5.7 Quantity Relationships in Chemical Reactions (Stoichiometry)

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1 5.7 Quantity Relationships in Chemical Reactions (Stoichiometry) We have previously learned that atoms combine in simple whole number ratios to form compounds. However, to perform a chemical reaction, we can t count or weigh atoms or compounds at the particulate level. We had to scale up to macroscopic numbers of atoms, and the mole was chosen as a reference number of atoms or molecules to work with, i.e x particles. Mole Mole Relationships Stoicheion is a Greek word meaning element. In chemistry, the term stoichiometry refers to measuring the quantitative relationships involved in chemical reactions. Reactions occur between molecules or atoms in specific whole number ratios. Stoichiometric ratios are the mole/mole ratios by which chemicals combine. They are employed to predict amounts of reactants required or products produced in chemical reactions. When reactants are combined in the correct mole ratio, we say they are mixed in stoichiometric quantities, i.e. the correct number of atoms or molecules. In the previous chapter, we introduced the Haber-Bosch reaction, which converts hydrogen and nitrogen gas into ammonia: H 2 + N 2 NH 3 The law of conservation of mass tells us that we must account for all atoms in a chemical reaction. That is, a chemical reaction merely rearranges the atoms; therefore the number of atoms we finish a chemical reaction with must exactly equal the number of atoms we started with. A balanced equation does just that, telling us the relative numbers of molecules of reactants and products in a chemical reaction, e.g 3H 2 + N 2 2NH 3 I will need 3 molecules of hydrogen gas (6 H atoms) for every molecule of nitrogen gas (2 N atoms). Note that the coefficients are the number of molecules, not the amount (mass) of each.

2 3H 2 + N 2 2NH x10 6 1x10 6 2x x x x moles 1 mole 2 mole In another example, the synthesis of methanol (methyl alcohol) is given by the equation: CO(g) + H 2 (g) CH 3 OH(l) We add coefficients to equalize and account for all the atoms on both sides of the arrow in the reaction. CO(g) + 2H 2 (g) CH 3 OH(l) When the coefficients account for all the atoms, we say the equation is balanced. Moreover, these molecules always react in the 1:2:1 ratio shown, no matter how many you start out with (Table 9.1). Returning to the Haber Bosch equation: 3H 2 + N 2 2NH 3 3 moles 1 mole 2 moles 6 moles 2 moles 4 moles 9 moles 3 moles 6 moles The atoms always combine in the same fixed number ratios: 3H 2 /1N 2 1N 2 /3H 2 and 1N 2 /2NH 3 3H 2 /2NH 3

3 Consider: What if you had 5.8 moles of N 2? How much H 2 would you need? How much NH 3 could you make? Because H 2 reacts with N 2 in a 3:1 ratio, you can use this relationship as follows: (3H 2 / 1N 2 ) x 5.8 N mol N 2 (2NH 3 / 1N 2 ) x 5.8 N mole H mol N mol NH 3 Example: The combustion of propane. C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g) If you are going to burn 4.3 mol C 3 H 8, how much oxygen will you need? First you must balance the equation: C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g) C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) Five mole of O 2 are required for each mole of C 3 H 8. A conversion factor of 5O 2 / C 3 H 8 is employed: (5O 2 / 1C 3 H 8 ) x 4.3 mol C 3 H 8 = 21.5 mol O 2. What if you wanted to know how much CO 2 would be produced? Use another conversion factor: (3CO 2 / C 3 H 8 ) x 4.3 mol C 3 H 8 = 12.9 mol CO Mass Mass Stoichiometry Of course, a mole is a number, a count of the uncountable. Moles are useful in representing the relative numbers of molecules that combine in a reaction, but usually you are not given the number of moles to work with. You are

4 given the grams of each reactant and asked to calculate the amount of product that might be made. The solution is to follow a three step mass-to-mass path: Mass of Reagent Moles of Reagent Moles of Product Mass of Product Example: Returning to the Haber Bosch reaction: 3H 2 + N 2 2NH 3 3 moles 1 mole 2 moles If you want to make 34 g of NH 3, how many grams of H 2 will you need? We know the mole ratios that these molecules combine in. 3H 2 /1N 2 1N 2 /3H 2 and 1N 2 /2NH 3 3H 2 /2NH 3 Before we can use these mole ratios, we must convert grams to moles using the appropriate conversions factors. Step 1: 34 g NH 3 x (1mol NH 3 /17g) = 2 mol NH 3 Step 2: 2 mol NH 3 x (3H 2 /2NH 3 ) = 3 mol H 2 Step 3: 3 mol H 2 x ( g H 2 /mol H 2 ) = 6.0 g H 2 Summary: To make 34 g. NH 3 (2 mol), you will need 6.0 g H 2 (3 mol). Example: The combustion of propane. Returning to the combustion of propane: C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) If you are going to burn 96.1 g of propane, how many grams of O 2 will you need?

