Answer to Question #136979 in General Chemistry for Miça

Question

A 0.100g sample of a compound C,H, and O is burned in oxygen producing 0.1783g of CO2 and 0.0734g of H2O. Determine the empirical formula of the compound, listing elements in the order of CnHnOn

Answer 1

Solution:

We need to find out the percent composition of hydrogen in water (H₂O) and of carbon in carbon dioxide (CO₂). This will allow to find out how many grams of hydrogen and carbon, respectively, were produced by the combustion reaction.

One mole of water (H₂O) contains 2 moles of hydrogen.

The percent composition of hydrogen in water is:

(2 × 1.00794 g/mol) / (18.015 g/mol) = 0.1119 or 11.19%

This means that every 100 grams of water contain 11.19 grams of hydrogen.

In our case, a 0.0734 g sample of water contains:

0.0734 g H₂O × (11.19 g H / 100 g H₂O) = 0.00821 g H

One mole of carbon dioxide (CO₂) contains one mole of carbon.

The percent composition of carbon in carbon dioxide is:

(1 × 12.011 g/mol) / (44.01 g/mol) = 0.2729 or 27.29%

This means that every 100 grams of carbon dioxide contain 27.29 grams of carbon.

In our case, a 0.1783 g sample of carbon dioxide contains:

0.1783 g CO₂ × (27.29 g C / 100 g CO₂) = 0.04866 g C

The initial mass of the sample is equal to 0.100 g.

This suggests that the sample also contains:

m_oxygen = m_sample - m_hydrogen - m_carbon

m_oxygen = 0.100 g - 0.00821 g - 0.04866 g = 0.04313 g

Using the molar masses and masses of each elements to convert their to moles:

For C: 0.04866 g / 12.011 g mol^-1 = 0.00405 moles C

For H: 0.00821 g / 1.00794 g mol^-1 = 0.008145 moles H

For O: 0.04313 g / 15.9994 g mol^-1 = 0.002696 moles O

Coefficients for the tentative empirical formula are derived by dividing each molar amount by the lesser of the three:

0.00405 moles C / 0.002696 = 1.5

0.008145 moles H / 0.002696 = 3.0

0.002696 moles O / 0.002696 = 1.0

This means that: C_1.5H₃O₁

Since we are looking for the smallest whole number ratio that exists between the elements, multiply each subscript by 2 to get:

(C_1.5H₃O₁)₂ ⇒ C₃H₆O₂

Answer: The empirical formula of the compound is C₃H₆O₂