24- Complete combustion of 2.384 g of an organic acid containing C, H and O prod

Question

24- Complete combustion of 2.384 g of an organic acid containing C, H and O produced 5.735 g CO: and 2.013 g H:O. Calculate the Empirical formula of the compound 25- 2.766 g aluminum oxide is produced from the reaction of 1.722 g aluminum with 1.722 g oxygen. (a) Calculate the percent vield of the reaction. (b) Calculate the mass of the remaining reagent(s) at the end of the reaction. 26-An element has three isotopes: A (25.675 amu, 24.3% abundance): B (24.703 amu) and C (26.788 amu). If the average atomie mass of the element is 25.316 amu, calculate the abundance of the isotope B and C. 27- An organic compound containing C, H, N and O. In a complete reaction with oxygen 3.547 g of the on, 2.048 g of the produced 3.302 g water and 7.176 g carbon dioxide . In another reacti compound compound were treated with sodium hydroxide and produced 0.4001 g NHs. Calculate the empirical formula of the compound. (a) The electron capacity of the fourth shell is (b) The electron capacity of the 4f subshell is. (c) The four quantum numbers of the outermost electron in Al are: ,., (d) The number of unpaired electrons in Ni ion is (e) The electron capacity of a 3d orbital i (f) The subshell represented by (n 5 and is . Subshell. (g) The number of orbitals in the third shell 28-Complete: 29- Write equations representing the following processes: (a) The second ionization of copper. (b) The electron affinity of Br (c) The ionization of F (d) The electron affinity of Sn 30-Use Hess's law to Determine H° for the reaction: From the following data: N2(g) H2O2 31-A compound has the following composition: 36.14% C ; 25.30% N and 38.55% O. Calculate 1° =-622 kJ/mol 1° =-286 kJ/mol AH188 kJ/mol (a) N2H4) +02(g) + 2H20() Al (c) H2(g) O2(g) + its Empirical formula.

Explanation / Answer

24.

% of H = Atomic weight of H * mass of H2O * 100 / (molecular weight of H2O * mass of organic acid)

% of H = 2 * 2.013 * 100 / (18 * 2.384)

% of H = 9.38 %

Similarly

% of C = atomic weight of C * mass of CO2 * 100 / (Molecular weight of CO2 * mass of organic compound)

= 12 * 5.735 * 100 / (44 * 2.384)

= 65.6 %

Now, % of O = 100 - 65.6 - 9.38 = 25.02 %

Element Mass Molar mass No.of moles division by small number nearest whole number

C 65.6 12 65.6 / 12 = 5.47 5.47 / 1.56 = 3.50 * 2 7

H 9.38 1 9.38 / 1 = 9.38 9.38 / 1.56 = 6.00 * 2 12

O 25.02 16 25.02 / 16 = 1.56 1.56 / 1.56 = 1.00 * 2 2

Therefore, the emperical formula of the compound is: C7H12O2

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