Empirical and Molecular Formulas CK12 Editor Say Thanks to the Authors Click http://www.ck12.org/saythanks (No sign in required)
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www.ck12.org Concept 1. Empirical and Molecular Formulas CONCEPT 1 Empirical and Molecular Formulas Lesson Objectives The student will reduce molecular formulas to empirical formulas. Given either masses or percent composition of a compound, the student will determine the empirical formula. Given either masses or percent composition of a compound and the molar mass, the student will determine the molecular formula. Introduction The empirical formula is the simplest ratio of atoms in a compound. Formulas for ionic compounds are always empirical formulas but for covalent compounds, the empirical formula is not always the actual formula for the molecule. Molecules such as benzene, C 6 H 6, would have an empirical formula of CH. Finding Empirical Formula from Experimental Data Empirical formulas can be determined from experimental data or from percent composition. Consider the following example. Example 11 We find that a 2.50 gram sample of a compound contains 0.900 grams of calcium and 1.60 grams of chlorine. The compound contains only these two elements. We can calculate the number of moles of calcium atoms and the number of moles of chlorine atoms in the compound. We can then find the ratio of moles of calcium atoms to moles of chlorine atoms and from this; we can determine the empirical formula. Solution First, we convert the mass of each element into moles. moles of Ca = 0.900 g = 0.0224 mole Ca 40.1 g/mol moles of Cl atoms = 1.60 g = 0.0451 mole Cl 35.5 g/mol At this point, we have the correct ratio for the atoms in the compound, Ca 0.0224 Cl 0.0451, except that this isn t an acceptable formula. We need to find the simplest whole number ratio. To find a simple whole number ratio for these numbers, we divide each of them by the smallest of them. 1
www.ck12.org moles of Ca = 0.0224 = 1.00 Ca 0.0224 moles of Cl = 0.0451 = 2.01 Cl 0.0224 Now, we can see the correct empirical formula for this compound is CaCl 2. It is important to note that when solving for empirical formulas, we are determining the number of atoms of each element in the compound. Therefore, those substances which occur in nature as diatomic molecules such as Cl 2, O 2, H 2, N 2, and so on, are dealt with as atoms in this procedure. Finding Empirical Formula from Percent Composition When finding the empirical formula from percent composition, the first thing we do is to convert the percentages into masses. For example, suppose we are given the percent composition of a compound as 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen. Since every sample of this compound regardless of size will have the same composition in terms of ratio of atoms, we could choose a sample of any size. Suppose we choose a sample size of 100. grams. The masses of each of the elements in this sample will be 40.0 grams of carbon, 6.71 grams of hydrogen, and 53.3 grams of oxygen. These masses can then be used to find the empirical formula. You should note that you could use any size sample. You could choose a sample size of 167.8 grams and take the percentages of this sample to get the masses of the individual elements. We choose a sample size of 100. grams because it makes the arithmetic simple. Example 12 Find the empirical formula of a compound whose percent composition is 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen. Solution We choose a sample size of 100. grams and multiply this 100. gram sample by each of the percentages to get masses for each element. This would yield 40.0 grams of carbon, 6.71 grams of hydrogen, and 53.3 grams of oxygen. The next step is to convert the mass of each element into moles. moles of C = 40.0 g = 3.33 moles C 12.0 g/mol moles of H = 6.71 g = 6.64 moles H 1.01 g/mol moles of O = 53.3 g = 3.33 mole Ca 16.0 g/mol Then, we divide all three numbers by the smallest one to get simple whole number ratios: C = 3.33 3.33 = 1 H = 6.64 3.33 = 2 O = 3.33 3.33 = 1 and finally, we can write the empirical formula, CH 2 O. Sometimes, the technique of dividing each of the moles by the smallest to get a whole number ratio does not yield whole numbers. Whenever the subscript for any element in the empirical formula is 1, dividing each of the moles by the smallest will yield a simple whole number ratio but if none of the elements in the empirical formula has a 2
