Question

A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.25g of this compound was mixed with $N{{a}_{2}}C{{O}_{3}}$ to convert all Ca into 0.16g $CaC{{O}_{3}}$. A 0.115g sample of compound was carried through a series of reaction until all its S was changed into $S{{O}_{4}}^{2-}$ and precipitated as 0.344g of $BaS{{O}_{4}}$. A 0.712g sample was processed to liberate all of its N as $N{{H}_{3}}$ and 0.155g $N{{H}_{3}}$ was obtained. The formula mass was found to be 156. Determine the empirical formula of the compound.

Answer 1

Empirical formula is the simplest formula where the number of atoms of each element in a molecule are converted into real numbers. For this, the molar ratio needs to be determined, and then divide it by the smallest number.

Complete step by step solution:
First, according to the data given in the question, we will find the mass percentage of individual atoms. There are 4 elements i.e Ca, S, N and C. So,
Mass % of Ca = $(\dfrac{0.16}{100})\times 40\times (\dfrac{100}{0.25})=25.6$
Mass % of S = $(\dfrac{0.334}{233})\times 32\times (\dfrac{100}{0.115})=41$
Mass % of N = $(\dfrac{0.155}{17})\times 14\times (\dfrac{100}{0.712})=17.9$
Mass % of C = $100-(25.6-17.9-41)=15.48$
On simplifying, we obtain the molar ratios of the given individual elements as:
Ca:0.64
S: 1.28
N: 1.28
C: 1.29
We can observe that 0.64 is the smallest mol ratio among the 4. Now we will divide every number with 0.64 and round off to the nearest positive integer. Then the result is as follows:
Ca : 1
S: 2
N: 2
C: 2

This is the simplest ratio, so the empirical formula of the compound comes out to be $Ca{{S}_{2}}{{N}_{2}}{{C}_{2}}$.

It is not the actual formula and such a compound with these numbers of atoms does not exist in real life.

Note: It is to be noted that the number of atoms that have been rounded off to the nearest integer cannot be negative. Empirical formula is the simplest formula and every empirical formula might not be correct, in terms of real life. For example, the compound $Ca{{S}_{2}}{{N}_{2}}{{C}_{2}}$ does not exist in real life, but a similar compound with some alterations in the number of atoms might exist, according to the question.