Chapter 3 - Chemical Reactions and Reaction Stoichiometry

Today we're going to discuss chapter three, "Chemical Reactions and Reaction Stoichiometry." The learning objectives for today are: 1. Explain the mole as a relative mass unit. You will be able to determine the molar masses of atoms and molecules. 2. Convert between mass, moles, and number of atoms or molecules. 3. Calculate the mass percent of each element given a chemical formula. 4. Use percent composition data to calculate the empirical formula of a compound. 5. Describe how combustion analysis is used to determine the percent composition of carbon, hydrogen, and oxygen in a compound. 6. Use coefficients to balance chemical reactions. 7. Calculate the quantities of substances involved in chemical reactions. 8. Determine limiting reagents, theoretical yields, and percent yields of chemical reactions. So, stoichiometry is basically the study of the mass relationship in chemistry, based on the law of conservation of mass. As Antoine Lavoisier mentioned in 1789, "nothing is created, an equal amount of matter exists before and after the experiment." This principle forms the basis of performing chemical experiments. Matter cannot be created or destroyed, only transformed. Now, let's introduce what a chemical equation is. A chemical equation is a representation of a chemical reaction. For example, 2 moles of hydrogen plus 1 mole of oxygen will produce 2 moles of water. The arrow represents the difference between what we have at the beginning (the reactants) and what we have after the reaction (the products). Here's another chemical equation: methane plus 2 oxygen will produce carbon dioxide and 2 moles of water. On the left side of the equation, we have the reactants. On the right side, we have the products. We need to make sure that we have the same amount of atoms on each side of the equation to conserve mass. In the first equation, we have 1 carbon, 4 hydrogen, and 4 oxygen on both sides, so it is a balanced chemical equation. We may also mention the state of the reactants and products using abbreviations like (G) for gas, (L) for liquid, (S) for solid, and (aq) for aqueous solution. In conclusion, chemical reactions and stoichiometry are essential in understanding the mass relationships in chemistry. It is important to balance chemical equations and consider the states of the reactants and products. By following the law of conservation of mass, we can accurately calculate quantities and determine the limiting reagents, theoretical yields, and percent yields of chemical reactions. Conservation of mass tells us that the coefficients in a balanced equation represent the number of molecules or atoms involved. For example, in the equation 2H2 + O2 -> 2H2O, the coefficients are 2, 1, and 2. These coefficients can vary depending on the balance needed to have the same amount of atoms on each side. It's important to place the coefficients in front of the molecules because changing subscripts will alter the identity and properties of the compounds. For instance, if we change the subscript of O2 from 2 to 4, we would end up with hydrogen peroxide instead of water. Thus, it is better to add coefficients rather than change subscripts when balancing equations. Let's take a look at a couple of examples. The equation 2H2 + O2 -> 2H2O shows that two moles of hydrogen react with one mole of oxygen to produce two moles of water. The states of matter are also indicated - hydrogen and oxygen are both gases, while water is a liquid. Another example is the equation H2 + O2 -> H2O2, which produces hydrogen peroxide, a pure liquid. The coefficients remain the same, but the identity of the compound changes. Now, let's discuss three different types of chemical reactions - combination reactions, decomposition reactions, and combustion reactions. Combination reactions occur when two or more substances react to form one product. For example, magnesium + oxygen -> magnesium oxide. In this case, two reactants combine to produce one product. Other examples include nitrogen + 3H2 -> 2NH3 and C3H6 + Br2 -> C3H6Br2. Decomposition reactions, on the other hand, involve the breakdown of a reactant into two or more substances. For instance, calcium carbonate (CaCO3) decomposes to form calcium oxide (CaO) and carbon dioxide (CO2). Similarly, potassium perchlorate (KClO4) decomposes into potassium chloride (KCl) and oxygen (O2). Another example is the decomposition of sodium azide (NaN3) to produce sodium (Na) and nitrogen (N2). Finally, combustion reactions are characterized by their rapid nature and the production of a flame. These reactions often involve oxygen from the air as a reactant. For example, methane (CH4) + O2 -> CO2 + H2O is a combustion reaction that produces carbon dioxide and water. Similarly, propane (C3H8) + 5O2 -> 3CO2 + 4H2O is another combustion reaction where heat is released, causing the water to be in vapor form. In conclusion, the coefficients in a balanced equation represent the number of molecules or atoms involved, and it is important to add coefficients rather than change subscripts to maintain the identity of the compounds. Different types of chemical reactions include combination reactions, decomposition reactions, and combustion reactions. So, we have this water that is going to be in a gas state. We can have apples of carbon dioxide and water, and the methane or propane also as a gas. Now let's talk about the formula weight. The formula weight is the sum of the atomic weight for the atoms in a chemical formula. This is the quantitative significance of a formula. The formula weight of calcium chloride (CaCl2) would be: calcium (Ca) - 1 times the atomic mass of calcium, and chloride (Cl) - 2 times the atomic mass of chlorine. The total is 110.99, which is the formula weight