Answer:
C₃H₈O
Explanation:
The following data were obtained from the question:
Mass of compound = 0.255 g
Mass of CO2 = 0.561 g
Mass of H2O = 0.306 g
Empirical formula =?
Next, we shall determine the mass of carbon (C), hydrogen (H) and oxygen (O) present in the compound. This can be obtained as follow:
For Carbon (C):
Mass of CO2 = 0.561 g
Molar mass of CO2 = 12 + (2×16) = 12 + 32 = 44 g/mol
Mass of C = 12/44 × 0.561
Mass of C = 0.153 g
For Hydrogen (H):
Mass of H2O = 0.306 g
Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol
Mass of H = 2/18 × 0.306
Mass of H = 0.034 g
For Oxygen (O):
Mass of compound = 0.255 g
Mass of C = 0.153 g
Mass of H = 0.034 g
Mass of O = Mass of compound – (Mass of C + Mass of H)
Mass of O = 0.255 – ( 0.153 + 0.034)
Mass of O = 0.255 – 0.187
Mass of O = 0.068 g
Finally, we shall determine the empirical formula of the compound. This can be obtained as follow:
C = 0.153 g
H = 0.034 g
O = 0.068 g
Divide by their molar mass
C = 0.153 / 12 = 0.0128
H = 0.034 / 1 = 0.034
O = 0.068 / 16 = 0.0043
Divide by the smallest
C = 0.0128 / 0.0043 = 3
H = 0.034 /0.0043 = 8
O = 0.0043 / 0.0043 = 1
Thus, the empirical formula of the compound is C₃H₈O
