11. Calculate the molecular formula for the compound found to have the following elemental composition by mass: 40.0% C, 6.71% H, 53.3% O. The molar mass of the compound is 90.08 g/mol. 1 M. F. = (E.F.) (n)

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Calculate the molecular formula for the compound found to have the following elemental composition by mass: 40.0% C, 6.71% H, 53.3% O. The molar mass of the compound is 90.08 g/mol.

 

I'm confused what to do next after acquiring the empirical formula mass.

11. Calculate the molecular formula for the compound found to have the following
elemental composition by mass: 40.0% C, 6.71% H, 53.3% O. The molar
mass of the compound is 90.08 g/mol.
MOLECULAR FORMULA =
(EMPIRICAL FORMULA). (n) | M.F. = (E.F.) (n)
400 4 ( 120012
чод
mol
6.71g
-
mol
H (1 m² ² )
-)
0.083 mol C
0.083mol
C₁
C (H₂00) 40
n
·) = 0.083 mal
2
= 0.083 mel CSMALLEST
VALVE
53.350 (mol) = 3.33 mol O
(+
= 6.64 mol H
6.64 mol H
0.083 mel
H ₂0
C: Riel
H: 80(1.01)
0:40 (16.00)
, 3.33 mol O
0.083
5
чо
-732.819
MASS
MOLAR MASS
M.F. = EMPIRICAL
M. F. = 732.812 (8)
90.09
M. F. =
s) (n)
Page 3 of 3
mol
Transcribed Image Text:11. Calculate the molecular formula for the compound found to have the following elemental composition by mass: 40.0% C, 6.71% H, 53.3% O. The molar mass of the compound is 90.08 g/mol. MOLECULAR FORMULA = (EMPIRICAL FORMULA). (n) | M.F. = (E.F.) (n) 400 4 ( 120012 чод mol 6.71g - mol H (1 m² ² ) -) 0.083 mol C 0.083mol C₁ C (H₂00) 40 n ·) = 0.083 mal 2 = 0.083 mel CSMALLEST VALVE 53.350 (mol) = 3.33 mol O (+ = 6.64 mol H 6.64 mol H 0.083 mel H ₂0 C: Riel H: 80(1.01) 0:40 (16.00) , 3.33 mol O 0.083 5 чо -732.819 MASS MOLAR MASS M.F. = EMPIRICAL M. F. = 732.812 (8) 90.09 M. F. = s) (n) Page 3 of 3 mol
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