# Introductory Chemistry Online/Quantitative Relationships
## Chapter 6. Quantitative Relationships in Chemistry

With the background that we have built in this course, thus far, we can now approach a more quantitative view of chemistry. While the notion of chemistry and math (together in the same room) may make you want to scream, we will see in this chapter that concepts such as chemical stoichiometry and mass balance are not overwhelming and can be approached using the same problem-solving algorithms that we have mastered in previous chapters. Once the concept of stoichiometric balance has been mastered, we will finally tackle limiting reactants. Limiting reactant problems may appear challenging, but we will see it is the same calculation that we do routinely… we simply have to do the calculations twice.

## 6.1 An Introduction to Stoichiometry

Stoichiometry… what a wonderful word! It sounds so complex and so chemical. In fact, it’s a fairly simple concept; stoichiometry is the relationship between the molar masses of chemical reactants and products in a given chemical reaction. In Chapter 5 we learned to balance chemical equations by inserting numerical coefficients in front of reactants or products so that there were the same number and types of atoms on each side of the equation. Thus, in the reaction between sodium metal and chlorine gas, the balanced chemical equation is: Figure 6.1
This equation tells us that two atoms of sodium react with one molecule of chlorine gas to give two sodium chlorides. The coefficients in front of the sodium and the sodium chloride are called the stoichiometric coefficients for this reaction. If we were to totally react a single molecule of chlorine gas in this reaction, this stoichiometry tells us that two atoms of sodium (atomic mass 22.99) having a total mass of 45.98 amu, would react with one molecule of chlorine gas (having a mass 70.90 amu) to give two sodium chlorides (formula mass 137.94), for a total of 275.9 amu of product. Because chemists usually don’t speak of chemical reactions in terms of individual atoms or molecules, it is much more common to describe the stoichiometry of a reaction in terms of grams of reactants and products, or more conveniently in terms of moles. Molar stoichiometry is simply the expression of the coefficients of a reaction in terms of moles of reactants and products.

## 6.2 An Molar Stoichiometry in Chemical Equations

If we were to describe the reaction of sodium metal with chlorine gas in molar terms, we would say that two moles of sodium metal combine with one mole of Cl$_2$ to give two moles of sodium chloride. In terms of mass, two moles of sodium, having a total mass of 45.98 grams, would react with one mole of chlorine gas (a mass 70.90 grams) to give two moles of sodium chloride, for a total of 275.9 grams of product. Likewise, the molar stoichiometry for the decomposition of hydrogen peroxide (H$_2$O$_2$) to form oxygen and water, can be described simply as two moles of H$_2$O$_2$ decompose to form one mole of oxygen gas and two moles of water.
The stoichiometric coefficients for this reaction gives us the key information about the relationship between molar quantities of reactants and products, but in the real world, we will not always be working with exactly two moles of hydrogen peroxide. What if you want to know how much oxygen gas will be formed when 0.28 moles of H$_2$O$_2$ decompose? One way to solve this type of problem is to utilize a tool that we will call a reaction pathway. The reaction pathway is a kind of simple map of the stoichiometry of a reaction, which uses arrows to show the relationship between reactants and products. For any simple reaction, a pathway can be drawn linking the reactants and products as follows: Scheme 1
Using the given-find-ratio algorithm that we introduced back in Chapter 1, if we were given mol reactant and we wanted to find mol product, we could set up a simple equation as follows: Scheme 2
$\left(molproduct\right)=\left(molreactant\right)\left({\frac {molproduct}{molreactant}}\right)$
The units mol reactant cancel to give the solution in mol product. If we were given mol product and we wanted to find mol reactant, we would set up the equation as follows in order for the units to properly cancel:
$\left(molreactant\right)=\left(molproduct\right)\left({\frac {molreactant}{molproduct}}\right)$
This basic approach can be used to solve for any molar (mass, or gas) conversion based on a balanced chemical equation, as long as you are careful to set up the ratios so that the units cancel, giving you the desired solution with the proper units. As an example, return to the question of the decomposition of H$_2$O$_2$. If 0.28 moles of H$_2$O$_2$ decompose, according to the equation given below, how many moles of oxygen gas (O$_2$) will be formed? Scheme 3
$\left({\frac {1{\text{ mol O}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}{\text{O}}_{2}}}\right)$
We set up the problem to solve for mol product; the general equation is:
$\left(molproduct\right)=\left(molreactant\right)\left({\frac {molproduct}{molreactant}}\right)$
The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting, Scheme 4
$\left(x{\text{mol O}}_{\text{2}}\right)=\left({\text{0}}{\text{.28 mol H}}_{\text{2}}{\text{O}}_{\text{2}}\right)\times \left({\frac {1{\text{ mol O}}_{\text{2}}}{{\text{2 mol H}}_{\text{2}}{\text{O}}_{\text{2}}}}\right)$ = 0.14 mol O$_2$
Thus, the decomposition of 0.28 mol of H$_2$O$_2$ will produce 0.14 mol of the product, oxygen gas (O$_2$).
Example 6.1 Using Molar Stoichiometry
Exercise 6.1 Molar Stoichiometry
Ammonia is produced industrially from nitrogen and hydrogen according to the equation:

