1. Determining Empirical Formula from Combustion Analysis 1

2. Empirical Formulas The simplest whole number ratio of elements in a compound The simplest whole number ratio of elements in a compound 2 Ethyne (C 2 H 2 ) is a gas used in welder’s torches. Styrene (C 8 H 8 ) is used in making polystyrene. Both have the same empirical formula (CH) but different molecular formulas.

3. Molecular Formulas The molecular formula of a compound is: The actual number of atoms of each element in a substance. Three compounds that have the same empirical formula (CH 2 O): Formaldehyde (CH 2 O) acetic acid (C 2 H 4 O 2 ) glucose (C 6 H 12 O 6 )

4. Combustion Analysis 4

5. Determining the Empirical Formula From Combustion Analysis Section 3.5 (p. 97) Another test to determine empirical formulas is the Combustion Analysis. The sample is combusted and the quantity of CO 2 and H 2 O produced is measured.

6. Combustion Analysis 6 Example 1: Combustion of 0.250 g of a hydrocarbon produces 0.686 g CO 2 and 0.562 g H 2 O. What is the empirical formula?

7. Combustion Analysis 7 Example 2: Combustion of 0.255 g isopropyl alcohol produces 0.561 g CO 2 and 0.306 g H 2 O. What is the empirical formula?

8. Combustion Analysis Combustion: when fuel burns, water and carbon dioxide are the products. Sometimes is a hydrocarbon fuel (contains carbon and hydrogen) while other times there are other components to the fuel source. Combustion Analysis: the technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion. 8 Ex: Methanol combusts in the presence of oxygen 2CH 3 OH + 3O 2  2 CO 2 + 4 H 2 0 Ex: Methane combusts in the presence of oxygen CH 4 + 2O 2  CO 2 + 2 H 2 0

9. 9 Combustion Analysis Combustion Analysis: the technique of finding the mass composition of an unknown sample (X) by examining the products of its combustion. X + O 2 → CO 2 + H 2 O 0.250 g of compound X produces: 0.686 g CO 2 and 0.562 g H 2 O

10. 10 Combustion Analysis X + O 2 → CO 2 + H 2 O Step 1. Find the mass of C & H that must have been present in X (multiply masses of products by percent composition of the products). = 0.187 g C C: 0.686 g CO 2 x (1 mole CO 2 /44.01 g CO 2 ) X (1 mole C/1 mole CO 2 ) X (12g C/1 mole C) = 0.187 g C = 0.063 g H H: 0.562 g H 2 O x (1 mole H 2 O /18.02 g H 2 O) X (2 moles H/ 1 mole H 2 O) X (2.002 g H /2 moles of H) = 0.063 g H

11. 11 Combustion Analysis X + O 2 → CO 2 + H 2 O Find out if there’s oxygen or unknown compounds in X. Subtract the mass of C and H from the original mass 0 g 0.250 g of compound X – (0.187 g C + 0.063 g H) = 0 g So compound X must contain only C and H only and NO oxygen !! Step 2. Find the number of moles of C and H 0.0156 moles C C: 0.187 g x (1 mole C /12.01 g of C) = 0.0156 moles C 0.063 moles H H: 0.063 g x (1 mole H /1.008 g of H) = 0.063 moles H

12. 12 Combustion Analysis X + O 2 → CO 2 + H 2 O 3. Find the RELATIVE number of moles of C and H in whole number units (divide by smallest number of moles) C: 0.0156 moles of C /0.0156 = 1 H: 0.063 moles of H /0.0156 = 4 NOTE: If these numbers are fractions, multiply each by the same whole number.

13. 13 Combustion Analysis X + O 2 → CO 2 + H 2 O 3. Write the Empirical Formula (use the relative numbers as subscripts) CH 4

14. 14 Combustion Analysis Summary 1. Find the mass of C and H in the sample. 2. Find the actual number of moles of C and H in the sample. 3. Find the relative number of moles of C and H in whole numbers. 4. Write the empirical formula for the unknown compound.

15. 15 Combustion Analysis NOTE! In step # 1 always check to see if the total mass of C and H adds up to the total mass of X combusted. If the combined mass of C and H is less than the mass of X, then the remainder is an unknown element (unless instructed otherwise). If a third element is known, calculate the mass of that element by subtraction (at the end of step 1), and include the element in the remaining steps.

16. 16 Combustion Analysis Combustion Analysis provides the Empirical Formula. If a second technique provides the molecular weight, then the molecular formula may be deduced. 1. Calculate the empirical formula weight. 2. Find the number of “formula units or (n)” by dividing the known molecular weight by the formula weight. 3. Multiply the number of atoms in the empirical formula by the number of formula units.

17. 17 Combustion Analysis The molecular weight of glucose is 180 g/mole and its empirical formula is CH 2 O. Deduce the molecular formula. 1. Formula weight for CH 2 O is 30.03 g/mole 2. # of “formula units or (n)” = (180 g/mole)/ (30.03g/mole) = 6 3. (CH 2 O)n  where n=6 Molecular formula = C 6 H 12 O 6

18. Determining the Empirical Formula From Combustion Analysis Section 3.5 (p. 97) Combustion of 0.255 g isopropyl alcohol produces 0.561 g CO 2 and 0.306 g H 2 O. What is the empirical formula? 1.Calculate the mass of each element (C, H, O): mass O = Mass of sample – (mass of C + mass of H) = 0.255 g – (0.153 g + 0.0343 g) = 0.068 g O 0.561 g CO 2 1 mole CO 2 1 mol C12.0 g C = 0.153 g C 44.0 g CO 2 1 mol CO 2 1 mol C 0.306 g H 2 O 1 mole H 2 O 2 mol H1.0 g H = 0.0343 g H 18.0 g H 2 O 1 mol H 2 O 1 mol H

19. Empirical Formula 4.Convert to whole numbers. (not needed) Empirical Formula: C 3 H 8 O ÷ 0.0043 = 3 ÷ 0.0043 = 8 ÷ 0.0043 = 1

20. Molecular Formulas The molecular formula of a compound is: The actual number of atoms of each element in a substance. Three compounds that have the same empirical formula (CH 2 O): Formaldehyde (CH 2 O) acetic acid (C 2 H 4 O 2 ) glucose (C 6 H 12 O 6 )

21. Molecular Formula of a Compound Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH 4 N. Step 1: Calculate the molar mass of CH 4 N30.0 g/mol Step 2: Divide the molar mass of the compound by the molar mass of the empirical formula: Step 3: Multiply the subscripts in the empirical formula by the number calculated in Step 2: CH 4 N x 2.0 = C 2 H 8 N 2