Respuesta :
CxHy + (x+0.25)O₂ → xCO₂ + 0.5yH₂O
m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}
0.2845/{44.01x}=0.1451/{9.01y}
x/y=0.4=2:5
C₂H₅ is the empirical formula
m(CO₂)/{xM(CO₂)}=m(H₂O)/{0.5yM(H₂O)}
0.2845/{44.01x}=0.1451/{9.01y}
x/y=0.4=2:5
C₂H₅ is the empirical formula
The empirical formula of the unknown compound is [tex]C_2H_5.[/tex]
Given:
The 0.2845 g of carbon dioxide and 0.1451 g of water were produced on combustion of an unknown compound.
To find:
The empirical formula of the compound.
Solution
Mass of carbon dioxide produced = 0.2845 g
Moles of carbon dioxide produced:
[tex]=\frac{0.2845 g}{44.01 g/mol}=0.006464 mol[/tex]
In one mole of carbon dioxide there is 1 mole of carbon, then in 0.006464 moles of carbon dioxide:
[tex]=1\times 0.006464 mol=0.006464\text {mol of C}[/tex]
Mass of water produced = 0.1451 g
Moles of water produced:
[tex]=\frac{0.1451 g}{18.01528 g/mol}=0.008054 mol[/tex]
In one mole of water there are 2 moles of hydrogen, then in 0.008054 moles of water:
[tex]=1\times 0.008054 mol=0.01611\text {mol of H}[/tex]
The moles of carbon and hydrogen are coming from the combustion of an unknown compound
The empirical formula of the compound = [tex]C_xH_y[/tex]
Moles of carbon =0.006464 mol
Moles of hydrogen = 0.01611mol
[tex]x=\frac{0.006464 mol}{0.006464 mol}=1\\\\y=\frac{ 0.01611mol}{0.006464 mol}=2.5[/tex]
The empirical formula of the compound:
[tex]C_xH_y=C_1H_{2.5}=C_{10}H_{25}=C_2H_5[/tex]
The empirical formula of the unknown compound is [tex]C_2H_5.[/tex]
Learn more about empirical formula here:
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