Answer:
[tex]C_4H_{10}O[/tex]
Explanation:
Hello,
In this case, the first step is to compute the moles of carbon in the sample that are contained in CO2 only at the products as shown below:
[tex]n_C=8.612gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.196molC[/tex]
Next the moles of hydrogen contained in the H2O only:
[tex]n_H=4.406gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.490molH[/tex]
Now, we compute the mass of oxygen in the sample, by subtracting mass of both carbon and hydrogen from the 3.626 g of sample:
[tex]m_O=3.626g-0.196molC*\frac{12gC}{1molC}-0.490molH*\frac{1gH}{1molH} =0.784gO[/tex]
And the moles:
[tex]n_O=0.784gO*\frac{1molO}{16gO} =0.049molO[/tex]
Now, the mole ratios by considering the moles of oxygen as the smallest:
[tex]C=\frac{0.196mol}{0.049mol}= 4\\\\H=\frac{0.49mol}{0.049mol}=10\\\\O=\frac{0.049mol}{0.049mol}=1[/tex]
Thus, empirical formula is:
[tex]C_4H_{10}O[/tex]
Regards.