Hmm I'll need help with this one 30. Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen, andoxygen produced 2.265 g of CO2 and 1.236 g of H20. What is the empirical formulaof this compound?a) C3H80b) C3H50c) C6H1602d) C3H9O3e) C3H603

On combustion of the given sample all hydrogen atoms present in the sample become the part of water and all the carbon atoms present in sample are become the part of carbon dioxide. Oxygen atom is there in water as well as in carbon dioxide.Now we can calculate the no of moles of water and carbon dioxide formed on combustion of the given sample. Moles of water=1.236 g18 g/mol=0.069 moleMoles of CO2 =2.265 g44 g/mol=0.052 mole In 1 mole water we have 2 mole hydrogen. In 0.069 mole water we have 0.069×2=0.138 mole hydrogen. In 1 mole CO2 we have 1 mole carbon. In 0.052 mole CO2 we have 0.052 mole carbon.No of moles of any substance is the mass per molar mass of that substance. Mass=no of moles × molar massMass of H=0.138 mol×1 g/mol=0.138 gMass of C=0.052 mol×12 g/mol=0.624 gMass of O=Mass of sample-mass of H+mass of C=1.031 g-0.138 g+0.624 g=0.269 g Now we can calculate the no of moles of oxygen in 0.2756 g. Moles of O=massmolar mass=0.269 g16 g/mol=0.017 mol For deducing the empirical formula we have to find the simplest whole number ratio between C,H and O. C:H:O=0.0520.017:0.1370.017:0.0170.017=3:8:1 Hence the empirical formula of the given sample is C3H8O. Option a is correct.