Question

A new pesticide is analyzed to consist of 19.526% carbon, 3.8172% hydrogen, and 76.657% chlorine by mass basis. Based on this information, calculate the empirical formula of this pesticide.

Answer 1

criven Pesticide :- of % mass % 19:520 Carbon > 3.8172 % hydrogen => chlorine → 76.657 % Elements in almol Empirical formula on molar mass of Carbon » 12.0 g/mol Hydrogen lood gl mol chlorine 35-45 g/mol as that is So mass let total mass is Loo grams follow:- ① Carbon 19.526 g Hydrogen 3.8722 g 2 ② Chlorine a 76.657 g mass moles into ทจรร Convert by molars Carbon * I mol 19.526 8 x 1.6271 mol 12 g Hydrogen 3.87218 x I mol 3.9794 moi (3-7794) loof g 76.6578x 2.1623 mol chlorine a mol 35.45 g Divide each mol value by smallest Number of mole. Carbon » 16275 mol 2.62 71 1 mol Hydrogen » 3.77 94 mol 2.33 1.0271 mol ① chlorthe 2.1623 mol L.6271 mol → 1.33

Round the nearest whole Number multiply each Number by lowest whole Number here we multiply by 3. Carbon I X3 3 hydrogen » 2.33 X 3 6.99 m 7 Chlorine 1.33x3 ñ 3.99 N? 4 So, Empirical formula of Pesticide is :- C3 H7 clq