Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of co2 and 0.2835 g of h2o. what is the empirical formula of the compound?

First, we have to get how many grams of C & H & O in the compound:

- the mass of C on CO2 = mass of CO2*molar mass of C / molar mass of CO2

= 0.5213 * 12 / 44 = 0.142 g

- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O

=0.2835 * 2 / 18 = 0.0315 g

- the mass of O = the total mass - the mass of C atom - the mass of H atom

= 0.3 - 0.142 - 0.0315 = 0.1265 g

Convert the mass to mole by divided by molar mass

C (0.142/12) H (0.0315/2) O (0.1265/16)

C (0.0118) H (0.01575) O (0.0079) by dividing by the smallest value 0.0079

C1.504 H3.99 O1 by rounding to the nearst fraction

C3/2 H4/1) 1/1 multiply by 2

∴ the emprical formula C3H8O2