Answer to Question #136345 in General Chemistry for Zahidul Islam

Question

A 3.77 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This produced 2.12 g CO2 and 1.30 g H2O . A second sample of this compound with a mass of 5.56 g produced 5.70 g SO3 . A third sample of this compound with a mass of 8.11 g produced 3.27 g HNO3 . Determine the empirical formula of the compound. Enter the correct subscripts on the given chemical formula.

Answer 1

QUESTION # 136345

A 3.77 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This produced 2.12 g CO2 and 1.30 g H2O . A second sample of this compound with a mass of 5.56 g produced 5.70 g SO3 . A third sample of this compound with a mass of 8.11 g produced 3.27 g HNO3 . Determine the empirical formula of the compound. Enter the correct subscripts on the given chemical formula.

ANSWER

From combustion of 3.77 g sample;

Mass of carbon from co₂ is 12/44×2.12= 0.578g

Mass of hydrogen from H₂O is 2/18×1.30 =0.144g

From combustion of 5.56 g sample;

Mass of sulphur from so₃ is 32/80×5.70=2.28g

If 5.56g compound gives 2.28g sulphur, 3.77g compound gives X

X=(3.77×2.28)÷5.56 =1.546g

From combustion of 8.11 g sample;

Mass of nitrogen from HNO₃ is 14/63×3.27=0.727g

If 8.11g compound gives 0.727g nitrogen, 3.77g compound gives X

X=(3.77×0.727)÷8.11= 0.338g

The mass of oxygen from sample =3.77-(0.578+0.144+1.546+0.338)=1.164

Elements       carbon           hydrogen      oxygen           nitrogen        sulphur

Mass              0.578              0.144              1.164              0.338              1.546

RAM               12                   1                     16                   14                   32

Moles             0.0482            0.144              0.0728            0.0239            0.0483

Mole ratio(0.0482/0.0239) (0.144/0.0239) (0.0728/0.0239) (0.0239/0.0239)( 0.0483/0.0239

                       =2                               =6          =3                         =1                      =2

The empirical formula is C₂H₆O₃NS₂