1. % Composition & Empirical Formula

2. V.4 PERCENTAGE COMPOSITION Percentage composition is the percentage (by mass) of the species in a chemical formula.

3. How to calculate it… For MgO 1) Find the total molar mass of your compound. 2) Find the specific molar mass of the each element you have. 3) Divide the specific molar mass / total mass 4) Multiply by 100%

4. Example #1: What is the percent composition of ? a) MgO (what % of the mass is magnesium, what % of the mass is oxygen?)

5. b) FeCl 2

6. c) (NH 4 ) 3 PO 4

7. Example #2: What is the percent composition of the bold species? NiSO 4  7H 2 0

8. Empirical Formula

9. V.5 EMPIRICAL & MOLECULAR FORMULA Empirical Formula is the SMALLEST ‘whole number ratio’ of atoms which represents the molecular make-up of a compound. CH 2, C 2 H 4, C 3 H 6, C 4 H 8, C 5 H 10 --- all contain twice as many H’s as C’s Therefore, the empirical formula (or simplest ratio) is CH 2 In this case, all the formulae are whole-number multiples of CH 2 C 2 H 4 = 2 x CH 2 C 3 H 6 = 3 x CH 2 C 4 H 8 = 4 x CH 2

10. The empirical formula is the simplest whole number ratio between atoms in a compound. It is determined experimentally by measuring the mass of the elements that combine to form a compound. Example: A compound was found to be composed of 38.7 g C, 9.68 g H and 51.6 g O. Calculate the empirical formula.

11. A compound was found to be composed of 38.7 g C, 9.68 g H and 51.6 g O. Calculate the empirical formula. CH 3 O = 1 = 3 = 1 38.7 g C x 1 mole= 3.225 mol 12.0 g 9.68 g H x 1 mole= 9.584 mol 1.01 g 51.6 g O x 1 mole= 3.225 mol 16.0 g 3.225 mol Change grams to moles first ! Divide by the smallest number to get whole numbers! Write the formula!

12. A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 0.5879 mol = 1.765 mol = 0.5890 mol 28.24 g O x 1 mole 16.0 g AgNO 3 = 3 = 1 63.55 g Ag x 1 mole 107.9 g 8.23 g N x 1 mole 14.0 g = 0.5879 mol Assume that you have 100 g- that means 63.55 g Ag, 8.23 g N, and 28.24 g O Change grams to moles using molar mass!

13. A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula. Change % into grams first, then into moles. Only do this for ionic compounds- rearrange it as a polyatomic ion Not for covalent! Cu 3 (PO 4 ) 2 0.5255 mol = 4 = 1.5 = 1 = 2.103 mol = 0.7885 mol 50.07 g Cu x 1 mole 63.5 g = 0.5255 mol 16.29 g P x 1 mole 31.0 g 33.64 g O x 1 mole 16.0 g 0.5255 mol Cu 3 P 2 O 8 Double = 3 = 2 = 8 Rearrange

14. What is the empirical formula of a compound containing 39.0 % Si and 61.0 % O?

15. Page 93 –Common Fractions → Decimal Conversions –Helpful to be able to recognize these! Saves you time! 2.67, 1.33, 5.67, 3.33, etc involve THIRDS ( x 3 to clear fraction) 1.75, 2.25, 3.75, etc involve QUARTERS ( x 4 to clear fraction) IMPORTANT: Don’t round intermediate values, keep 3 or 4 decimals...

16. Homework Percentage composition questions Page 91 #44 a, d, g, I, k, n #45 a, b, f Empirical Formula questions Page 93 # 46 (odd)