Respuesta :
Answer: The empirical formula for the given compound is [tex]CCl_{2}[/tex]
Explanation:
The chemical equation for the combustion of compound having carbon and chlorine follows:
[tex]C_xCl_y+O_2\rightarrow CO_2+Cl_2[/tex]
where, 'x' and 'y' are the subscripts of carbon and chlorine respectively.
We are given:
Mass of [tex]CO_2=1.37g[/tex]
We know that:
Molar mass of carbon dioxide = 44 g/mol
- For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 1.37 g of carbon dioxide, [tex]\frac{12}{44}\times 1.37=0.374g[/tex] of carbon will be contained.
- Mass of chlorine in the compound = (2.58) - (0.374) = 2.206 g
To formulate the empirical formula, we need to follow some steps:
- Step 1: Converting the given masses into moles.
Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.374g}{12g/mole}=0.031moles[/tex]
Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{2.206g}{35.45g/mole}=0.062moles[/tex]
- 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 0.031 moles.
For Carbon = [tex]\frac{0.031}{0.031}=1[/tex]
For Chlorine = [tex]\frac{0.062}{0.031}=2[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of C : Cl = 1 : 2
Hence, the empirical formula for the given compound is [tex]C_1Cl_2=CCl_2[/tex]
