Determining Chemical Formulas

Determining Chemical Formulas Scientists estimate that new chemical compounds are being discovered at a rate of two every minute, which is about one million new compounds every year! Many compounds, such as the recently invented cancer drug temozolomide, are tested for possible medical benefits. Before a compound can be approved for use as a drug ianada, however, the Food and Drug Act requires the drug company to conduct extensive tests to ensure that the drug is safe and effective. As part of the approval process, the drug company must submit the molecular formula of the compound to the appropriate department of the Ministry of Health. How is the molecular formula of a new compound determined? Analytical laboratories have sophisticated instruments that can determine almost any characteristic of an element or compound. All of these instruments operate on the basis of fundamental chemical principles, some of which were discovered hundreds of years ago. In 1799, Joseph Proust (Figure 1) proposed the law of constant composition. This law states that a compound contains elements in certain fixed proportions (ratios) and in no other combinations, regardless of how the compound is prepared or where it is found in nature. Therefore, one molecule of methane (natural gas), CH 4(g), always contains one atom of carbon and four atoms of hydrogen. One molecule of water always contains two atoms of hydrogen and one atom of oxygen. Adding or removing atoms from a compound changes the compound to a completely different substance. For example, adding an additional oxygen atom to a water molecule changes the compound to hydrogen peroxide, H 2 O 2(l). Whereas water is safe to consume, hydrogen peroxide is not. Hydrogen peroxide is a powerful bleaching agent and a potent disinfectant. Since chemists cannot see individual molecules, they cannot directly determine the types and numbers of atoms in molecules. In the nineteenth century, chemists used several complex and time-consuming laboratory procedures to determine the molecular formula of a new compound. Today, most of the procedures are automated and efficiently performed by computerized equipment. Two key instruments that are used to determine the formula of a compound are the mass spectrometer and the combustion analyzer. The Mass Spectrometer A mass spectrometer (Figure 2) is used to measure the molar mass of a compound. A small sample of the compound is vaporized (turned into gas) and bombarded by a beam of electrons. The electrons cause the molecules to become electrically charged (ionized) and possibly broken up into a number of smaller fragments (Figure 3, on the next page). The charged fragments are accelerated by an electric field and deflected by a magnetic field. The amount of deflection depends on the mass and charge of the fragments. The molar mass of the original molecule can be determined from the molar mass of the 2.3 Figure 1 Joseph Proust (1754 1826) law of constant composition A compound contains elements in certain fixed proportions (ratios), regardless of how the compound is prepared or where it is found in nature. mass spectrometer laboratory instrument that is used to measure the molar mass of a compound Figure 2 A mass spectrometer is used to identify the drugs that an athlete may have used before a race. NEL Quantities ihemistry 107

Intensity + C 2 H 3 (27) Mass Spectrograph of Lighter Fluid + C 3 H 7 (43) + C 4 H 10 (58) 0 10203040 5060708090100 Molar mass of fragments (g/mol) Figure 4 A mass spectrograph of lighter fluid is used to determine the molar mass of the chemical in the fluid. Expert analysis of the mass spectrograph provides the basis for the molecular model shown here. combustion analyzer laboratory instrument that is used to determine the percentages of carbon, hydrogen, oxygen, and nitrogen in a compound accelerating plates gas inlet (isobutane, C 4 H 10 ) + electron gun (ionizing gun) + + + + + + + slit magnet beam of C 3 H 7 + ions beam of C 4 H 10+ ions slit detector Figure 3 A mass spectrometer is used to determine the masses of ionized fragments by measuring the amount of deflection in the path of the fragments as they pass through a magnetic field. largest fragment. Figure 4 shows a mass spectrograph (a printout from a mass spectrometer) of butane (lighter fluid), C 4 H 10(l). The Combustion Analyzer A combustion analyzer is used to determine the percentages of carbon, hydrogen, oxygen, and possibly nitrogen in a compound. An accurately known mass of a pure compound is burned in a stream of pure oxygen gas at a temperature of approximately 980 o C. Oxygen is used because it is required for the combustion of carbon-based compounds. All the carbon in the sample is incorporated into carbon dioxide gas. All the hydrogen is incorporated into water vapour (steam). For example, if the sample contains methane, the following combustion reaction occurs: CH 4(g) 2O 2(g) CO 2(g) 2 H 2 O (g) Notice that every hydrogen atom (H) in the original sample ends up in an H 2 O (g) molecule; every carbon atom (C) in the original sample ends up in a CO 2(g) molecule; all the atoms, and therefore all the mass of the original sample, are accounted for in the products. As the combustion products (carbon dioxide and water) pass through the system (Figure 5), the water vapour is absorbed by a water trap. The water trap is a tube that is packed with a compound, such as calcium chloride, CaCl 2(s), that absorbs water. To determine the mass of water that is produced in the combustion reaction, the mass of the water trap before it has absorbed water vapour is subtracted from the mass of the water trap after it has absorbed water vapour. 108 Unit 2 NEL

