Respuesta :
Answer:
The empirical formula of the compound = [tex]C_4H_8S_1O_1[/tex]
Explanation:
Mass of carbon dioxide gas = 59.060 mg = 0.059060 g
1 mg = 0.001 g
Moles of carbon dioxide = [tex]\frac{0.059060 g}{44 g/mol}=0.0013 mol[/tex]
Moles of carbon in 0.0013 moles of carbon dioxide gas = 1 × 0.0013 mol = 0.0013 mol
Mass of 0.0013 moles of carbon = [tex]12 g/mol\times 0.0013 mol=0.0156 g[/tex]
Mass of water = 24.176 mg = 0.024176
Moles of water = [tex]\frac{0.024176 g}{18 g/mol}=0.0013 mol[/tex]
Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol
Mass of 0.0013 moles of hydrogen= [tex]1 g/mol\times 0.0013 mol=0.0013 g[/tex]
Mass of sulfur dioxide = 20.326 mg = 0.020326 g
Moles of sulfur dioxide = [tex]\frac{0.020326 g}{64 g/mol}=0.00032 mol[/tex]
Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol
Mass of 0.00032 moles of sulfur = [tex]32 g/mol\times 0.00032 mol=0.01024 g[/tex]
Mass of oxygen in the sample = x
Mass of sample = 33.153 mg = 0.033153 g
0.033153 g = 0.0156 g + 0.0013 g + 0.01024 g + x
x = 0.006013 g
Moles of oxygen = [tex]\frac{0.006013 g}{16 g/mol}=0.00038 mol[/tex]
For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:
Carbon : [tex]\frac{0.0013 mol}{0.00032 mol}=4[/tex]
Hydrogen: [tex]\frac{0.0026 mol}{0.00032 mol}=8[/tex]
Sulfur : [tex]\frac{0.00032 mol}{0.00032 mol}=1[/tex]
Oxygen : [tex]\frac{0.00038 mol}{0.00032 mol}=1[/tex]
The empirical formula of the compound = [tex]C_4H_8S_1O_1[/tex]
