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MOLECULAR FORMULAS Chemistry 10 2014-2015. MOLECULAR FORMULAS  If the empirical formula is different from the molecular formula, the molecular formula.

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Presentation on theme: "MOLECULAR FORMULAS Chemistry 10 2014-2015. MOLECULAR FORMULAS  If the empirical formula is different from the molecular formula, the molecular formula."— Presentation transcript:

1 MOLECULAR FORMULAS Chemistry

2 MOLECULAR FORMULAS  If the empirical formula is different from the molecular formula, the molecular formula will always be a simple multiple of the empirical formula.  EX: The empirical formula for hydrogen peroxide is HO; the molecular formula is H 2 O 2.  In both formulas, the ratio of oxygen to hydrogen is 1:1

3 DETERMINE THE EMPIRICAL FORMULAS FOR EACH OF THE FOLLOWING MOLECULAR FORMULAS. 1. C 6 H 18 …..______________ 2. H 2 O 2 ……______________ 3. Hg 2 Cl 2 …..______________ 4. C 3 H 6 …….______________ 5. Na 2 C 2 O 4..._____________ 6. H 2 O......________________ 7. C 4 H 8 ….._______________ 8. C 4 H 6 …..________________ 9. C 7 H 12 …._______________ 10. CH 3 COOH….___________

4 REVIEW: CALCULATING EMPIRICAL FORMULAS  Use the following poem to remember the steps: Percent to mass Mass to moles Divide by small Multiply ‘til whole

5 EXAMPLE PROBLEM  Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16% hydrogen, and 43.20% oxygen.

6 STEP ONE: PERCENT TO MASS  Let’s assume we have a 100. g sample of methyl acetate. This means that each element’s percent is also the number of grams of that element.  48.64% C = g C  8.16% H = 8.16 g H  43.20% O = g O

7 STEP TWO: MASS TO MOLES  Convert each mass into moles using the molar mass of each element.  g C x 1 mol C = mol C g  8.16 g H x 1 mol H = 8.10 mol H g  g O x 1 mol O = mol O g

8 STEP THREE: DIVIDE BY SMALL  Oxygen accounts for the smallest number of moles in the formula, so divide each element by oxygen’s number of moles: mol  Carbon: mol / mol = = 1.5  Hydrogen: 8.10 mol / mol = 3.00 = 3  Oxygen: mol / mol = = 1  Remember, we will want whole-number ratios

9 STEP FOUR: MULTIPLY ‘TIL WHOLE  In the previous slide, the ratio of C:H:O is 1.5:3:1  We need a whole-number ratio, so we can multiply everything by 2 to get rid of the 1.5  C  3, H  6, O  2 So, the final empirical formula is C 3 H 6 O 2

10 CALCULATING MOLECULAR FORMULAS

11 EXAMPLE PROBLEM  Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 52.24% oxygen and has a molar mass of g/mol. Determine the empirical and molecular formulas for succinic acid.

12 STEP ONE: FIND EMPIRICAL FORMULA  % C = g C x 1mol C = mol C g  5.08 % H = 5.08 g H x 1 mol H = 5.04 mol H g  54.24% O = g O x 1 mol O = mol O g C: H:O mol : 5.04 mol : mol mol mol mol 1 : 1.48: 1  1 : 1.5 : 1, multiply by 2  C 2 H 3 O 2

13 STEP TWO: DIVIDE MOLAR MASSES  Molar mass empirical formula (2 x 12.0 g/mol) + (3 x 1.01 g/mol) + (2 x 16.0 g/mol) = 59.0 g/mol  Given (experimental) molar mass = g/mol  Multiplier = g/mol = g/mol

14 STEP THREE: USE MULTIPLIER  Empirical Formula = C 2 H 3 O 2  x 2 from step two  Molecular formula = C 4 H 6 O 4

15 PRACTICE QUESTION 2 What is the empirical and molecular formula of Vitamin D 3 if it contains 84.31% C, 11.53% H, and 4.16% O, with a molar mass of g/mol?

16 THINKING QUESTIONS 1) What information must a chemist obtain in order to determine the empirical formula of an unknown compound? 2) What information must a chemist have to determine the molecular formula of a compound?


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