for calcium chloride. When you calculate the formula weight, you use the atomic mass of each component of the molecular formula. And always remember, if you have more than one, you need to multiply the molar atomic mass by the number of times you have it in the molecule. For example, in the molecule of ethane (C2H6), we have 2 carbon atoms, so it is 2 times the atomic mass of carbon. And we have 6 hydrogen atoms, so it is 6 times the atomic mass of hydrogen. This gives us a total of 30.07 atomic mass units, which is the molecular weight for that compound (C2H6). For ionic compounds, empirical formulas and formula weight are used, not molecular weight. Ionic compounds exist with a three-dimensional order of ions, and there is no simple group of atoms to call a molecule. One can find the percentage of the mass of the atoms in the compound that comes from each element in the compound by using this equation: percent of element = number of atoms times atomic weight divided by formula weight of the compound, times 100. For example, to find the percentage of carbon in ethane, we multiply 2 (number of carbon atoms) by the atomic mass of carbon, then divide it by the molecular weight of ethane. This gives us 24.22 divided by 30.07, times 100, which equals 79.887% of carbon. Since this compound only contains carbon and hydrogen, the difference between 100 and 79.887% will be the percentage of hydrogen. Now, Avogadro's number should easily be understood in the same way as other numbers that represent a quantity or amount. For example, a dozen means 12, and a six-pack means 6. Similarly, Avogadro's numbers represent an amount in the lab. In the lab, we work with individual molecules, which are really small. 6.02 times 10 to the power of 23 atoms or molecules is an amount that we call one mole. It's like when you have one dozen of something, it represents 12. In the same way, one mole represents 6.02 times 10 to the power of 23 atoms, molecules, or particles. These can be atoms, molecules, or particles of any kind. For example, just like you can have one dozen of elephants, cats, cars, or students, you can have 6.02 times 10 to the power of 23 atoms, molecules, or ions, and that amount represents one mole. So, one mole of carbon has a mass of 12.00 grams, and one mole of water has a mass of 18.00 grams. In this case, Avogadro's number represents the amount, and one mole represents that amount. This amount of molecules of water, for example, moles of water, will be eighteen point zero grams. Okay, so for one mole of water, that represents 6.02 times 10^23 molecules of water. The mass of a mole and the amount of molecules is eighteen point zero grams. In the square, you'll find the molar mass. The molar mass is the mass divided by one mole. So, the molar mass of an element is the atomic weight for the element from the periodic table. If it's diatomic, it's twice that atomic weight. The formula weight, which is most of the time in atomic mass units, will be the same number as the molar mass. The units will always be grams per mole. We have one mole of oxygen inside this red balloon, and that mole of oxygen has a mass of 32.0 grams. Here we have one mole of water, and this mole of water will have a mass of eighteen point zero grams. And this is one mole of sodium chloride. This has a mass of 58.45 grams. Okay, so this is basically the representation of different compounds and their different amounts. Even though they are one mole, you can see that they have different masses. They can take up more or less space depending on the state of the compound. But you can see that when you have different compounds, you're going to have a different mass for the same amount. As I mentioned before, it's just like if we said one dozen of elephants will have a specific weight, and one dozen of pencils will have a specific weight. Both of them have the same amount of particles, twelve and twelve, but the mass is going to be different. Okay, so that's what happens here. One mole of oxygen, one mole of water, one mole of sodium chloride, with different masses. Moles are very useful because they are the only way that we can compare one compound to another compound or one compound to an atom. We need to use the units of moles to do that type of conversion or ratio or relationship. So, if we have grams, we can go from grams to moles by using a conversion factor that is the molar mass. We can also go from moles to mass by using the same molar mass but flipped. The same way, if you are in moles and you want to determine the formula units or the amount of particles, remember we can have formula units, ions, or molecules. One mole is going to determine the amount of particles. And as we have the amount of particles, we can calculate how many moles that amount represents. We could have a number of species with specific numbers of moles. Moles provide a bridge from the molecular scale to the real-world scale. That is the purpose of the mole. Here we have a table with different mole relationships. We have atomic nitrogen and we have molecular nitrogen. The atomic nitrogen has a molar mass and a formula weight of 14. The formula weight and molar mass of nitrogen is 28 because it's double. We have two particles of nitrogen. Now, when you're talking about some nitrogen, 6.02 times 10^23 atoms of nitrogen will have this mass. And for N2, we're going to have the same concept. 