A 33.153 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis apparatus, yielding 59.060 mg of carbon dioxide and 24.176 mg of water. In another experiment, 47.029 mg of the compound is reacted with excess oxygen to produce 20.326 mg of sulfur dioxide. Add subscripts below to correctly identify the empirical formula of this compound (use this order of elements: CHSO)

The empirical formula of the compound =

Explanation:

Mass of carbon dioxide gas = 59.060 mg = 0.059060 g

1 mg = 0.001 g

Moles of carbon dioxide =

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 =

Mass of water = 24.176 mg = 0.024176

Moles of water =

Moles of hydrogen in 0.0013 moles of water = 2 × 0.0013 mol = 0.0026 mol

Mass of 0.0013 moles of hydrogen=

Mass of sulfur dioxide = 20.326 mg = 0.020326 g

Moles of sulfur dioxide =

Moles of sulfur in 0.00032 moles of sulfur dioxide = 1 × 0.00032 mol = 0.00032 mol

Mass of 0.00032 moles of sulfur =

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 =

For empirical formula divide the lowest number of moles of elemnt from all the moles of the all the elements:

Carbon :

Hydrogen:

Sulfur :

Oxygen :

The empirical formula of the compound =