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?

Question

12908 views around the world

You can reuse this answer
Creative Commons License

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#