6.02 times 10^23 molecules of nitrogen. So, this amount that is the same as this but from nitrogen diatomic molecule will have a molar mass of 28. Silver as an atom will have a molar mass of 107.9. And this will be the representation of the mass of 6.02 times 10^23 atoms of silver. If you have a mole of silver, it's going to be the same mass as the atoms of silver. You're going to have the same molar mass and also the same number of particles, but instead of atoms, now you're talking about ions. The number is going to be the same because it's one mole by Avogadro's number. For barium chloride, it's going to depend. If you're talking just about barium chloride, the molar mass will be 208.2. And you can represent the number of formula units as 6.02 times 10^23 formula units of barium chloride. But if you want to be separated by ions, you have just one ion of barium, so that means that you're going to have 6.02 times 10^23 ions of barium. While you have two chlorines, so you multiply two times this value, and this will be the representation of the number of ions that you will have in the dissociation of barium chloride. Basically, as you can see here, for example, nitrogen as an atom, nitrogen as a molecule, the Avogadro's number is going to be the same, 6.02 x 10^23, as well as atoms and ions of the same molecule. That's why it's really important to understand this ratio, this amount of particles. Those particles can be atoms, ions, or molecules. Atoms can be molecules, formula units, or ions. A mole of molecules or formula units contains Avogadro's number of particles. The molecular formula shows the number of each type of atom in a molecule, while the empirical formula is the simplest ratio of atoms in a compound. Sometimes, the molecular formula is also the empirical formula, such as in the case of H2O. The ratio of atoms in the empirical formula must be integers. To determine the empirical formula for a compound, we need to convert the percent composition to grams assuming a 100 gram sample. Then, we divide by the molar mass of each element to get the moles. Finally, we divide by the smallest number of moles to obtain the empirical formula. Let's apply this method to the compound para-aminobenzoic acid, often listed as PABA on sunscreen bottles. PABA is composed of 40.31% carbon, 5.14% hydrogen, 10.21% nitrogen, and 23.33% oxygen. To start, we convert the percentages to grams by assuming a 100 gram sample. We then divide by the molar mass of each element: 12.01 g/mol for carbon, 1.01 g/mol for hydrogen, 14.01 g/mol for nitrogen, and 16.00 g/mol for oxygen. Next, we divide each amount by the smallest value obtained. For carbon, the smallest value is 0.7288, so we divide 40.31 by 0.7288 to get 55.24. We round this to 55. For hydrogen, the smallest value is 0.7288, so we divide 5.14 by 0.7288 to get 7.04. We round this to 7. For nitrogen, the smallest value is 0.7288, so we divide 10.21 by 0.7288 to get 14.01. We round this to 14. For oxygen, the smallest value is 0.7288, so we divide 23.33 by 0.7288 to get 32.00. We round this to 32. Therefore, the empirical formula for PABA is C7H7NO2. This method allows us to determine the empirical formula from the percent composition. By dividing the empirical formula back by the molar mass, we can also determine the molecular formula, if different from the empirical formula. In summary, to determine the empirical formula from percent composition:1. Convert percent composition to grams using a 100 gram sample.2. Convert grams to moles using the molar mass of each element.3. Divide the moles by the smallest value obtained to obtain the empirical formula. The molecular formula is a multiple of the number of atoms in an empirical formula. We can calculate the molecular formula from the empirical formula by finding that multiple. To do this, we need to know the molar mass of the molecular compound and the empirical formula. For example, let's say the empirical formula of a compound is CH and the molar mass is 78 grams per mole. To find the molecular formula, we calculate the multiple by dividing the molar mass by the molar mass of the empirical formula. In this case, it would be 78/13, which is equal to 6. So, the multiple is 6, which means the compound is C6H6. This is because we multiply each element in the empirical formula by the multiple. If we had, for example, C2H as the empirical formula, we would multiply C6 by 2 and H6 by 2, resulting in C12H6. In general, the multiple is used to multiply the coefficient of each element in the empirical formula. Keep in mind that the coefficient in the empirical formula remains the same for each element. Now, let's discuss combustion analysis. This is a method used to determine the empirical formula of a compound that contains carbon, hydrogen, and oxygen. In combustion analysis, a sample of the compound is burned in the presence of oxygen. By trapping the resulting products, we can determine the amounts of water and carbon dioxide produced. The mass gained by each absorber corresponds to the mass of carbon dioxide and water produced. Absorbers are parts of the instrument used to collect these compounds. Carbon is determined from the mass of carbon dioxide produced, while hydrogen is determined from the mass of water produced. Oxygen, on the other hand, is determined by calculating the difference. Since matter cannot be created or destroyed, the difference in mass must be oxygen. For example, if we have the equation 2H2 + O2 -> 2H2O, we can interpret it as two moles of hydrogen reacting with one mole of oxygen to produce two moles of water. To convert this to grams, we need to use the molar mass. The molar mass of hydrogen is 4 grams, one mole of oxygen is 32 grams, and two moles of water is 36 grams. Remember that mass is conserved in chemical reactions. So, if we start with 36 grams of reactant, we will also have 36 grams of product. This follows the law of conservation of mass. The coefficients in the balanced equation represent the relative numbers of molecules of reactants and products. For example, four moles of hydrogen reacting with one mole of oxygen always produces two moles of water. So, in summary, the molecular formula can be