Skills Worksheet Problem Solving Empirical Formulas Suppose you analyze an unknown compound that is a white powder and find that it is composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur. You can use those percentages to determine the mole ratios among sodium, sulfur, and oxygen and write a formula for the compound. To begin, the mass percentages of each element can be interpreted as grams of element per 100 grams of compound. To make things simpler, you can assume you have a 100 g sample of the unknown compound. The unknown compound contains 36.5% sodium by mass. Therefore 100.0 g of the compound would contain 36.5 g of sodium. You already know how to convert mass of a substance into number of moles, so you can calculate the number of moles of sodium in 36.5 g. After you find the number of moles of each element, you can look for a simple ratio among the elements and use this ratio of elements to write a formula for the compound. The chemical formula obtained from the mass percentages is in the simplest form for that compound. The mole ratios for each element, which you determined from the analytical data given, are reduced to the smallest whole numbers. This simplest formula is also called the empirical formula. The actual formula for the compound could be a multiple of the empirical formula. For instance, suppose you analyze a compound and find that it is composed of 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen. If you determine the formula for this compound based only on the analytical data, you will determine the formula to be CH 2 O. There are, however, other possibilities for the formula. It could be C 2 H 4 O 2 and still have the same percentage composition. In fact, it could be any multiple of CH 2 O. It is possible to convert from the empirical formula to the actual chemical formula for the compound as long as the molar mass of the compound is known. Look again at the CH 2 O example. If the true compound were CH 2 O, it would have a molar mass of 30.03 g/mol. If you do more tests on the unknown compound and find that its molar mass is 60.06, you know that CH 2 O cannot be its true identity. The molar mass 60.06 is twice the molar mass of CH 2 O. Therefore, you know that the true chemical formula must be twice the empirical formula, (CH 2 O) 2, or C 2 H 4 O 2. Any correct molecular formula can be determined from an empirical formula and a molar mass in this same way. Holt ChemFile: Problem-Solving Workbook 85 Empirical Formulas
General Plan for Determining Empirical Formulas and Molecular Formulas 1 Percentage of element expressed as grams of element per 100 g unknown Convert using the molar mass of each element. 2 Amount of each element per 100 g of unknown Use the amount of the least-abundant element to calculate the simplest wholenumber ratio among the elements. 4 Empirical formula of the compound Convert using the experimental molar mass of the unknown and the molar mass of the simplest formula. The calculated ratio is the simplest formula. 3 Calculated whole-number ratio among the elements 5 Molecular formula of the compound Holt ChemFile: Problem-Solving Workbook 86 Empirical Formulas
Sample Problem 1 Determine the empirical formula for an unknown compound composed of 36.5% sodium, 38.1% oxygen, and 25.4% sulfur by mass. Solution ANALYZE What is given in the problem? What are you asked to find? the percentage composition of the compound the empirical formula for the compound Items The percentage composition of the unknown subtance The molar mass of each element* Amount of each element per 100.0 g of the unknown? mol Simplest mole ratio of elements in the unknown? * determined from the periodic table Data 36.5% sodium 38.1% oxygen 25.4% sulfur 22.99 g Na/mol Na 16.00 g O/mol O 32.07 g S/mol S PLAN What steps are needed to calculate the amount in moles of each element per 100.0 g of unknown? State the percentage of the element in grams and multiply by the inverse of the molar mass of the element. What steps are needed to determine the whole-number mole ratio of the elements in the unknown (the simplest formula)? Divide the amount of each element by the amount of the least-abundant element. If necessary, multiply the ratio by a small integer that will produce a whole-number ratio. 1 Mass of Na per multiply by the inverse of the molar mass of Na 2 Amount Na in mol per percent of Na stated as grams 1 Na per 100 g unknown molar mass Na 36.5 g Na 1 mol Na 22.99 g Na mol Na Holt ChemFile: Problem-Solving Workbook 87 Empirical Formulas
