Respuesta :
Answer: The empirical formula is C3H8S2NO.
Explanation: To determine the formula, find the porcentage of mass of each element in the sample:
1) For C (Molar Mass = 12u)
Note: molar mass of CO2 = 44g/mol
1 mol of CO2 = [tex]\frac{4,06}{44}[/tex] = 0,092g/mol
In 1 mol of CO2, 1 mol of C, so mass of C is
mC = 0,092 . 12 = 1,104g
[tex]\frac{mC}{msample}[/tex] . 100 = [tex]\frac{1,104}{3,76}[/tex] . 100 = 29,4%
2) For H (Molar Mass = 1u)
Note: molar mass of H20 = 18g/mol
1 mol of H2O = [tex]\frac{2,22}{18}[/tex] = 0,124 g/mol
In 1 mol of H2O, 2 mols of H, so mass is
mH = 0,124 . 2 . 1 = 0,248g
[tex]\frac{mH}{msample}[/tex] . 100 = [tex]\frac{0,248}{3,76}[/tex] . 100 = 6,6%
3) For S (Molar Mass = 32u)
Note: molar mass of SO3 = 80g/mol
1 mol of SO3 = [tex]\frac{3,73}{80}[/tex] = 0,046g/mol
In 1 mol of SO3, 1 mol of S, so its mass is:
mS = 0,046 . 32 = 1,472g
[tex]\frac{mS}{msample}[/tex] . 100 = [tex]\frac{1,472}{3,73}[/tex] . 100 = 39,5%
4) For N (Molar Mass = 14u)
Note: molar mass of HNO3 = 63g/mol
1 mol of HNO3 = [tex]\frac{4,40}{63}[/tex] = 0,07
In 1 mol of HNO3, 1 mol of N, so the mass is
mN = 0,07 . 14 = 0,98g
[tex]\frac{mN}{msample}[/tex] . 100 = [tex]\frac{0,98}{8,53}[/tex] . 100 = 11,5%
5) For O (Molar Mass = 16u).
O = 100% - (29,4%+6,6%+39,5%+11,5%)
O = 13%
It's determined the porcentage of the composition. Assuming that 100g of the compound, we calculte the mols of each:
Carbon = [tex]\frac{29,4}{12}[/tex] = 2,45 mol
Hidrogen = 6,6 mol
Sulfur = 1,25 mol
Nitrogen = 0,82 mol
Oxygen = 0,81 mol
Divide each by 0,81, the empirical formula of the compound is C3H8S2NO
