Determine the MF of a compound with. a. a molar mass of 110 g/mol that consists

Question

Determine the MF of a compound with. a. a molar mass of 110 g/mol that consists of 65.45 % C, 5.492 % H and 29.06 % O. b. a molar mass of 34.01 g/mol that consists of 5.93 % H and 94.09 % O: write the name of the compound c. a molar mass of 60.05 g/mol that consists of 40.00 % C, 6.714 % H and 53.29 % O: write the name of the compound d. a molar mass of 525.19 g/mol that consists of 76.29 % Hg, 5.33 % N and 18.28 % O: write the name of the compound e. a molar mass of 224.24 g/mol that consists of 57.19 % S and 42.81 % O: write the name of the compound f. a molar mass of 92.01 g/mol that consists of 30.45 % N and 69.56 % O: write the name of the compound a 2.78 mg sample of a particular compound that contains carbon, and oxygen produced 6.32 mg of carbon dioxide and 2.58 mg of water upon complete combustion. Determine the EF of the compound.

Explanation / Answer

It can be solved easily by using a table.

b)

Thus empirical formula = HO

Empirical formula mass = (1.01 +16.00) = 17.01g/mol

Molecular mass = 34.01

n = Molecular mass / Empirical formula mass = 34.01/17.01 = 2

Molecular formula = (empirical formula )n = (HO)2

Thus, Molecular formula = H2O2

e)

Thus empirical formula = S2O3

Empirical formula mass = (32.06×2)+(16.00×3) = 112.12 g/mol

Molecular mass =224.24

n = Molecular mass / Empirical formula mass = 224.24/112.12 = 2

Molecular formula = (empirical formula )n = (S2O3)2

Thus, Molecular formula = S4O6

Element percentage Relative number of atoms Dividing by smallest factor whole number ratio of atoms H 5.93 5.93/1.01 =5.87 5.87/5.87 = 1.00 1 O 94.09 94.09/16.00=5.88 5.88/5.87 = 1.00 1

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