Answer: Empirical formula = C3H6O
C6H1202---Molecular formula
Explanation:
we first find the masses of C, H and O contained in the sample
mass of carbon
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
Mass Hydrogen
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
a. The empirical formula of caproic acid is [tex]C_3H_6O[/tex]
b. The molecular formula of caproic acid is [tex]C_6H_{12}O_2[/tex]
Given the following data:
a. To find the empirical formula of caproic acid:
First of all, we would determine the mass of each chemical element in the products.
For carbon:
Molar mass of carbon = 12 g/mol
Molar mass of carbon dioxide ([tex]CO_2[/tex]) = [tex](12 + [16\times2]) = 44 \;g/mol[/tex]
[tex]Mass\;of\;carbon = \frac{12}{44} \times 0.512\\\\Mass\;of\;carbon = \frac{6.144}{44}[/tex]
Mass of carbon = 0.1396 grams
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{0.1396}{12}[/tex]
Number of moles = 0.0116 moles
For hydrogen:
Molar mass of hydrogen ([tex]H_2[/tex]) = [tex]1 \times 2 = 2\;g/mol[/tex]
Molar mass of water ([tex]H_2O[/tex]) = [tex]( [1\times2] +16) = 18 \;g/mol[/tex]
[tex]Mass\;of\;hydrogen = \frac{2}{18} \times 0.209\\\\Mass\;of\;hydrogen = \frac{0.418}{18}[/tex]
Mass of hydrogen = 0.0232 grams
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{0.0232}{1}[/tex]
Number of moles = 0.0232 moles
For oxygen:
[tex]Mass\;of\;oxygen = Total\;mass\;of\;compound - [Mass\;of\;C + Mass\;of\;H]\\\\Mass\;of\;oxygen = 0.225 - [0.1396 + 0.0232]\\\\Mass\;of\;oxygen = 0.225 - 0.1628[/tex]
Mass of hydrogen = 0.0622 grams
[tex]Number\;of\;moles = \frac{mass}{molar\;mass}\\\\Number\;of\;moles = \frac{0.0622}{16}[/tex]
Number of moles = 0.0039 moles
Next, we would divide by the smallest number of moles:
[tex]Carbon = \frac{0.0116}{0.0039} = 2.97 =3[/tex]
[tex]Hydrogen = \frac{0.0232}{0.0039} = 5.95 =6[/tex]
[tex]Oxygen = \frac{0.0039}{0.0039} = 1[/tex]
Empirical formula = [tex]C_3H_6O[/tex]
b. To find the molecular formula of caproic acid:
[tex](C_3H_6O)n = 116\\\\([12\times3] + [1\times6] +16)n = 116\\\\(36+6+16)n=116\\\\58n =116\\\\n = \frac{116}{58} \\\\n = 2[/tex]
[tex]Molecular\;formula = (C_3H_6O)2\\\\Molecular\;formula = C_6H_{12}O_2[/tex]
