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Section 6.2 Page 268 Empirical and Molecular Formulas.

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1 Section 6.2 Page 268 Empirical and Molecular Formulas

2 Empirical vs. Molecular Formula Empirical Formula (E f ) A formula that gives the simplest whole-number ratio of the atoms of each element in a compound.

3 Empirical vs. Molecular Formula Empirical Formula (E f ) A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula (M f ) A formula that gives the actual number of atoms of each element in a compound.

4 CH 2 O CH 3 OOCH = C 2 H 4 O 2 CH 3 O Molecular Formula (actual) Empirical Formula (simplest) H2O2H2O2H2O2H2O2HO C 6 H 12 O 6 CH 2 O

5 Steps to determining the empirical formula. Steps to determining the empirical formula. Step 1. Find mole amounts.

6 Steps to determining the empirical formula. Steps to determining the empirical formula. Step 1. Find mole amounts. Step 2. Divide each mole amount by the smallest mole amount.

7 Example 1: Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.

8 1.Find mole amounts.

9 n Cl = 2.128 g = 0.0600 mol Cl 35.45 g/mol 35.45 g/mol

10 1.Find mole amounts. n Cl = 2.128 g = 0.0600 mol Cl 35.45 g/mol 35.45 g/mol n Ca = 1.203 g = 0.0300 mol Ca 40.08 g/mol 40.08 g/mol

11 1.Find mole amounts. n Cl = 2.128 g = 0.0600 mol Cl 35.45 g/mol 35.45 g/mol n Ca = 1.203 g = 0.0300 mol Ca 40.08 g/mol 40.08 g/mol

12 2.Divide each mole by the smallest mole.

13 Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300 0.0300

14 2.Divide each mole by the smallest mole. Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300 0.0300 Ca = 0.0300 mol Ca = 1.00 mol Ca 0.0300 0.0300 Ratio – 1 Ca : 2 Cl Empirical Formula = CaCl 2

15 Example 2: A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula?

16 Always assume that you have a 100g sample. That way you can convert percent directly to grams... In this case you would have 50.81g of Zn, 16.04g of P, and 33.15g of O.

17 Rhyme “Percent to mass Mass to mole Divide by small Multiply ‘til whole”

18 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol P 30.97g/mol O 16g/mol

19 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81g P 30.97g/mol 16.04g O 16g/mol 33.15g

20 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol P 30.97g/mol 16.04g O 16g/mol 33.15g

21 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol O 16g/mol 33.15g

22 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol

23 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol

24 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol 0.518 = 1 O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol

25 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol 0.518 = 1 O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol 2.072 0.518 = 4

26 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 1.5 x 2 = 3 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol 0.518 = 1 O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol 2.072 0.518 = 4

27 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 1.5 x 2 = 3 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol 0.518 = 1 1 x 2 = 2 O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol 2.072 0.518 = 4

28 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 1.5 x 2 = 3 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol 0.518 = 1 1 x 2 = 2 O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol 2.072 0.518 = 4 4 x 2 = 8

29 A compound consists of 50.81% zinc, and 16.04% phosphorus, and 33.15% oxygen. What is the empirical formula? AtomMassMoleMole Ratio Whole # Ratio Zn 65.38g/mol 50.81gn Zn = 50.81g 65.38g/mol = 0.777 mol 0.777 0.518 = 1.5 1.5 x 2 = 3 P 30.97g/mol 16.04gn P = 16.04g 30.97g/mol = 0.518 mol 0.518 = 1 1 x 2 = 2 O 16g/mol 33.15gn O = 33.15g 16g/mol = 2.072 mol 2.072 0.158 = 4 4 x 2 = 8 Therefore empirical formula would be Zn 3 P 2 O 8

30 Example 3: The percentage composition of a compound is 48.63% carbon, 21.59% oxygen, 18.90% nitrogen, and the rest hydrogen. Find the empirical formula for the compound.

31 Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound.

32 Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Molar Mass. (M Ef ) 3. Divide the Molecular Formula Molar Mass (M Mf ) by the “M Ef ”. 4. Multiply empirical formula by factor.

33 Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Molar Mass. (M Ef ) 3. Divide the Molecular Formula Molar Mass (M Mf ) by the “M Ef ”. 4. Multiply empirical formula by factor. Note: The molecular formula molar mass (M Mf ) is always given in the question.

34 Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH 2 O 3.

35 Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH 2 O 3. Step 2.“M Ef ” = 62.03 g

36 Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH 2 O 3. Step 2.“M Ef ” = 62.03 g Step 3.M Mf /M Ef = 124.06/62.03 = 2

37 Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH 2 O 3. Step 2.“M Ef ” = 62.03 g Step 3.M Mf /M Ef = 124.06/62.03 = 2 Step 4.2(CH 2 O 3 ) = C 2 H 4 O 6

38 Example 1: Find the molecular formula for a compound whose molar mass is ~124.06 g/mol and empirical formula is CH 2 O 3. Step 2.“M Ef ” = 62.03 g Step 3.M Mf /M Ef = 124.06/62.03 = 2 Step 4.2(CH 2 O 3 ) = C 2 H 4 O 6  Therefore the Molecular formula (Mf) is C 2 H 4 O 6

39 Example 2: A compound containing carbon, hydrogen, iodine and oxygen is found to be 18.0% carbon, 2.5% hydrogen, 63.5% iodine, and 16.0% O. The molar mass of this compound is known to be 400 g/mol. What is its molecular formula? 1.Find the empirical formula. 2.Then you can find the molecular formula.

40 Mass Spectrometry Demo!

41

42 Poison Identification. Acetone: C 3 H 6 O Hydrogen cyanide: HCN Ethylene Glycol: C 2 H 6 O 2 Isopropanol: C 3 H 8 O Methanol: CH 4 O Sodium hypochlorite: NaOCl Acetaminophen: C 8 H 9 NO 2

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