EXPERIMENT 6 – Empirical Formula of a Compound, Exams of Chemistry

Download EXPERIMENT 6 – Empirical Formula of a Compound and more Exams Chemistry in PDF only on Docsity! CHEM 1405 Experiment 6 1 EXPERIMENT 6 – Empirical Formula of a Compound INTRODUCTION Chemical formulas indicate the composition of compounds. A formula that gives only the simplest ratio of the relative number of atoms in a compound is the empirical formula or simplest formula. The ratio usually consists of small whole numbers. We call a formula that gives the actual numbers of each type of atom in a compound the molecular formula. The numbers in a molecular formula will be whole number multiples of the numbers in an empirical formula. To determine the molecular formula of a compound, we need to know both the empirical formula and the molar mass of the compound. Benzene, for example, has an empirical formula of CH. In a molecule of benzene, the number of carbon atoms (C) and hydrogen atoms (H) are the same. The molar mass of benzene is 78.11 g/mol. We can determine the molecular formula of benzene by first calculating the molar mass of the empirical formula, which is 13.02 g/mol. We then determine the number of empirical units in a molecule by dividing the molar mass of benzene by the empirical molar mass: Number of empirical units = (78.11 g/mol) / (13.02 g/mol) = 6 empirical units Multiplying the empirical formula by 6 gives the molecular formula of benzene, (CH) X 6 or C6H6. Experimentally, we can determine the empirical formula of a compound by first finding the mass of each element in a sample of the compound. We then convert the mass of each element to the equivalent number of moles of that element. To find the simplest formula of a compound, we will combine the elements in the compound under conditions that allow us to determine the mass of each element. From these data, the moles of atoms of each element may be calculated. By dividing the numbers to the smallest number of moles, you obtain quotients that are in a simple ratio of integers or are readily converted to such a ratio. The ratio of moles of atoms of the elements in a compound is the same as the ratio of individual atoms that is expressed in an empirical formula. Example: Suppose that we want to determine the empirical formula of the oxide that formed when we ignited 0.175 g of aluminum (Al) in an open container to produce a compound of Al and O that weighed 0.331 g. The gain in mass is due to the presence of oxygen atoms that combined with the aluminum atoms in the reaction. First, we find the number of grams of each element in the sample of the compound. We know we started with 0.175 grams of Al, so we can calculate the grams of O in the compound by subtracting the mass of Al from the mass of compound: Mass of O = 0.331 g – 0.175 g = 0.156 g Next, we find the number of moles of each element in the compound: CHEM 1405 Experiment 6 2 Moles of Al = (mass of Al) / (molar mass of Al) = (0.175 g) / (26.98 g/mol) = 0.00649 mol Moles of O = (mass of O) / (molar mass of O) = (0.156 g) / (16.00 g/mol) = 0.00975 mol At this stage, our formula is “Al0.00649O0.00975”. Now we need to convert the subscripts to the simplest whole numbers. Use the following steps to do this: 1) Start by dividing each number of moles by smallest number of moles in the formula. The smallest number of moles is the moles of Al, 0.00649 mol. Al0.00649O0.00975 = Al1.00O1.50 0.00649 0.00649 2) Usually the subscripts at this stage are very near whole numbers, but in this case we still have a fractional subscript (1.5). To correct this, we will multiply both subscripts by a whole number (2, 3, 4, etc.) in order to obtain the simplest whole number subscripts. In this example, if we multiply 1.5 by 2, we will get a whole number, 3 for the O subscript. Therefore, multiplying both subscripts by 2, we obtain, Al2.00O3.00 which gives the empirical formula Al2O3. You may find it convenient to organize your calculations by arranging both data and results in a table as follows: Element Convert Grams to Moles Divide Each by the Smallest Number of Moles Convert to Whole Numbers Al 0.175 g / 26.98 g/mol = 0.00649 mol 0.00649 mol / 0.00649 mol = 1.00 1.00 X 2 = 2.00 = 2 O 0.156 g / 16.00 g/mol = 0.00975 mol 0.00975 mol / 0.00649 mol = 1.50 1.50 X 2 = 3.00 = 3 In this experiment, you will react a known mass of magnesium (Mg) with hydrochloric acid, HCl (aq), to form a compound containing only the elements Mg and Cl (magnesium chloride). The mass of Cl reacting with the Mg will be found from the difference in the mass of the product and the mass of Mg used. By following the same sequence of calculations used in the Al2O3 example, you will be able to experimentally verify the empirical formula of magnesium chloride. The reaction of magnesium with hydrochloric acid is an example of a single replacement reaction. Can you write the balanced equation for the reaction using the known formula of magnesium chloride? CHEM 1405 Experiment 6 5 EXPERIMENT 6 Name Pre-Laboratory Questions and Exercises Due before lab begins. Answer in the space provided. 1. a) How many moles of copper atoms are in 150 g of copper metal? b) How many copper atoms are in this amount of copper? 2. Write the empirical formula for the following compounds containing, a) 0.0200 mole of Al and 0.0600 mole of Cl. b) 0.0800 mole of Ba, 0.0800 mole of S, and 0.320 mole of O. 3. When 0.424 g of iron powder is burned in an oxygen atmosphere, 0.606 g of a reddish brown oxide is obtained. Determine the empirical formula of the oxide. 4. What safety precautions are cited in this experiment? CHEM 1405 Experiment 6 6 EXPERIMENT 6 Name Post-Laboratory Questions and Exercises Due after completing the lab. Answer in the space provided. 1. Why was an evaporating dish more suitable for this lab procedure, rather than using a beaker? 2. How would your experimental formula of magnesium chloride “MgClx” have been affected if your product was not dried completely before weighing it? Would “x” be too high or two low? 3. When 6.25 grams of pure iron are allowed to react with oxygen, a black oxide forms. If the product weighs 8.15 g, what is the empirical formula of the oxide? 4. A compound of nitrogen and oxygen is 30.46% by mass N and 69.54% by mass O. The molar mass if the compound was determined to be 92 g/mol. a) What is the empirical formula of the compound? b) What is the molecular formula of the compound?