Repeat this step for the remaining elements. 2 Amount of Na in mol per divide by the amount of the least-abundant element 4 Empirical formula 3 Whole-number ratio among the elements COMPUTE 36.5 g Na 38.1 g O 25.4 g S Divide the amount of each element by the amount of the least-abundant element, which in this example is S. This can be accomplished by multiplying the amount of each element by the inverse of the amount of the least abundant element. 1.59 mol Na 2.38 mol O 0.792 mol S 1 mol Na 22.99 g Na 1 mol O 16.00 g O 1 mol S 32.07 g S 0.792 mol S 0.792 mol S 0.792 mol S 1.59 mol Na 2.38 mol O 0.792 mol S 2.01 mol Na 1 mol S 3.01 mol O 1 mol S 1.00 mol S 1 mol S From the calculations, the simplest mole ratio is 2 mol Na:3 mol O:1 mol S. The simplest formula is therefore Na 2 O 3 S. Seeing the ratio 3 mol O:1 mol S, you can use your knowledge of chemistry to suggest that this possibly represents a sulfite group, SO 3 and propose the formula Na 2 SO 3. EVALUATE Are the units correct? Yes; units canceled throughout the calculation, so it is reasonable to assume that the resulting ratio is accurate. Is the number of significant figures correct? Yes; ratios were calculated to three significant figures because percentages were given to three significant figures. Holt ChemFile: Problem-Solving Workbook 88 Empirical Formulas
Is the answer reasonable? Yes; the formula, Na 2 SO 3 is plausible, given the mole ratios and considering that the sulfite ion has a 2 charge and the sodium ion has a 1 charge. Practice 1. Determine the empirical formula for compounds that have the following analyses: a. 28.4% copper, 71.6% bromine ans: CuBr 2 b. 39.0% potassium, 12.0% carbon, 1.01% hydrogen, and 47.9% oxygen ans: KHCO 3 c. 77.3% silver, 7.4% phosphorus, 15.3% oxygen ans: Ag 3 PO 4 d. 0.57% hydrogen, 72.1% iodine, 27.3% oxygen ans: HIO 3 Holt ChemFile: Problem-Solving Workbook 89 Empirical Formulas
Sample Problem 2 Determine the empirical formula for an unknown compound composed of 38.4% oxygen, 23.7% carbon, and 1.66% hydrogen. Solution ANALYZE What is given in the problem? What are you asked to find? the percentage composition of the compound the empirical formula for the compound PLAN What steps are needed to calculate the amount in moles of each element per 100.0 g of unknown? State the percentage of the element in grams and multiply by the inverse of the molar mass of the element. What steps are needed to determine the whole-number mole ratio of the elements in the unknown (the simplest formula)? Divide the amount of each element by the amount of the least-abundant element. If necessary, multiply the ratio by a small integer to produce a wholenumber ratio. 1 Mass of K in g per multiply by the inverse of the molar mass of K 2 Amount of K in mol per divide by the amount of the least-abundant element, and multiply by an integer that will produce a wholenumber ratio 4 Empirical formula 3 Whole-number ratio among the elements COMPUTE 38.4 g K 1 mol K 39.10 g K 0.982 mol K Proceed to find the amount in moles per 100.0 g of unknown for the elements carbon, oxygen, and hydrogen, as in Sample Problem 1. When determining the formula of a compound having more than two elements, it is usually advisable to put the data and results in a table. Holt ChemFile: Problem-Solving Workbook 90 Empirical Formulas
