Question

An inorganic salt gave the following percentage composition sodium is 29.1% ,sulphur is 40.51%, oxygen is 30.38%. Calculate the emperical formula of salt.

Answer 1

therefore empirical formula is Na2S2O3

Answer 2

Answer:

The empirical formula for the inorganic salt given is $\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$

Explanation:

Empirical Formula of a salt is the simplest positive integer ratio of all the components of the salt or the compound. We are given Sodium=29.1%, Sulphur=40.51% and Oxygen=30.38%.

We first calculate the mass in grams of each of the elements. For this, we consider the mass of each element= percentage given. So we get:

Sodium= 29.1 gm

Sulphur= 40.51 gm

Oxygen= 30.38 gm

We now calculate the molar mass of each of the elements. We know that the molar weight of sodium= 23 gm, sulphur= 32 gm and oxygen= 16 gm.

We now calculate the molar mass of the three elements:

Sodium or Na= (29.1× 1mol Na)÷23= 1.26 mol Na

Sulphur or S= (40.51×1mol S)÷32 = 1.26 mol S

Oxygen or O=(30.38 ×1mol O)÷16=1.89 mol O

Now, we divide the molar mass of each element with the smallest molar weight calculated to get an integer value for all the elements.

Na= 1.26/1.26 mol Na= 1

S= 1.26/1.26mol S= 1

O= 1.89/1.26 mol O=1.5  

We now have to factorize to get the lowest whole number. Since one solution is 1.5, we will multiply all the solutions by 2.

So we get, Na= 1×2= 2

S= 1×2= 2

O=1.5×2=3

The ratio of the elements Na:S:O= 2:2:3

So we get the empirical formula as $\bold{\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}}$