What is the empirical formula of a compound containing 5.03 grams carbon, 0.42 grams hydrogen, and 44.5 grams chlorine?

c2h3cl
c2h2cl
chcl4
chcl3

CHCl₃

Explanation:

Given parameters:

Carbon = 5.03g

Hydrogen = 0.42g

Chlorine = 44.5g

The empirical formula shows the simplest formula of a compound.

To deduce the empirical, we need two pieces of information:

> Mass of the elements or the percentage composition of the compound

>The relative atomic masses of the elements

In order to derive the empirical formula from these parameters,

>>> find the number of moles of elements by dividing the mass given by the relative atomic mass of the respective atom

>>> Divide through by the smallest mole

>>> Approximate or multiply by a factor that would make it possible for whole numbers to be obtained

From the question, we have been given the mass of each element.

Now using the period table, we can obtain the relative atomic masses of each atom:

Carbon = 12gmol⁻¹

Hydrogen = 1gmol⁻¹

Chlorine = 37.5gmol⁻¹

C H Cl

Mass(in g)5.03 0.42 44.5

Moles 5.03/12 0.42/1 44.5/37.5

0.42 0.42 1.19

Dividing

by

smallest 0.42/0.42 0.42/0.42 1.19/0.42

Mole ratio 1 1 2.83

Approximate 1 1 3

The empirical formula is CHCl₃

CHCl₃

Explanation:

Given parameters:

Carbon = 5.03g

Hydrogen = 0.42g

Chlorine = 44.5g

The empirical formula shows the simplest formula of a compound.

To deduce the empirical, we need two pieces of information:

> Mass of the elements or the percentage composition of the compound

>The relative atomic masses of the elements

In order to derive the empirical formula from these parameters,

>>> find the number of moles of elements by dividing the mass given by the relative atomic mass of the respective atom

>>> Divide through by the smallest mole

>>> Approximate or multiply by a factor that would make it possible for whole numbers to be obtained

From the question, we have been given the mass of each element.

Now using the period table, we can obtain the relative atomic masses of each atom:

Carbon = 12gmol⁻¹

Hydrogen = 1gmol⁻¹

Chlorine = 37.5gmol⁻¹

C H Cl

Mass(in g)5.03 0.42 44.5

Moles 5.03/12 0.42/1 44.5/37.5

0.42 0.42 1.19

Dividing

by

smallest 0.42/0.42 0.42/0.42 1.19/0.42

Mole ratio 1 1 2.83

Approximate 1 1 3

The empirical formula is CHCl₃

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.