CH-180 The Empirical Formula of an Ionic Compound A compound’s empirical formula is the simplest whole-number ratio of the atoms. The empirical formula's subscripts indicate the ratio of the elements. For example, the empirical formula of hydrogen peroxide is HO: in hydrogen peroxide, the ratio of hydrogen to oxygen is 1:1. Don't confuse the empirical formula with the molecular (or true) formula. The true formula gives not just the ratios, but the actual number of atoms of each element in one molecule of the compound. For hydrogen peroxide, the true formula is H2O2, since each molecule contains two hydrogens and two oxygens. Do you see how the empirical formula information is contained within the true formula? The hydrogen-to-oxygen ratio, according to the true formula, is 2 to 2, which is the same as the 1 to 1 ratio expressed by the empirical formula. Knowledge of the empirical formula can't allow us to narrow down a compound's identity to a single possibility, but it can help us to rule out all but a very few of the millions of known compounds. The empirical formula is therefore very important information about a compound. Fortunately, the empirical formula is rather easy to obtain: even the early chemists, such as Dalton, were able to get the empirical formula for many compounds. In this experiment, you will convert a weighed amount of zinc (Zn) to zinc chloride by reaction with hydrochloric acid (HCl). As the Zn reacts with the HCl, hydrogen gas (H2) is released. The procedure that you will follow will give enough data to permit you to determine the empirical formula of the zinc chloride. The remarkable thing about this experiment is that you will be able to draw conclusions about the composition of a single formula unit of zinc chloride, even though we make our measurements on a huge number of zinc chloride formula units. We don't need to see an individual formula unit in order to make significant discoveries about it! Mole Concept In our everyday lives, we use the term “dozen”, which is a counting unit, to refer to 12 objects. In the same way, chemists also use counting units to refer to the number of atoms, ions, or molecules in a sample of matter. This is done to simplify things because even a small amount of any sample of matter contains a tremendously large number of particles. Chemists use a counting unit referred to as the mole to refer to the amount of particles in any sample. A mole is defined as 6.02 x 1023 of any type of particle. Thus, one mole of ping pong balls would be 6.02 x 1023 ping pong balls. As it turns out, this number equals the number of atoms in exactly 12 g of pure carbon-12. Thus, 12 g of carbon-12 equals one mole. Because the carbon-12 atom does not have the same mass as an iron-56 atom, 1 mole of iron-56 will have a different mass than 1 mole of carbon-12. So if we have more than 12 g of carbon-12, i.e. say 28 g, how do we determine the number of moles of this element that we have? Interestingly, it has been found that the mass of a single atom of an element, i.e. the atomic weight (in atomic mass units, amu), is numerically equal to the mass (in grams) of 1 mole of that element: 2 1 atom of C-12 has a mass of 12 amu so 1 mole of C-12 has a mass of 12 grams 1 atom of Fe-56 has a mass of 56 amu so 1 mole of Fe-56 has a mass of 56 grams 1 atom of N-14 has a mass of 14 amu so 1 mole of N-14 has a mass of 14 grams Thus, to calculate the number of moles contained in 28 g of carbon-12: 28 g x 1 mole/12 g = 2.3 moles = 1.4 x 1024 atoms When we look in the periodic table, we notice that the atomic mass of an element is not a whole number, but rather is a number that contains decimals. Recall that the atomic masses found in the periodic table are weighted averages of the atomic masses of the naturally occurring isotopes for an element. Thus, carbon’s atomic mass or atomic weight is listed as 12.0107 g. As a result, if one has 28 grams of carbon containing all isotopes as found in nature, the mass per mole used in the calculation would be 12.0107 grams: 28 g x 1 mole/12.0107 g = 2.3 moles Little difference is observed in the answers in these two examples so in most cases, the atomic mass of the most abundant form of the element is used for these types of calculations. What about compounds? We can also calculate the number of moles of a compound from its mass provided that we know the mass of one mole of the compound. To arrive at the number of moles for a compound, we need to first know its formula weight. The formula weight (FW) for a compound is simply the sum of the atomic weights (in amu) of each atom in its chemical formula. For example, the formula weight for water, H2O, is calculated from the atomic masses if hydrogen and oxygen: FW = 1 + 1 + 16 = 18 amu or more correctly 1.00794 + 1.00794 + 15.9994 = 18.01528 amu, which for convenience we can round to 18.01 amu. Second, we need to know that the same numerical relationship exists between a compound’s formula weight (in amu) and the mass (in grams) of one mole of that substance. So as an example, one mole of water molecules will have a mass of approximately 18 grams. Let’s say we have 50 grams of water. How many moles of water does this represent? 