## 6.3 Mass Calculations

The methods described in the previous section allow us to express reactants and products in terms of moles, but what if we wanted to know how many grams of a reactant would be required to produce a given number of grams of a certain product? This logical extension is, of course, trivial! In Chapter 4, we learned to express molar quantities in terms of the masses of reactants or products. For example, the reduction of iron (III) oxide by hydrogen gas, produces metallic iron and water. If we were to ask how many grams of elemental iron will be formed by the reduction of 1.0 grams of iron (III) oxide, we would simply use the molar stoichiometry to determine the number of moles of iron that would be produced, and then convert moles into grams using the known molar mass. For example, one gram of Fe$_2$O$_3$ can be converted into mol Fe$_2$O$_3$ by remembering that moles of a substance is equivalent to grams of that substance divided by the molar mass of that substance: Figure 6.3
$moles=\left({\frac {grams}{molar\ mass}}\right)=\left({\frac {grams}{^{grams}\!\!\diagup \!\!_{mol}\;}}\right)=\left(grams\right)\times \left(^{mol}\!\!\diagup \!\!_{grams}\;\right)$
Using this approach, the mass of a reactant can be inserted into our reaction pathway as the ratio of mass-to-molar mass. This is shown here for the reduction of 1.0 gram of Fe$_2$O$_3$. Scheme 5
Given: $\left({\frac {{\text{1}}{\text{.0 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 }}{\frac {{\text{g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}}}\right){\text{ }}$ Find: x mol Fe
We set up the problem to solve for mol product; the general equation is:
$\left(molproduct\right)=\left(molreactant\right)\times \left({\frac {molproduct}{molreactant}}\right)$
The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting,
$x{\text{ mol Fe=}}\left({\frac {{\text{1}}{\text{.0 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 }}{\frac {{\text{g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}}}\right)\times \left({\frac {\text{2 mol Fe}}{{\text{1 mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)$
It is often simpler to express the ratio (mass)/(molar mass) as shown below,
$\left({\frac {{\text{1}}{\text{.0 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 }}{\frac {{\text{g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{{\text{mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}}}\right)=\left({\text{1}}{\text{.0 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}\right)\times \left({\frac {1{\text{ mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)$
Doing this, and rearranging,
$x{\text{ mol Fe=}}\left({\text{1}}{\text{.0 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}\right)\times \left({\frac {1{\text{ mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)\times {\text{ }}\left({\frac {\text{2 mol Fe}}{{\text{1 mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)$ = 0.013 mol
That is, the reduction of 1.0 grams of Fe$_2$O$_3$ by excess hydrogen gas will produce 0.013 moles of elemental iron. All of these calculations are good to two significant figures based on the mass of iron (III) oxide in the original problem (1.0 grams). Note that we have two conversion factors (ratios) in this solution; one from mass to molar mass and the second, the stoichiometric mole ratio from the balanced chemical equation. Knowing that we have 0.013 moles of Fe, we could now convert that into grams by knowing that one mole of Fe has a mass of 55.85 grams; the yield would be 0.70 grams.
We could also modify our basic set-up so that we could find the number of grams of iron directly. Our modified pathway would look like this: Scheme 6
Here we have simply substituted the quantity (moles  molar mass) to get mass of iron that would be produced. Again, we set up the problem to solve for mol product;
$\left(molproduct\right)=\left(molreactant\right)\left({\frac {molproduct}{molreactant}}\right)$
In place of mol product and mol reactant, we use the expressions for mass and molar mass, as shown in the scheme above. The stoichiometric mole ratio is set up so that mol reactant (the given) will cancel, giving a solution in mol product. Substituting,
$\left(x{\text{ mol Fe}}\right)\left({\text{ }}{\frac {5{\text{5}}{\text{.85g Fe}}}{\text{1 mol Fe}}}\right){\text{=}}\left({\text{1 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}{\text{ }}\right)\left({\frac {{\text{1 mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)\times {\text{ }}\left({\frac {\text{2 mol Fe}}{{\text{1 mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)$
Rearranging and canceling units,
$x{\text{ g Fe=}}\left(5{\text{5}}{\text{.85 }}{\frac {\text{g Fe}}{\text{mol Fe}}}\right)\left({\frac {{\text{1 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}{\text{ }}\times {\text{ mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{1{\text{59}}{\text{.70 g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)\left({\frac {\text{2 mol Fe}}{{\text{1 mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}}\right)$ = 0.70 g
Example 6.2 Mass Calculations in Stoichiometry
Aqueous solutions of silver nitrate and sodium chloride react in a double-replacement reaction to form a precipitate of silver chloride, according to the balanced equation shown below. Figure 6.4
Example 6.3 Mass Calculations in Stoichiometry
Aluminum and chlorine gas react to form aluminum chloride according to the balanced equation shown in the Scheme below.
Exercise 6.2 Molar Stoichiometry
Ammonia, NH$_{3,}$ is also used in cleaning solutions around the house and is produced from nitrogen and hydrogen according to the equation: Figure 6.5

