Empirical Formulas and Molar Mass: Part 1-Determination of an Empirical Formula

1. Empirical Formulas and Molar Mass:Part 1-Determination of an Empirical Formula

2. Objectives • -Explain what empirical formulas are • -Be able to determine empirical formulas using charges or using experimental data

3. Empirical Formulas • Empirical formula (“formula unit”) -the lowest whole number ratio of atoms in an ionic compound • There are two ways to determine the empirical formula for a compound • Using charges (use your ions and this is easy) C CRISS O SS • Mathematically (yippee!!!)

4. Method #1 – Charges EX: Write the empirical formula for the compound formed by Na & P c. K & Ne Sr & Cl d. Cu & Cl No compound Na3P CuCl or CuCl2 SrCl2

5. Method #2 – Mathematically • Step 1:Use the info given in the problem and DA with the atomic mass of the element (from periodic table- round to 3 SDs) to find moles • Step 2:Take all the mole values and divide them by the SMALLEST one to figure out a ratio • Step 3: Use the answers as subscripts in the empirical formula

6. Law of Conservation of Mass Chemical Equation Na + Cl NaCl Reactants  Products Mass of Reactants = Mass of Products Example: Na+ Cl NaCl 5.0g + 5.0g = ? 10.0g

7. Atomic Mass of an Element = Mole • Each element has a unique atomic mass and this is a standard for each element • Atomic mass = mass of 1 mole of an element • The atomic masses are from the periodic table and we use grams 1 mole O = 6.02 x 1023 atoms O = 15.999 g O 1 mole Mg= 6.02 x 1023 atoms Mg= 24.305 g Mg

8. Empirical Formulas7.06 g of silver combine with an excess of fluorine to produce 8.30 g of a compound Silver + Fluorine  Ag?F? 7.06g 1.24g 8.30g Found by subtracting! = 1.00 or 1 X 7.06 g Ag 1.24 g F = .0654 moles Ag Ag1F1 = 1.00 or 1 X = .0653 moles F ANSWER = AgF

9. Empirical FormulasA compound contains 24.58% K, 35.81% Mn, and 40.50% O. Find the empirical formula (assume working with 100 grams of the compound and change percentages to grams) X = 1.00 or 1 = .629 mole K 24.58 g K 35.81 g Mn 40.50 g O X = 1.04 or 1 = .652 moles Mn = 4.02 or 4 X = 2.53 moles O ANSWER = KMnO4

10. Uneven Empirical Formulas • When figuring empirical formulas mathematically, sometimes the resulting numbers don’t come out so clean • You can’t just assume and round how you choose

11. .05 Rule • This 0.05 rule allows for experimental error that occurs causing varied number values: • If avalue is within .05 of a whole number (+0.05 or - 0.05), then the value may be rounded • Ex: 1.96 can be rounded to 2 • 1.07 cannot be rounded to 1 • 3.02 could be rounded to 3 • 1.93 cannot be rounded to 2 • If one value is not within .05 of a whole number, all the values must be multiplied by an integer so all values fall within .05 of whole numbers

12. Uneven Empirical Formulas 4.35 g sample of zinc is combined with an excess of the element phosphorus. 5.72 g of compound are formed. Calculate the empirical formula. Zinc + Phosphorous  Zn?P? 4.35g 1.37g 5.72g Found by subtracting! = 1.50 X = .0665 moles Zn X 2 = 3.00 or 3 4.35 g Zn 1.37 g P Not within .05 of a whole number X = 1.00 = .0442 moles P X 2 = 2.00 or 2 ANSWER = Zn3P2

13. Let’s Do It!!! A compound is found to contain 72.3% Fe and 27.7% O by weight. Calculate the empirical formula. Assume in 100 g of compound there would be 72.3 g Fe and 27.7 g O

14. 1 mole Fe X —————— 55.8 g Fe 72.3g Fe = 1.30 mole Fe 1.30 mole 1.30 mole = 1.00 X 3 = 3.00 = 3 Fe3O4 1 mole O X —————— 16.0 g O 27.7g O = 1.73 mole O 1.73 mole 1.30 mole =1.33 X 3 = 3.99 = 4

15. Objectives • -Explain what empirical formulas are • -Be able to determine empirical formulas using charges or using experimental data