Mass per 100.0 g Amount in mol per Element of unknown Molar mass 100.0 g of unknown Potassium 38.4 g K 39.10 g/mol 0.982 mol K Carbon 23.7 g C 12.01 g/mol 1.97 mol C Oxygen 36.3 g O 16.00 g/mol 2.27 mol O Hydrogen 1.66 g H 1.01 g/mol 1.64 mol H Again, as in Sample Problem 1, divide each result by the amount in moles of the least-abundant element, which in this example is K. You should get the following results: Amount in mol of element Amount in mol of element Element per 100.0 g of unknown per mol of potassium Potassium 0.982 mol K 1.00 mol K Carbon 1.97 mol C 2.01 mol C Oxygen 2.27 mol O 2.31 mol O Hydrogen 1.64 mol H 1.67 mol H In contrast to Sample Problem 1, this calculation does not give a simple whole-number ratio among the elements. To solve this problem, multiply by a small integer that will result in a whole-number ratio. You can pick an integer that you think might work, or you can convert the number of moles to an equivalent fractional number. At this point, you should keep in mind that analytical data is never perfect, so change the number of moles to the fraction that is closest to the decimal number. Then, choose the appropriate integer factor to use. In this case, the fractions are in thirds so a factor of 3 will change the fractions into whole numbers. Amount in mol of element per Fraction nearest the Integer Whole-number mole of potassium decimal value factor mole ratio 1.00 mol K 1 mol K 3 3 mol K 2.01 mol C 2 mol C 3 6 mol C 2.31 mol O 2 1/3 mol O 3 7 mol O 1.67 mol H 1 2/3 mol H 3 5 mol H Thus, the simplest formula for the compound is K 3 C 6 H 5 O 7, which happens to be the formula for potassium citrate. Holt ChemFile: Problem-Solving Workbook 91 Empirical Formulas
EVALUATE Is the answer reasonable? Yes; the formula, K 3 C 6 H 5 O 7 is plausible, considering that the potassium ion has a 1 charge and the citrate polyatomic ion has a 3 charge. Practice 1. Determine the simplest formula for compounds that have the following analyses. The data may not be exact. a. 36.2% aluminum and 63.8% sulfur ans: Al 2 S 3 b. 93.5% niobium and 6.50% oxygen ans: Nb 5 O 2 c. 57.6% strontium, 13.8% phosphorus, and 28.6% oxygen ans: Sr 3 P 2 O 8 or Sr 3 (PO 4 ) 2 d. 28.5% iron, 48.6% oxygen, and 22.9% sulfur ans: Fe 2 S 3 O 12 or Fe 2 (SO 4 ) 3 Holt ChemFile: Problem-Solving Workbook 92 Empirical Formulas
Sample Problem 3 A compound is analyzed and found to have the empirical formula CH 2 O. The molar mass of the compound is found to be 153 g/mol. What is the compound s molecular formula? Solution ANALYZE What is given in the problem? What are you asked to find? the empirical formula, and the experimental molar mass the molecular formula of the compound Items Data Empirical formula of unknown CH 2 O Experimental molar mass of unknown 153 g/mol Molar mass of empirical formula 30.03 g/mol Molecular formula of the compound? PLAN What steps are needed to determine the molecular formula of the unknown compound? Multiply the experimental molar mass by the inverse of the molar mass of the empirical formula. The subscripts of the empirical formula are multiplied by the whole-number factor obtained. 4 Empirical formula of unknown multiply the experimental molar mass by the inverse of the molar mass of the empirical formula, and multiply each subscript in the empirical formula by the resulting factor 5 Molecular formula of unknown COMPUTE given 153 g 1 mol unknown 153 g 1 mol unknown 1 molar mass of empirical formula 1 mol CH 2 O 30.03 g 1 mol CH 2 O 30.03 g factor that shows the number of times the empirical formula must be multiplied to get the molecular formula mol CH 2 O 1 mol unknown Allowing for a little experimental error, the molecular formula must be five times the empirical formula. Molecular formula (CH 2 O) 5 C 5 H 10 O 5 5.09 mol CH 2 O 1 mol unknown Holt ChemFile: Problem-Solving Workbook 93 Empirical Formulas
EVALUATE Is the answer reasonable? Yes; the calculated molar mass of C 5 H 10 O 5 is 150.15, which is close to the experimental molar mass of the unknown. Reference books show that there are several different compounds with the formula C 5 H 10 O 5. Practice 1. Determine the molecular formula of each of the following unknown substances: a. empirical formula CH 2, experimental molar mass 28 g/mol ans: C 2 H 4 b. empirical formula B 2 H 5, experimental molar mass 54 g/mol ans: B 4 H 10 c. empirical formula C 2 HCl, experimental molar mass 179 g/mol ans: C 6 H 3 Cl 3 d. empirical formula C 6 H 8 O, experimental molar mass 290 g/mol ans: C 18 H 24 O 3 e. empirical formula C 3 H 2 O, experimental molar mass 216 g/mol ans: C 12 H 8 O 4 Holt ChemFile: Problem-Solving Workbook 94 Empirical Formulas
