12.915 g of a chemical compound was burned in an atmosphere containing 50.123 g of oxygen. Subsequent analysis of the gaseous result yielded 18.942 g carbon dioxide, 7.749 g water, and 36.347 g of oxygen.
a) Determine the empirical formula of the substance
18.942/44.0 CO2; 7.749/18.0 H2O; 36.347/32.0 O2; Original O2 50.123/32.0
This gives you the number of moles of each substance.
double the moles of H2O for moles of H. For moles of O you need to take 2x final O2 value from (2xCO2 value + H2O value +2xO2 value).
To get empirical formula you now have ratio C:H:O; divide all these numbers by the smallest one and hopefully this gives approximate whole numbers that can be rounded. If any are .5 or .333 or .6667, then double all or 3x all gives the whole number ratios. Write as formula.