Question

Complete combustion of 3.90 g of a hydrocarbon produced 12.5 g of CO2 and 4.28 g of H2O. What is the empirical formula for the hydrocarbon?

Answer 1

`44g \ CO_(2) \ \ \ \ \rightarrow \ \ \ 12g \ C\\ 12,5g \ CO_(2) \ \ \rightarrow \ \ \ x\\\\ x=(12,5g*12g)/(44g)=3,41g \ \ \ \Rightarrow \ \ \ n=(3,41g)/(12(g)/(mol))=0,28mol\\\\\\ 18g \ H_(2)O \ \ \ \ \ \ \rightarrow \ \ \ 2g \ \ H\\ 4,28g \ H_(2)O \ \ \ \rightarrow \ \ \ x\\\\ x=(4,28g*2g)/(18g)=0,48g \ \ \ \Rightarrow \ \ \ n=(0,48g)/(1(g)/(mol))=0,48mol\\\\\\\\ n_(c):n_(x)=0,28:0,48\approx1:2\\\\\\ empirical \ formula: \ CH_(2)`

Answer 2

Empirical formula of the hydrocarbon is
`C_1H_2`.

Step-by-step explanation:

Mass of carbon dioxide produced = 12.5 g

Moles of carbon dioxide =
`(12.5 g)/(44 g/mol)=0.2841 mol`

Moles of carbon atom in 0.2841 moles of carbon dioxide:

1 × 0.2841 mol = 0.2841 mol

Mass of water produced = 4.28 g

Moles of water =
`(4.28 g)/(18 g/mol)= 0.2378 mol`

Moles of hydrogen in water = 2 × 0.2378 mol = 0.4756 mol

For empirical formula divide the moles of element which are is less amount with all the moles of every element.

Carbon =
`(0.2841 mol)/(0.2841 mol)=1`

Hydrogen =
`(0.4756 mol)/(0.2841 mol)=1.67\approx 2`

Empirical formula of the hydrocarbon is
`C_1H_2`.