Section 2.3 O 2 sample furnace H 2 O absorber CO 2 absorber N 2 absorber Figure 5 When a substance is burned in a combustion analyzer, oxides are produced. The oxides are captured by absorbers in chemical traps. The initial and final masses of each trap indicate the masses of the oxides produced. These masses are then used to calculate the percentage composition of the substance burned. The remainder of the combustion gas now contains only carbon dioxide gas. The carbon dioxide travels through a carbon dioxide trap, which absorbs all the gas. The carbon dioxide trap is a tube that is packed with a compound that reacts completely with carbon dioxide, such as solid sodium hydroxide, NaOH (s). To determine the mass of carbon dioxide produced in the combustion reaction, the mass of the carbon dioxide trap before it has absorbed carbon dioxide is subtracted from the mass of the trap after it has absorbed carbon dioxide. The masses of hydrogen and carbon in the sample compound are calculated as follows: m H m H2 O O2 The masses of hydrogen and carbon are then used to calculate the percentage, by mass, of each element in the sample compound: m H msample 1 mol H 2 O 18.02 g H2 O 1 mol CO 2 44.01 g CO2 % H 100% 2.02 g H 1 mol H2 O 12.01 g C 1 mol CO2 m C msample % C 100% If the compound is a hydrocarbon (containing only hydrogen and carbon), the sum of % H and % C will be 100%. If the compound also contains oxygen, the sum of % H and % C will be less than 100%, with the difference equal to the percentage of oxygen in the compound. The percentage of oxygen in the compound is determined by subtraction because the oxygen atoms in water and carbon dioxide do not all come from the sample compound. Some oxygen atoms come from the oxygen gas that is used in the combustion process. The percentage, by mass, of each element in a compound is called the percentage composition of the compound. percentage composition the percentage, by mass, of each element in a compound NEL Quantities ihemistry 109

TRY THIS activity What Makes Popcorn Pop? In each kernel of popping corn, there is a small drop of water in a circle of soft starch. When heated, the water expands and builds up pressure against the hard outer surface, eventually exploding and turning the kernel inside out. Materials: popping corn, hot-air popcorn popper, balance 1. Measure the mass of some unpopped popping corn. 2. Pop the popping corn. 3. Allow the popcorn to cool, and measure the mass again. (Assume that any difference in masses is caused by loss of water from the kernels.) 4. Calculate the percentage of water in the sample of popcorn. 5. Repeat the activity with kernels of popping corn that have been cut in half, either lengthwise or crosswise. 6. Record the percentage of popped kernels from cut and uncut popping corn. (a) Do your results confirm the given reason why popcorn pops? Explain. Modern combustion analyzers are equipped with computers that perform these calculations automatically. Table 1 illustrates a typical printout from a combustion analyzer for a compound that contains carbon, hydrogen, and oxygen. Ten samples of the compound were run through the combustion analyzer. The average percentage composition was calculated for each element in the compound. Note that the total of the average percentage composition values equals 100%. Table 1 Combustion Analyzer Data Sample % C% H % O 1 38.70 9.71 51.60 2 38.69 9.71 51.58 3 38.71 9.71 51.58 4 38.71 9.73 51.59 5 38.73 9.72 51.58 6 38.72 9.71 51.57 7 38.71 9.71 51.58 8 38.71 9.70 51.58 9 38.71 9.70 51.56 10 38.71 9.70 51.58 Average % composition 38.71 9.71 51.58 110 Unit 2 NEL

