Respuesta :
Answer: The molecular formula for the given compound is [tex]C_3H_6[/tex]
Explanation : Given,
Percentage of C = 85.7 %
Percentage of H = 14.3 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 85.7 g
Mass of H = 14.3 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{85.7g}{12g/mole}=7.14moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{14.3g}{1g/mole}=14.3moles[/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 7.14 moles.
For Carbon = [tex]\frac{7.14}{7.14}=1[/tex]
For Hydrogen = [tex]\frac{14.3}{7.14}=2.00\approx 2[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 2
The empirical formula for the given compound is [tex]C_1H_2=CH_2[/tex]
Mass of empirical formula = [tex]CH_2[/tex] = 1(12) + 2(1) = 14 g/eq.
Now we have to determine the molar mass of compound.
As, 112 mL of volume contains 0.21 g of compound
So, 22400 mL of volume contains [tex]\frac{22400}{112}\times 0.21=42g[/tex] of compound
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 = 42 g/mol
Mass of empirical formula = 14 g/mol
Putting values in above equation, we get:
[tex]n=\frac{42}{14}=3[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]CH_2=(CH_2)_n=(CH_2)_3=C_3H_6[/tex]
Thus, the molecular formula for the given compound is [tex]C_3H_6[/tex]
The molecular formula of the compound containing 85.7% C and 14.3% is C₃H₆
We'll begin by obtaining the the empirical formula of the compound. This can be obtained as follow:
C = 85.7%
H = 14.3%
Divide by their molar mass
C = 85.7 / 12 = 7.142
H = 14.3 / 1 = 14.3
Divide by the smallest
C = 7.142 / 7.142 = 1
H = 14.3 / 7.142 = 2
Thus, the empirical formula of the compound is CH₂
Next, we shall determine the number of mole of compound.
At STP,
22400 mL = 1 mole of the gas
Therefore,
112 mL = 112 / 22400 = 0.005 mole of the gas
Next, we shall determine the molar mass of the compound.
Mass of compound = 0.21 g
Mole of compound = 0.005 mole
Molar mass =?
Molar mass = mass / mole
Molar mass of compound = 0.21 / 0.005
Molar mass of compound = 42 g/mol
Finally, we shall determine the molecular formula of the compound.
Empirical = CH₂
Molar mass of compound = 42 g/mol
Molecular formula =?
Molecular formula = Empirical × n
[CH₂]n = 42
[12 + (2×1)]n = 42
14n = 42
Divide both side by 14
n = 42 / 14
n = 3
Molecular formula = [CH₂]ₙ
Molecular formula = [CH₂]₃
Molecular formula = C₃H₆
Thus, the molecular formula of the compound is C₃H₆
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