Calculating Empirical and Molecular Formulas in Chemistry, Lecture notes of Chemistry

Download Calculating Empirical and Molecular Formulas in Chemistry and more Lecture notes Chemistry in PDF only on Docsity! Simple vs. True 454 Laying the Foundation in Chemistry 17 Simple vs. True Calculating Empirical and Molecular Formulas OBJECTIVE Students will learn to calculate empirical and molecular formulas and practice applying logical problem- solving skills. LEVEL Chemistry NATIONAL STANDARDS UCP.1, UCP.2, UCP.3, B.2 CONNECTIONS TO AP AP Chemistry: III. Reactions B. Stoichiometry 3. Mass and volume relations with emphasis on the mole concept, including empirical formulas TIME FRAME 45 minutes MATERIALS TEACHER NOTES This lesson is designed to be an integral part of the classroom unit involving the mole concept. It is best placed after students have mastered percent composition calculations. The student notes provide the basis for your lecture concerning empirical formula and molecular formulas. The examples easily serve as guided practice. If student white boards are available, it is a good idea to use them to engage and monitor the students. Be sure to explain each problem-solving step of the example problems; the solutions follow. Once students have mastered the concept of converting the given quantities to moles in search of a mole:mole ratio, encourage the use of the graphing calculator to allow students to visualize the data and better understand the simplified, yet whole number mole:mole ratio. calculator periodic table student white boards (optional) T E A C H E R P A G E S 17 Simple vs. True Laying the Foundation in Chemistry 455 EXAMPLE PROBLEM 1: Many of the biochemicals in our body consist of the elements carbon, hydrogen, oxygen and nitrogen. One of these chemicals, norepinephrine, is often released during stressful times and serves to increase our metabolic rate during the “fight or flight” response. The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O, and 8.28% N. Calculate the simplest formula for this biological compound. Step 1: Convert the percent composition data of each element to moles of each element. (Assume a 100. gram sample) 1molC 56.8 g C = 4.729 mol C 12.01 g C  1mol H 6.56g H = 6.495 mol H 1.01 g H  1mol O 28.4 g O = 1.775mol O 16.00g O  1mol N 8.28g N = 0.5910 mol N 14.01g N  Step 2: Calculate the mole:mole ratio. The smallest number of moles calculated corresponds to nitrogen so, divide all of the moles calculated by nitrogen’s number of moles, 0.5910 to obtain a simplified ratio. 4.729molC = 8 0.5910mol N 6.495mol H = 11 0.5910mol N Empirical Formula = 8 11 3C H O N 1.775molO = 3 0.5910mol N 0.5910mol N = 1 0.5910mol N T E A C H E R P A G E S Simple vs. True 458 Laying the Foundation in Chemistry 17 EXAMPLE PROBLEM 4: An organic alcohol was quantitatively found to contain the following elements in the given proportions: C = 64.81%; H = 13.60%; O = 21.59%. Given that the molecular weight of this alcohol is 74 g/mol, determine the molecular formula and name this alcohol. Step 1: Convert the percent composition data of each element to moles of each element. (Assume a 100. gram sample) 1mol C 64.81g C = 5.396 mol C 12.01g C  1mol H 13.60 g H = 13.47 mol H 1.01g H  1molO 21.59 g O = 1.349 molO 16.00 g O  Step 2: Calculate the mole:mole ratio. The smallest number of moles calculated corresponds to oxygen so, divide all of the moles calculated by oxygen’s number of moles, 1.349 to obtain a simplified ratio. 5.396 mol C = 4 1.349 mol O 13.47 mol H = 10 1.349 molO 1.349molO = 1 1.349molO The empirical formula is 4 10C H O but, this is an alcohol thus written as 4 9C H OH . Step 3: First, calculate the empirical mass for 4 9C H OH .     g 4 12.01 10 1.01 16.00 74.14 mol    Next, simplify the ratio of the molecular mass:empirical mass. molecular mass 74 1 empirical mass 74.14   Since the empirical and molecular masses are the same, the alcohol is butanol. T E A C H E R P A G E S 17 Simple vs. True Laying the Foundation in Chemistry 459 ANSWERS TO THE ANALYSIS QUESTIONS 1. A 2.676 gram sample of an unknown compound was found to contain 0.730 g of sodium, 0.442 g of nitrogen and 1.504 g of oxygen. Calculate the empirical formula for this compound and name it. Step 1: Convert the given data for each element to moles of each element. 1mol Na 0.730 g Na = 0.03175 mol Na 22.99 g Na  1mol N 0.442 g N = 0.03155mol N 14.01 gN  1molO 1.504 g O = 0.9400molO 16.00 g O  Step 2: Calculate the mole:mole ratio. The smallest number of moles calculated corresponds to nitrogen so, divide all of the moles calculated by nitrogen’s number of moles, 0.03155 to obtain a simplified ratio. 