Question

Complete combustion of 2.10 g of a hydrocarbon produced 6.75 g of CO2 and 2.30 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

$44g \ \ CO_{2} \ \ \ \ \ \rightarrow \ \ 12g \ \ C\\ 6,75g \ \ CO_{2} \ \ \rightarrow \ \ x \ \ C\\\\ x=\frac{6,75g*12g}{44g}=1,84g \ \ \ \Rightarrow \ \ n=\frac{1,84g}{12\frac{g}{mol}}=0,15mol\\\\\\ 18g \ \ H_{2}O \ \ \ \ \rightarrow \ \ 2g \ H\\ 2,3g \ \ H_{2}O \ \ \ \rightarrow \ \ x \ H\\\\ x=\frac{2,3g*2g}{18g}=0,26g \ \ \ \ \Rightarrow \ \ n=\frac{0,26g}{1\frac{g}{mol}}=0,26mol\\\\\\ n_{C}:n_{H}=0,15:0,26\approx1:2\\\\ empirical \ formula: \ \ CH_{2}$