www.ck12.org Concept 1. Empirical and Molecular Formulas subscript of 1, then this technique will not yield a simple whole number ratio. In those cases, a little more work is required. Example 13 Determine the empirical formula for a compound that is 66.0% calcium and 34.0% phosphorus. Solution We choose a sample size of 100. grams and multiply the 100. grams by the percentage of each element to get masses. This yields 66.0 grams of calcium and 34.0 grams of phosphorus. We then divide each of these masses by their molar mass to convert to moles. We then divide each of these moles by the smallest. moles of Ca = 66.0 g = 1.65 moles Ca 40.1 g/mol moles of P = 34.0 g = 1.10 moles P 31.0 g/mol Ca = 1.65 1.10 = 1.50 P = 1.10 1.10 = 1.00 In this case, dividing each of the numbers by the smallest one does not yield a simple whole number ratio. In such a case, we must multiply both numbers by some factor that will produce a whole number ratio. If we multiply each of these by 2, we get a whole number ratio of 3Ca to 2 P. Therefore, the empirical formula is Ca 3 P 2. Finding Molecular Formulas Empirical formulas show the simplest whole number ratio of the atoms in a compound. Molecular formulas show the actual number of atoms of each element in a compound. When you find the empirical formula from either masses of elements or from percent composition, as demonstrated in the previous section, for the compound N 2 H 4, you will get an empirical formula of NH 2 and for C 3 H 6, you will get CH 2. If we want to determine the actual molecular formula, we need one additional piece of information. The molecular formula is always a whole number multiple of the empirical formula. That is, in order to get the molecular formula for N 2 H 4, you must double each of the subscripts in the empirical formula. Since the molecular formula is a whole number multiple of the empirical formula, the molecular mass will be the same whole number multiple of the formula mass. The formula mass for NH 2 is 14 g/mol and the molecular mass for N 2 H 4 is 28 g/mol. When we have the empirical formula and the molecular mass for a compound, we can divide the formula mass into the molecular mass and find the whole number that we need to multiply by each of the subscripts in the empirical formula. Example 14 Suppose we have the same problem as in example 12 except that we are also given the molecular mass of the compound as 180 grams/mole and we are asked for the molecular formula. In example 12, we determined the empirical formula to be CH 2 O. This empirical formula has a formula mass of 30.0 g/mol. In order to find the molecular formula for this compound, we divide the formula mass into the molecular mass (180 divided by 30) and find the multiplier for the empirical formula to be 6. As a result, the molecular formula for this compound will be C 6 H 12 O 6. Example 15 3
www.ck12.org Find the molecular formula for a compound with percent composition of 85.6% carbon and 14.5% hydrogen. The molecular mass of the compound is 42.1g/mol. Solution We choose a sample size of 100. g and multiply each element percentage to get masses for the elements in this sample. This yields 85.6 g of C and 14.5 g of H. Dividing each of these by their atomic mass yields 7.13 moles of C and 14.4 moles of H. Dividing each of these by the smallest yields a whole number ratio of 1 carbon to 2 hydrogen. Thus, the empirical formula will be CH 2. The formula mass of CH 2 is 14 g/mol. Dividing 14 g/mol into the molecular mass of 42.1g/mol yields a multiplier of 3. The molecular formula will be C 3 H 6. Lesson Summary The empirical formula of a compound indicates the simplest whole number ratio of atoms present in the compound. The empirical formula of a compound can be calculate from the masses of the elements in the compound or from the percent composition. The molecular formula of a compound is some whole number multiple of the empirical formula. Review Questions 1. What is the empirical formula for C 8 H 18? 2. What is the empirical formula for C 6 H 6? 3. What is the empirical formula for WO 2? 4. A compound has the empirical formula C 2 H 8 N and a molar mass of 46 grams/mol. What is the molecular formula of this compound? 5. A compound has the empirical formula C 2 H 4 NO. If its molar mass is 116.1 grams/mole, what is the molecular formula of the compound? 6. A sample of pure indium chloride with a mass of 0.5000 grams is found to contain 0.2404 grams of chlorine. What is the empirical formula of this compound? 7. Determine the empirical formula of a compound that contains 63.0 grams of rubidium and 5.90 grams of oxygen. 8. Determine the empirical formula of a compound that contains 58.0%Rb, 9.50%N, and 32.5%O. 9. Determine the empirical formula of a compound that contains 33.3%Ca, 40.0%O, and 26.7%S. 10. Find the molecular formula of a compound with percent composition 26.7%P, 12.1%N, and 61.2%Cl and with a molecular mass of 695 g/mol. 4