5 Five mole of O 2 are required for each mole of C 3 H 8. A conversion factor of 5O 2 / C 3 H 8 will be employed to calculate the moles of O 2 required: (5O 2 / 1C 3 H 8 ) x mol C 3 H 8 = mol O 2. Before you can employ the conversion factor however, you must first convert the grams of C 3 H 8 to moles of C 3 H g C 3 H 8 x (1mol C 3 H 8 / 42 g C 3 H 8 ) = g C 3 H 8 Complete the calculation. Calculate moles of O 2 required. Convert the moles of O 2 to grams of O 2 What would be the volume of the O 2 at STP? Al + I 2 AlI 3 If you react 35 g of Al, how much I 2 will you need? Limiting Reagents When chemicals are mixed in stoichiometric quantities, each will be completely consumed or run out at the same time. In many chemical reactions, however, one of the reagents is present in excess such that one reagent runs out before the other. The significance of this is that the reagent that runs out first defines or limits the amount of product made. We call the reagent that runs out first the limiting reagent. The one that does not run out is called the excess reagent. If the reagents are mixed in nonstoichiometric quantities, one must calculate the amount of product that each could theoretically produce to determine which reagent is limiting. The maximum amount of product that can be made is called the theoretical yield.

6 For example, the reaction of carbon + oxygen, starting with three carbon atoms and two oxygen molecules: Since they react on a one-to-one basis, a carbon atom will be left over. We say that carbon was in excess. The following figure illustrates the following reaction: Zn(s) + HCl(aq) H 2 (g) + ZnCl 2 (aq) To fixed amount of HCl(aq), three different amounts of Zn(s) were added. Flask 1 Flask 2 Flask 3 6.1g 2.37g 0.41g

7 In Flask 2 and 3, all the Zn(s) was consumed, but not Flask 1 Returning once more to the Haber Bosch equation: 3H 2 + N 2 2NH 3 3 moles 1 mole 2 moles If you mixed 3 moles of H 2 and 3 moles of N 2, it should be apparent from the balanced equation that H 2 would be completely consumed long before the N 2 was, and only 2 mol of NH 3 would be produced. Indeed, there is an excess of 2 mol of N 2. H 2 is the limiting reagent. Example: 20 g of N 2 gas and 5 g of H 2 gas are mixed and reacted in the Haber Bosch reaction. How many grams of ammonia will be produced? We know the mole ratios that these molecules combine in to form product,

8 so we can use these to predict the product from each reagent. 2NH 3 /3H 2 2NH 3 /1N 2 Before we can use these mole ratios to predict product, we must first convert grams of each reagent to moles. There is more than one approach to take. This is one approach. 3H 2 + N 2 2NH 3 5g 20 g g? Hydrogen: How much NH 3 can be produced? Step 1: 5 g H 2 x (1mol H 2 /2.016 g) = 2.48 mol H 2 Step 2: How much NH 3 can be produced from 2.48 mol H 2? 2.48 mol H 2 x (2NH 3 /3H 2 ) = 1.65 mol NH 3 Hence, there is enough H 2 to make 1.65 mol of NH 3 Nitrogen: How much NH 3 can be produced? Step 1: 20 g N 2 x (1 mol N 2 / g N 2 ) = mol N 2 Step 2: mol N 2 x (2NH 3 /1N 2 ) = 1.45 mol NH 3 Hence, there is enough N 2 to make 1.45 mol of NH 3 Ammonia: The reagent that produces the least amount of NH 3 is the limiting reagent. In this example, the limitation to the amount of NH 3 produced is determined by the amount of N 2. Therefore, even though there are 5 times more grams of N 2 than grams of H 2, N 2 is the limiting reagent because on a molecule to molecule basis, or mole to mole basis, it is used up before the H 2. The amount of NH 3 produced, in grams, would be: 1.45 mol NH 3 x (17.03 g / mol NH 3 ) = g NH 3 To illustrate the need for moles, calculate the amount of product that can be made in the problem below. Mole-Mole: Given 10 moles of each reagent, how much product can be made? What is the limiting reagent?

9 3H 2 + N 2 2NH 3 10 mol 10 mol mole? Gram Gram: Given 10 g of each reagent, how much product can be made? What is the limiting reagent? 3H 2 + N 2 2NH 3 10 g 10 g g? How do these two examples compare, and what do they tell us? Aluminum reacts with Cl 2 to form aluminum chloride. If 3.5 g of Al are reacted with 5.6 g of Cl 2, how many grams of aluminum chloride could be made? 1. Write the balanced equation 2. Determine the amount of aluminum chloride that could be produced by each reagent separately 3. Decide which reagent is the limiting reagent (which one runs out first) g of Al(s) are added to 63.0 g of HBr(aq) to form hydrogen gas and aluminum bromide. What is the limiting reagent? Theoretical Yield, Actual Yield, and Percent Yield

