## Introduction

Stoichiometry is a quantitatve process. Given an initial mass or volume of reactant or product, the molar relationships between reactants and products in a chemical reaction are used to calculate a specific mass or volume of another reactant or product.

## Problems - Mass/Mass

Essentially, all stoichiometry problems can be broken down into three steps:
1. Take the given quantity (i.e. mass or volume) and convert to moles
2. Use a mole to mole ratio to find the number of moles of the desired compound
3. Answer the question - convert the moles of the desired compound to the appropriate quantity (i.e. mass, volume). When unsure of what the question is asking, look for clues in the wording. If a phrase such as "find the number of grams" is used, the unit grams indicates that the mass should be found.

Consider the following reaction and problem: Determine the mass of iron that is produced from 25.36 g of iron(III) oxide.
This problem is often referred to as a mass-mass problem since you are given the mass of a compound in the problem and asked to find the mass of another compound. The three step method described above can be applied in the setup shown below: In the first conversion factor above, the molar mass of Fe2O3 is determined since it is needed to convert to moles (step one). When given a mass (as in this problem), dividing by molar mass always converts to moles. In the second conversion factor (step two), a ratio of Fe to Fe2O3 is determined from the coefficients in the reaction. It is worth noting that this is the only time the coefficients of the reaction are used when solving a stoichiometry problem. Finally, the molar mass of Fe is used to convert moles of Fe to grams of Fe (step three). It's worth noting that in a mass-mass problem, the first and third steps are opposites of each other. That is, in step one you convert grams to moles (divide by molar mass) and in the third step you convert moles to grams (multiply by molar mass). Alternatively, this may visually represented in a simplified manner:
25.36 g Fe2O3 ÷ 159.70 g/mol Fe2O3 × 2 Fe:1 Fe2O3 × 55.85 g/mol Fe = 17.74 g Fe

## Problems - Mass/Volume

The next problem is a mass-volume problem. In this case, either the mass of a compound will be given and the volume of another is asked, or the volume of a gas will be given and the mass of another compound will be asked. Reconsider the equation from above: Determine the volume of carbon dioxide gas that will be produced from 112.5 grams of iron at STP.
First, it's important to understand the concept of STP, standard temperature and pressure. Standard temperature and pressure is a set of conditions (273.15 K and 1 atm) at which 1 mole of any ideal gas will occupy 22.414 L. As a conversion factor, 1 mole gas = 22.414 L.
Although many authors will assume STP unless otherwise specified, it is important to determine the conditions of the reaction. If the reaction is not occuring at STP, the conversion factor given above cannot be used. The problem can still be solved, but the ideal gas law must be incorporated. Note the usage of the aforementioned conversion factor in the solution: ## Problems - Volume/Volume

A volume-volume problem concerns only the gaseous compounds of a reaction. Again, the relationship between 1 mole of gas at STP and the molar volume of 22.414 L is important. Consider the reaction below: What is the volume of ammonia gas will react with 22.5 L of oxygen gas?
Note that the first and last step in a volume-volume problem will cancel each other. This is because the first step, converting to liters of oxygen to moles, requires a division by 22.414. In the third step, conversion of moles of ammonia to liters, requires multiplication by 22.414. These two steps cancel each other and render step two (mole to mole ratio) the only important step. It needs to be stressed that this only happens in a volume-volume problem. ## Problems - Volumes Not at STP

Let's reconsider the mass/volume problem from earlier. What if the question had asked to determine the mass of carbon monoxide produced from 112.5 g of iron at 35.5°C and 855 torr of pressure? This problem now becomes a little longer because the 1 mol of gas = 22.414 L cannot be used. By clearly stating a temperature and/or pressure (in this case it's both) that are not at STP, the ideal gas law must be utilized in this problem. However, the first two steps of the problem remain unchanged. This is because the first step requires converting mass to moles. A mass-mol conversion is only reliant on molar mass, and the pressure/temperature of the reaction is irrelevant. The second step involves a mol-mol ratio, once again pressure and temperature are immaterial. The final step involves calculating a volume of gas. It is at this point that the ideal gas law is used. After these first two steps, the following can be determined: The ideal gas law, PV=nRT, must be used to finish this problem. The variable P represents pressure, and must be in atm. The variable V is the volume, and is what we are solving for. The variable n represents moles, and 0.4764 will be substituted into the equation here. The variable R is the gas law constant and has a value of 0.0821. Its units are atm • L • mol-1 • K-1; note how the unit incorporates the units of all the other variables. It is for this reason that pressure must be in atmospheres. The temperature T must be in kelvin. First, let's make the necessary conversions for temperature and pressure.

For temperature, to convert degrees Celsius to kelvin, add 273.15. Therefore, 35.5 + 273.15 = 308.65 K. For pressure we must setup a set of proportions using the conversion factor 1 atm = 760 torr. Solving the above proportion gives a value of 1.13 atm for "x." Now that these conversions are complete, the pressure, temperature, moles of carbon dioxide (solved earlier), and gas law constant can be plugged into the ideal gas law: ## Problems - Volume of Liquids

On occasion, a liquid reactant may be used and the mass is not given. Instead, the volume of the liquid is given as the starting quantity. Be careful with this as 22.414 L/mol cannot be used since that is only useful for gases. If lucky, the density of the liquid will be given in the problem. If not, then it must be found in literature. Using the density formula, the mass of the substance can be found (mass equals volume multiplied by density) and from there, the moles of the substance can be found. One last reminder: densities are given in g/mL, so make sure the volume of liquid given in the problem is also given in milliliters.

