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Empirical Formula and Percent Composition • For Lawn Fertilizer Amount of each nutrient put on lawn at different times of the year is very important. • High percent nitrogen content in Spring to stimulate green leaf growth • High percent potassium content in Fall to stimulate root growth • Relative amount of each element in a compound expressed as Percent Composition. Example: Calculating Percent??? Relative amount of each element in a compound expressed as Percent Composition. K2CrO4 Potassium Chromate 40.3 % Potassium 26.8% Chromium 32.9% Oxygen 40.3 + 26.8 + 32.9 = 100.0 How do you calculate this weight percent for each element??? Elemental Percent % Mass of Element E = Mass of E x 100 Mass of Compound Percent Composition in a Formula Just like finding the % you get on a test… For CaCO3 first find the mass…but do it like this… Then divide each individual part by the whole… then times by 100 to get %! Make sure to write % to the hundredth place (2 after the decimal). Ca = 40.078 C = 12.011 O = 3(15.999) 100.086 / 100.086 (100) = 40.04 % Ca / 100.086 (100) = 12.00 % C / 100.086 (100) = 47.96 % O They need to add to 100.00 Elemental % in Potassium Chromate Report to two after the decimal. Using atomic masses: K2CrO4 K = 39.098 Cr = 51.996 O = 15.999 2(39.098) + 51.996 + 4(15.999 )= 194.188 Total Molar Mass of K2CrO4 % K = 2(39.098) / 194.188 x100 = 40.26% K % Cr = 51.996 / 194.188 x100 = 26.78 % Cr % O =4(15.999 )= 194.188 x100= 32.96% O Elemental % in Potassium Dichromate K2Cr2O7 (Potassium Dichromate) Molar Mass = 2(39.098) + 2(51.996) + 7(15.999) = 294.18 g 2(39.098) g 100 = 26.58% K 294.18 g 2(51.996)g 100 = 35.35 % Cr 294.18 g 7(15.999) g 100 = 38.07 % O 294.18 g How do the two compounds differ even though they are composed of the same three elements? MOLE SONG (Used for Empirical Formulas) Mass over formula weight, That’s the way you get the mole. Then divide the largest by the smallest. Press the little buttons as you go. Multiply it, Multiply it, Multiply it, Till it’s whole! Empirical Formula • Gives the lowest whole number ratio of atoms in a chemical compound • HO would be empirical formula for H2O2 • % Composition can be used to figure out Empirical Formula of compound. Empirical Formula • The empirical formula may be the same as the molecular formula ( CO2) • If empirical formula is not the same as molecular formula it is just a whole number ratio of the molecular formula empirical formula molecular formula NH2 N2H4 CH2O C6H12O6 Find Empirical Formula • Elemental Analysis: 25.9 % N , 74.1% O • Assume 100 g sample and calculate moles • Get empirical formula in lowest whole number ratio 25.9 % N , 74.1% O 25.9 g N and 74.1 g O, change to moles. 25.9 g N | 1 mole N = 1.85 mole N | 14.007 g N 74.1 g O | 1mole O = 4.63 mole O | 15.999 g O Then divide all the moles by the smallest mole to get the ratio. 1.85 mole N = 1.0 N 1.85 mole 4.63 mole O = 2.5 O 1.85 mole If the number is not a whole number multiply all the numbers till they are all a whole number. 1.0 N(2) = 2 N 2.5 O(2) = 5 O Then you have the empirical formula ratio: N2O5 Empirical Formula (Simplest Formula) This is basically reverse of what we just did with percent composition. Let’s find the formula for saccharin, an artificial sweetener. It has the composition 45.90% C, 2.75% H, 26.20% O, 17.50% S and 7.65% N. The easiest way to do this is to assume a 100.00g sample. So . . .45.90 g C, 2.75 g H, 26.20 g O, 17.50 g S and 7.65 g N. Then convert each into moles… 3.021 45.90 g C | 1 mole C = | 12.011 g C 2.75 g H | 1 mole H = 2.728 | 1.0079 g H 26.20 g O | 1 mole O = 1.638 | 15.999 g O 17.50 g S | 1mole S = 0.5458 | 32.065 g S 7.65 g N | 1 mole N = 0.5462 | 14.007 g N Then divide each of the moles by the smallest mole (ratio) / 0.5462 = 5.5 = 11 C This is the ratio of / 0.5462 = each. If it isn’t a whole 5.0 = 10 H number… you must multiply all the numbers 3.0 = 6 O until all are whole numbers! 1.0 = 2 S / 0.5462 = 1.0 / 0.5462 = / 0.5462 = =2N C11H10O6S2N2 Molecular Formula • You can find the Molecular Formula of a Compound if you know the empirical formula and the molar mass. • Calculate the empirical formula mass, divide into molar mass and then multiply this number by each atom in the empirical formula… Given the empirical formula is CH2 and the molecular formula mass is 112.21 g/mole find the molecular formula. CH2 has a formula mass of 14.027 g/mole. 112.21 g/mole = 8 14.027 g/mole therefore the molecular formula is C8H16 C(1 x 8) H(2 x 8) Your turn to try… Make sure to ask questions if you don’t understand…and go back and look at slides to help you through your problems!