Question

A 3.76 g sample of a compound consisting of carbon, hydrogen, oxygen, nitrogen, and sulfur was combusted in excess oxygen. This produced 4.06 g CO 2 and 2.22 g H 2 O . A second sample of this compound with a mass of 5.69 g produced 3.73 g SO 3 . A third sample of this compound with a mass of 8.53 g produced 4.40 g HNO 3 . Determine the empirical formula of the compound. Enter the correct subscripts on the given chemical formula.?

Answer 1

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 = $\frac{4,06}{44}$ = 0,092g/mol

In 1 mol of CO2, 1 mol of C, so mass of C is

mC = 0,092 . 12 = 1,104g

$\frac{mC}{msample}$ . 100 = $\frac{1,104}{3,76}$ . 100 = 29,4%

2) For H (Molar Mass = 1u)

Note: molar mass of H20 = 18g/mol

1 mol of H2O = $\frac{2,22}{18}$ = 0,124 g/mol

In 1 mol of H2O, 2 mols of H, so mass is

mH = 0,124 . 2 . 1 = 0,248g

$\frac{mH}{msample}$ . 100 = $\frac{0,248}{3,76}$ . 100 = 6,6%

3) For S (Molar Mass = 32u)

Note: molar mass of SO3 = 80g/mol

1 mol of SO3 = $\frac{3,73}{80}$ = 0,046g/mol

In 1 mol of SO3, 1 mol of S, so its mass is:

mS = 0,046 . 32 = 1,472g

$\frac{mS}{msample}$ . 100 = $\frac{1,472}{3,73}$ . 100 = 39,5%

4) For N (Molar Mass = 14u)

Note: molar mass of HNO3 = 63g/mol

1 mol of HNO3 = $\frac{4,40}{63}$ = 0,07

In 1 mol of HNO3, 1 mol of N, so the mass is

mN = 0,07 . 14 = 0,98g

$\frac{mN}{msample}$ . 100 = $\frac{0,98}{8,53}$ . 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 = $\frac{29,4}{12}$ = 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