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An organic compound containing carbon hydrogen and chlorine has 10% of carbon, 0.84 % of hydrogen and rest chlorine. what is the empirical formula of the compound.​

Answer 1

Answer:

An organic compound contains 10% of carbon 0.84% of hydrogen ... Chlorine: 88.76gfind the no. of moles of each (no. of moles= total mass/ atomic mass): ... Mass of C + Mass of H + Mass of 3 atoms of Cl= empirical formula mass ... Explanation: c is 10. h is 0.84. but cl is rest. we know.

molecular mass is 49.

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Answer 2

$\mathcal{\gray{\underline{\underline{\blue{GIVEN:-}}}}}$

• An organic compound containing,

1. Carbon = 10%
2. Hydrogen = 0.84%
3. Chlorine = rest of percentage .

So, % of Chlorine

=> % Chlorine = 100% - [10% - 0.84%]

=> % Chlorine = 100% - 10.84%

=> % Chlorine = 89.16%

$\mathcal{\gray{\underline{\underline{\blue{TO\:FIND:-}}}}}$

• The empirical formula of the compound .

$\mathcal{\gray{\underline{\underline{\blue{SOLUTION:-}}}}}$

✍️ To calculate the empirical formula of the Organic compound, we must follow these following steps :-

STEP - 1 :-

☃️ First of all we can pick a “100.0 g” sample of the given organic compound and use it's percentage concentration by mass, to find that it contains,

• Carbon = 10g.

• Hydrogen = 0.84g.

• Chlorine = 89.16g.

STEP - 2 :-

☃️ Next use the molar masses of the three elements, to find how many moles of each we have in this sample .

$\orange\bigstar\:\rm{\pink{\overbrace{\underbrace{\purple{No.\:of\:moles\:=\:\dfrac{Total\:mass}{Atomic\:mass}\:}}}}}$

For “Carbon” :-

=> No. of moles of Carbon = 10g/12g

=> No. of moles of Carbon = 0.83 .

FOR “Hydrogen” :-

=> No. of moles of “H” = 0.84g/1g

=> No. of moles of “H” = 0.84 .

FOR “Chlorine” :-

=> No. of moles “Cl” = 89.16g/35.5g

=> No. of moles “Cl” = 2.51 .

STEP - 3 :-

☃️ To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one .

☞ Here,

• the smallest one = 0.83

FOR “Carbon” :-

• $\bf{\implies\:\dfrac{0.83\:moles}{0.83\:moles}}$

• $\bf\green{\implies\:1}$

FOR “Hydrogen” :-

• $\bf{\implies\:\dfrac{0.84\:moles}{0.83\:moles}}$

• $\bf\green{\implies\:1.01\:\approx\:1}$

FOR “Chlorine” :-

• $\bf{\implies\:\dfrac{2.51\:moles}{0.83\:moles}}$

• $\bf\green{\implies\:3.02\:\approx\:3}$

STEP - 4 :-

☃️ The compounds empirical formula, which tells us what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

$\green\bigstar\:\bf{\pink{\overbrace{\underbrace{\purple{C_1\:H_1\:Cl_3\:\implies\:CHCl_3\:}}}}}$

$\rule{200}2$

✍️ If we calculate all these above steps in tabular format then the output comes following .

$\rm\begin{array}{| c | c | c | c | c |}\cline{1 - 5} Elements & given \% & By mass & No. of moles & Mole ratio \\ \\ \cline{1 - 5} C & 10\% & 10 g & \dfrac{10}{12} = 0.83 & \dfrac{0.83}{0.83} = 1 \\ \\ \cline{1 - 5} H & 0.84\% & 0.84 g & \dfrac{0.84}{1} = 0.84 & \dfrac{0.84}{0.83} = 1.01 \approx 1 \\ \\ \cline{1 - 5} Cl & 89.16\% & 89.16 g & \dfrac{89.16}{35.5} = 2.51 & \dfrac{2.51}{0.83} = 3.02 \approx 3 \\ \\ \cline{1 - 5} \end{array}$

$\bf\red{\therefore}$ The empirical formula of the compound is “ $\bf\pink{CHCl_3}$ ” .