Respuesta :
Answer: The empirical formula for the given compound is [tex]C_3HF_2[/tex] and molecular formula for the given compound is [tex]C_{24}H_8F_{16}[/tex]
Explanation : Given,
Mass of C = 18.24 g
Mass of H = 0.51 g
Mass of F = 16.91 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{18.24g}{12g/mole}=1.52moles[/tex]
Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.51g}{1g/mole}=0.51moles[/tex]
Moles of Fluorine = [tex]\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{16.91g}{19g/mole}=0.89moles[/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 0.51 moles.
For Carbon = [tex]\frac{1.52}{0.51}=2.98\approx 3[/tex]
For Hydrogen = [tex]\frac{0.51}{0.51}=1[/tex]
For Fluorine = [tex]\frac{0.89}{0.51}=1.74\approx 2[/tex]
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H : F = 3 : 1 : 2
The empirical formula for the given compound is [tex]C_3H_1F_2=C_3HF_2[/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:
Molar mass = 562.0 g/mol
Mass of empirical formula = 3(12) + 1(1) + 2(19) = 75 g/eq
Putting values in above equation, we get:
[tex]n=\frac{562.0}{75}=7.49\approx 8[/tex]
Multiplying this valency by the subscript of every element of empirical formula, we get:
[tex]C_3HF_2=(C_3HF_2)_n=(C_3HF_2)_8=C_{24}H_8F_{16}[/tex]
Thus, the molecular formula for the given compound is [tex]C_{24}H_8F_{16}[/tex]
The proportionate of the atom found in a given compound is called the empirical formula while the molecular formula represents the actual digits of the atoms found in the given molecule of the compound.
The empirical formula for the given compound is [tex]\rm C_{3}HF_{2}[/tex] and molecular formula for the given compound is [tex]\rm C_{24}H_{8}F_{16}[/tex].
The empirical formula can be determined as:
Given,
- Mass of C = 18.24 g
- Mass of H = 0.51 g
- Mass of F = 16.91 g
Step 1: For determining the empirical formula, convert mass into moles:
For Carbon:
[tex]\begin{aligned}&\rm Moles \;of \;Carbon = \dfrac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}\\\\&= \dfrac{18.24 \rm g}{ 12 \rm g/mol}\\\\&=1.52\text{ moles}\end{aligned}[/tex]
For Hydrogen:
[tex]\begin{aligned}&\rm Moles \;of \;Hydrogen= \dfrac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}\\\\&= \dfrac{0.51\rm g}{ 1\rm g/mol}\\\\&=0.51\text{ moles}\end{aligned}[/tex]
For Fluorine:
[tex]\begin{aligned}&\rm Moles \;of \;Fluorine= \dfrac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}\\\\&= \dfrac{16.91\rm g}{ 19\rm g/mol}\\\\&=0.89\text{ moles}\end{aligned}[/tex]
Step 2: Calculation of mole ratio for the assigned elements:
For this divide the moles of elements with the smallest number of moles calculated above.
[tex]\begin{aligned}\text{For Carbon} & = \dfrac{1.52 }{0.51}\\&= 2.98\end{aligned}[/tex]
Moles of C = 3
[tex]\begin{aligned}\text{For Hydrogen}& = \dfrac{0.51}{ 0.51}\\&= 1\end{aligned}[/tex]
Moles of Hydrogen = 1
[tex]\begin{aligned}\text{For Fluorine}& = \dfrac{0.89}{ 0.51}\\&= 1.74\end{aligned}[/tex]
Moles of Fluorine = 2
Step 3: The calculated mole ratio will be the subscript of the elements.
C : H : F = 3 : 1 : 2
Therefore, the empirical formula will be,
[tex]\rm C_{3}H_{1}F_{2} \;or\; C_{3}HF_{2}[/tex]
The molecular formula can be determined as:
Determining the valency (n):
[tex]n = \dfrac {\text{molecular mass} }{\text{empirical mass}}[/tex]
Given,
- Molar mass = 562.0 g/mol
- Mass of empirical formula = [tex]3(12) + 1(1) + 2(19)[/tex]
= 75 g/eq
Substituting values in the above equation, we get:
[tex]\begin{aligned}n &= \dfrac{562 }{ 75}\\\\&= 7.49 \\\\&= 8\end{aligned}[/tex]
Multiply the valency calculated with the empirical formula:
[tex]\begin{aligned}\rm C_{3}HF_{2} &=\rm (C_{3}HF_{2})n\\&= \rm (C_{3}\rm HF_{2})8\\\\\\\\&= \rm (C_{24}H_{8}F_{16})\end{aligned}[/tex]
Therefore, the molecular formula is [tex]\rm C_{24}H_{8}F_{16}[/tex].
Thus, the empirical formula is [tex]\rm C_{3}HF_{2}[/tex] and the molecular formula is [tex]\rm C_{24}H_{8}F_{16}[/tex].
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