Section 2.3 Determining the Empirical Formula Percentage composition values can be used to calculate the empirical formula of a compound. The empirical formula tells you the lowest ratio of atoms in a compound, but it does not necessarily tell you the exact number of each type of atom. The following Sample Problem and Example show how an empirical formula can be calculated using percentage composition data. empirical formula a formula that gives the lowest ratio of atoms in a compound SAMPLE problem 1 Determining the Empirical Formula of a Compound (a) Use the average percentage composition values given in Table 1 to determine the empirical formula of the compound. The empirical formula of a compound is based on its percentage composition. From Table 1, the percentage composition of the compound is 38.71% carbon, 9.71% hydrogen, and 51.58% oxygen. Step 1: List Given Values C 38.71% H 9.71% O 51.58% Step 2: Calculate Mass (m) of Each Element in 100-g Sample 38.71 100 100 g 38.71 g m H 9.71 100 100 g m H 9.71 g 51.58 100 100 g 51.58 g Step 3: Convert Mass (m) into Amount (n) 38.71 g C 3.22 mol C 1 mol H n H 9.71 g H 1.01 g H n H 9.61 mol H 51.58 g O 3.22 mol O 1 mol C 12.01 g C 1 mol O 16.00 g O NEL Quantities ihemistry 111

Step 4: State Amount Ratio : 3.22 : 9.61 : 3.22 Step 5: Calculate Lowest Whole-Number Amount Ratio Divide all the amounts by the smallest amount. Here the smallest amount is 3.22 mol O. 3.22 9.61 3.22 : : : 3.22 3.22 3.22 : 1.00 : 2.98 : 1.00 Notice that the value for hydrogen is 2.98. Round 2.98 to the closest whole number, which is 3. : 1 : 3 : 1 Therefore, the empirical formula of the compound is CH 3 O. (b) The percentage composition of a compound is 69.9% iron and 30.1% oxygen. What is the empirical formula of the compound? Step 1: List Given Values Fe 69.9% O 30.1% Step 2: Calculate Mass (m) of Each Element in 100-g Sample m Fe 69.9 100 100 g m Fe 69.9 g 30.1 100 g 100 30.1 g Step 3: Convert Mass (m) into Amount (n) 1 mol Fe n Fe 69.9 g Fe 55.85 g Fe n Fe 1.25 mol Fe 1 mol O 30.1 g O 16.00 g O 1.88 mol O Step 4: State Amount Ratio n Fe : 1.25 : 1.88 Step 5: Calculate Lowest Whole-Number Amount Ratio Divide all the amounts by the smallest amount. Here the smallest amount is 1.25 mol Fe. 1.25 1.88 n Fe : : 1.25 1.25 n Fe : 1.00 : 1.50 112 Unit 2 NEL

Section 2.3 Notice that this ratio has one number that is not a whole number. To obtain a whole-number ratio, multiply both numbers by 2. n Fe : 2(1.00) : 2(1.50) 2.00 : 3.00 n Fe : 2 : 3 Therefore, the empirical formula of the compound is Fe 2 O 3. Example The percentage composition of a compound is 21.6% sodium, 33.3% chlorine, and 45.1% oxygen. What is the empirical formula of the compound? Solution Na 21.6% Cl 33.3% O 45.1% 21.6 m Na 100 g 100 m Na 21.6 g l 33.3 100 100 g l 33.3 g 45.1 100 g 100 45.1 g 1 mol Na n Na 21.6 g Na 22.99 g Na n Na 0.940 mol Na 1 mol Cl l 33.3 g Cl 35.45 g Cl l 0.939 mol Cl 1 mol O 45.1 g O 16.00 g O 2.82 mol O n Na : l : 0.940 : 0.939 : 2.82 0.940 0.939 2.82 : : 0.939 0.939 0.939 1.00 : 1.00 : 3:00 n Na : l : 1 : 1 : 3 The empirical formula of this compound is NaClO 3. NEL Quantities ihemistry 113

