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

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

How to find Percent Composition Percent composition can be determined in 2 ways. 1.Determined using formulas % mass of element = molar mass of element x 100 molar mass of compound 2.Found experimentally % mass of element = given mass of element x 100 given mass of compound

What is the % Composition of Potassium Chromates? Determined using formulas Potassium Chromate K 2 CrO 4 % K = (2 x 39.1)/ % % Cr = 52.0/ % % O = (4 x 16.0)/ % 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/ % % Cr = 1.77/ % % O = 1.91/ %

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 Fertilizers

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

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: CH 2 O 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 : C 6 H 12 O 6

CH 2 O & CH 2 O C 2 H 4 O 2 & CH 2 O C 4 H 8 O 4 & CH 2 O C 3 H 6 O 3 & CH 2 O What’s its formula? (CH 2 O) C 5 H 10 O 5 & CH 2 O C 6 H 12 O 6 & CH 2 O

Name Molecular Formula MultipleMolar massUse or function FormaldehydeCH 2 O130 Disinfectant, biology preservative Acetic AcidC2H4O2C2H4O2 260Polymers, vinegar Lactic AcidC3H6O3C3H6O3 390 Sours milk, product of exercise ErythroseC3H6O3C3H6O Forms during sugar metabolism RiboseC 5 H 10 O Nuclei acids RNA & vitamin B 12 GlucoseC 6 H 12 O Nutrient for energy Empirical Formula CH 2 O

Going from % Comp to empirical formula 1.Assume you have a 100g. 100 g means % directly coverts to grams 2.Convert elements to moles. 3.Find a whole number ratio. Divide each element by the smallest mole value to create a 1. 4.Write the empirical formula using the molar ratios as subscripts %C = g C %H = g H g C x 1 mol C = 6.65 mol C 12.0 g C g H x 1 mol H = mol H 1.0 g H C 1 H 3 = CH 3 _________ = 1 C 6.65 mol ________ = 3 H 6.65 mol

What if it doesn’t come out to a whole # 1.Assume you have a 100g. 100 g means % directly coverts to grams 2.Convert elements to moles. 3.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 or ⅓ x or ¼ x or ⅕ x 5 4.Write the empirical formula using the molar ratios as subscripts % P = g P 56.34% O = g O g P x 1 mol P = 1.41 mol P 31.0 g P g O x 1 mol O = 3.52 mol O 16.0 g O P2O5P2O5 _________ = 1 P 1.41 mol _________ = 2.5 O 1.41 mol X 2 = 2 P X 2 = 5 O

Going from Empirical Formula to Molecular Formulas 1.Calculate the empirical formula mass 2.Divide the molar mass of the compound by the empirical formula mass. 3.Multiply this number by all subscripts in the empirical formula to get the molecular formula. CH 4 N = (4 x 1.0) = 30.0 g 60.0g / 30.0 g = 2 CH 4 N x2 = C 2 H 8 N 2 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 CH 4 N.