Answer : The molecular formula of a compound is, [tex]C_6H_5Cl_1[/tex]
Solution : Given,
Mass of C = 64.03 g
Mass of H = 4.48 g
Mass of Cl = 31.49 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{64.03g}{12g/mole}=5.34moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{4.48g}{1g/mole}=4.48moles[/tex]
Moles of Cl = [tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{31.49g}{35.5g/mole}=0.887moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{5.34}{0.887}=6.02\approx 6[/tex]
For H = [tex]\frac{4.48}{0.887}=5.05\approx 5[/tex]
For Cl = [tex]\frac{0.887}{0.887}=1[/tex]
The ratio of C : H : Cl = 6 : 5 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_6H_5Cl_1[/tex]
The empirical formula weight = 6(12) + 5(1) + 1(35.5) = 112.5 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{112.5}{112.5}=1[/tex]
Molecular formula = [tex](C_6H_5Cl_1)_n=(C_6H_5Cl_1)_1=C_6H_5Cl_1[/tex]
Therefore, the molecular of the compound is, [tex]C_6H_5Cl_1[/tex]
