Answer:
[tex]C_{3}H_4O_2[/tex]
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:
[tex]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[/tex]
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:
[tex]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[/tex]
In such a way, the mass of oxygen comes from the mass of the compound minus the mass of carbon and oxygen:
[tex]m_O=50.1g-25.0g-2.79g=22.31gO[/tex]
And the moles:
[tex]n_O=22.31gO*\frac{1molO}{16gO}=1.39molO[/tex]
Then, we compute the subscripts by diving the moles of C, H and O by the moles of oxygen as the smallest moles:
[tex]C:\frac{2.09}{1.39}=1.5 \\\\H:\frac{2.79}{1.39}=2\\ \\O:\frac{1.39}{1.39} =1[/tex]
After that, we write:
[tex]C_{1.5}H_2O[/tex]
Which must be shown in whole number only, thereby we multiply the subscripts by 2, so the empirical formula turns out:
[tex]C_{3}H_4O_2[/tex]
Best regards.
