Question

an organic compound contains 78.6% carbon, 8.4% hydrogen and 12.95% nitrogen the molecular weight of the compound is 107. calculate the empirical and molecular formula of the compound​

Answer 1

Answer:

imperical formula is C7H9N2 and molecular formula is also the same..

Answer 2

Need to FinD :-

• The empirical Formula of Compound .

$\red{\frak{Given}}\begin{cases}\textsf{ An organic compound contains 78.6 \% carbon.}\\\textsf{ Also , it has 8.4 \% hydrogen and 12.95 \% nitrogen .}\\\textsf{ The Molecular weight is 107 .}\end{cases}$

• By Question :-

• The different percentage of the elements in the compound are given. We will divide those percentages , by thier Atomic weight to obtain the Relative no. of Atoms .
• Thereafter we will divide all Relative nos. with the smallest one b/w them .

$\rule{200}2$

• Here's the table :-

$\boxed{\begin{array}{c|c|c|c|c}\underline{\sf \red{ Element}} & \underline{\sf \red{ Percentage}} & \underline{\sf \red{ At. \ mass }} & \underline{\sf \red{ Relative\ no.} }& \underline{\sf \red{ Simplest\ Ratio }} \\\\\sf Carbon &\sf 78.6\% & \sf 12 &\sf \dfrac{78.6}{12}= 6.55 &\sf \dfrac{6.55}{0.925}= 7 \\\\ \sf Hydrogen &\sf 8.4\% & \sf 1 &\sf \dfrac{8.4}{1}= 8.4 &\sf \dfrac{8.4}{0.925}= 9 \\\\\sf Nitrogen &\sf \sf 12.95\% &\sf 14 &\sf \dfrac{12.95}{14} &\sf \dfrac{0.925}{0.925}= 1\end{array}}$

$\rule{200}2$

• Therefore :-

$\sf \to The \ Empirical\ Formula \ = \ \red{ C_7H_9N}$

• Now here the Empirical Formula weight , will be equal to 12(7)+(1)9+14 = 107
• According to Question , the molecular weight is also 107.

$\sf\dashrightarrow n = \dfrac{Molecular\ wt.}{Empirical\ wt.}\\\\\sf\dashrightarrow n =\dfrac{107}{107}\\\\\sf\dashrightarrow \boxed{\purple{\sf n = 1 }}$

$\rule{200}2$

• Hence the Compound is :-

$\sf \dashrightarrow Compound = (Empirical \ Formula)_n \\\\\sf \dashrightarrow Compound = (C_7H_9N)_1 \\\\\sf\dashrightarrow \boxed{\pink{\frak{Compound = C_7H_9N }}}$

$\rule{200}2$