Question

Complete combustion of 6.50 g of a hydrocarbon produced 21.2 g of co2 and 6.49 g of h2o. what is the empirical formula for the hydrocarbon?

Answer 1

Answer:- Empirical formula is $C_2H_3$ .

Solution:-  Masses of the product carbon dioxide and water are given. These are converted to moles and then using mol ratio the moles of C and H are calculated as one mole of carbon dioxide has one mole of carbon and one mole of water has two moles of hydrogen. Then we calculate their ratio that gives empirical formula.

Let's do the calculations as:

Calculations for the moles of C:

$21.2gCO_2(\frac{1molCO_2}{44gCO_2})(\frac{1molC}{1molCO_2})$

= $0.482molC$

Calculations for the moles of H:

$6.49gH_2O(\frac{1molH_2O}{18gH_2O})(\frac{2molH}{1molH_2O})$

= $0.721molH$

For the ratio if moles of C and H we divide the moles of H by C since moles of C are less than moles of H.

H = $\frac{0.721}{0.482}$  = 1.5

C = $\frac{0.482}{0.482}$  = 1

For empirical formula, we need the whole number ratio, since 1.5 is not a whole number, we need to multiply both the numbers by 2 to make it a whole number.

So, H = 3  and C = 2

Hence, the empirical formula is $C_2H_3$ .