2. Empirical & Molecular Formulas Law of Constant Composition a compound contains elements in a certain fixed proportions (ratios), regardless of how the compound is prepared or where it is found in nature. If you have one molecule of methane gas, you will always have 1 carbon atoms and 4 hydrogen atoms.

3. 1. Mass Spectrometer This machine measure the molar mass of a compound. A small sample of the compound is vaporized and hit with a beam of electrons The fragments are put through an electric field and the amount of deflection determines molar mass

4. 2. Combustion Analyzer Is an instrument that can determine the percentages of carbon, hydrogen, oxygen & nitrogen in a compounds A combustion reaction occurs and the individual parts of the products are captured and measured Using mass of products and individual atom mass, one can determine the percent composition

5. Empirical Formula Empirical Formula is the formula that gives the lowest ratio of atoms in a compound. It does not necessarily tell you the exact number of each type of atom. Example 1: The percent composition of a compound is 69.9 % iron and 30.1% oxygen. What is the empirical formula of a compound?

6. Step 1: List the given values Fe=69.9% and O = 30.1% Step 2: Calculate the mass (m) of each element in a 100g sample. m Fe = 69.9 x 100g = 69.9g 100 m O = 30.1 x 100g = 30.1g 100

7. Step 3: Convert Mass (m) into moles (n) n Fe = m/M = 69.9g/55.86g/mol = 1.25 mol Fe n O = m/M = 30.1g/16.00g/mol = 1.88 mol O Step 4: State the Amount Ratio n Fe : n O 1.25mol : 1.88 mol Step 5: Calculate lowest whole number ratio 1.25mol : 1.88 mol 1.25mol 1.25 mol 1 : 1.5 2 : 3 Empirical Formula is Fe 2 O 3 When you don’t get a whole number, multiply entire ratio by 2, 3, 4 etc. until you get a whole number

8. Example 2: The percent composition of a compound is 21.6% sodium, 33.3% chlorine, and 45.1% oxygen. What is the empirical formula of the compound?

9. Step 1: List the given values Cl=33.3%, Na = 21.6% and O = 45.1% Step 2: Calculate the mass (m) of each element in a 100g sample. m Cl = 33.3 x 100g = 33.3g Cl 100 m Na = 21.6 x 100g = 21.6g Na 100 m O = 45.1 x 100g = 45.1g O 100

10. Step 3: Convert Mass (m) into moles (n) n Cl = m/M = 33.3g/35.5g/mol = 0.94 mol Cl n Na = m/M = 21.6g/23.0g/mol = 0.94 mol Na n O = m/M = 45.1g/16.00g/mol = 2.82 mol O Step 4: State the Amount Ratio n Fe : n Na : n O 0.94mol : 0.94mol : 2.82 mol Step 5: Calculate lowest whole number ratio 0.94mol : 0.94mol : 2.82 mol 0.94mol : 0.94mol : 0.94 mol 1 : 1: 3 Empirical Formula is NaClO 3

11. Molecular Formula Molecular Formula of a compound tells you exact number of atoms in one molecule of a compound. This formula may be equal to the empirical formula or may be a multiple of this formula. To determine, you need: –The empirical formula –The molar mass of the compound

12. Molecular Formula - shows the actual number of atoms Example: C 6 H 12 O 6 Empirical Formula - shows the ratio between atoms Example: CH 2 O

13. The empirical formula of a compound is CH 3 O and its molar mass is 93.12g/mol. What is the molecular formula? Step 1: List given values Empirical Formula=CH 3 O M compound = 93.12 g/mol Step 2: Determine the molar mass for the empirical formula, CH 3 O. M Empirical = 12.01g/mol + 3(1.01g/mol) + 16.00g/mol = 31.04 g/mol

14. Step 3. Divide the molar mass by the empirical formula molar mass. = = 3 Step 4. Calculate Molecular Formula by multiplying this number by the empirical formula. Molecular formula = x (empirical formula) 3 x CH 3 O Therefore, the molecular formula is C 3 H 9 O 3 Molecular formula molar mass Empirical formula molar mass 93.12 g/mol 31.04 g/mol

15. Example 2: The percent composition of a compound is determined by a combustion and analyzer is a 40.03% carbon, 6.67% hydrogen, & 53.30% oxygen. The molar mass is 180.18g/mol. What is the molecular formula Step 1: List given values C= 40.03%, O=53.30%, H=6.67% M compound = 180.18 g/mol Step 2: Calculate the mass of each element in a 100g sample m C =40.03g m O =53.30g m H =6.67g

16. Step 3: Convert Mass (m) into moles (n) n C = m/M = 40.03g/12.01g/mol = 3.33 mol C n H = m/M = 6.67g/1.01g/mol = 6.60 mol H n O = m/M = 53.30g/16.00g/mol = 3.33 mol O Step 4: State the Amount Ratio n C : n H : n O 3.33mol : 6.60mol : 3.33 mol Step 5: Calculate lowest whole number ratio 3.33mol : 6.60mol : 3.33 mol 3.33mol : 3.33mol : 3.33 mol 1 : 2: 1 Empirical Formula is CH 2 O

17. Step 6: Determine the molar mass for the empirical formula M Empirical = 12.01g/mol + 2(1.01g/mol) + 16.00g/mol = 30.03 g/mol Step 7. Divide the molar mass by the empirical formula molar mass. = = 6 Step 8. Calculate Molecular Formula by multiplying this number by the empirical formula. Molecular formula = x (empirical formula) 6 x (CH 2 O) Therefore, the molecular formula is C 6 H 12 O 6 Molar mass Empirical formula molar mass 180.18 g/mol 30.03 g/mol

18. Example 3: The percent composition of a compound is determined by a combustion and analyzer is a 32.0% carbon, 6.70% hydrogen, 42.6% oxygen & 18.7% nitrogen. The molar mass is 75.08g/mol. What is the molecular formula? Calculate the mass of each element in a 100g sample m C =32.0g m O =42.6g m H =6.70g m N =18.7g Convert Mass (m) into moles (n) n C = m/M = 32.0g/12.01g/mol = 2.66 mol C n H = m/M = 6.70g/1.01g/mol = 6.65 mol H n O = m/M = 42.6g/16.00g/mol = 2.66 mol O n N = m/M = 18.7g/14.01g/mol = 1.33 mol N

19. State the Amount Ratio n C : n H : n O : n N 2.66mol : 6.65mol : 2.6 mol:1.33mol Step 5: Calculate lowest whole number ratio 2.66mol : 6.65mol : 2.6 mol:1.33mol 1.33mol : 1.33mol : 1.33 mol:1.33mol 2 : 5: 2: 1 Empirical Formula is C 2 H 5 O 2 N

20. Determine the molar mass for the empirical formula M Empirical = 75.08g Divide the molar mass by the empirical formula molar mass. = = 1 Calculate Molecular Formula by multiplying this number by the empirical formula. Molecular formula = x (empirical formula) 1 x ( C 2 H 5 O 2 N ) Therefore, the molecular formula is C 2 H 5 O 2 N Molar mass Empirical formula molar mass 75.08 g/mol