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Combustion analysis of a compound known to contain justC,H,N, and Odemonstrated that it is 46.21wt%C,9.02wt% H, 13.74wt N and, by difference, 100 - 46.21 - 9.02 -13.74 = 31.03% O. This means that of unknown would contain46.21gof C,9.02g ofH, and so on. Find the atomic ratio C : H : N : Oand express it as the lowest reasonable integer ratio.

Short Answer

Expert verified

The atomic ratio C : H : N : O is 3.992 : 9.120 : 1.000 : 1.978 .

Step by step solution

01

Derivation of atomic ratio.

The atomic ratio is a measure of the proportion of one type of atom I to another type of atom (ii) (j). The atomic percent (or at. percent) is a closely related concept that expresses the percentage of one type of atom relative to the total number of atoms. The molar fraction or molar percent are the molecular equivalents of these concepts.

02

Rewriting the compounds.

Compound known to contain just C,H,N and O with 46.21wt \% \mathrm{C}, 9.02 wt%H ,13.74 wt%N and, by difference, 100 - 46.21 - 9.02 - 13.74 = 31.03%O .

First consider that $100 \mathrm{~g}$ of compound would contain:

46.21gofC9.02gofH13.74gofN31.03gofO

03

Determining the atomic ratio.

Calculating the atomic ratio of C : H : N : O :

mM(C):mM(H):mM(N):mM(O):46.21g12.0107g/mol:9.02g1.00794g/mol:13.7414.00674g/mol:31.03g15.9994g/mol

Considering that 0.981 is the smallest factor we will divide all of the factors above with 0.981 which will then give us the ratio:

3.992 : 9.120 : 1.000 : 1.978

The empirical formula would then be C4H9NO2

Therefore the atomic ratio is 3.992 : 9.120 : 1.000 : 1.978 .

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