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

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}}= (46.47g)/(12g/mole)=3.87moles`

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

Moles of Cl =
`\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= (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 =
`(3.87)/(1.28)=3.02\approx 3`

For H =
`(7.80)/(1.28)=6.09\approx 6`

For Cl =
`(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}}=(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`

Answer 2

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