50 g x 1 mole/18 g = 2.8 moles water The mass in grams of one mole of any substance is called the molar mass of the substance. The molar mass of any substance is simply equal numerically to its formula weight (atomic weight if an element) in atomic mass units. Alternatively, if we know the number of moles of a substance that we have, we can also calculate the number of grams we have provided we know the molar mass. Assume we have 0.50 moles of water. How many grams of water do we have? 0.50 moles x 18 g/1mole = 9.0 g water 3 Now let’s consider chemical reactions briefly. For the reaction between hydrogen and oxygen: 2H2(g) + O2(g) → 2H2O(l) 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to form 2 molecules of water. Since the mole is simply a counting unit that represents the number of atoms, molecules, or ions there are in a sample of a substance, we can also say that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water. And since we can use the number of moles to determine masses, we can determine the masses of reactants and products that react. So for the reaction between hydrogen and oxygen: 2 moles H2 x 2 g/1 mole = 4 g H2 1 mole O2 x 32 g/1 mole = 32 g O2 2 mole H2O x 18 g/1mole = 36 g H2O 4 g of hydrogen gas react with 32 g of oxygen gas to form 36 g of water, as discovered originally by Joseph Proust in the 1790’s when he proposed the law of definite proportions. We can also observe the law of mass conservation here as well, since the total mass is the same on both sides of the chemical equation. (Remember Lavoisier?) The same concepts apply to all chemical equations. Calculating Empirical Formulas: An Example The mole concept provides us with the easiest way to calculate an empirical formula. All we need is mass data (either in a recognized mass unit - g, kg, lb, etc. - or as percents). We simply convert the mass data for each element to moles, then calculate the ratios that relate the elements. Finally, a math trick can help us to recognize simple whole-number ratios. Watch it work: A compound with the daunting name of thiopropionaldehyde-S-oxide is responsible for our tears when we chop raw onions. A 7.243 g sample of the compound was found to contain 2.895 g C, 0.486 g H, and 1.286 g O. The rest of the compound was Sulfur. What was the empirical formula of the sample? First, we need to know the mass of each element. C, H, and O are easy; they’re given in the problem statement. S is a little tougher. Fortunately, we have the Law of Conservation of Matter to help: the mass of the entire sample equals the sum of the masses of each element, or, mass of sample = mass of C + mass of H + mass of O + mass of S We have numbers for every term except “mass of S”; we can rearrange (by algebra) and solve: or, or, mass of S = mass of sample - (mass of C + mass of H + mass of O) mass of S = 7.243 g - (2.895 g + 0.486 g + 1.286 g) mass of S = 2.576 g Now convert each mass into moles. C: 2.895 g 1 mole C = 0.2410 moles C 12.011 g C H: 0.486 g 1 mole H = 0.482 moles H 1.00794 g H O: 1.286 g 1 mole O = 0.08038 moles O 15.9994 g O S: 2.576 g 1 mole S = 0.08035 moles S 32.06 g S 4 Since moles count atoms (just like dozens count eggs), the answers represent a count of the number of atoms of each element present. But, a chemical formula also counts the number of atoms present (in a molecule), so the answers at left are related to the empirical formula! By converting the moles to whole numbers, we can easily see the relationship. Here’s a math trick to help us see the wholenumber ratios: Just divide all the answers by the smallest (0.08035, in this case). That ensures that all of the answers are at least “1”: 0.2410 moles C = 3.000 0.08035 moles S 0.08038 moles O = 1.000 0.08035 moles S 0.482 moles H = 6.00 0.08035 moles S 0.08035 moles S = 1.000 0.08035 moles S So, the empirical formula is C3H6OS. We’re done! 5 First Lab Period's Procedure 1. You may work in teams of two (not three, or four, etc.) 2. Record your partner's name, just in case