Question

An organic compound containing only C, H, and possibly O was subjected to combustion analysis. A
sample weighing 0.4119 g yielded 0.666 g CO2 and 0.327 g H2O. What is the empirical formula of the
compound?

Answer 1

.666g CO2 / (molar mass of CO2) = moles of CO2 = moles of C since there is 1 mole of C per 1 mole of CO2 (no subscript)

.327g H2O / (molar mass of H2O) = moles of H2O (multiply this by 2 to get moles of H)

.666g CO2 / (molar mass of CO2) = moles of CO2 (multiply by 2 again for moles of O)
.327g H2O / (molar mass of H2O) = moles of H2O = moles of O
***add the two values above to get total moles of O

divide the number of moles of each element by the number of moles of the element with the lowest number of moles (Ex. if O had the least number of moles, divide moles of H by moles of O, divide moles of C by moles of O, and divide moles of O by moles of O). Then round to closest whole number.

*mass of sample is not needed unless you need to find the molecular formula*