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

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