Question

Determine the number of atoms of each element in the empirical formula of a compound with the following composition: 54.95 percent C, 8.455 percent H, 36.60 percent O.

Answer 1

Assuming that the total mass of compound is 100 g. The mass of the elements will be as follows: Mass of caibon = 54.95g Mass of hydrogen = 8.455g Mass of oxygen = 36.60g Calculate the number of moles of C as follows: Mass Moles = Molar Mass 54.95g 12g/mol = 4.58mol Calculate the number of moles of H as follows: Mass Moles = Molar Mass 8.455g 1g/mol = 8.455 mol

Calculate the number of moles of O as follows: Mass Moles = Molar Mass 36.60g 16g/mol = 2.28mol Calculate the least ratio as follows: 4.58 2.28 H 8.455 = 2.28 = 3.708

Multiply the ratio by 3. Rounding all the three ratios, the ratio obtained is as follows: C:H:0 :: 6:11:3 Therefore, the empirical formula of the given compound is C,H,Oą. Calculate the number of atoms of carbon as follows: Number of atoms = Moles Avogadro's constant = 4.58moles 6.022x1023 = 2.76x1024 atoms Calculate the number of atoms of hydrogen as follows: Number of atoms = Moles x Avogadro's constant = 8.455 moles 6.022x1023 = 5.09x1024 atoms Calculate the number of atoms of oxygen as follows: Number of atoms = Moles Avogadro's constant = 2.28moles x 6.022 x 1023 =1.37x1024 atoms