Respuesta :
Answer: The empirical formula for the given compound is [tex]PbCl_2[/tex]
Explanation:
We are given:
Mass of beaker, [tex]m_1[/tex] = 204.35 g
Mass of beaker and lead before the reaction, [tex]m_2[/tex] = 214.71 g
Mass of beaker and lead chloride after the reaction, [tex]m_3[/tex] = 218.26 g
Mass of lead = [tex]m_2-m_1[/tex] = [214.71 - 204.35] = 10.36 g
Mass of chlorine = [tex]m_3-m_2[/tex] = [218.26 - 214.71] = 3.55 g
To formulate the empirical formula, we need to follow some steps:
- Step 1: Converting the given masses into moles.
Moles of Lead =[tex]\frac{\text{Given mass of Lead}}{\text{Molar mass of Lead}}=\frac{10.36g}{207.2g/mole}=0.05moles[/tex]
Moles of Chlorine = [tex]\frac{\text{Given mass of Chlorine}}{\text{Molar mass of Chlorine}}=\frac{3.55g}{35.5g/mole}=0.1moles[/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 0.05 moles.
For Lead = [tex]\frac{0.05}{0.05}=1[/tex]
For Chlorine = [tex]\frac{0.1}{0.05}=2[/tex]
- Step 3: Taking the mole ratio as their subscripts.
The ratio of Pb : Cl = 1 : 2
Hence, the empirical formula for the given compound is [tex]PbCl_2[/tex]