Additional Problems 1. Determine the empirical formula for compounds that have the following analyses: a. 66.0% barium and 34.0% chlorine b. 80.38% bismuth, 18.46% oxygen, and 1.16% hydrogen c. 12.67% aluminum, 19.73% nitrogen, and 67.60% oxygen d. 35.64% zinc, 26.18% carbon, 34.88% oxygen, and 3.30% hydrogen e. 2.8% hydrogen, 9.8% nitrogen, 20.5% nickel, 44.5% oxygen, and 22.4% sulfur f. 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine 2. Sometimes, instead of percentage composition, you will have the composition of a sample by mass. Use the same method shown in Sample Problem 1, but use the actual mass of the sample instead of assuming a 100 g sample. Determine the empirical formula for compounds that have the following analyses: a. a 0.858 g sample of an unknown substance is composed of 0.537 g of copper and 0.321 g of fluorine b. a 13.07 g sample of an unknown substance is composed of 9.48 g of barium, 1.66 g of carbon, and 1.93 g of nitrogen c. a 0.025 g sample of an unknown substance is composed of 0.0091 g manganese, 0.0106 g oxygen, and 0.0053 g sulfur 3. Determine the empirical formula for compounds that have the following analyses: a. a 0.0082 g sample contains 0.0015 g of nickel and 0.0067 g of iodine b. a 0.470 g sample contains 0.144 g of manganese, 0.074 g of nitrogen, and 0.252 g of oxygen c. a 3.880 g sample contains 0.691 g of magnesium, 1.824 g of sulfur, and 1.365 g of oxygen d. a 46.25 g sample contains 14.77 g of potassium, 9.06 g of oxygen, and 22.42 g of tin 4. Determine the empirical formula for compounds that have the following analyses: a. 60.9% As and 39.1% S b. 76.89% Re and 23.12% O c. 5.04% H, 35.00% N, and 59.96% O d. 24.3% Fe, 33.9% Cr, and 41.8% O e. 54.03% C, 37.81% N, and 8.16% H f. 55.81% C, 3.90% H, 29.43% F, and 10.85% N Holt ChemFile: Problem-Solving Workbook 95 Empirical Formulas
5. Determine the molecular formulas for compounds having the following empirical formulas and molar masses: a. C 2 H 4 S; experimental molar mass 179 b. C 2 H 4 O; experimental molar mass 176 c. C 2 H 3 O 2 ; experimental molar mass 119 d. C 2 H 2 O, experimental molar mass 254 6. Use the experimental molar mass to determine the molecular formula for compounds having the following analyses: a. 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen; experimental molar mass 116.07 b. 54.53% carbon, 9.15% hydrogen, and 36.32% oxygen; experimental molar mass 88 c. 64.27% carbon, 7.19% hydrogen, and 28.54% oxygen; experimental molar mass 168.19 7. A 0.400 g sample of a white powder contains 0.141 g of potassium, 0.115 g of sulfur, and 0.144 g of oxygen. What is the empirical formula for the compound? 8. A 10.64 g sample of a lead compound is analyzed and found to be made up of 9.65 g of lead and 0.99 g of oxygen. Determine the empirical formula for this compound. 9. A 2.65 g sample of a salmon-colored powder contains 0.70 g of chromium, 0.65 g of sulfur, and 1.30 g of oxygen. The molar mass is 392.2. What is the formula of the compound? 10. Ninhydrin is a compound that reacts with amino acids and proteins to produce a dark-colored complex. It is used by forensic chemists and detectives to see fingerprints that might otherwise be invisible. Ninhydrin s composition is 60.68% carbon, 3.40% hydrogen, and 35.92% oxygen. What is the empirical formula for ninhydrin? 11. Histamine is a substance that is released by cells in response to injury, infection, stings, and materials that cause allergic responses, such as pollen. Histamine causes dilation of blood vessels and swelling due to accumulation of fluid in the tissues. People sometimes take antihistamine drugs to counteract the effects of histamine. A sample of histamine having a mass of 385 mg is composed of 208 mg of carbon, 31 mg of hydrogen, and 146 mg of nitrogen. The molar mass of histamine is 111 g/mol. What is the molecular formula for histamine? 12. You analyze two substances in the laboratory and discover that each has the empirical formula CH 2 O. You can easily see that they are different substances because one is a liquid with a sharp, biting odor and the other is an odorless, crystalline solid. How can you account for the fact that both have the same empirical formula? Holt ChemFile: Problem-Solving Workbook 96 Empirical Formulas