Answers 1. HCO 2 2. Al 2 S 3 Practice Understanding Concepts 1. Calculate the empirical formula of a compound that, on analysis, is found to contain 2.2% hydrogen, 26.7% carbon, and 71.1% oxygen. 2. The percentage composition of a compound is 35.9% aluminum and 64.1% sulfur. What is the empirical formula of the compound? SUMMARY Steps Used to Determine an Empirical Formula Step 1: List Given Values Step 2: Calculate Mass (m) of Each Element in 100-g Sample Step 3: Convert Mass (m) into Amount (n) Step 4: State Amount Ratio Step 5: Calculate Lowest Whole-Number Amount Ratio molecular formula a formula that indicates the actual numbers of atoms in one molecule of a compound Determining the Molecular Formula The molecular formula of a compound tells you the exact number of atoms in one molecule of the compound. The molecular formula may be equal to the empirical formula, or it may be a multiple of this formula. The molecular formula of water, H 2 O (l), cannot be reduced any further, so it is also the empirical formula. The molecular formula of hydrogen peroxide is H 2 O 2(l), which can be reduced to its empirical formula of HO. To determine the molecular formula of a compound, you need to know the empirical formula and the molar mass of the compound. As explained earlier, the molar mass can be obtained using a mass spectrometer. SAMPLE problem 2 Calculating the Molecular Formula of a Compound (a) The empirical formula of a compound is CH 3 O, and its molar mass is 93.12 g/mol (as determined by a mass spectrometer). What is the molecular formula of the compound? Step 1: List Given Values empirical formula of compound CH 3 O 93.12 g/mol Step 2: Determine Molar Mass of Empirical Formula M CH3 O 1(12.01 g/mol) 3(1.01 mol/l) 1(16.00 mol/l) 31.04 g/mol M CH3 O 114 Unit 2 NEL

Section 2.3 Step 3: Determine Ratio of Molar Mass of Compound to Molar Mass of Empirical Formula The ratio gives the factor by which the molar mass of the Mempirical formula compound is greater than the molar mass of the empirical formula. MCH3 O MCH3 O 3 Step 4: Calculate Molecular Formula Use the ratio determined in step 3 to calculate the molecular formula. The molar mass of the compound is three times greater than the molar mass of its empirical formula. Multiply the subscripts of the empirical formula by 3 to obtain the subscripts of the molecular formula. molecular formula 3(empirical formula) 3(CH 3 O) molecular formula C 3 H 9 O 3 The molecular formula of the compound is C 3 H 9 O 3. (b) The percentage composition of a compound, as determined by a combustion analyzer, is 40.03% carbon, 6.67% hydrogen, and 53.30% oxygen. Using a mass spectrometer, the molar mass of the compound is found to be 180.18 g/mol. What is the molecular formula of the compound? Notice that the empirical formula is not given. You can use the percentage composition values, however, to determine the empirical formula of the compound. Then you can use the empirical formula and the molar mass of the compound to determine the molecular formula. The solution to this problem includes the calculations for determining the empirical formula of a compound (Part 1) and the calculations for determining the molecular formula (Part 2). Part 1: Determine Empirical Formula Step 1: List Given Values C 40.03% H 6.67% O 53.30% 180.18 g/mol 93.12 g/mol 31.04 g/mol Step 2: Calculate Mass (m) of Each Element in 100-g Sample 40.03 100 100 g 40.03 g NEL Quantities ihemistry 115

m H 6.67 100 100 g m H 6.67 g 53.30 100 100 g 53.30 g Step 3: Convert Mass (m) into Amount (n) 40.03 g C 3.33 mol C 1 mol H n H 6.67 g H 1.01 g H n H 6.60 mol H 53.30 g O 3.33 mol O Step 4: State Amount Ratio : 3.33 : 6.60 : 3.33 Step 5: Calculate Lowest Whole-Number Amount Ratio To obtain the lowest whole-number ratio of elements, divide all the amounts by the smallest amount. Here the smallest amount is 3.33 mol O. 3.33 6.60 3.33 : : : 3.33 3.33 3.33 : 1.00 : 1.98 : 1.00 Notice that the value for hydrogen is 1.98. Round 1.98 to the closest whole number, which is 2. : 1 : 2 : 1 Therefore, the empirical formula of the compound is CH 2 O. Part 2: Determine Molecular Formula Step 6: Determine Molar Mass of Empirical Formula M CH2 O 1(12.01 g/mol) 2(1.01 g/mol) 1(16.00 g/mol) 30.03 g/mol M CH2 O Step 7: Determine Ratio of Molar Mass of Compound to Molar Mass of Empirical Formula MCH2 O MCH2 O 6 1 mol C 12.01 g C 1 mol O 16.00 g O 180.18 g/mol 30.03 g/mol 116 Unit 2 NEL