www.ck12.org Concept 1. Empirical and Molecular Formulas Further Reading / Supplemental Links Zumdahl, Steven S. and Zumdahl, Susan A., Chemistry, Fifth Edition, Chapter 3: Stoichiometry, Houghton Mifflin Company, New York, 2000. Website with lessons, worksheets, and quizzes on various high school chemistry topics. Lesson 5-5 is on Empirical Formulas. http://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson55.htm http://misterguch.brinkster.net/molarmass.html http://www.ausetute.com.au/percentc.html http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm Vocabulary empirical formula The formula giving the simplest ratio between the atoms of the elements present in a compound. molecular formula A formula indicating the actual number of each kind of atom contained in a molecule. Labs and Demonstrations for Teacher s Pages for Empirical Formula of Magnesium Oxide Investigation and Experimentation Objectives In this activity, the student will collect and interpret data, use mathematics to determine a solution, and communicate the result. Lab Notes It is not crucial that the magnesium ribbon be exactly 35 cm, but it should be clean. A good way to clean the ribbon is to dip it in 0.1 MHCl for a couple of seconds, then rinse it in distilled water, and dry it in alcohol or acetone. The more finely divided the ribbon is, the faster it will react. 0.5 cm to 1.0 cm pieces seems to work best. The crucibles often react with the magnesium during this process. This can cause greenish-black discoloration to the crucibles. This does not really affect chemical behavior for later reactions, as long as they are cleaned. However, the crucibles will often crack during this procedure. If a crucible cracks, discard it. Stress to the students that the 5
www.ck12.org crucibles get extremely hot. A hot crucible can cause a very serious burn. Show the students how to handle a crucible properly by using crucible tongs. A common source of error in this experiment is not to react the nitride. Make sure the students do this portion of the lab it is not washing. Answers to Pre-Lab Questions 1. Assume you have 100 grams of the compound. This then changes the percent composition to grams of each element. Find the number of moles of each element present. (Divide the grams of each element by its molar mass.) Divide each of these answers by the smallest answer. This will give the empirical formula. 2. The molecular weight of the compound is needed. 3. Yes. The charge of magnesium increases from 0 to +2, and the charge of the oxygen is reduced from 0 to 2. 4. 3.93 g for MgO,3.28 g for Mg 3 N 2 Lab - Empirical Formula of Magnesium Oxide Background Information In this lab, magnesium metal (an element) is oxidized by oxygen gas to magnesium oxide (a compound). Magnesium reacts vigorously when heated in the presence of air. The Mg O 2 reaction is energetic enough to allow some Mg to react with gaseous N 2. Although there is a higher percentage of N 2 gas in the atmosphere than O 2, O 2 is more reactive, and the magnesium oxide forms in a greater amount than the nitride. The small amount of nitride that forms can be removed with the addition of water, which converts the nitride to magnesium hydroxide and ammonia gas. Heating the product again causes the loss of water and conversion of the hydroxide to the oxide. The unbalanced equations are: Mg (s) + N 2(g) + O 2(g) MgO (s) + Mg 3 N 2(s) MgO (s) + Mg 3 N 2(s) + H 2 O (L) MgO (s) + Mg(OH) 2(s) + NH 3(g) MgO (s) + Mg(OH) 2(s) MgO (s) + H 2 O (g) Pre-Lab Questions a. If the mass percent of each element in a compound is known, what steps are taken to determine the compound s empirical formula? b. If the empirical formula of a compound is known, what additional information is required to determine the molecular formula of the compound? c. Is the reaction of magnesium metal and oxygen gas an oxidation-reduction reaction? If so, what is the change in oxidation number of each type of atom? d. What is the theoretical yield in grams of MgO if 2.37 gmgmetal reacts with excess O 2? What is the theoretical yield of Mg 3 N 2 if the same amount of Mg reacts with excess N 2? Purpose To determine the empirical formula of magnesium oxide, and to reinforce the concepts of the law of mass conservation and the law of multiple proportions. Apparatus and Materials 6 Safety goggles
