Respuesta :
Answer : The molecular formula of a compound is, [tex]CHCl_3[/tex]
Solution :
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 0.1006 g
Mass of H = 0.0084 g
Mass of Cl = 0.8910 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{0.1006g}{12g/mole}=0.0084moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.0084g}{1g/mole}=0.0084moles[/tex]
Moles of Cl = [tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{0.8910g}{35.5g/mole}=0.0251moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{0.0084}{0.0084}=1[/tex]
For H = [tex]\frac{0.0084}{0.0084}=1[/tex]
For Cl = [tex]\frac{0.0251}{0.0084}=2.98\approx 3[/tex]
The ratio of C : H : Cl = 1 : 1 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_1H_1Cl_3=CHCl_3[/tex]
The empirical formula weight = 1(12) + 1(1) + 3(35.5) = 119.5 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}}[/tex]
[tex]n=\frac{119.6}{119.5}=1[/tex]
Molecular formula = [tex](CHCl_3)_n=(CHCl_3)_1=CHCl_3[/tex]
Therefore, the molecular of the compound is, [tex]CHCl_3[/tex]