Section 2.3 Step 8: Calculate Molecular Formula The molar mass of the compound is six times greater than the molar mass of the empirical formula. Therefore, multiply the subscripts of the empirical formula by 6 to obtain the subscripts of the molecular formula. molecular formula 6(empirical formula) 6(CH 2 O) molecular formula C 6 H 12 O 6 The molecular formula of the compound is C 6 H 12 O 6. Example A combustion analyzer determines the percentage composition of a compound to be 32.0% carbon, 6.70% hydrogen, 42.6% oxygen, and 18.7% nitrogen. Using a mass spectrometer, the molar mass of the compound is found to be 75.08 g/mol. What is the molecular formula of the compound? Solution C 32.0% H 6.70% O 42.6% N 18.7% 75.08 g/mol 32.0 42.6 100 g 100 g 100 100 32.0 g 42.6 g 6.70 18.7 m H 100 g m N 100 g 100 100 m H 6.70 g 1 mol C 32.0 g C 12.01 g C 2.66 mol C m N 18.7 g 42.6 g O 2.66 mol O 1 mol O 16.00 g O 1 mol H n H 6.70 g H n N 18.7 g N 1.01 g H n H 6.63 mol H n N 1.33 mol N : 2.66 : 6.63 : 2.66 : 1.33 2.66 6.63 2.66 1.33 : : : 1.33 1.33 1.33 1.33 2.00 : 4.98 : 2.00 : 1.00 : 2 : 5 : 2 : 1 The empirical formula of the compound is C 2 H 5 O 2 N. 1 mol N 14.01 g N NEL Quantities ihemistry 117

M C2 H 5 O 2 N 2(12.01 g/mol) 5(1.01 g/mol) 2(16.00 g/mol) 1(14.01 g/mol) M C2 H 5 O 2 N 75.00 g/mol C 2 H 5 O 2 N MC2 H 5 O 2 N MC2 H 5 O 2 N 1 75.00 g/mol 75.08 g/mol molecular formula 1(empirical formula) 1(C 2 H 5 O 2 N) molecular formula C 2 H 5 O 2 N The molecular formula of the compound is C 2 H 5 O 2 N. Answers 3. C 8 H 12 O 2 4. C 45 H 85 O 5 5. K 2 Cr 2 O 7 6. C 10 H 14 N 2 Practice Understanding Concepts 3. A combustion analyzer determines the percentage composition of a compound to be 68.54% carbon, 8.63% hydrogen, and 22.83% oxygen. A mass spectrometer determines its molar mass to be 140.20 g/mol. What is the molecular formula of the compound? 4. A fat that is used to make soap contains 76.5% carbon, 12.2% hydrogen, and 11.3% oxygen by mass. Determine the molecular formula of the fat if its molar mass is 706.3 g/mol. 5. A substance contains 26.65% potassium, 35.33% chromium, and 38.02% oxygen. Its molar mass is 294.20 g/mol. Determine the molecular formula of the compound. 6. The percentage composition of nicotine is 74.0% carbon, 8.7% hydrogen, and 17.3% nitrogen. Its molar mass is 162.26 g/mol. What is the molecular formula of nicotine? Calculating Percentage Composition by Mass When the chemical formula of a compound is known, its percentage composition may be calculated using the atomic masses of its elements and the molecular mass (or formula unit mass) of the compound. SAMPLE problem 3 Calculating the Percentage Composition of a Compound Calculate the percentage composition of carbon dioxide, CO 2(g). Step 1: Calculate Total Mass of Each Element iompound 12.01 u 2(16.00 u) 32.00 u 118 Unit 2 NEL

