1. EMPIRICAL FORMULAS

2. Methanal vs. acetic acid Both substances have C:H:O in a 1:2:1 ratio, but they are completely different substances.

3. Methanal vs. acetic acid The EMPIRICAL FORMULA for both these substances is CH 2 O. The MOLECULAR FORMULA for acetic acid is C 2 H 4 O 2.

4. EMPIRICAL VS. MOLECULAR FORMULAS EMPIRICAL FORMULAS show the simplest whole number ratio of elements in a compound. MOLECULAR FORMULAS show the element symbols and the exact no. of each type of atom in a molecular compound. **formulas for ionic compounds are ALWAYS empirical.

5. Comparing Empirical & Molecular Formulas Sometime the empirical and molecular formulas are the same and sometimes they are different

6. Finding the Empirical Formula from Percent Composition 1. Use a chart to organize your work\ 1. Assume you have 100 g sample so that your mass % is the same as your mass in grams. 2. To get a whole number at the end…multiply. AtomAssume 100g: Mass (g) MM (g/mol)n (mol)Smallest Ratio Whole Number A rhyme to help you remember what goes in the chart: Percent to mass Mass to mole Divide by small Multiply til whole If the decimal ends in….FractionMultiply by…...33 1/3 3 …25 1/4 4... 1.5 ½ 2 …,67 ….75

7. EXAMPLE 1: Find the empirical formula of a compound with 35.4% sodium and the remainder nitrogen. STEP 1: Always assume you are working with 100 g of the compound. AtomAssume 100g: Mass (g) MM (g/mol)n (mol)Smallest Ratio Whole Number Na35.4g22.99 N64.6g14.01

8. EXAMPLE 1: Find the empirical formula of a compound with 35.4% sodium and the remainder nitrogen. STEP 2: Calculate the amount of each element in moles using n=m/MM AtomAssume 100g: Mass (g) MM (g/mol)n (mol)Smallest Ratio Whole Number Na35.4g22.99n=m/MM =35.4/22.99 =1.539… N64.6g14.01n=4.61…

9. EXAMPLE 1: Find the empirical formula of a compound with 35.4% sodium and the remainder nitrogen. STEP 3: Determine the whole number ratio by dividing each element mole value by the lowest number. AtomAssume 100g: Mass (g) MM (g/mol)n (mol)Smallest Ratio Whole Number Na35.4g22.99n=m/MM =35.4/22.99 =1.539… 1 N64.6g14.01n=4.61… 3

10. EXAMPLE 1: Find the empirical formula of a compound with 35.4% sodium and the remainder nitrogen. STEP 4: Multiply each number by an integer to obtain all whole numbers. Empirical Formula is NaN 3 **not needed this time AtomAssume 100g: Mass (g) MM (g/mol) n (mol)Smallest Ratio Whole Number Na35.4g22.99n=m/MM =35.4/22.99 =1.539… 1 N64.6g14.01n=4.61… 3

11. EXAMPLE 2: Determine the empirical formula of a compound that contains 69.9% iron and 30.1% oxygen by mass. ***Assume 100g of substance*** STEP 1: Calculate the amount of each element in 100g.

12. EXAMPLE 2: Determine the empirical formula of a compound that contains 69.9% iron and 30.1% oxygen by mass. STEP 2: Divide the amount of each element by the smallest amount.

13. EXAMPLE 2: Determine the empirical formula of a compound that contains 69.9% iron and 30.1% oxygen by mass. STEP 3: Find an equivalent whole number ratio and write the empirical formula.

14. Tips for Solving Empirical Formula Problems Don’t round until the very end. If the value is 0.95 round to 1, if the value is 0.05 round to 0, if between 0.45 to 0.55 round to 0.5.

15. Determining the Empirical Formula by Experiment You can calculate a molecular formula using the periodic table, for example you know that pure copper will oxidize to become CuO. But, as a scientist you need to confirm this by experimentation. This can be done using a variety of methods such as a synthesis reaction.

16. HOMEWORK: pp. 292 #1 (practice) pp. 293 #1-5, 7

17. Molecular Formula The molecular formula (actual formula) can only be found if you have both the: o Molar mass of the compound & o Empirical Formula (or percent composition)

18. Steps to Finding Molecular Formula 1. Find the molar mass of the empirical formula 2. Find the ratio of the molar mass of the compound to the empirical formula using: 3. Multiply the empirical formula by the ratio to obtain the molecular formula.

19. Example 1 The empirical formula for Vitamin C is C 5 H 4 O 3 and its molar mass is 176.17 g/mol. Find the molecular formula MM EF = 88.07 g/mol Ratio = MM MF = 176.17g/mol = 2 MM EF 88.07g/mol Molecular Formula = C 5 x 2 H 4 x 2 O 3 x 2 = C 10 H 8 O 6

20. Learning Check A compound with an empirical formula of C 2 OH 4 and a molar mass of 88g/mol. What is the molecular formula of the compound? C4O2H8C4O2H8

21. Exit Card A sample of indium chloride with a mass of 0.5000g is found to have 0.2404g of chlorine. What is the empirical formula of the indium compound?