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A 0.3832-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7120 g of CO2 and 0.1458 g of H2O. What is the empirical formula of the compound

User Kpalser
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Answer:


C_3H_3O_2

Step-by-step explanation:

Hello,

In this case, since the law of conservation of mass allows us to notice that the mass of carbon in the burned compound is also present in the resulting carbon dioxide, we can compute such moles as shown below:


n_C=0.7120gCO_2*(1molCO_2)/(44gCO_2) *(1molC)/(1molCO_2) =0.0162molC

Next, hydrogen in the sample is present at the products in the water only, and in one mole of water, two moles of hydrogen are present, thereby:


n_H=0.1458gH_2O*(1molH_2O)/(18gH_2O)*(2molH)/(1molH_2O)=0.0162molH

Nevertheless, the amount of oxygen in the sample must be computed by subtracting both the mass of carbon and hydrogen from the previously computed moles:


m_O=0.3832g-0.0162molC*(12gC)/(1molC) -0.0162molH*(1gH)/(1molH) =0.1726gO

And the moles:


n_O=0.1726gO*(1molO)/(16gO) =0.0108molO

Next, we compute the mole ratios with respect to the element having the smallest number of moles (oxygen) to obtain the subscripts in the empirical formula:


C=(0.0162)/(0.0108)=1.5\\ \\H=(0.0162)/(0.0108)=1.5\\\\O=(0.0108)/(0.0108)=1

Then, we have:


C_(1.5)H_(1.5)O_1

Finally, by multiplying by two, we obtain the smallest whole subscripts:


C_3H_3O_2

Best regards.

User Avin Kavish
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