Question

A 4.215 g sample of a compound containing only carbon, hydrogen, and oxygen is burned in an excess of oxygen gas, producing 9.582 g CO2 and 3.922 g H2O. What percent by mass of oxygen is contained in the original sample?

Answer 1

Answer:

$\% O=27.6\%$

Explanation:

Hello,

In this case, for the sample of the given compound, we can compute the moles of each atom (carbon, hydrogen and oxygen) that is present in the sample as shown below:

- Moles of carbon are contained in the 9.582 grams of carbon dioxide:

$n_C=9.582gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2} =0.218molC$

- Moles of hydrogen are contained in the 3.922 grams of water:

$n_H=3.922gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.436molH$

- Mass of oxygen is computed by subtracting both the mass of carbon and hydrogen in carbon dioxide and water respectively from the initial sample:

$m_O=4.215g-0.218molC*\frac{12gC}{1molC} -0.436molH*\frac{1gH}{1molH} =1.163gO$

Finally, we compute the percent by mass of oxygen:

$\% O=\frac{1.163g}{4.215g}*100\% \\\\\% O=27.6\%$

Regards.