calculated from the empirical formula by finding the multiple. Combustion analysis is a method used to determine the empirical formula for compounds containing carbon, hydrogen, and oxygen. It involves burning the compound in the presence of oxygen and analyzing the resulting products. Amounts are also known as conversion factors because we can go from one mole of oxygen to two moles of hydrogen, two moles of water to two moles of oxygen, two moles of water to one mole of oxygen. So those are different conversion factors that can show the relationship between those two different compounds. And the only way is through the moles, the relative number of moles of reactants and products, which can be converted to mass eventually. So let's talk about the stoichiometric calculations. Here, we're going to start with grams of substance A to know how much we're going to have of substance B. So to do that, we need to change the mass of substance A to moles of substance A, and by that, we're going to use the molar mass of A. Now that we have moles of A, we need to calculate how many moles of B will be produced by using the moles of A. And to determine that, we're going to use the conversion factor produced from the coefficients of the balanced equation. That's going to be a ratio of how many moles of A react with B or how many moles of A will produce moles of B. This depends on what A and B are. A and B could both be reactants, A and B could be one reactant and the other product, or A and B could also be both products. The important thing is that the only way we can change from one substance to another is through moles. And once we have the moles of B, we can use the molar mass of B to determine the mass of B. We have already seen in this chapter how to convert from grams to moles or moles to grams. The new calculation is how to compare two different materials using the mole ratio from the balanced equation. So that's why we need to use moles because moles calculate the amount between two different materials. Let's do an example of stoichiometric calculation: How many grams of water can be produced from 1 gram of glucose? This is the equation: glucose + oxygen ‚Üí carbon dioxide + water. We're going to look for the ratio from one reactant to two products. In this case, A is glucose and B is water. We can also calculate how many grams of carbon dioxide will be produced. So we can go from A to B or A to C or A to D. We have three different possibilities. By using glucose as A, we can obtain the ratio or compare both reactants, or compare one reactant with another product, or one product with the other product. In this case, we have 1 gram of glucose to start. So, we need to calculate B grams of water. If we have the grams of glucose, we need to convert that to moles of glucose by using the molar mass of glucose. So, we have 1 gram divided by 180 grams per mole, which will give us the moles of glucose. Now that we have the moles of glucose, we can use the ratio between glucose and water, which is 1:6. We multiply the moles of glucose by 6/1 because that's the ratio. Now we have moles of water that will be produced by one mole of glucose. Next, we need to convert from moles to mass because the question is asking for the mass of water that will be produced. We're going to use the molar mass of water, which is 18. So, we multiply 18 by the number of moles of water, and that will give us the mass of water that will be produced by using 1 gram of glucose. Sometimes, in chemical reactions, when there is more than one reactant, one of them is going to be the limiting reactant. This means that it is the reactant present in the smallest stoichiometric amount. In other words, it is the reactant that you run out of first. The limiting reactant is the one that limits the amount of product that can be formed. For example, before the reaction, we have 10 moles of hydrogen and the same number of moles of oxygen. After the reaction, you can see that we have an excess of oxygen. This means that there are no more molecules of hydrogen to react with. So, the molecule of hydrogen was the limiting reactant. Limiting reactant in this reaction. Okay, oxygen was the excess reagent, so the limiting reactant is used in chemistry calculations to determine the amounts of products and the amounts of any other reactant using the reaction. So, if you need to calculate the amount of product, you need to determine which of the reactants is your limiting reactant. By using the limiting reactant, you can calculate your products or the amounts of other reactants that will be left in the reaction. The theoretical yield is the maximum amount of product that can be made, and you can calculate that by using the limiting reactant. In other words, it is the amount of product possible as calculated through the stoichiometry problem. This is different from the actual yield, which is the amount actually produced and measured. So, the theoretical yield is the yield that you calculate using pencil and paper to obtain that theoretical yield. The actual yield is the experimental one. You obtain that by weighing it on a scale, or measuring its volume using a graduated cylinder or similar equipment. The theoretical yield is the one that you calculate, while the actual yield is the one that you work for to obtain that product. The percent yield is found by comparing the actual yield to the theoretical yield, and it is possible to make the expected theoretical yield. In other words, the percent yield is the actual yield divided by the theoretical yield, times 100. That's the percent yield for any chemical reaction, and also non-chemical reactions. It's never going to be 100% yield due to different reasons. So, that will be the end of Chapter 3: Chemical Reactions and Reaction Stoichiometry.