Respuesta :
Answer: The molecular formula for the compound is [tex]C_6H_6[/tex]
Explanation:
We are given:
Percentage of C = 92.26 %
Percentage of H = 7.74 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 92.26 g
Mass of H = 7.74 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{92.26g}{12g/mole}=7.68moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{7.74g}{1g/mole}=7.74moles[/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 7.68 moles.
For Carbon = [tex]\frac{7.68}{7.68}=1[/tex]
For Hydrogen = [tex]\frac{7.74}{7.68}=1[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 1
The empirical formula for the given compound is [tex]CH[/tex]
- Calculating the molar mass of the compound:
To calculate the molecular mass, we use the equation given by ideal gas equation:
PV = nRT
Or,
[tex]PV=\frac{m}{M}RT[/tex]
where,
P = pressure of the gas = 820 torr
V = Volume of gas = 250 mL = 0.250 L (Conversion factor: 1 L = 1000 mL )
m = mass of gas = 0.6883 g
M = Molar mass of gas = ?
R = Gas constant = [tex]62.3637\text{ L. torr }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]100^oC=(100+273)K=373K[/tex]
Putting values in above equation, we get:
[tex]820torr\times 0.250L=\frac{0.6883g}{M}\times 62.3637\text{ L torr }mol^{-1}K^{-1}\times 373K\\\\M=\frac{0.6883\times 62.3637\times 373}{820\times 0.250}=78.10g/mol[/tex]
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 = 78.10 g/mol
Mass of empirical formula = 13 g/mol
Putting values in above equation, we get:
[tex]n=\frac{78.10g/mol}{13g/mol}=6[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_{(1\times 6)}H_{(1\times 6)}=C_6H_6[/tex]
Hence, the molecular formula for the compound is [tex]C_6H_6[/tex]
