A sample of a compound containing only carbon and hydrogen is burned, producing 157 mg CO2 and 42.8 mg H2O. What is the empirical formula?

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Answer 1 · 2017-09-06T00:35:06

Combustion analysis!

#157mg * (10^-3g)/(mg) * (CO_2)/(44.01g) approx 3.57*10^-3mol#

#42.8mg * (10^-3g)/(mg) * (H_2O)/(18.02g)* (2H)/(H_2O) approx 4.75*10^-3mol#

If you divide the largest molar mass by the smallest...

#C_1.33H#

Annoying... multiply by three so integers are the subscripts:

#C_4H_3#