Question

Complete combustion of 5.90 g of a hydrocarbon produced 19.2 g of CO2 and 5.89 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

`44g \ CO_(2) \ \ \ \rightarrow \ \ 12g \ C\\ 19,2g \ CO_(2) \rightarrow \ \ \ x\\\\ x=(19,2g*12g)/(44g)\approx5,24g \ \ \ \Rightarrow \ \ n=(5,24g)/(12(g)/(mol))\approx0,44mol\\\\\\ 18g \ H_(2)O \ \ \ \rightarrow \ \ \ 2g \ H\\ 5,89g \ H_(2)O \ \rightarrow \ \ y\\\\ y=(5,89g*2g)/(18g)\approx0,65g \ \ \ \ \Rightarrow \ \ n=(0,65g)/(1(g)/(mol))=0,65mol\\\\\\ n_(C):n_(H)=0,44:0,65\approx1:1\\\\ CH`