10 Who hasn t experienced this? You start with 20 kernels of popping corn, but you end up with only 16 pieces of popcorn. In other words, not all the kernels popped. What is the theoretical number of popcorn that we could expect? What is the actual percent of the kernels popped? Note that in all the examples of chemical reactions given so far, it is assumed that each reaction works to perfection, and that no product is lost in collecting it, washing it, drying it, transferring it, and weighing it. Under the assumption of perfection in all aspects, the calculated amount of product produced is called the theoretical yield. This potential yield of product is called theoretical because in reality perfection can never achieved. For instance, the chemical reaction does not occur completely, and some product is lost along the way. The amount we actually end up with is always less than the theoretical maximum amount. We call this amount the actual yield.

11 One measure of how well the chemical reaction went, and how good you were in collecting the product, is to calculate a ratio of the actual yield over the theoretical yield. Expressed as a percentage, we call this the percent yield. In one of the above examples, 3H 2 + N 2 2NH 3 5g 20 g g? We found that N 2 was the limiting reagent, with the 20 g N 2 producing a calculated amount of g NH 3. This is the theoretical amount. If the NH 3 was actually weighed and the amount was found to be 20.0 g of NH 3, we would call this the actual yield. The percent yield = [20.0 g NH 3 (actual yield) / g NH 3 (theoretical yield)] x100 = 82.3% yield g of mercury(ii) oxide decompose to form 39.0 g of mercury. What is the percent yield? Phosphorus and Bromine gas react vigorously to form PBr 3 1. If 5 g of phosphorus and 35 g of bromine react, which reagent is the limiting reagent? 2. How many grams of PBr 3 will theoretically be produced? 3. If the actual amount of PBr 3 produced is 30 g, what is the percent yield?

12 For a tutorial on stoichiometry: Energy in Chemical Reactions A collision between molecules is necessary for a chemical reaction to take place. In addition, the molecules must collide in a specific orientation, and particularly with sufficient energy, or they will simply bounce off each other. The amount of energy required is called the activation energy. For many reactions, the sum of the energy (or enthalpy) contained in the reactants is greater than the total energy contained in the resulting products. This difference in energy is given off as heat during the reaction (heat of reaction). Reactions that produce heat are called exothermic reactions. For reactions whose products contain a greater amount of energy than the reagents, the input of energy is required for a chemical reaction to take place. Reactions requiring the absorption of energy are called endothermic reactions.

13 Rate of Reactions The activation energy represents a hurdle which all reactions must overcome for a chemical reaction to take place. The height of that hurdle, or more correctly, the amount of activation energy required, can determine the speed or rate of the chemical reaction. The reaction rate can be accelerated by increasing the temperature or increasing the concentration of the reagents. The rate can also be increased by lowering the activation energy with something called a catalyst.

14 Thermochemical Stoichiometry The study of energy is called thermodynamics. The first law of thermodynamics (the law of conservation of energy) is: energy like matter is neither created nor destroyed. That can also be stated as: the energy of the universe is constant. Enthalpy: A special function, enthalpy (H), was created to represent energy. It represents the internal potential energy possessed by a chemical system at a given temperature and pressure. Reactants Products H R H P In a chemical reaction, when heat is transferred (without any work being done), the change in energy ( H) equals the change in energy that flows as heat: H = H R - H P H = heat (or enthalpy) released or consumed by the reaction, in Joules (J) Note: When heat is released (an exothermic reaction), the heat is expressed with a minus sign (heat is removed from the system).

15 For an endothermic reaction, where heat is consumed, the heat is expressed as a positive number (heat is added to the system0. Enthalpy example: When 1.0 moles of methane (CH 4 ) are burned, a total of 890 kj of energy is released as heat. In other words, CH 4 + 2O 2 CO 2 + 2H 2 O At constant pressure, Q = H = - 890kJ Note: the -890kJ is understood to correlate to either 1 mole of CH 4, or 2 moles of O 2. Questions: How many moles of each reactant are consumed, and each product are produced, to cause the release of 890 kj of heat? Suppose you reacted half as many moles of each reactant and produced half as many moles of product. How much heat would be released in this reaction? Compute the mass of CO 2 produced if the above reaction generated 89kJ of heat. Calculate the change in enthalpy ( H) when 5.8 g of methane are burned/ Enthalpy Stoichiometry: 2KClO 3 (s) 3O 2 (g) + 2KCl(s) H = 89.5 kj (energy consumed)

16 How many grams of KClO 3 can be decomposed by 225 kj of heat? C 8 H 18 (l) + O 2 (g) CO 2 (g) + H 2 O(g) H = x 10 3 kj What is the H for 175 g of C 8 H 18? Mg(s) + O 2 (g) MgO(s) H = x 10 3 kj Compute the heat released when 2.4 g of Mg(s) reacts at atmospheric pressure (i.e. at constant pressure).