Question

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2},\) and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Short answer

(a) The empirical formula of the hydrocarbon is C2H3O. (b) 1.6 g of O2 were used in the reaction. (c) 1.87 g of O2 would have been required for complete combustion.

Step-by-step solution

Calculate the amount of moles of CO, CO2, and H2O formed

Using the molar masses of CO (\(28.01\mathrm{~g/mol}\)), CO2 (\(44.01\mathrm{~g/mol}\)), and H2O (\(18.02\mathrm{~g/mol}\)), let's calculate the amount of moles of each compound formed. moles of CO: \(\dfrac{0.467\mathrm{~g}}{28.01\mathrm{~g/mol}} = 0.0167\mathrm{~mol}\) moles of CO2: \(\dfrac{0.733\mathrm{~g}}{44.01\mathrm{~g/mol}} = 0.0167\mathrm{~mol}\) moles of H2O: \(\dfrac{0.450\mathrm{~g}}{18.02\mathrm{~g/mol}} = 0.025\mathrm{~mol}\)

Determine the amount of C, H, and O in the hydrocarbon and its empirical formula

There are two sources of carbon in the products: CO and CO2. Both of the carbon moles in CO and CO2 come from the hydrocarbon. Carbon: \(0.0167+0.0167=0.0334\mathrm{~mol}\) The moles of hydrogen in the hydrocarbon come from the water produced: Hydrogen: \(2\times0.025=0.050\mathrm{~mol}\) Oxygen: Total oxygen used in the reaction - Oxygen used to form CO and CO2 Total oxygen used in the reaction comes from the moles of CO and CO2 in the products and their oxygen content: \(0.0167+2\times0.0167=0.05\mathrm{~mol}\) So the empirical formula is C\(_{0.0334}\)H\(_{0.050}\)O\(_{0.0167}\). To obtain the simplest whole-number ratio of elements in the compound, we will divide each subscript by the smallest subscript: C: \(\dfrac{0.0334}{0.0167} = 2\) H: \(\dfrac{0.050}{0.0167} = 3\) O: \(\dfrac{0.0167}{0.0167} = 1\) The empirical formula of the hydrocarbon is C2H3O. (a) Empirical formula: C2H3O

Calculate the grams of O2 used in the reaction

To find the grams of O2 used, lets first find the moles of O2: moles of O2: \(0.05\mathrm{~mol}\) (from the previous calculations) Now, using the molar mass of O2 (\(32.00\mathrm{~g/mol}\)), we can find the grams of O2 used: grams of O2: \(0.05\mathrm{~mol} \times 32.00\mathrm{~g/mol} = 1.6\mathrm{~g}\) (b) Grams of O2 used in the reaction: 1.6 g

Calculate grams of O2 required for complete combustion

For complete combustion, the hydrocarbon would react with oxygen to produce only CO2 and H2O. The equation for the complete combustion would be: C2H3O + \(x\)O2 \(\rightarrow\) 2CO2 + \(1.5\)H2O To find the value of x, we'll balance the oxygen atoms: 2x = 4 + 3 x = 3.5 The balanced equation for complete combustion would be: C2H3O + 3.5O2 \(\rightarrow\) 2CO2 + \(1.5\)H2O Now, we can calculate the grams of O2 required for complete combustion: moles of O2: \(3.5 \times 0.0167\mathrm{~mol} = 0.05845\mathrm{~mol}\) grams of O2: \(0.05845\mathrm{~mol} \times 32.00\mathrm{~g/mol} = 1.87\mathrm{~g}\) (c) Grams of O2 required for complete combustion: 1.87 g

Key concepts

Combustion Reaction

Combustion reactions are chemical processes in which a substance combines with oxygen to release energy. When hydrocarbons, which are compounds made mostly of hydrogen and carbon, undergo combustion, they react with oxygen to produce carbon dioxide (\(\mathrm{CO_2}\)) and water (\(\mathrm{H_2O}\)). These reactions are exothermic, meaning they release energy in the form of heat and light. Incomplete combustion occurs when there is a limited supply of oxygen, leading to the production of carbon monoxide (\(\mathrm{CO}\)) in addition to \(\mathrm{CO_2}\) and \(\mathrm{H_2O}\).

• In complete combustion, enough oxygen is provided to fully oxidize the carbon and hydrogen into \(\mathrm{CO_2}\) and \(\mathrm{H_2O}\).
• Incomplete combustion occurs in low oxygen environments, producing \(\mathrm{CO}\) and potentially hazardous smoke.
• Combustion reactions are important for understanding energy release in fuels and environmental impact due to emissions.

Understanding these reactions is crucial for calculating the empirical formula of compounds resulting from hydrocarbons burning in insufficient oxygen supply.

Moles Calculation

The concept of moles is fundamental in chemistry for quantifying the amount of substance. A mole is the amount of a chemical substance that contains the same number of particles as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's number, approximately \(6.022 \times 10^{23}\).

• It allows chemists to count atoms and molecules by weighing them.
• In calculations, the formula used is \(\mathrm{moles} = \dfrac{\mathrm{mass \; in \; grams}}{\mathrm{molar \; mass \; in \; g/mol}}\)

In the initial problem, you calculated the moles of \(\mathrm{CO}\), \(\mathrm{CO_2}\), and \(\mathrm{H_2O}\) formed from known masses. Correctly determining these values is key to discovering how many atoms of each element were involved, ultimately aiding in finding the hydrocarbon's empirical formula.
Accurate calculations of moles ensure the empirical formula represents the simplest whole-number ratio of atoms in a compound, providing foundational insight into chemical composition.

Hydrocarbon Chemistry

Hydrocarbon chemistry focuses on compounds that consist entirely of hydrogen and carbon. These compounds can be found in several structures and forms, primarily as alkanes, alkenes, and alkynes. Hydrocarbons play a significant role because they are the primary sources of energy in fuels and raw materials in the chemical industry.

• Alkanes are saturated hydrocarbons, consisting only of single bonds between carbon atoms.
• Alkenes and alkynes are unsaturated hydrocarbons, containing double or triple bonds, respectively.

In the combustion process, hydrocarbons undergo reactions to produce energy. By understanding the type of hydrocarbon, you can predict the nature of the reaction and the resulting products.
In the given problem, the challenge was to determine the empirical formula of an unknown hydrocarbon by examining its combustion products. The empirical formula gives the simplest ratio of carbon, hydrogen, and oxygen; a necessary step in analyzing and utilizing a hydrocarbon in further applications.