0.03175mol Na 1 0.03155mol N  0.03155mol N = 1 0.03155mol N 0.09400molO 3 0.03155mol N  Therefore, the empirical formula is 3NaNO , sodium nitrate. T E A C H E R P A G E S Simple vs. True 460 Laying the Foundation in Chemistry 17 2. A mysterious white powder was found at a crime scene. It was quantitatively analyzed and its percent composition was determined to be 27.37% sodium, 1.20% hydrogen, 14.30% carbon with the remainder being oxygen. Calculate the empirical formula for this compound and name it. Step 1: Convert the percent composition data of each element to moles of each element. (Assume a 100. gram sample) 1mol Na 27.37g Na = 1.191mol Na 22.99 g Na  1mol H 1.20 g H = 1.188mol H 1.01g H  1molC 14.30 g C = 1.191molC 12.01g C  Calculate the percent of oxygen present.  % oxygen 100 27.37 1.20 14.30 57.13 %     1molO 57.13 g O = 3.571molO 16.00 g O  Step 2: Calculate the mole:mole ratio. The smallest number of moles calculated corresponds to hydrogen so, divide all of the moles calculated by hydrogen’s number of moles, 1.188 to obtain a simplified ratio. 1.191mol Na = 1 1.188mol H 1.188mol H 1 1.188mol H  1.191molC = 1 1.188mol H 3.571molO = 3 1.188mol H Therefore, the empirical formula is 3NaHCO , and the compound is known as either sodium bicarbonate or sodium hydrogen carbonate. T E A C H E R P A G E S 17 Simple vs. True Laying the Foundation in Chemistry 463 Simple vs. True Calculating Empirical and Molecular Formulas How do chemists determine the true chemical formula for a newly synthesized or unknown compound? In this lesson we will explore some of the mathematics chemists apply to experimental evidence to quickly and accurately determine the true chemical formula of a compound. PURPOSE In this lesson you will learn problem-solving strategies that will enable you to calculate empirical and molecular formulas given experimental data. MATERIALS CLASS NOTES The simplest formula or empirical formula for a compound represents the smallest whole number ratio of atoms present in a given chemical substance. The molecular formula represents the true ratio of atoms actually present in a molecular compound. Sometimes the empirical formula and the molecular formula are identical. For example, the formula for water, H2O, is both the simplest ratio of atoms contained per molecule of water as well as the true ratio. In other instances, the molecular formula is a whole number multiple of the empirical formula. For example, the formula for butane is C4H10. This formula represents the molecular formula, which is the true ratio of atoms present in a molecule of butane. The empirical formula for this compound is easily determined by reducing the subscripts to the simplest whole number ratio possible. This is accomplished by dividing all of the subscripts by the greatest common factor, which is 2, to yield C2H5. For ionically-bonded substances, the empirical formula is the representation of the smallest formula unit. For example, in the formula NaCl, Na and Cl are in a 1:1 ratio, however, sodium chloride crystals are actually arranged in a crystal lattice that is face- centered cubic. One unit cell requires many more ions yet maintains the 1:1 ion to ion ratio. When a new substance is discovered, the formula is unknown until some qualitative and quantitative analyses are performed on the compound. First, qualitative analysis reveals which elements are in the compound. Next, quantitative analysis determines the amounts of those elements in the compound. Chemists use this type of experimental data to determine the empirical formula. Additional data must be collected in order to determine the molecular formula. calculator periodic table student white boards (optional) Simple vs. True 464 Laying the Foundation in Chemistry 17 CALCULATING EMPIRICAL FORMULAS 1. Convert the grams given for each element into moles. If the data is given as percent composition data, it is simplest to assume a 100g sample so that each percentage is converted directly to grams. For example, if a compound contains 20.0% Na, then convert this directly to 20.0 grams of sodium and then convert the quantity into moles of sodium. Record the number of moles to at least four significant figures. Rounding early is not recommended. 