Percent Composition & Chemical Formulas (empirical to molecular) Chapter 10.3

1. Percent Composition & Chemical Formulas (empirical to molecular)Chapter 10.3

2. Percent Composition • The percent by mass of each element in the compound

3. How to find Percent Composition • Percent composition can be determined in 2 ways. • Determined using formulas % mass of element =molar mass of element x 100 molar mass of compound • Found experimentally % mass of element =givenmass of element x 100 given mass of compound

4. What is the % Composition ofPotassium Chromates? • Determined using formulas • Potassium Chromate K2CrO4 % K = (2 x 39.1)/194.2 40.3% % Cr = 52.0/194.2 26.8% % O = (4 x 16.0)/194.2 32.9% • Found experimentally • Upon analysis of a 5.00 g sample of a potassium chromate it was found to contain 1.33g K, 1.77g Cr & 1.91g O. % K = 1.32/5.00 26.6% % Cr = 1.77/5.00 35.4% % O = 1.91/5.00 38.1%

5. Fertilizers Nitrogen (N) Key element in turf grass nutrition Promotes vigorous leaf and stem growth to improve the overall quality of the turf Essential component of the chlorophyll molecule which gives turf its dark green color Involved in regulating the uptake of other key elements Phosphorous (P) Used in the formation and transfer of energy within the plant Influences early root development and growth Encourages plant establishment Potassium (K) Used by the plant in large quantities, second only to nitrogen Key component in the formation of carbohydrates, or food for the plant Encourages rooting and wear tolerance Enhances drought and cold tolerance Key component in cell wall strength and resistance to disease

6. What’s the right mix? What factors do you consider?

8. Empirical & Molecular Formulas • Empirical Formula • A formula that gives the lowest whole-number ratio of elements in the compound. • To get empirical formula from molecular, divide all subscripts by the greatest common multiple. • Example: CH2O • Molecular Formula • A formula that gives the actual ratio of elements in the compound. • Molecular formula is a simple multiple of the empirical formula • Example : C6H12O6

9. What’s its formula? (CH2O) 1. 3. 2. 4. CH2O & CH2O C3H6O3 & CH2O C2H4O2 & CH2O 5. C4H8O4 & CH2O C5H10O5 & CH2O 6. C6H12O6 & CH2O

10. Empirical Formula CH2O

11. Going from % Comp to empirical formula • Assume you have a 100g. 100 g means % directly coverts to grams • Convert elements to moles. • Find a whole number ratio. Divide each element by the smallest mole value to create a 1. • Write the empirical formula using the molar ratios as subscripts 79.85 %C = 79.85 g C 20.15 %H = 20.15 g H 79.85 g C x 1 mol C = 6.65 mol C 12.0 g C 20.15 g H x 1 mol H = 20.15 mol H 1.0 g H C1H3 = CH3 _________ =1 C6.65 mol ________ = 3 H 6.65 mol

12. What if it doesn’t come out to a whole # • Assume you have a 100g. 100 g means % directly coverts to grams • Convert elements to moles. • Find a whole number ratio. Divide each element by the smallest mole value to create a 1. Multiply all elements to generate a whole number 0.5 or ½ x 2 0.33 or ⅓ x 3 0.25 or ¼ x 4 0.2 or ⅕ x 5 • Write the empirical formula using the molar ratios as subscripts 43.66 % P = 43.66 g P 56.34% O = 56.34 g O 43.66 g P x 1 mol P = 1.41 mol P 31.0 g P 56.34 g O x 1 mol O = 3.52 mol O 16.0 g O P2O5 _________ =1 P1.41 mol _________ = 2.5 O 1.41 mol X 2 = 2 P X 2 = 5 O

13. Going from Empirical Formula to Molecular Formulas You can determine the molecular formula if you know empirical formula and molar mass. ? Find molecular formula with a molar mass of 60.0g and empirical formula of CH4N. • Calculate the empirical formula mass • Divide the molar mass of the compound by the empirical formula mass. • Multiply this number by all subscripts in the empirical formula to get the molecular formula. CH4N = 12.0 + (4 x 1.0) + 14.0 = 30.0 g 60.0g / 30.0 g = 2 CH4N x2 = C2H8N2