Question

An unknown compound with a molar mass of 155.06 g/mol consists of 46.47% c, 7.80% h, and 45.72% cl. find the molecular formula for the compound.

Answer 1

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

Answer 2

Answer : The molecular of the compound is, $C_6H_{12}Cl_2$

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 = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{46.47g}{12g/mole}=3.87moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{7.80g}{1g/mole}=7.80moles$

Moles of Cl = $\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{45.72g}{35.5g/mole}=1.28moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.87}{1.28}=3.02\approx 3$

For H = $\frac{7.80}{1.28}=6.09\approx 6$

For Cl = $\frac{1.28}{1.28}=1$

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 = $C_3H_6Cl_1$

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 :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{155.06}{77.5}=2$

Molecular formula = $(C_3H_6Cl_1)_n=(C_3H_6Cl_1)_2=C_6H_{12}Cl_2$

Therefore, the molecular of the compound is, $C_6H_{12}Cl_2$