- Firstly, we should write the equation of organic compound's combustion:
CxHyOz + O2 → xCO2 + y/2H2O
- According to the equation, we can calculate, for the beginning, amounts of substances of C and H:
ν (CO2) = m/M = 29.3(g) / 44(g/mol) = 0.67 mol = ν (C)
ν (H2O) = m/M = 11.7(g) / 18(g/mol) = 0.65 mol → ν (H) = 2 ⋅ ν(H2O) = 2 ⋅ 0.65(mol) = 1.3 mol
- Then, we can calculate the masses of these substances:
m(C) = ν⋅M = 0.67(mol) ⋅ 12(g/mol) = 8.04 g
m(H) = ν⋅M = 1.3(mol) ⋅ 1(g/mol) = 1.3 g
summary mass of m(C) and m(H) = 8.04(g) + 1.3(g) = 9.34(g)
- then m(O) will be calculated this way:
m(O) = m(compound)−m(summary) = 20(g) - 9.34(g) = 10.66(g) → ν(O) = m/M = 10.66(g) / 16(g/mol) = 0.67 mol
- And now we can determine the empirical formula of the compound, according to this ratio:
x:y:z=ν(C):ν(H):ν(O)
x : y : z = 0.67(mol) : 1.3(mol) : 0.67(mol) ∣ :0.67(mol)
x : y : z = 1 : 2 : 1
- So, the empirical formula of the compound is H2CO, which corresponds to formaldehyde
Answer: the empirical formula of the compound is H2CO (formaldehyde)
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