Complete combustion of 2.00 g of a hydrocarbon produced 6.39 g of CO2 and 2.29 g of H2O. What is the empirical formula for the hydrocarbon?

Question

anonymous · Answer

$$2C_nH_{2n+2} + (3n+1)O_2\rightarrow (2n+2)H_2O+2nCO_2$$
for example if $n =1$
$$2CH_{4} +4O_2\rightarrow 4H_2O+2CO_2$$

and you need the molair mass for all the atoms to calculate the molair mass of all the molecules

apromimately
1 mole of C = 12 g
1 mole of O = 16 g
1 mole of H = 1 g

so
1 mole of $CO_2$ = 44 g
1 mole  of $H_2O$ = 18 g
1 mole of $O_2$= 32 g

hence
6.39g of $CO_2$ is about 0.145 mole of $CO_2$  corresponds to 2$n$
2.29g of $H_2O$ is about 0.127 mole of $H_2O$ corresponds to 2$n$+2
2g      of $O_2$    is about  0.125 or $\frac{1}{8}$  corresponds to 3$n$+1