Respuesta :
Answer: The molecular formula for the given organic compound is [tex]C_{12}H_{22}O_4[/tex]
Explanation:
We are given:
Percentage of C = 62.58 %
Percentage of H = 9.63 %
Percentage of O = 27.79 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 62.58 g
Mass of H = 9.63 g
Mass of O = 27.79 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{62.58g}{12g/mole}=5.215moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{9.63g}{1g/mole}=9.63moles[/tex]
Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{27.79g}{16g/mole}=1.737moles[/tex]
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.737 moles.
For Carbon = [tex]\frac{5.215}{1.737}=3.00\approx 3[/tex]
For Hydrogen = [tex]\frac{9.63}{1.737}=5.5[/tex]
For Oxygen = [tex]\frac{1.737}{1.737}=1[/tex]
The ratio of C : H : O = 3 : 5.5 : 1
To make the ratio in a whole number we are multiplying ratio by 2, we get:
The ratio of C : H : O = 6 : 11 : 2
Step 3: Taking the mole ratio as their subscripts.
The empirical formula for the given compound is [tex]C_6H_{11}O_2[/tex]
Mass of empirical formula = [tex]C_6H_{11}O_2[/tex]
= 6(12) + 11(1) + 2(16) = 115 g/eq.
For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.
The equation used to calculate the valency is :
[tex]n=\frac{\text{molecular mass}}{\text{empirical mass}}[/tex]
We are given:
Mass of molecular formula = 230 g/mol
Mass of empirical formula = 115 g/mol
Putting values in above equation, we get:
[tex]n=\frac{230}{115}=2[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_6H_{11}O_2=(C_6H_{11}O_2)_n=(C_6H_{11}O_2)_2=C_{12}H_{22}O_4[/tex]
Thus, the molecular formula for the given compound is [tex]C_{12}H_{22}O_4[/tex]
