Question

A compound contains C, H, Cl and O. Combustion of 1.962 g of the compound gave 2.200 g CO2 and 0.676 g H2 O. In a separate analysis 1.208 g of the compound was converted into 2.205 g AgCl. The approximate molecular weight is 160. What is the empirical formula?

Answer 1

Answer: The empirical formula for the given compound is $C_{3}H_{5}O_{3}Cl$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, iron and oxygen follows:

$Cl_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'w', 'x', 'y' and 'z' are the subscripts of chlorine, carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=2.200g$

Mass of $H_2O=0.676g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 2.200 g of carbon dioxide, $\frac{12}{44}\times 2.200=0.6g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.676 g of water, $\frac{2}{18}\times 0.676=0.075g$ of hydrogen will be contained.

For calculating the mass of chlorine:

Now we have to calculate the moles of AgCl.

Moles of AgCl = $\frac{\text{Given mass of }AgCl}{\text{Molar mass of }AgCl}=\frac{2.205g}{143.32g/mole}=0.0154moles$

Now we have to calculate the moles of chlorine.

As we know that, 1 mole of AgCl dissociate to give 1 mole of silver ion and 1 mole of chloride ion.

So, the moles of chloride ion = Moles of AgCl = 0.0154 mole

Now we have to calculate the mass of chlorine.

$\text{Mass of chlorine}=\text{Moles of chlorine}\times \text{Molar mass of chlorine}$

$\text{Mass of chlorine}=(0.0154mole)\times (35.5g/mole)=0.5467g$

Now we have to calculate the mass percent of chlorine.

$\%\text{ mass of chlorine}=\frac{\text{Given amount}}{\text{Total amount}}\times 100=\frac{0.5467}{1.962}\times 100=27.86\%$

So, the amount of chlorine present in 1.962 g of compound = $0.2786\times 1.962=0.5466g$

Mass of oxygen in the compound = (1.962) - (0.6 + 0.075 + 0.5466) = 0.7404 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =$\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.6g}{12g/mole}=0.05moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.075g}{1g/mole}=0.075moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.7404g}{16g/mole}=0.0463moles$

Moles of Chlorine = $\frac{\text{Given mass of chlorine}}{\text{Molar mass of chlorine}}=\frac{0.5466g}{35.5g/mole}=0.0154moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{0.05}{0.0154}=3.24\approx 3$

For Hydrogen  = $\frac{0.075}{0.0154}=4.87\approx 5$

For Oxygen  = $\frac{0.0463}{0.0154}=3.006\approx 3$

For Chlorine  = $\frac{0.0154}{0.0154}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Cl : C : H : O = 1 : 3 : 5  : 3

Hence, the empirical formula for the given compound is $C_{3}H_{5}O_{3}Cl_1=C_{3}H_{5}O_{3}Cl$