Determination of Empirical Formula

DETERMINATION OF EMPIRICAL FORMULA The empirical formula of a compound represents the simplest whole number ratio of the elements in that compound. A real-life example of an empirical formula is glucose (C₆H₁₂O₆), a sugar found in many foods. The molecular formula for glucose is C₆H₁₂O₆, which shows that for every 6 carbon atoms, there are 12 hydrogen atoms and 6 oxygen atoms. However, the empirical formula simplifies this ratio to CH₂O, which shows that for every carbon atom, there are two hydrogen atoms and one oxygen atom in the simplest ratio. So, even though glucose has the molecular formula C₆H₁₂O₆, its empirical formula is CH₂O. EXAMPLE # 1 A sample of a compound contains 5.30 g of nitrogen (N) and 14.70 g of oxygen (O). What is the empirical formula of the compound? STEP 1: OBTAIN THE MASS OF EACH ELEMENT Mass of nitrogen (N) = 5.30 g Mass of oxygen (O) = 14.70 g STEP 2: CONVERT THE MASS OF EACH ELEMENT TO MOLES • moles of nitrogen: 5.30 g/ 14.01 g/mol / = 0.378mol of N • moles of oxygen: 14.70g/ 16.00 g/mol / Data from periodic table = 0.919mol of O Notice that grams (g) were canceled because they appear in both the numerator and the denominator. STEP 3: DETERMINE THE SIMPLEST MOLE RATIO Now, divide the number of moles of each element by the smallest number of moles (which is 0.378 mol for nitrogen): • Ratio of nitrogen: 0.378 0.378 = 1 • Ratio of Oxygen: 0.919 0.378 = 2.43 EXAMPLE # 2 A compound is found to contain 40.0% carbon (C), 6.7% hydrogen (H), and 53.3% oxygen (O). Determine the empirical formula of the compound. STEP 1: ASSUME A 100 G SAMPLE If the total mass of the compound is 100 g, the percentage values can be treated as grams: STEP 4: ROUND TO THE NEAREST WHOLE NUMBER (OR ADJUST) • Mass of carbon = 40.0 g Since 2.43 is close to 2.5, we multiply both ratios by 2 to get whole numbers: • Mass of oxygen = 53.3 g • Mass of hydrogen = 6.7 g • Ratio of nitrogen = 1 × 2 = 2 STEP 2: CONVERT THE MASS OF EACH ELEMENT TO MOLES • Ratio of oxygen = 2.43 × 2 = 4.86 or 5 Use the molar masses of each element: STEP 5: WRITE THE EMPIRICAL FORMULA The simplest mole ratio is: Nitrogen (N) = 2 Oxygen (O) = 5 Thus, the empirical formula of the compound is N₂O₅ • Carbon (C): 40.0 g/ = 3.33 mol 12.01 g/mol / • Hydrogen (H): 6.7 g/ 6.65mol = 1.008 /g/mol • Oxygen (O): 53.3 g/ = 3.33 mol 16.00 g/mol / STEP 3: DETERMINE THE SIMPLEST MOLE RATIO Divide each number of moles by the smallest value (3.33 mol): • carbon (C): 33.3 = 1 33.3 • hydrogen (H): 6.65 33.3 = 2 • oxygen (O): 33.3 33.3 = 1 STEP 4: WRITE THE EMPIRICAL FORMULA The simplest mole ratio is: • Carbon (C) = 1 • Hydrogen (H) = 2 • Oxygen (O) = 1 Thus, the empirical formula of the compound is CH₂O . PRACTICE 1. A compound is found to contain 92.3% carbon (C) and 7.7% hydrogen (H) by mass. Determine its empirical formula. 2. A compound contains 38.7% chlorine (Cl), 1.6% hydrogen (H), and 59.7% oxygen (O) by mass. Determine the empirical formula of the compound. 3. A compound is analyzed and found to contain 28.03% magnesium (Mg), 21.60% silicon (Si), 1.16% hydrogen (H), and 49.21% oxygen (O) by mass. Determine the empirical formula of the compound. 4. A sample of an organic compound contains 54.5% carbon, 9.1% hydrogen, and 36.4% oxygen. Assume a 100 g sample of the compound and determine its empirical formula. Note: In the chemical formula CH₂O, we don't write the number "1" next to carbon (C) or oxygen (O) because it's implied that there is only one atom of each. However, when there are two or more atoms, we must include the subscript to indicate the quantity. For example, if there were two carbons, the formula would be written as C₂H₂O.