## 6.4 Percentage Yield

When we use stoichiometric calculations to predict quantities in a reaction, our results are based on the assumption that everything happens in reality exactly as described by the chemical equation. Unfortunately, that is not always the case. When you work in the laboratory, things often go wrong. When you are weighing out reactants and transferring the materials to reaction vessels, some material will often remain on the spatula or in the weighing vessel. As you collect product, some may spill, or in a vigorous reaction, material may escape from the reaction vessel. Stoichiometric calculations will give you a theoretical yield for a reaction; the yield that you should obtain assuming that the reaction proceeds with 100% efficiency and that no material is lost in handling. The amount of material that you isolate from a given reaction is called the actual yield and it is always less than the theoretical yield. The percentage of the theoretical yield that you actually isolate is called the percentage yield.
Consider the reaction between silver nitrate and sodium chloride to form solid silver chloride. If we react 10.00 grams silver nitrate with excess sodium chloride, we would predict that we would obtain: Scheme 6.7
We set up the problem to solve for mol product; the general equation is:
$\left(molproduct\right)=\left(molreactant\right)\times \left({\frac {molproduct}{molreactant}}\right)$
For mol product and mol reactant, we use the expressions for (mass)/(molar mass), as shown in the scheme above. The stoichiometric mole ratio is set up so that mol reactant will cancel, giving a solution in mol product. Substituting,
$\left(x{\text{ g AgCl}}\right)\left({\frac {\text{1 mol AgCl}}{1{\text{4}}3.32{\text{ g AgCl}}}}\right){\text{ =}}\left(1{\text{0}}{\text{.00 g AgNO}}_{\text{3}}\right)\left({\frac {{\text{1 mol AgNO}}_{\text{3}}}{{\text{169}}{\text{.88 g AgNO}}_{\text{3}}}}\right)\times \left({\frac {1{\text{ mol AgCl}}}{{\text{1 mol AgNO}}_{\text{3}}}}\right)$
or,
$\left(x{\text{ g AgCl}}\right){\text{=}}\left(1{\text{0}}{\text{.00 g AgNO}}_{\text{3}}\right)\left({\frac {{\text{1 mol AgNO}}_{\text{3}}}{{\text{169}}{\text{.88 g AgNO}}_{\text{3}}}}\right)\times \left({\frac {1{\text{ mol AgCl}}}{{\text{1 mol AgNO}}_{\text{3}}}}\right)\times \left({\frac {1{\text{4}}3.32{\text{ g AgCl}}}{\text{1 mol AgCl}}}\right){\text{ }}$ = 8.440 g
This mass, calculated from the masses of starting materials and the stoichiometry of the equation is the theoretical yield. Because the silver chloride is a precipitate from an aqueous solution, however, we must filter it, dry it, transfer it to our balance and weigh it, before we can measure how much product we obtain as our actual yield. As we filter it, a small amount of solid is likely to remain stuck to the sides of the flask. When it is dry and we transfer it to the balance, some solid will remain on the filter paper, some on the spatula, and (most likely) your lab partner will sneeze at an inopportune moment and blow some of it all over the desktop. Considering all of this, it is very unlikely that we will end up with an actual yield of 8.044 grams of solid AgCl.
Let’s assume that we have done all of these operations (including the sneeze) and when we weigh our solid AgCl, we actually obtain 7.98 grams of solid. We know that we obtain 7.98 grams of product (our actual yield) and we calculated that we should obtain 8.440 grams of product (the theoretical yield). The percentage of the theoretical yield that we obtain is called the percentage yield, and is calculated as (actual yield)/(theoretical yield)  100. In the present case:
$\left(^{7.98{\text{ g}}}\!\!\diagup \!\!_{{\text{8}}{\text{.440 g}}}\;\right)\times 100=94.5\%$
The percentage yield of solid AgCl that we obtained in this reaction is therefore 94.5% (not bad, actually, considering your lab partner). The concept of percentage yield is generally applied to all experimental work in chemistry.
Example 6.4 Percentage Yield
Exercise 6.3 Percentage Yield
The Harber process is used making ammonia from nitrogen and hydrogen according to the equation shown below. The yield of the reaction, however, is not 100%.