Section 2.3 Step 2: Calculate Molecular Mass (or Formula Unit Mass) of Compound O2 12.01 u 32.00 u O2 44.01 u Step 3: Calculate Percentage Composition by Mass of Compound m C mco2 m O mco2 % C 100% % O 100% 12.01 u 32.00 u 100% 100% 44.01 u 44.01 u % C 27.29% % O 72.71% The percentage composition by mass of carbon dioxide is 27.29% carbon and 72.71% oxygen. Example Calculate the percentage composition by mass of potassium sulfate, K 2 SO 4(s). Solution m K 2(39.10 u) m S 32.06 u 4(16.00 u) m K 78.20 u = 64.00 u m K2 SO 4 2(m K ) (m S ) 4( ) 2(39.10 u) (32.06 u) 4(16.00 u) 78.20 u 32.06 u 64.00 u m K2 SO 4 174.26 u 78.20 u % K 100% 174.26 u 32.06 u % S 100% 174.26 u % K 44.87% % S 18.40% 64.00 u % O 100% 174.26 u % O 36.73% The percentage composition by mass of potassium sulfate is 44.87% potassium, 18.40% sulfur, and 36.73% oxygen. Practice 7. Calculate the percentage composition by mass of each of the following compounds: (a) C 6 H 8 O 6(s) (b) Al 2 O 3(s) (c) Zn(NO 3 ) 2(s) Answers 7. (a) 40.91% C, 4.59% H, 54.50% O (b) 52.92% Al, 47.08% O (c) 34.52% Zn, 14.79% N, 50.69% O NEL Quantities ihemistry 119

SUMMARY Steps Used to Determine a Molecular Formula Part 1: Determine Empirical Formula Step 1: List Given Values Step 2: Calculate Mass (m) of Each Element in 100-g Sample Step 3: Convert Mass (m) into Amount (n) Step 4: State Amount Ratio Step 5: Calculate Lowest Whole-Number Amount Ratio Part 2: Determine Molecular Formula Step 6: Determine Molar Mass of Empirical Formula Step 7: Determine Ratio of Molar Mass of Compound to Molar Mass of Empirical Formula Step 8: Calculate Molecular Formula SUMMARY Steps Used to Calculate Percentage Composition by Mass Step 1: Calculate Total Mass of Each Element iompound Step 2: Calculate Molecular Mass (or Formula Unit Mass) of Compound Step 3: Calculate Percentage Composition by Mass of Compound Section 2.3 Questions Understanding Concepts 1. State the law of constant composition, in your own words. 2. (a) What is a mass spectrometer used for? Briefly describe how it works. (b) What is a combustion analyser used for? Briefly describe how it works. 3. (a) What information do you need to determine the empirical formula of a compound? (b) What information, in addition to the empirical formula, do you need to determine the molecular formula of a compound? 4. Write the empirical formula for each of the following molecular formulas: (a) C 2 H 4 O 2 (b) NH 3 (c) C 6 H 6 5. What is the empirical formula of a compound that contains 26.6% potassium, 35.4% chromium, and 38.1% oxygen? 6. The percentage composition values of two antibiotics are given. Determine the empirical formula of each antibiotic. (a) chloromycetin: 40.87% carbon, 3.72% hydrogen, 8.67% nitrogen, 24.77% oxygen. The rest is chlorine. (b) sulfanilamide: 41.86% carbon, 4.65% hydrogen, 16.28% nitrogen, 18.60% oxygen. The rest is sulfur. 7. Calculate the percentage composition by mass of each of the following compounds: (a) H 2 O (l) (b) Ca(OH) 2(s) Applying Inquiry Skills 8. The analysis of a compound shows that it contains 21.9% sodium, 45.7% carbon, 1.9% hydrogen, and 30.5% oxygen. A mass spectrometer determines its molar mass to be 210 g/mol. What is the molecular formula of the compound? 9. The analysis of a newly discovered compound produced the following data: combustion analyzer: 49.38% carbon, 3.55% hydrogen, 9.40% oxygen, 37.67% sulfur mass spectrometer: molar mass 170.2 g/mol Determine the molecular formula of the compound. 120 Unit 2 NEL