Empirical Formulas, Lecture notes of Engineering Chemistry

Download Empirical Formulas and more Lecture notes Engineering Chemistry in PDF only on Docsity! Page 1 Percent Composition & Empirical Formulas A pure compound always consists of the same elements combined in the same proportions by weight. Therefore we can express molecular, composition as PERCENT BY WEIGHT Ethanol, C2H6O 52.13% C 13.15% H 34.72% O Percent Composition Consider NO2, Molar mass = ? What is the weight percent of N and of O? %30.45=100%•N g 14.01=N%.Wt Wt. % O = 2 (16 .00g O per mole ) 46 .01 g x 100 % = 69 .55% NO g 46.01 2 What are the weight percentages of N and O in NO? 46.68 % N and 53.32 % O Determining Formulas In chemical analysis we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula. PROBLEM: A compound of B and H is 81.10% B. What is its empirical formula? Page 2 • Because it contains only B and H, it must contain 18.90% H. • In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H. A compound of B and H is 81.10% B. What is its empirical formula? • Calculate the number of moles of each constituent. Calculate the number of moles of each element in 100.0 g of sample. 81.10 g B • 1 mol 10 81 g = 7.502 mol B A compound of B and H is 81.10% B. What is its empirical formula? . 18.90 g H • 1 mol 1.008 g = 18.75 mol H Take the ratio of moles of B and H. Always divide by the smaller number. 18.75 mol H 7 502 mol B = 2.499 mol H 1 000 mol B = 2.5 mol H 1 0 mol B A compound of B and H is 81.10% B. What is its empirical formula? But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B2H5 . . .