## 6.5 Limiting Reactants

Gloves will typically come is right- and left-handed models. In order to make a pair of gloves, you need one that is designed to fit each hand. If you had a box containing 50 left-handed gloves and another box containing 40 right-handed gloves, you could make 40 proper pairs and you would have ten left-handed gloves left over. The number of pairs of gloves that you could assemble is limited by the glove in the smallest number (the right-handed glove). The other glove in this example, the left, is present in excess. Figure 6.6
The same sort of logic applies to chemical reactions in which there are two or more reactants. In Example 6.4, we carefully weighed out 0.010 mole of solid zinc and solid sulfur in order to react them to form 0.010 mole of the product, ZnS. If instead, we had reacted 0.010 mole of Zn with 0.020 moles of sulfur, how much ZnS would have (theoretically) formed? The answer is still 0.010 mole of ZnS. What happens to the leftover sulfur? It just sits there! When the reaction is complete, there is (theoretically) 0.010 mole of ZnS mixed with the remaining 0.010 mole of sulfur. This is shown graphically in Figure 6.7; 10 atoms of Zn reacting with 20 atoms of S yield 10 molecules of ZnS with 10 atoms of S remaining.
You may have noticed that, in many of the problems in this chapter, we stated that one reactant reacted with an excess of a second reactant. In all of these cases, the theoretical yield of product is determined by the limiting reactant in the reaction, and some of the excess reactant is left over. If aluminum and chlorine gas, a diatomic gas, react to form aluminum chloride according to the equation shown below,
and there are 2.0 moles of aluminum and 14 moles of chlorine present, two moles of aluminum chloride are formed and 11 moles of chlorine gas remain in excess. Our stoichiometry is:
$\left(^{3{\text{ mol Cl}}_{\text{2}}}\!\!\diagup \!\!_{{\text{2 mol AlCl}}_{\text{3}}}\;\right)$
If three moles of chlorine gas are used in the reaction, (14 – 3) = 11 moles of chlorine must remain. When a problem is presented and one reactant is labeled as excess, the theoretical yield of product is equal to the moles of the limiting reagent, adjusted for the stoichiometry of the reaction.
Although it would be easier if reactants were routinely labeled as “limiting” or “excess”, more commonly problems are written in such a way that it is not always trivial to identify the limiting reactant. For example, if you were told that 6.0 grams of aluminum was reacted with 3.8 grams of chlorine gas and you were asked to calculate the mass of AlCl$_3$ that would be formed, there would be no simple way to identify the limiting- and excess reactants. In a case like this what you want to do is to simply solve the problem twice. First you would calculate the number of moles of aluminum in 6.0 grams, and then calculate how many moles of AlCl$_3$ could be formed. Next, you calculate how many moles of chlorine are present in 3.8 grams of chlorine gas and again, calculate how many moles of AlCl$_3$ could be formed. Whichever reagent produces the smallest number of moles of product must be limiting and the other reagent must be in excess.
To work this example we can expand our standard scheme to show both reactants: Scheme 6.8
Find: x moles of AlCl$_3$
Find: x moles of AlCl$_3$
We set up the problem to solve for mol product for each reactant. The general equation is:
$\left(molproduct\right)=\left(molreactant\right)\times \left({\frac {molproduct}{molreactant}}\right)$
The solutions for both reactants are:
$x{\text{ mol AlCl}}_{\text{3}}{\text{=}}\left({\text{6}}{\text{.0 g Al}}\right)\left({\frac {\text{1 mol Al}}{{\text{26}}{\text{.98 g Al}}}}\right)\left({\frac {\text{1 mol AlCl}}{\text{1 mol Al}}}\right)$ , and
$x{\text{ mol AlCl}}_{\text{3}}{\text{=}}\left({\text{3}}{\text{.8 g Cl}}_{\text{2}}\right)\left({\frac {{\text{1 mol Cl}}_{\text{2}}}{{\text{70}}{\text{.90 g Cl}}_{\text{2}}}}\right)\left({\frac {\text{2 mol AlCl}}{{\text{3 mol Cl}}_{\text{2}}}}\right)$
Solving these equations, we see that, beginning with 6.0 grams of aluminum, 0.22 moles of AlCl$_3$ can be formed, and that, beginning with 3.8 grams of chlorine, 0.036 moles of AlCl$_3$ can be formed.
Keeping score, 6.0 grams of Al yields 0.22 moles of AlCl$_3$, and 3.8 grams of chlorine gas yields 0.036 moles of AlCl$_3$. The lowest yield comes from the chlorine gas, therefore it must be limiting and aluminum must be in excess. The reaction in the problem will therefore produce 0.036 moles of product, which is equivalent to:
$0.036{\text{ mol AlCl}}_{\text{3}}{\text{ }}\left(133.33{\text{ }}^{\text{g}}\!\!\diagup \!\!_{\text{mol}}\;\right)=4.8{\text{ grams of AlCl}}_{\text{3}}$
Example 6.5 Limiting Reactant Problems
Although “limiting reactant problems” may be tedious, they are not difficult. When you are faced with a limiting reactant problem, just remember, you do the simple molar yield calculations twice, one for each reactant. The reactant that yields the lowest molar quantity is your limiting reagent and the molar value you calculate determines the theoretical yield in the problem.
Exercise 6.4 Limiting Reactants
Ammonia, which is the active ingredient in “smelling salts”, is prepared from nitrogen and hydrogen according to the equation shown below.

