1. Chemical Formulas NaCl H2OH2O C 6 H 12 O 6 NaHCO 3

2. What information is in a chemical formula? A ratio of ATOMS. In 1 molecule of H 2 O there are 2 atoms of H and 1 atom of O. A ratio of MOLES. In 1 mole of H 2 O there are 2 moles of H and 1 mole of O. Formulas are NOT ratios of grams. In 1 mole H 2 O there are NOT 2 g of H for every g of O.

3. How do chemists determine the formula of a compound? Step 1: Elemental Analysis Identify ELEMENTS present Determine % COMPOSITION by MASS for each element.

4. How do chemists determine the formula of a compound? Step 2: Determination of empirical formula Empirical Formula = SIMPLEST Ratio of atoms in a formula

5. IONIC COMPOUND (+ metal ion/ nonmetal – ion or polyatomic ion) Empirical Formula = “Final” Formula Ions bond together in a lattice – giant “molecule” Actual formula would be to difficult write, e.g. Na 10 23 Cl 10 23 So formula is written as simplest ratio, NaCl

6. Covalent Compound (2 or more nonmetals sharing pairs of electrons exist as molecules. Empirical Formulas (simplest ratio) may or may not be true ratio True or Actual Formula is called MOLECULAR Formula Example: H 2 O: empirical = molecular Benzene : empirical ≠ molecular Empirical = CH Molecular C 6 H 6

7. MOLECULAR FORMULA DETERMINATION Requires empirical formula + MOLAR MASS Molar Mass is Determined by Experiment (in problems, molar mass must be provided).

8. Empirical Formula vs Molecular Formula Empirical Formula = SIMPLEST ratio of atoms in a compound. Molecular Formula = ACTUAL ratio of atoms in a compound. Ionic compound → Empirical Formula is “actual” formula Molecular compound (covalent bonds) → Molecular formula is usually different than empirical formula. Determining empirical formula is still a key step in the process of the determining the formula of a Molecular Compound.

9. Examples of Empirical and Molecular Formulas for covalently bonded compounds: Carbon dioxide: Empirical = CO 2 Molecular = CO 2 Benzene: Empirical CH Molecular = C 6 H 6 Glucose: Molecular = C 6 H 12 O 6 Empirical = CH 2 O

10. % Composition by Mass Definition of %: Part out of ; To calculate % : x 100 % Definition of Mass % = 100 Part Total Mass of Element x 100% Total Mass of Compound Mass % of each element must add up to 100

11. Example #1: % from mass data What is the % by mass of a compound containing 28.0 g Fe and 8.00 g O? % Fe = % O = Mass of Fe x 100 % Total Mass of Compound (Fe + O) 28.0 g Fe 28.0 + 8.00 = 36.0 x 100% = 77.8% 8.00 g O 36.0 g Fe + O x 100% = 22.2%

12. Example #2: % from a chemical formula What is the % by mass of each element in water, H 2 O? Step 1 – Calculate grams of each element and molar mass: H: 2 mol x 1.01 g/mol = 2.02 g O: 1 mol x 16.00 g/mol = 16.00 g 18.02 g Step 2 – Calculate % H: 2.02 / 18.02 x 100 % = 11.2 % O: 16.00/ 18.02 x 100 % = 88.8 % 2 nd method to calculate % O = 100 – 11.2 = 88.8% HW Self check 8.7

13. Determining Empirical Formula from Elemental Analysis Data Example #1: Elemental Analysis (% by mass) for a compound is: Al: 32.13 % F: 67.87% Recall = Simplest Ratio of Atoms: Al x F y, where x,y = integers Step 1: Find masses of elements. Since data in mass %, assume 100 g sample; then can replace % with g. Assume 100 g sample: 32.13% of 100 = 32.13 g Al 67.87% of 100 = 67.87 g F

14. Step #2: Convert to moles Al: 32.13 g F: 67.87 g ( 26.98 g Al 1 mole Al ) = 1.191 moles Al ( 19.00 g F 1 mole F ) = 3.572 moles F Can we leave final Answer as: Al 1.911 F 3.572 ? No! Final Answer must be simplest ratio!

15. Step 3: Divide by smallest # of moles: Al: 1.191 moles F: 3.572 moles F 3.572 > 1.191 moles so divide both by 1.191 Al: 1.191 ÷ 1.191 = 1 F: 3.572 ÷ 1.191 = 3 AlF 3 Final Answer: AlF 3 HW: Self check 8-8

16. Determining Empirical Formula from Elemental Analysis Data Example #2: An oxide of aluminum is formed by the reaction of 4.151 g of al with 3.692 g of oxygen. Calculate the empirical formula of the compound. Al: 4.151 g O: 3.692 g Recall = Simplest Ratio of Atoms: Al x O y, where x,y = integers Step 1: Find masses of elements: Given here

17. Step #2: Convert to moles Al: 4.151 g O: 3.692 g ( 26.98 g Al 1 mole Al ) = 0.1539 moles Al ( 16.00 g O 1 mole O ) = 0.2308 moles O Can we leave final Answer as: Al 01539 O 0.2308 ? No! Final Answer must be simplest ratio!

18. Step 3: Divide by smallest # of moles: Al: 0.1539 moles O: 0.2308 moles O 0.2308 > 0.1539 moles so divide both by 0.1539 Al: 0.1539 ÷ 0.1539 = 1 O: 0.2308 ÷ 0.1539 = 1.5 AlO 1.5 Can’t leave answer as: AlO 1.5

19. Step 4: Multiply the numbers from step 3 by the smallest integer that convert them both to whole numbers. 1.5 = 3/2 so x 2 will get rid of 2 in denominator 1.500 moles O x 2 = 3 moles O 1.000 moles Al x 2 = 2 moles Al Al 2 O 3 Final Answer: Al 2 O 3 HW: Self check 8-9 and 8.10

20. Common Decimals/Fractions in our problems FractionDecimal 1/4 1/3 1/2 2/3 3/4 0.25 0.33 0.50 0.67 0.75

21. Moles XMoles Y Multiply byFinal Answer 1.00 2.00 NA 1.25 1.00 4X5Y4X5Y4 1.33 1.00 1.50 1.00 1.67 1.00 1.751.00 1.99 1.00 1.891.00 3 2 3 4 Round to 2 Error check your math X4Y3X4Y3 X3Y2X3Y2 X5Y3X5Y3 X7Y4X7Y4 X2YX2Y XY 2