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 compound is C2H3. (b) 2.40 g of O2 were used in the reaction. (c) 2.93 g of O2 would have been required for complete combustion.

Step-by-step solution

Calculate moles of products and oxygen used

First, we need to convert the given mass of each product into moles. We will use the molar mass of each component (C = 12.01 g/mol, O = 16.00 g/mol, H = 1.008 g/mol). Moles of CO: \(\frac{0.467 \mathrm{~g}}{28.01 \mathrm{~g/mol}} = 0.0167 \mathrm{~mol}\) Moles of CO2: \(\frac{0.733 \mathrm{~g}}{44.01 \mathrm{~g/mol}} = 0.0167 \mathrm{~mol}\) Moles of H2O: \(\frac{0.450 \mathrm{~g}}{18.02 \mathrm{~g/mol}} = 0.0250 \mathrm{~mol}\) Since the hydrocarbon contains only carbon and hydrogen, we can calculate the moles of carbon (C) and hydrogen (H) in the hydrocarbon based on the moles of CO and CO2 and moles of H2O, respectively. Moles of C = moles of CO + moles of CO2 = 0.0167 + 0.0167 = 0.0334 moles Moles of H = 2 * moles of H2O = 2 * 0.0250 = 0.0500 moles Oxygen is present in CO, CO2, and H2O. Now, let's calculate the moles of O used in the reaction. Moles of O = (moles of CO + 2 * moles of CO2) + moles of H2O = (0.0167 + 2 * 0.0167) + 0.0250 = 0.0751 moles

Determine the empirical formula of the hydrocarbon

Now, let's determine the empirical formula by finding the mole ratio of C and H in the hydrocarbon. Mole ratio of C and H = \(\frac{0.0334}{0.0500} = 0.668\) Since the ratio is close to 2:3, the empirical formula of the hydrocarbon is C2H3.

Calculate mass of oxygen used and required for complete combustion

(a) The empirical formula of the compound is C2H3. (b) To calculate the mass of oxygen used, multiply the moles of oxygen used by the molar mass of oxygen: Mass of O2 used = 0.0751 moles * 32.00 g/mol = 2.40 g (c) To calculate the mass of oxygen required for complete combustion, we need to calculate the theoretical O2 requirement to completely oxidize the hydrocarbon. A completely oxidized hydrocarbon would produce CO2 and H2O as the only products: C2H3 + O2 -> 2CO2 + 3/2H2O Now, we calculate the required moles of O2: Moles of O2 required = moles of C * 2 + moles of H * 1/2 = (0.0334 * 2) + (0.0500 * 1/2) = 0.0668 + 0.0250 = 0.0918 moles Mass of O2 required = 0.0918 moles * 32.00 g/mol = 2.93 g So, 2.93 g of O2 would have been required for complete combustion.

Key concepts

Understanding Combustion Reactions

Combustion reactions are a type of chemical reaction where a substance combines with oxygen and releases energy in the form of heat and light. This process typically occurs in hydrocarbons, which are compounds comprised of hydrogen and carbon atoms. When hydrocarbons combust, they produce carbon dioxide, water, and sometimes carbon monoxide if there isn't enough oxygen present.
The basics involve a hydrocarbon reacting with oxygen (\[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \] ). However, if the reaction is incomplete due to limited oxygen, carbon monoxide (CO) might also form, as seen in the following incomplete combustion reaction:
\[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{CO} + \text{H}_2\text{O}. \]
Understanding these reactions is crucial for calculating the products formed during combustion, as seen in exercises requiring the determination of empirical formulas.

The Role of Mole Calculations

Mole calculations are fundamental for understanding chemical reactions, especially those involving combustion. The mole is a unit that measures the amount of a substance. It helps chemists count particles like atoms and molecules by relating mass to molecular weight through Avogadro's number. During combustion reactions, mole calculations allow us to convert the mass of reactants and products to moles, facilitating the empirical formula determination.
By using the molar mass of substances, chemists can find out how many moles of each element are involved in the reaction. For example, in the problem, calculating the moles of carbon monoxide, carbon dioxide, and water produced from the original hydrocarbon lets us understand how many moles of carbon and hydrogen are present. This, in turn, allows the determination of the stoichiometry of the reaction:
- Use the molar mass to determine moles (e.g., CO has a molar mass of 28.01 g/mol). - Calculate the moles of elements in products to find their ratios. - From there, deduce the empirical formula (e.g., determine moles of C and H to find the mole ratio and the empirical formula).
Such calculations are key to understanding how reactants convert to products in combustion reactions.

Hydrocarbons and Their Combustion

Hydrocarbons are a major focus in studies of combustion, as they represent molecules solely made up of carbon and hydrogen. Their combustion, particularly in chemistry problems, often requires balancing between complete and incomplete combustion scenarios. Complete combustion refers to a situation where adequate oxygen leads to all carbon atoms forming carbon dioxide and hydrogen forming water, as in the equation:
\[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}. \]
Incomplete combustion occurs when there isn't enough oxygen available, hence forming carbon monoxide alongside carbon dioxide and water:
\[ \text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{CO} + \text{H}_2\text{O}. \]
These differing scenarios change the outcome of products substantially.

• In complete combustion, the hydrocarbon fully reacts, producing maximum possible energy release.
• In incomplete, some energy is lost as it forms less energy-efficient products like CO.

Understanding these principles helps chemists predict the kinds of products formed and balance equations accordingly, and it also plays a vital role in figuring out necessary oxygen quantities for full reactions.