Question

A newly discovered pollutant in fossil fuels was characterized to determine its composition. Elemental analysis shows the presence of carbon, hydrogen, and sulfur elements in the compound. When burned with excess oxygen 0.397 g carbon dioxide, 0.190 g water, and 0.193 g sulfur dioxide were found as products. What is the empirical formula of this compound? If 150 g of fuel contains 3.6 g of this pollutant what is the mass percent of the pollutant in fuel?

Answer 1

Moles

Carbon dioxide $=\frac{0.397}{44.01}=0.009moles$

Water $=\frac{0.190}{18.02}=0.01moles$

Sulfur oxide $=\frac{0.193} {64.066}=0.003$

Ratio $,\frac{0.009}{0.003}=3 ,\frac{0.01}{0.009}=3.3 ,\frac{0.003}{0.003}=1$