Answer: Empirical formula = C3H6O
C6H1202---Molecular formula
Explanation:
we first find the masses of C, H and O contained in the sample
<u>mass of carbon </u>
Mass of CO2 = 0.512g
molar mass of carbon = 12g/mol
Molar mass CO2 = 12+ (16X2)=44g/mol
Mass C = 12/44x 0.512 = 0.1396g of carbon
<u>Mass Hydrogen </u>
mass of H2O= 0.209g
Molar mass H2O = 18g/mol
Molar mass Hydrogen = 1.0079g x 2 = 2.0158g since Hydrogen gas is diatomic
Mass H = 2.0158/18 X 0.209 = 0.0234g Hydrogen
Mass of Oxygen in the sample = overall mass of Sample - mass of hydrogen and oxgen=
0.225 - ( 0.1396 + 0.0234) = 0.062g Oxygen
A) T o find Empirical formula
1ST STEP-- Divide through by each relative atomic mass:
C = 0.1396/12 = 0.01163
H = 0.0234/1.0079 = 0.0232
O = 0.0619/16 = 0.00387
2nd step Divide by smallest answer
C = 0.01163/0.00387 = 2.0998 = 3
H = 0.0232/0.00387 = 5.99 = 6
O = 0.00387/0.00387=1
Empirical formula = C3H6O
B)To find molecular formula
Given that molar mass of caproic acid as 116g,
we will use our empirical formulae to find molecular formulae with the equation
(C3H6O)n= 116g/mol
12x3 + 1x6 + 1 x16= 58g/mol
58n= 116g/mol
n = 116/58= 2
= (C3H6O)2= C(3X2) H(6X2) O(1X2)
C6H1202--- MOLECULAR FORMULA