www.ck12.org Concept 1. Empirical and Molecular Formulas Magnesium ribbon, Mg Balance (to 0.01 g or better) Ring stand Bunsen burner Ring support with clay triangle Crucible with lid Crucible Tongs Heat resistant tile or pad Safety Issues The crucible and all of the apparatus gets very hot. The ammonia emitted during the secondary reaction is irritating. Open flames can be dangerous. Do not place a hot crucible on an electronic balance. It can damage the electronics. In addition, a hot crucible causes the air above it to become buoyant. If placed on a balance, the buoyant air will cause a mass reading, which is less than the actual mass. Procedure 1. Heat the empty crucible and lid for about 3 minutes to remove water, oils, or other contaminants and to make sure there are no cracks. The bottom of the crucible should glow red-hot for about 20 seconds. Remove the flame and cool the crucible with lid. 2. Record the mass of crucible and lid once it has cooled. Handle the crucible with tongs. 3. Obtain about 0.3 g (35 cm) magnesium ribbon (do not handle the ribbon with your hands). Cut the magnesium into 0.5 1.0 cm pieces with scissors. 4. Record the mass of the magnesium ribbon, lid and crucible. 5. Place the crucible securely on the clay triangle. Set the lid slightly off-center on the crucible to allow air to enter but to prevent the magnesium oxide from escaping. 6. Place the Bunsen burner under the crucible, light it, and heat the bottom of the crucible with a gentle flame for about 1 minute; then, place the burner under the crucible and heat strongly. 7. Heat until all the magnesium turns into gray-white powder (around 10 minutes). 8. Stop heating and allow the crucible, lid and contents to cool. 9. Add about 1 ml (approx. 10 drops) of distilled water directly to the solid powder. Carefully waft some of the gas that is generated toward your nose, but be very careful. Record any odor. 10. Heat the crucible and contents, with the lid slightly ajar, gently for about 2 minutes and then strongly for about another 3 to 5 minutes. 11. Allow the crucible to cool and then record the mass of the crucible, lid and contents. 7
www.ck12.org 12. Follow instructions for oxide disposal given by your teacher. Clean all equipment thoroughly. Data Mass of crucible and lid = g Mass of the crucible, crucible lid, and the magnesium = g Mass of the crucible, crucible lid, and magnesium oxide = g Post-Lab Questions 1. Determine the mass of magnesium ribbon used in the experiment by subtracting the mass of the crucible and lid from the mass of the crucible, lid, and magnesium. Mass of magnesium = g 2. Determine the number of moles of magnesium used. Remember: mass atomic weight = number of moles. The atomic weight of magnesium is 24.3 g/mol. Number of moles of magnesium = mole 3. Determine the mass of magnesium oxide that was formed by subtracting the mass of the mass of the crucible and lid from the mass of the crucible, lid, and magnesium oxide. Mass of magnesium oxide formed = g 4. Determine the mass of oxygen that combined with the magnesium. Mass of oxygen = mass of magnesium oxide - mass of magnesium Mass of oxygen that combined with the magnesium = g 5. Determine the number of moles of oxygen atoms that were used. This is elemental oxygen so use 16.0 g/mol for the atomic weight. Number of moles of oxygen atoms that were used = mole 6. Calculate the ratio between moles of magnesium atoms used and moles of oxygen atoms used. Remember, this is simple division. Divide the number of moles of magnesium by the number of moles of oxygen. Round your answer to the nearest whole number, as we do not use part of an atom. This represents the moles (and also atoms) of magnesium. The moles (and also atoms) of oxygen, are represented by 1, because it was on the bottom of the division. Moles of Magnesium : Moles of Oxygen : Teacher s Pages for Water of Hydration Lab Investigation and Experimentation Objectives In this activity, the student will collect and interpret data, use mathematics to determine a solution, and communicate the result. Lab Notes This lab is fairly straightforward and simple in execution. The biggest hazard is with the crucibles. They become very, very hot enough to weld your skin on to it. Make sure to handle hot crucibles with crucible tongs, and never 8
