{\text{O}_{\text{2}}} \right)=\frac{\text{4 mol NH}_{3}}{\text{5 mol O}_{\text{2}}}      \text{S}\left( \frac{\text{O}_{\text{2}}}{\text{NO}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NO}}

\text{S}\left( \frac{\text{NH}_{3}}{\text{NO}} \right)=\frac{\text{4 mol NH}_{3}}{\text{4 mol NO}}      \text{S}\left( \frac{\text{O}_{\text{2}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{6 mol H}_{\text{2}}\text{O}}

\text{S}\left( \frac{\text{NH}_{3}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{4 mol NH}_{3}}{\text{6 mol H}_{\text{2}}\text{O}}      \text{S}\left( \frac{\text{NO}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{4 mol NO}}{\text{6 mol H}_{\text{2}}\text{O}}

There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.]

When any chemical reactionA process in which one or more substances, the reactant or reactants, change into one or more different substances, the products; chemical change involves rearrangement, combination, or separation of atoms. Also called chemical change. occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (1) as an example, this means that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the stoichiometric ratio S(O2/NH3):

\frac{n_{\text{O}_{\text{2}}\text{ consumed}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}}=\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{NH}_{3}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{3}}

Similarly, the ratio of the amount of H2O produced to the amount of NH3 consumed must be


\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}}=\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{3}} \right)=\frac{\text{6 mol H}_{\text{2}}\text{O}}{\text{4 mol NH}_{3}}

In general we can say that

\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{                         (3}\text{a)}

or, in symbols,

\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{                                                          (3}\text{b)}

Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent any reactantA substance consumed by a chemical reaction. or any productA substance produced by a chemical reaction. in the balanced chemical equation from which the stoichiometric ratio was derived. No matterAnything that occupies space and has mass; contrasted with energy. how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.

EXAMPLE 2 Find the amount of water produced when 3.68 mol NH3 is consumed according to Eq. (1).

Solution The amount of water produced must be in the stoichiometric ratio S(H2O/NH3) to the amount of ammonia consumed:

\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)=\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}}

Multiplying both sides nNH3 consumed, by we have

n_{\text{H}_{\text{2}}\text{O produced}}&=n_{\text{NH}_{\text{3}}\text{ consumed}}\times\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)
\&=\text{3}\text{.68 mol NH}_{\text{3}}\times \frac{\text{6 mol H}_{\text{2}}\text{O}}{\text{4 mol NH}_{\text{3}}}
\&=\text{5}\text{.52 mol H}_{\text{2}}\text{O}

This is a typical illustration of the use of a stoichiometric ratio as a conversion factorA relationship between two units of measure that is derived from the proportionality of one quantity to another; for example, the mass of a substances is proportional to its volume and the conversion factor from volume to mass is density.. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where densityThe ratio of the mass of a sample of a material to its volume. was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar massThe mass of a mole of substance; the same as molecular weight for molecular substances. were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form

\text{amount of X consumed or produced}\overset{\begin{smallmatrix} 
 \text{stoichiometric} \ 
 \text{   ratio X/Y} 
\end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}

or symbolically.

                                      n_{\text{X   consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y   consumed or produced}}

When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol NH3 cancels 1 mol NH3 but does not cancel 1 mol H2O.

The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.

EXAMPLE 3 Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation

4 FeS2 + 11 O2 → 2 Fe2O3 + 8 SO2

Solution The problem asks that we calculate the mass of SO2 produced. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of SO2 to the mass of SO2. Therefore this problem in effect is asking that we calculate the amount of SO2 produced from the amount of O2 consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio

\text{S}\left( \frac{\text{SO}_{\text{2}}}{\text{O}_{\text{2}}} \right)=\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}

The amount of SO2 produced is then

n_{\text{SO}_{\text{2}}}\text{ produced}&=n_{\text{O}_{\text{2}}\text{ consumed}}\text{ }\!\!\times\!\!\text{  conversion factor}
\&=\text{3}\text{.84 mol O}_{\text{2}}\times \frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}
\&=\text{2}\text{.79 mol SO}_{\text{2}}

The mass of SO2 is

\text{m}_{\text{SO}_{\text{2}}}&=\text{2}\text{.79 mol SO}_{\text{2}}\times \frac{\text{64}\text{.06 g SO}_{\text{2}}}{\text{1 mol SO}_{\text{2}}}
\&=\text{179 g SO}_{\text{2}}

With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as

n_{\text{O}_{\text{2}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/O}_{\text{2}}\text{)}}\text{ }n_{\text{SO}_{\text{2}}}\text{ }\xrightarrow{M_{\text{SO}_{\text{2}}}}\text{ }m_{\text{SO}_{\text{2}}}


\text{m}_{\text{SO}_{\text{2}}}=\text{3}\text{.84 mol O}_{\text{2}}\times \text{ }\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol SO}_{\text{2}}}=\text{179 g}

These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations.

4 FeS2 + 11 O2 2 Fe2O3 + 8 SO2
m (g) 168 123 111 179
M (g/mol) 120.0 32.0 159.7 64.06
n (mol) 1.40 3.84 0.698 2.79

The chemical reaction in this example is of environmental interest. Iron pyrite (FeS2) is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide (SO2), a major air pollutant.

Our next example also involves burning a fuel and its effect on the atmosphereA unit of pressure equal to 101.325 kPa or 760 mmHg; abbreviated atm. Also, the mixture of gases surrounding the earth..

EXAMPLE 4 What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?

Solution First, write a balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically

m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}}

m_{\text{O}_{\text{2}}}&=\text{3}\text{.3 }\times \text{ 10}^{\text{15}}\text{ g }\times \text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\times \text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\times \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}}
\&=\text{1}\text{.2 }\times \text{ 10}^{\text{16}}\text{ g }

Thus 12 Pg (petagrams) of O2 would be needed.

The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen.