Respuesta :
Explanation:
First, calculate the moles of [tex]CO_{2}[/tex] using ideal gas equation as follows.
PV = nRT
or, n = [tex]\frac{PV}{RT}[/tex]
= [tex]\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}[/tex] (as 1 bar = 1 atm (approx))
= 0.183 mol
As, Density = [tex]\frac{mass}{volume}[/tex]
Hence, mass of water will be as follows.
Density = [tex]\frac{mass}{volume}[/tex]
0.998 g/ml = [tex]\frac{mass}{3.26 ml}[/tex]
mass = 3.25 g
Similarly, calculate the moles of water as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{3.25 g}{18.02 g/mol}[/tex]
= 0.180 mol
Moles of hydrogen = [tex]0.180 \times 2[/tex] = 0.36 mol
Now, mass of carbon will be as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
0.183 mol = [tex]\frac{mass}{12 g/mol}[/tex]
= 2.19 g
Therefore, mass of oxygen will be as follows.
Mass of O = mass of sample - (mass of C + mass of H)
= 3.50 g - (2.19 g + 0.36 g)
= 0.95 g
Therefore, moles of oxygen will be as follows.
No. of moles = [tex]\frac{mass}{\text{molar mass}}[/tex]
= [tex]\frac{0.95 g}{16 g/mol}[/tex]
= 0.059 mol
Now, diving number of moles of each element of the compound by smallest no. of moles as follows.
C H O
No. of moles: 0.183 0.36 0.059
On dividing: 3.1 6.1 1
Therefore, empirical formula of the given compound is [tex]C_{3}H_{6}O[/tex].
Thus, we can conclude that empirical formula of the given compound is [tex]C_{3}H_{6}O[/tex].
