Question

A compound containing only carbon, hydrogen, and oxygen is analyzed using combustion analysis. When 50.1 g of the compound is burned, 91.8 g of carbon dioxide and 25.1 g of water are collected. In order to determine the moles of carbon in the compound, first determine the moles of carbon dioxide that were produced from the combustion.

Answer 1

Answer:

$C_{3}H_4O_2$

Explanation:

Hello,

In this case, since the carbon of the initial compound is present in the carbon dioxide product, we can compute the mass and moles of carbon in the compound:

$n_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =2.09molC\\\\m_C=91.8g CO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2}*\frac{12gC}{1molC} =25.0gC$

Next, the mass and moles of hydrogen in the compound, is contained in the yielded amount of water, thus, we compute the mass and moles of hydrogen in the compound:

$n_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =2.79molH\\\\m_H=25.1gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} *\frac{1gH}{1molH} =2.79gH$

In such a way, the mass of oxygen comes from the mass of the compound minus the mass of carbon and oxygen:

$m_O=50.1g-25.0g-2.79g=22.31gO$

And the moles:

$n_O=22.31gO*\frac{1molO}{16gO}=1.39molO$

Then, we compute the subscripts by diving the moles of C, H and O by the moles of oxygen as the smallest moles:

$C:\frac{2.09}{1.39}=1.5 \\\\H:\frac{2.79}{1.39}=2\\ \\O:\frac{1.39}{1.39} =1$

After that, we write:

$C_{1.5}H_2O$

Which must be shown in whole number only, thereby we multiply the subscripts by 2, so the empirical formula turns out:

$C_{3}H_4O_2$

Best regards.