your partner fails to show up next time: Name _ 3. Obtain four large (20 150) test tubes from the lab drawer. Clean the test tubes so that no deposits appear on the interior glass. The test tubes need not be dry, but they must be clean, in order for this lab to work well. Label the four test tubes by letter: A, B, C, D. 4. Use forceps to place Zn pieces into a weighing boat until the mass of the Zn is ALMOST to the required mass (this should be within about 0.1 g of the amounts listed below). If the actual mass is greater than the required mass, remove pieces of Zn until the actual mass is less than the required mass. Then, use your spatula to CAREFULLY add powdered Zn to the weighing boat until the required mass is obtained. Follow this procedure for each of the four test tubes: test tube A B C D mass of Zn (g) 1.000 2.000 3.000 4.000 5. Transfer the Zn to each test tube. Then, slowly add 10 mL of 6 M HCl (hydrochloric acid) to each test tube. You should observe the formation of hydrogen bubbles almost immediately. Make sure you add the HCl slowly enough that the bubbles don't overflow the mouth of the test tube. 6. Stand the four test tubes in a 400 mL beaker to save for next week. By then, the reaction will have ceased, leaving a product which you will isolate and study. 7. Because the first lab period’s procedure only takes about 30 minutes, we will use the extra time for an additional 1 or 1.5 hours of lecture. Second Lab Period's Procedure 1. Weigh a dry evaporating dish on the analytical balance. Can you think of a good reason why the dish must be dry before weighing? 6 2. Pour the liquid from your sample into the evaporating dish. Be careful here: the liquid may still contain hydrochloric acid! If any metallic Zn remains in the test tube, pour the solution in such a way as to keep the unreacted Zn in the test tube. 3. Wash the interior of the test tube with 5 mL distilled water and add the water to the evaporating dish. Be careful not to splash any water out of the dish while making the transfer, and keep any unreacted Zn in the test tube, as before. 4. Devise and carry out a procedure to dry and weigh any unreacted Zn. Calculate the mass of Zn used up in the reaction. 5. Place the evaporating dish on a ringstand as shown on the next page. Set a wire screen on the iron ring as a platform to hold the evaporating dish. Carefully heat the dish with a Bunsen burner so that the solution dries to a solid residue without spattering. If the solution begins to spatter, remove the burner until spattering subsides, then resume heating by waving the burner back and forth under the dish. 6. Continue heating until the residue is dry. After the residue has dried, keep heating until the solid melts and a small pool of liquid forms in the bottom of the dish. STOP HEATING AS SOON AS THE LIQUID BEGINS TO APPEAR. 7. Allow the dish to cool by setting it on a wire screen until you can touch it without risk of a burn. (You can tell if the dish is too hot to touch by cupping your hands over the dish.) 8. After the dish has cooled, weigh the dish and its contents. 9. Design and perform calculations that will give you the empirical formula of the compound in the evaporating dish. See the example on the last page of the handout. Repeat steps 1 - 9 for the other three samples. Of course, you may wish to begin your work with one of the other samples as soon as your equipment becomes available, rather than waiting until the first sample is completely worked up. A particularly useful strategy is to heat two of your samples at the same time (by setting up two burners and ringstands), in order to cut your heating time in half. Your lab report for this experiment is a formal report that will be collected the week after the experiment is completed. 7 Data Attach this section to your formal lab report. Write your data and calculations for this lab in the space below. Putting your data in table format is strongly recommended. (Be sure to include this table of data in your results section of the formal report.) Based on your data, what should the empirical formula of the Zn/Cl compound be? 8 List your observations of the reaction in the large test tubes in the space below.: How would the % Zn change if the 5 mL wash water was not added to the evaporating dish? What error in the % Zn would result if the solution in the evaporating dish had spattered while boiling? What error in the % Zn would result if the unreacted Zn was not completely dried before weighing? 9 What error in the % Zn would result if extra water were added to the evaporating dish during heating?