Question #108578
When 20gof compound that contains carbon, hydrogen and oxygen only was burnt in air,29.3gof carbon (iv) oxide and 11.7g of water were produced . calculate the empirical formula of the compound
1
Expert's answer
2020-04-08T13:26:44-0400
  • Firstly, we should write the equation of organic compound's combustion:

CxHyOz + O2 \to xCO2 + y/2H2O

  • According to the equation, we can calculate, for the beginning, amounts of substances of C and H:

ν\nu (CO2) = m/Mm/M = 29.3(g) / 44(g/mol) = 0.67 mol = ν\nu (C)

ν\nu (H2O) = m/Mm/M = 11.7(g) / 18(g/mol) = 0.65 mol \to ν\nu (H) = 2 ⋅ ν\nu(H2O) = 2 ⋅ 0.65(mol) = 1.3 mol

  • Then, we can calculate the masses of these substances:

m(C) = νM\nu ⋅ M = 0.67(mol) ⋅ 12(g/mol) = 8.04 g

m(H) = νM\nu ⋅ 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)m(compound) - m(summary) = 20(g) - 9.34(g) = 10.66(g) \to ν\nu(O) = m/Mm/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 = \nu(C): \nu(H):\nu(O)

x : y : z = 0.67(mol) : 1.3(mol) : 0.67(mol) \mid :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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