Question

The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 3.626 g sample of ether was combusted in an oxygen rich environment to produce 8.612 g of CO2(g) and 4.406 g of H2O(g). Insert subscripts to complete the empirical formula of ether.

Answer 1

Answer:

$C_4H_{10}O$

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:

$n_C=8.612gCO_2*\frac{1molCO_2}{44gCO_2} *\frac{1molC}{1molCO_2} =0.196molC$

Next the moles of hydrogen contained in the H2O only:

$n_H=4.406gH_2O*\frac{1molH_2O}{18gH_2O} *\frac{2molH}{1molH_2O} =0.490molH$

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:

$m_O=3.626g-0.196molC*\frac{12gC}{1molC}-0.490molH*\frac{1gH}{1molH} =0.784gO$

And the moles:

$n_O=0.784gO*\frac{1molO}{16gO} =0.049molO$

Now, the mole ratios by considering the moles of oxygen as the smallest:

$C=\frac{0.196mol}{0.049mol}= 4\\\\H=\frac{0.49mol}{0.049mol}=10\\\\O=\frac{0.049mol}{0.049mol}=1$

Thus, empirical formula is:

$C_4H_{10}O$

Regards.