For the problem involving calculations, in the spaces provided below every probl

Question

For the problem involving calculations, in the spaces provided below every problem on the exam paper. Specifically, every problem requiring the full work to be shown is marked with an asterisk, * If space is insufficient for a particular problem, you may use the back of the sheet on which the problem appears Omitting the full work on a problem will result in major deductions. * I. A certain inorganic compound consists only of sodium (Na), sulfur (S) and oxygen (O). Elemental analysis of this compound indicates that it contains 29.08 %Na and 40.56 %S by weight. The empirical formula of this compound is: NaSO Na_2S_3 Na_3S_4 Na_2SO_3 Na_2SO_4 Na_2S_2O_3 Na_2S_3O_2 Na_2S_3O_3 Na_2S_4O_6 Na_3S_4O_3 A certain organic compound consists only of carbon(C), hydrogen (H) and oxygen (O). Elemental analysis of a sample of this compound weighing 0.25000 g shows that it contains 0.10000 g C and 0.01678 g H. The molecular weight of this compound is determined to be 120.11 g mol^-1. The molecular formula of this compound is: CH_2O C_2H_4O_2 C_3H_4O_5 C_3H_6O_4 C_3H_6O_5 C_4H_4O_4 C_4H_8O_6 C_4H_8O_4 C_8H_16O_8

Explanation / Answer

Q1.

assume a basis of 100 g

so

29.0 g are Na

40.56 g of S

mass of O = 100-29.08-40.56 = 30.36

mol of Na = mass/MW = 29/23 = 1.260

mol of S= mass/MW = 40.56 /32 = 1.2675

mol of O = mass/MW = 30.36/16 = 1.8975

Na:O = 1.8975/1.260 = 1.5

so

Na2S2O3 is the empirical formula (F)

Q2.

m C = 0.1

m H = 0.01678

m = (0.25-0.1-0.01678) = 0.13322 g

mol of C = 0.1/12 = 0.0083333

mol of H = 0.01678/1 = 0.01678

mol of O = 0.13322 /16 = 0.008326

ratios

C:O = 0.0083333/0.008326 = 1

C:H = 0.01678/0.0083333 = "

so

C1H2O1

nearest is A

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