www.ck12.org Concept 1. Empirical and Molecular Formulas to weigh a hot crucible. Teach your students how to gauge the temperature of a piece of glassware by having them approach the hot item with the back of their hand. If they can bring their hand to within a centimeter of the piece of glassware and it is hot, it is too hot to handle! If it is too hot for the back of your hand, it is certainly too hot for the front. Instruct the students on the proper procedure for using a desiccator. Water of Hydration Lab Background Information When ionic crystals form, they often incorporate water molecules within their structure. This water within the crystal is called the water of hydration, and the compounds themselves are called hydrates. This water of hydration can often be removed by simply heating the hydrate, because the water molecules are only weakly attracted to the ions present. When this is done, the resulting leftover substance is called the anhydrous form of the crystal. The amount of water present is often in a whole-number stoichiometric amount relative to the anhydrous form. Examples include barium chloride-2-hydrate, BaCl 2 2H 2 O, and cobalt nitrate-6-hydrate, Co(NO 3 ) 2 6H 2 O. Hydration numbers are most often integers, but in calcium compounds they are often fractional. Portland cement is an example of an important hydrate. When water is introduced to the anhydrous form, the water incorporates into the structure, and the cement hardens. Since the removal of water requires an input of heat, it should not be surprising that adding water to a hydrate gives off heat, and it gives off as much heat as was put into the system to remove the water in the first place. This is a problem for civil engineers who pour large amounts of cement: the heat given off by the hardening cement can be so great as to break the cement that has already hardened, due to heat stress. Steps must be taken to remove this heat. The Hoover Dam is such a large piece of concrete (cement + aggregate) that the dam is still cooling and the last of the cement was poured in 1935. Purpose To determine the hydration number and empirical formula of copper(ii) sulfate hydrate. Apparatus and Materials Ring Stand and Ring Crucible Clay Triangle Bunsen burner Wash bottle Matches Electronic Balance Copper(II)sulfate hydrate (approximately 3 g per lab group) desiccator watch glass Safety Issues Always handle crucibles with crucible tongs. Never place a hot crucible on a balance. It can damage the electronics and give a measurement, which is less than the actual mass. Procedure 1. Clean a porcelain crucible with soap and water. Rinse and dry the crucible by placing the crucible and cover on a clay triangle over a laboratory burner and heating until red-hot. 2. Carefully remove the crucible and cover with crucible tongs and let it cool. Handle the crucible and cover with tongs for the remainder of the experiment. 3. Measure the mass of the empty crucible and cover to the nearest 0.01 gram. 9
www.ck12.org 4. Add about 3 g of CuSO 4 hydrate crystals to the crucible, replace the cover, and measure the mass to the nearest 0.01 g. 5. Begin heating slowly. Increase the heat until you have heated the crucible strongly for about 10 minutes. 6. Remove the crucible from the triangle support, let it cool in a desiccator, and measure the mass. 7. Reheat with a hot flame for a few minutes, cool, and measure the mass again. If the mass is different from that recorded in Step 6, continue to heat and measure until the masses agree. 8. Remove the button of anhydrous copper sulfate by tipping it into a watch glass. Add a few drops of water to the anhydrous copper sulfate, and record your observations below. TABLE 1.1: Data Number Object Mass (grams) 1. Mass of Crucible + Cover grams 2. Mass of Crucible + Cover + Copper grams Sulfate Hydrate 3. Mass of Crucible + Cover + Anhydrous grams Copper Sulfate 4. Mass of Water (2 3) grams 5. Mass of Anhydrous Copper Sulfate grams (3 1) 6. Moles of Anhydrous Copper Sulfate moles 7. Moles of water driven off moles 8. Ratio of moles of water to moles of anhydrous CuSO 7 4 6 9. Describe the behavior of anhydrous copper sulfate when water is added. Post-Lab Questions a. Compare the number of moles of anhydrous CuSO 4 to the number of moles of water in the hydrate. Use the ratio of these two values to predict a formula for the hydrated CuSO 4. b. Why is it necessary to let the crucible cool before measuring mass? Why should the mass of the crucible be measured immediately after the crucible cools, and not later? c. In this experiment, you cooled your crucible in a desiccator. What is a desiccator? How does a desiccator work? d. How would your experimental results be affected if you did not use a desiccator when cooling the crucible and contents? e. How can you account for the behavior of the anhydrous form of the copper sulfate when the water was added? What do you think the new substance is? 10