Question

Combustion analysis of 2.400 g of an unknown compound containing carbon, hydrogen, and oxygen produced 4.171 g of CO2and 2.268 g of H2O. What is the empirical formula of the compound

Answer 1

Answer:

$C_3H_8O_2$

Explanation:

Hello,

In this case, for the given combustion process, we can compute the present moles of carbon atoms by considering that the only product containing carbon is CO2, thus, we have:

$n_C=4.171gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2} =0.095molC$

Next, we compute the moles of hydrogen that is present at the water only as a product:

$n_H=2.268gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} =0.252molH$

Then, we compute the moles of oxygen by considering the oxygen present in both the CO2 and H2O and the initial sample:

$m_O=2.400g-0.095molC*\frac{12gC}{1molC} -0.252molH*\frac{1gH}{1molH} \\\\m_O=1.008gO\\\\n_O=1.008gO*\frac{1molO}{16gO} =0.063molO$

Now, we divide the moles of C, H and O by the moles of O since they are lower amount, in order to compute the subscript in the empirical formula:

$C=\frac{0.095}{0.063}=1.5\\ \\H=\frac{0.252}{0.063}=4\\ \\O=\frac{0.063}{0.063}=1$

But the smallest whole number is:

$C_3H_8O_2$

Regards.