Question

The combustion of $0.4196\mathrm{g}$ of a hydrocarbon releases $17.55\mathrm{kJ}$ of heat. The masses of the products are ${\mathrm{CO}}_2=1.419\mathrm{g}$ and ${\mathrm{H}}_2\mathrm{O}=0.290\mathrm{g}$. (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is $76\mathrm{g}/\mathrm{mol}$, calculate its standard enthalpy of formation.

Short answer

(a) The empirical formula is CH. (b) The standard enthalpy of formation is approximately -3178.26 kJ/mol.

Step-by-step solution

Calculate Moles of Products

First, find the moles of carbon and hydrogen in the products. Using the molar mass of carbon dioxide (CO₂), which is 44.01 g/mol, calculate the moles of CO₂:$$molesC{O}_2=\frac{1.419g}{44.01g/mol}=0.0322mol$$One mole of CO₂ contains one mole of carbon, so there are 0.0322 mol C.Next, find the moles of hydrogen in water. The molar mass of water (H₂O) is 18.02 g/mol:$$moles{H}_{2}O=\frac{0.290g}{18.02g/mol}=0.0161mol$$Since water has two hydrogen atoms, there are:$$molesH=2\times 0.0161mol=0.0322mol$$

Determine Mass of Carbon and Hydrogen

Using the moles calculated, determine the mass of C and H.For carbon:$$massC=0.0322mol\times 12.01g/mol=0.387g$$For hydrogen:$$massH=0.0322mol\times 1.008g/mol=0.0325g$$

Calculate Mass of Oxygen in Compound

To find the mass and composition of oxygen in the hydrocarbon, subtract the combined mass of carbon and hydrogen from the total mass of the hydrocarbon:$$massO=0.4196g-(0.387g+0.0325g)=0.0001g$$The negligible mass indicates an error or extremely low oxygen content, implying no oxygen in the compound given the hydrocarbon nature.

Find the Empirical Formula

Now, calculate the simplest mole ratio of carbon to hydrogen. The moles of carbon and hydrogen are both approximately 0.0322. Thus, the empirical formula is CH.

Calculate Molecular Formula

To determine the molecular formula, compare the formula mass of the empirical formula to the given molar mass:Empirical formula mass (CH):$$12.01+1.008=13.018g/mol$$Divide the molar mass by the empirical formula mass:$$\frac{76g/mol}{13.018g/mol}\approx 5.84$$The nearest whole number is 6, suggesting the molecular formula is C₆H₆.

Calculate Standard Enthalpy of Formation

The standard enthalpy of formation requires the energy released and the moles of compound.Calculate moles of compound combusted using molar mass:$$moles{C}_6{H}_6=\frac{0.4196g}{76g/mol}=0.00552mol$$Enthalpy change per mole:$$\mathrm{\Delta }{H}_f=\frac{-17.55kJ}{0.00552mol}$$This gives:$$\mathrm{\Delta }{H}_f\approx -3178.26kJ/mol$$

Key concepts

Combustion Reactions

Combustion reactions are a form of chemical reaction where a substance reacts with oxygen, releasing energy in the form of heat and light. These reactions are common in everyday life, such as burning wood or gasoline. For hydrocarbons, like the one mentioned in the exercise, combustion typically produces carbon dioxide (CO₂) and water (H₂O).
These reactions are exothermic, meaning they release more energy than they absorb. This energy release is often measured in joules or kilojoules. In the given problem, the combustion of a hydrocarbon releases energy, helping identify its empirical formula by examining its combustion products. Understanding combustion reactions is crucial in fields like energy production and environmental science.

• Observe the products formed, often CO₂ and H₂O for hydrocarbons
• Calculate the heat released, known as enthalpy change
• Utilize findings to gain insights into chemical properties of the substance

Molar Mass Calculation

Molar mass calculation is vital for understanding the relationship between moles of a substance and its mass. It is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Molar mass is determined by summing the atomic masses of all atoms present in a molecule, which are found on the periodic table.
In the exercise provided, the molar mass of CO₂ and H₂O was used to calculate their respective moles. This information helped determine the moles of carbon and hydrogen, which are crucial for identifying the empirical formula of the hydrocarbon.

• Identify chemical formula of a compound
• Use atomic masses from the periodic table
• Sum atomic masses to find molar mass

Understanding molar mass helps in performing stoichiometric calculations and converting between mass and moles, which is fundamental in chemical analysis and reactions.

Enthalpy of Formation

Enthalpy of formation refers to the heat change that occurs when one mole of a compound is formed from its elements in their standard states. It's expressed as $\mathrm{\Delta }{H}_f$ and is a crucial concept in thermodynamics, helping scientists understand energy changes during chemical reactions.
For hydrocarbons, the enthalpy of formation can be determined from combustion data. The heat released during combustion provides clues about the energy changes when the compound is formed. In this context, specific calculations using the heat released helped deduce the standard enthalpy of formation for the hydrocarbon in question.

• Standard enthalpy values can be positive or negative
• Negative values indicate an exothermic formation process
• Use enthalpy to predict reaction behavior and stability

Mastering the concept of enthalpy of formation equips you to evaluate the energetics of chemical processes effectively.

Mole Concept

The mole concept is a fundamental principle in chemistry that allows chemists to quantify substances and reactants precisely. A mole is defined as containing exactly 6.022 x 10^{23} entities (Avogadro's number) of particles such as atoms or molecules. This provides a bridge between the atomic and macroscopic worlds, relating the mass of substances to amounts in terms of atoms or molecules.
The exercise leverages the mole concept to convert mass into moles, essential for calculating empirical formulas and enthalpies of formation. By understanding the amount of substance present, it becomes easier to connect chemical equations with practical lab measurements.

• Converts atomic or molecular scales to practical use
• Assists in balancing chemical equations
• Is a pivotal concept in stoichiometry

Grasping the mole concept is crucial for succeeding in various branches of chemistry, ensuring accurate and meaningful computations in chemical analysis.