## Study Points

* Stoichiometry is the relationship between the masses of chemical reactants and products in a given chemical reaction. The coefficients placed in a chemical equation in order to balance it are called the stoichiometric coefficients. Molar stoichiometry is simply the expression of the coefficients of a reaction in terms of moles of reactants and products.
* In order to find the number of moles of a product that is produced in a chemical reaction when you are given moles of reactant, simply multiply the moles of reactant by the stoichiometric ratio relating that reactant and the desired product; i.e.,
$\left(molreactant\right)\times \left({\frac {molproduct}{molreactant}}\right)$.
* In order to find the number of moles of a product that are produced in a chemical reaction when you are given mass of reactant, simply divide the mass of reactant by the molar mass (to get moles reactant) and then multiply by the stoichiometric ratio relating that reactant and the desired product; i.e.,
$\left({\frac {grams}{^{grams}\!\!\diagup \!\!_{mol}\;}}\right)\times \left({\frac {molproduct}{molreactant}}\right)$ .
* Always remember, mass divided by molar mass equals moles;
$\left({\frac {grams}{\left(^{grams}\!\!\diagup \!\!_{mol}\;\right)}}\right)=mol$.
* The mass or the number of moles that you calculate for a product based on reaction stoichiometry is called the theoretical yield for the reaction. The amount of material that you actually isolate from a given reaction is called the actual yield and it is always less than the theoretical yield. The ratio of the actual and theoretical yields, expressed as a percentage is called the percentage yield.
* If a reaction requires more than one reactant and if you are given the mass, or the number of moles of each reactant, you must approach the calculation as a limiting reactant problem. To solve a limiting reactant problem, simply perform the standard mass calculation for each reactant, noting the mass (or number of moles) of product formed in each calculation. The reactant that yields the smallest amount or product from these calculations is called the limiting reactant. Reactants that yield larger amounts of products in these calculations are called excess reactants. The theoretical yield in the reaction will be based solely on the calculated amount for the limiting reactant.
* If a reactant in a chemical reaction is said to be “in excess”, you assume that you have unlimited amount of the reactant, and that it will never be the limiting reactant.

## Supplementary Problems

Please follow this link...