## Problems - Heat of Reaction (Enthalpy)

A thermochemical reaction shows the value for ΔHreaction. When included on the products side, the reaction is exothermic. If included on the reactants side, the reaction in endothermic. Either way, a mole-enthalpy ratio can be generated to determine a relationship between enthalpy, mass, volume, or any other stoichiometric quantity. Consider the exothermic reaction shown below: Consider the following question: What mass of europium will yield 475 kJ of heat? Unlike previous stoichiometry problems that required three steps to solve, this one will only need two. This is because the step that integrates a mole-mole ratio will be replaced with a mole-enthalpy ratio. This allows for a unit conversion (moles to kilojoules) and a stoichiometric ratio based on the reaction equation to be completed in one step. The solution: ## Problems - Limiting Reactant

In the previous example, it was assumed that there was an unlimited supply of carbon monoxide to react with all of the iron. Sometimes this is not an appropriate or plausible assumption. Sometimes two distinct masses of reactants are given, and it cannot be assumed that they will consume each other completely. Imagine trying to bake a cake. The recipe states that two eggs are needed to make a cake. With a dozen eggs available, six cakes can be made. What if the recipe also states that a cup of sugar is necessary and only four cups of sugar are available? Regardless of the dozen eggs, only four cakes can be made, because after consuming four cups of sugar (with eight eggs), there will be no sugar remaining. At this point, no more cakes can be made. Sugar is considered the limiting reactant in this example. The eggs are the excess reactant. Recall the reaction from before: What mass of iron will be produced from 25.00 g of iron(III) oxide and 25.00 g of carbon monoxide?
The solution is similar to the mass-mass problem from before, except there are two problems being solved at the same time. The reactant that produces the smaller mass (or volume for a gas) of product is the limiting reactant. The limiting reactant always dictates the mass/volume of product that is produced, therefore the smaller quantity of product is always the solution to a limiting reactant problem.

## Problems - Excess Reactant

In the last problem it was imperative for us to calculate which reactant was consumed first because the reaction would stop at that point. The mass of products had to be determined from this substance, which was called the limiting reactant (Fe2O3). The other substance that was not fully consumed (CO) is the excess reactant. The mass of carbon monoxide that is consumed can be calculated, and the mass of carbon dioxide that remains unreacted can also be found. In order to do so, a stoichiometry problem must first be completed in which the limiting reactant is used to calculate the mass of excess reactant consumed. ## Stoichiometry Calculator

Utilize the list of reactions and find a specific reaction for mass/mass, mass/volume, and volume/volume calculations.

## Solution Stoichiometry

A solution stoichiometry problem will involve aqueous reactants for which you will need to calculate a molarity or volume. Calculations that are part of a titration experiment are guaranteed to be solution stoichiometry. First, let's look at a solution stoichiometry problem that also incorporates some concepts discussed earlier on this page:
Given the reaction that follows, find the volume of sulfur dioxide gas that will be produced from 25.36 mL of 0.966 M hydrochloric acid at STP. This problem is unique in that there are two numbers given within the problem, but since they are not values for different compounds, it is not a limiting reactant problem. The volume and molarity given must be used to find the number of moles of HCl. From there, the rest of the problem continues in the same manner as previous problems. Note that the volume must be converted to liters. The molarity formula does not need to be done separately as it can be included into the normal dimensional analysis setup. Notice the units of the first two ratios. The first value is strictly liters, the unit for volume. The next ratio's units are mol/L, the unit for molarity, and when multiplied by the first value it is the equivalent of n = M×V. ## Solution Stoichiometry - Titration

The following could be a problem from a textbook, or data from a lab experiment: 19.52 mL of 0.285 M sulfuric acid was needed to titrate 42.81 mL of sodium hydroxide. Find the molarity of the sodium hydroxide.
Since the problem explicitly states that the molarity of sodium hydroxide is the unknown, then we must have enough information to start with sulfuric acid. Upon inspecting the values again, it can be seen that both the molarity and volume of sulfuric acid are given, enough to find moles of sulfuric acid and complete the problem. ## Density

By now, it should be apparent that there are many variations to a stoichiometry problem. Virtually any quantifiable property given in a problem that can be converted to moles can serve as a starting point for a problem. Consider the problem:
What will be the volume of carbon dioxide produced at STP from the full combustion of 45.0 mL of ethanol? The density of alcohol is 0.789 g/mL. The volume of ethanol cannot be immediately converted to moles. That is because there is no volume to moles conversion factor for liquids. Interestingly, the final step of the problem, in which the volume of a gas is solved, requires a volume to moles conversion. In this case it is possible because carbon dioxide is a gas, and at STP, one mole of any gas is assumed to occupy 22.414 L: 