2. Examine your mole calculations and identify the least number of moles calculated. Divide all of the mole calculations by the smallest number of moles calculated to simplify the mole:mole ratio. This step may yield whole numbers or very close to whole numbers. If so, these whole numbers serve as the subscripts for the empirical formula. a. If the mole:mole ratio contains numbers other than whole numbers you may have to multiply all of the moles by the same factor to convert them to whole numbers. Try multiplying by 2 first, then by 3, etc. b. For example: If the mole:mole ratio comes out 1: 2.5: 1; multiplying each number in the ratio by 2 will yield the same proportion, but eliminate the ½. The ratio becomes 2: 5: 2. Remember that subscripts must be whole numbers so your calculated mole:mole ratio must be very near whole numbers. Also, remember you must multiply all of the calculated moles by the same number to keep them proportional. c. Watch for numbers that have the following terminal decimal values:  0.20 (multiply by 5 to yield 1.0)  0.25 (multiply by 4 to yield 1.0)  0.33 (multiply by 3 to yield 1.0)  0.50 (multiply by 2 to yield 1.0)  0.67 (multiply by 3 to yield  2.0)  0.75 (multiply by 4 to yield  3.0)  0.80 (multiply by 5 to yield  4.0) 3. Write the empirical or molecular formula with proper subscripts and name the compound if asked. (Usually the problem lists the elements in the order they appear in the formula.) EXAMPLE PROBLEM 1: Many of the biochemicals in our body consist of the elements carbon, hydrogen, oxygen and nitrogen. One of these chemicals, norepinephrine, is often released during stressful times and serves to increase our metabolic rate during the “fight or flight” response. The percent composition of this hormone is 56.8% C, 6.56% H, 28.4% O, and 8.28% N. Calculate the simplest formula for this biological compound. 17 Simple vs. True Laying the Foundation in Chemistry 465 EXAMPLE PROBLEM 2: A sample of a white, granular ionic compound having a mass of 41.764 grams was found in the photography lab. Analysis of this compound revealed that it was composed of 12.144 grams of sodium, 16.948 grams of sulfur, and the rest of the compound was oxygen. Calculate the empirical formula for this compound and provide its name. CALCULATING MOLECULAR FORMULAS It may be necessary for you to calculate the empirical formula first in order to determine the molecular formula. The molecular formula is simply a multiple of the empirical formula. 1. Calculate the empirical formula mass. 2. Determine the factor that the empirical formula will be multiplied by to determine the molecular formula. Simply divide the molecular mass by the empirical formula mass. This should yield a whole number. 3. Multiply all of the subscripts in the empirical formula by the whole number obtained from the previous step to get the true ratio of atoms in the molecular formula. EXAMPLE PROBLEM 3: Calculate the molecular formula for an organic compound whose molecular mass is 180. g mol and has an empirical formula of CH2O. Name this compound. EXAMPLE PROBLEM 4: An organic alcohol was quantitatively found to contain the following elements in the given proportions: C = 64.81%; H = 13.60%; O = 21.59%. Given that the molecular weight of this alcohol is 74 g/mol, determine the molecular formula and name this alcohol. Simple vs. True Name ______________________________________ Period _____________________________________ 468 Laying the Foundation in Chemistry 17 Simple vs. True Calculating Empirical and Molecular Formulas ANALYSIS Solve the following problems on this answer page. Be sure to show all work for full credit. 1. A 2.676 gram sample of an unknown compound was found to contain 0.730 g of sodium, 0.442 g of nitrogen and 1.504 g of oxygen. Calculate the empirical formula for this compound and name it. 2. A mysterious white powder was found at a crime scene. It was quantitatively analyzed and its percent composition was determined to be 27.37% sodium, 1.20% hydrogen, 14.30% carbon with the remainder being oxygen. Calculate the empirical formula for this compound and name it. 17 Simple vs. True Laying the Foundation in Chemistry 469 3. A common organic solvent has an empirical formula of CH and a molecular mass of 78 g/mole. Calculate the molecular formula for this compound and name it. 4. A gas was qualitatively analyzed and found to contain only the elements nitrogen and oxygen. The compound was further analyzed to it was determined the gas was composed of 30.43% nitrogen. Given that the molecular mass of the compound is 92.0 g/mole, calculate the molecular formula.