o. Find the empirical formula of a compound inat contains: 40.22 & Catbon, 731:
ID: 1026152 • Letter: O
Question
o. Find the empirical formula of a compound inat contains: 40.22 & Catbon, 731: Hydrogen, 1,82 Nitrogen, 4465 g Oxygen Cxmol 3.348mol C5.998o 12.o1 q 1 65581 0.558 0.5581 0.5581 N 0.5581 mol N1 4.01 9 Ox i mol_ 2.790 mol 4999 5 7. a. Determine the empirical formula for a compound that contains: 95.21 g Carbon, 4.79g o.00 12.01-9 51.668 x3-5 4.75 mol H 1.008 4.15 b. What is the molecular formula given that its the molar mass is 255 grams per mole? 2559 x602240% nol Determine the empirical ormula for a compound that contains: 36.25% Carbon, 4.82 26.75 % Chlorine, 32.19 % Oxygen 4x3-12 -6.33663-1 13 3 12.01 9 .82x 1 6.1545 07545 35.45 9 Handbook 2 © 2018 Cuesta College. All rights reserved o.00 9Explanation / Answer
6.Element mass A.Wt no of moles simple ratio
C 40.22 12 40.22/12 = 3.35 3.35/0.558 = 6
H 7.31 1 7.31/1 = 7.31 7.31/0.558 = 13
N 7.82 14 7.82/14 = 0.558 0.558/0.558 = 1
O 44.65 16 44.65/16 = 2.8 2.8/0.558 = 5
The empirical formula = C6H13NO5
7.
Element mass A.Wt no of moles simple ratio simplest ratio
C 95.21 12 95.21/12 = 7.93 7.93/4.79 = 1.655 1.655*3 = 5
H 4.79 1 4.79/1 = 4.79 4.79/4.79 = 1 1*3 = 3
The empirical formula C5H3
E.F.Wt = 12*5 + 1*3 = 63
molecular formula = ( empirical formula)n
n = M.Wt/E.F.wt
= 255/63 = 4
molecular formula = (C5H3)4
= C20H12
8.
Element % A.wt no of moles simple ratio simplest ratio
C 36.25 12 36.25/12 = 3 3/0.75 = 4 4*3 = 12
H 4.82 1 4.82/1 = 4.82 4.82/0.75 = 6.43 6.43*3 = 19
Cl 26.75 35.5 26.75/35.5 = 0.75 0.75/0.75 = 1 1*3 = 3
O 32.19 16 32.19/16 = 2 2/0.75 = 2.66 2.66*3 = 8
The empirical formula = C12H19Cl3O8
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