Respuesta :
the empirical formula should be first determined. Empirical formula is the simplest ratio of whole numbers of components in a compound.
for 100 g of the compound
C H Cl
mass 46.47 g 7.80 g 45.72 g
number of moles 46.47 g/12 g/mol 7.80 g/1 g/mol 45.72 g/35.5 g/mol
= 3.87 mol = 7.80 mol = 1.29 mol
divide all by the least number of moles
3.87/1.29 = 3 7.80/1.29 = 6.04 1.29/1.29 = 1
when they are all rounded off to the nearest whole number
number of atoms are as follows
C - 3
H - 6
Cl - 1
the empirical formula is C₃H₆Cl
molecular formula is the actual number of components in the compound
molar mass = 155.06 g/mol
we have to find the mass of the empirical unit
mass of C₃H₆Cl - (12 g/mol x 3) + (1 g/mol x 6) + (35.5 g/mol x 1) = 77.5
we have to then find how many empirical units are in the molecular formula
number of units = molecular mass / empirical unit mass
= 155.06 g/mol / 77.5 = 2.00
there are 2 empirical units
molecular formula = 2 (C₃H₆Cl)
molecular formula - C₆H₁₂Cl₂
for 100 g of the compound
C H Cl
mass 46.47 g 7.80 g 45.72 g
number of moles 46.47 g/12 g/mol 7.80 g/1 g/mol 45.72 g/35.5 g/mol
= 3.87 mol = 7.80 mol = 1.29 mol
divide all by the least number of moles
3.87/1.29 = 3 7.80/1.29 = 6.04 1.29/1.29 = 1
when they are all rounded off to the nearest whole number
number of atoms are as follows
C - 3
H - 6
Cl - 1
the empirical formula is C₃H₆Cl
molecular formula is the actual number of components in the compound
molar mass = 155.06 g/mol
we have to find the mass of the empirical unit
mass of C₃H₆Cl - (12 g/mol x 3) + (1 g/mol x 6) + (35.5 g/mol x 1) = 77.5
we have to then find how many empirical units are in the molecular formula
number of units = molecular mass / empirical unit mass
= 155.06 g/mol / 77.5 = 2.00
there are 2 empirical units
molecular formula = 2 (C₃H₆Cl)
molecular formula - C₆H₁₂Cl₂
Answer : The molecular of the compound is, [tex]C_6H_{12}Cl_2[/tex]
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the given percentage.
Mass of C = 46.47 g
Mass of H = 7.80 g
Mass of Cl = 45.72 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of Cl = 35.5 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{46.47g}{12g/mole}=3.87moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.80g}{1g/mole}=7.80moles[/tex]
Moles of Cl = [tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{45.72g}{35.5g/mole}=1.28moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.87}{1.28}=3.02\approx 3[/tex]
For H = [tex]\frac{7.80}{1.28}=6.09\approx 6[/tex]
For Cl = [tex]\frac{1.28}{1.28}=1[/tex]
The ratio of C : H : Cl = 3 : 6 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_3H_6Cl_1[/tex]
The empirical formula weight = 3(12) + 6(1) + 1(35.5) = 77.5 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{155.06}{77.5}=2[/tex]
Molecular formula = [tex](C_3H_6Cl_1)_n=(C_3H_6Cl_1)_2=C_6H_{12}Cl_2[/tex]
Therefore, the molecular of the compound is, [tex]C_6H_{12}Cl_2[/tex]
