Question

30 pts, A combustion analysis is done on 522.4g of an unknown substance. The analysis produces 499.4g of CO2 and 204.5g of H2O. There are no other products. What is the empirical formula of the unknown substance?

Answer 1

Below

Step-by-step explanation:

The oxygen comes form the air used to burn the sample and the carbon and hydrogen come from the sample

Using periodic table:

499.2 of CO2 is 12.011/ (12.011 + 2* 15.999) * 499.4 = 136.297 g of C

204.5 of H2 o is 1.008 *2 / (1.008*2 + 15.999) * 204.5 = 22.885 g of H2

136.297 gm of C is 11.35 moles of C

22.885 gm of H2 is also 11.35 moles of H2

so the substance empirical formula is CH2

11.35 CH2 ===> 11.35 CO2 + 11.35 H2O <====> you can use this to check