Respuesta :
The compound contain only carbon, hydrogen and oxygen, let the empirical formula of compound be [tex]C_{x}H_{y}O_{z}[/tex].
The mass of compound, carbon dioxide and water is 43.9 mg, 119 mg and 24.4 mg respectively . The molar mass of compound, carbon dioxide and water is 162 g/mol, 44.01 g/mol and 18 g/mol respectively.
Converting the mass into number of moles as follows:
[tex]n=\frac{m}{M}[/tex]
Here, m is mass and M is molar mass.
Number of moles of carbon dioxide [tex]CO_{2}[/tex] will be:
[tex]n=\frac{119\times 10^{-3}}{44.01 g/mol }=0.0027 mol[/tex]
1 mol of [tex]CO_{2}[/tex] has 1 mol of C thus, number of moles of C from 0.0027 mol of [tex]CO_{2}[/tex] will be 0.0027 mol.
Molar mass of C is 12 g/mol thus, mass of C will be:
[tex]m_{C}=0.0027 mol\times 12 g/mol=0.0324 g[/tex]
Number of moles of water [tex]H_{2}O[/tex] will be:
[tex]n=\frac{24.4\times 10^{-3}}{18 g/mol }=0.001356 mol[/tex]
1 mol of [tex]H_{2}O[/tex] has 2 mol of H thus, 0.001356 mol of [tex]H_{2}O[/tex] will have [tex]2\times 0.001356 mol=0.002712 mol[/tex] mole of H.
Molar mass of H is 1 g/mol thus, mass of H will be:
[tex]m_{H}=0.002712 mol\times 1 g/mol=0.002712 g[/tex]
Sum of mass of C and H will be:
[tex]m_{C+H}=0.0324+0.002712=0.035112 g[/tex]
Mass of compound is 43.9 mg or 0.0439 g thus, mass of oxygen will be:
[tex]m_{O}=m_{compound}-m_{C+H}=(0.0439-0.035112)g=0.008788 g[/tex]
Molar mass of oxygen is 16 g/mol thus, number of moles of oxygen will be:
[tex]n=\frac{0.008788 g}{16 g/mol}=0.00055 mol[/tex]
Taking the molar ratio:
C:H:O=0.0027:0.002712:0.00055=5:5:1
Therefore, empirical formula will be [tex]C_{5}H_{5}O[/tex].
The molar mass of [tex]C_{5}H_{5}O[/tex] is 81 g/mol
The molar mass of given compound is 162 that is 2 times the molar mass calculated above thus, molecular formula will be [tex]2\times C_{5}H_{5}O=C_{10}H_{10}O_{2}[/tex].
Therefore, empirical formula and molecular formula of the compound is [tex]C_{5}H_{5}O[/tex] and [tex]C_{10}H_{10}O_{2}[/tex] respectively.