Problem 42 - Chemistry and Chemical Reactivity

Question

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}_{1} \mathrm{O}_{2}\). You burn \(0.1523 \mathrm{g}\) of the compound and isolate \(0.3718 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.1522 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula of the compound? If the molar mass is \(72.1 \mathrm{g} / \mathrm{mol}\), what is the molecular formula? (See Exercise 4.9.)

Step-by-Step Solution

Answer

The empirical and molecular formulas are both \(\mathrm{C}_4\mathrm{H}_8\mathrm{O}\).

Step 1: Calculate Moles of Carbon

Given the mass of \(\mathrm{CO}_{2}\) produced is 0.3718 g. Since each molecule of \(\mathrm{CO}_{2}\) contains one carbon atom, first calculate the moles of \(\mathrm{CO}_{2}\) using its molar mass (44.01 g/mol):\[ \text{Moles of } \mathrm{CO}_{2} = \frac{0.3718}{44.01} = 0.00845 \text{ moles} \]Now, the moles of carbon are the same, 0.00845 moles.

Step 2: Calculate Moles of Hydrogen

Given the mass of \(\mathrm{H}_{2}\mathrm{O}\) produced is 0.1522 g. Each molecule of \(\mathrm{H}_{2}\mathrm{O}\) contains two hydrogen atoms, so first find the moles of \(\mathrm{H}_{2}\mathrm{O}\) using its molar mass (18.02 g/mol):\[ \text{Moles of } \mathrm{H}_{2}\mathrm{O} = \frac{0.1522}{18.02} = 0.00845 \text{ moles} \]The moles of hydrogen atoms are twice this value:\[ 0.00845 \times 2 = 0.0169 \text{ moles} \]

Step 3: Calculate Moles of Oxygen in the Compound

The compound's total mass is 0.1523 g. We know the masses of carbon and hydrogen have been converted entirely into \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) respectively.Mass of carbon: \( 0.00845 \times 12.01 = 0.1014 \text{ g} \)Mass of hydrogen: \( 0.0169 \times 1.008 = 0.0170 \text{ g} \)Now, the oxygen mass in the compound is:\[ \text{Mass of oxygen} = 0.1523 - 0.1014 - 0.0170 = 0.0339 \text{ g} \]Now, calculate the moles of oxygen in the compound using its atomic mass (16.00 g/mol):\[ \text{Moles of oxygen} = \frac{0.0339}{16.00} = 0.00212 \text{ moles} \]

Step 4: Calculate the Empirical Formula

The moles of each element are in the smallest whole number ratio:- Carbon: 0.00845 moles - Hydrogen: 0.0169 moles - Oxygen: 0.00212 molesTo find the ratio, divide each by the smallest number of moles, 0.00212:- Carbon: \( \frac{0.00845}{0.00212} = 3.99 \approx 4 \)- Hydrogen: \( \frac{0.0169}{0.00212} = 7.97 \approx 8 \)- Oxygen: \( \frac{0.00212}{0.00212} = 1 \)The empirical formula is \(\mathrm{C}_4\mathrm{H}_8\mathrm{O}\).

Step 5: Calculate the Molecular Formula

The empirical formula mass of \(\mathrm{C}_4\mathrm{H}_8\mathrm{O}\) is:\[ 4(12.01) + 8(1.008) + 16 = 72.1 \text{ g/mol} \]This matches the given molar mass of the compound, 72.1 g/mol. Therefore, the empirical and molecular formulas are the same:The molecular formula is \(\mathrm{C}_4\mathrm{H}_8\mathrm{O}\).

Key Concepts

Combustion Analysis

A combustion analysis is a powerful laboratory technique used to find the empirical formula of organic compounds. When a hydrocarbon or other organic compound completely burns, it converts into carbon dioxide (CO₂) and water (H₂O). By measuring the mass of these combustion products, we can deduce the original composition of the compound.

In our problem, we have an unknown compound that yields 0.3718 g of CO₂ and 0.1522 g of H₂O. These amounts help us calculate the moles of carbon and hydrogen in the compound.

• CO₂ gives us information about carbon: Each molecule contains one carbon atom.
• H₂O gives us hydrogen: Each molecule has two hydrogen atoms.

By analyzing these combustion products, we derive the proportions of carbon and hydrogen in the unknown compound. The remaining mass can be attributed to other elements like oxygen. This forms the basis for calculating an empirical formula.

Mole Calculation

Mole calculations are essential for determining the number of entities (like atoms or molecules) in a given sample. They help us convert between mass
The mole is based on Avogadro's number, which is approximately 6.022 x 10²³. For the problem, here’s how we use it:
To find the moles of carbon and hydrogen created during combustion, we start with the known mass of each product:

• Use the molar mass of CO₂ (44.01 g/mol) to find how many moles of carbon are present.
• With H₂O, using 18.02 g/mol, find the moles of hydrogen, knowing each molecule has two hydrogen atoms.

For any element in the compound, its moles can be found by dividing its mass by its atomic or molecular molar mass. This fundamental process allows us to quantify the composition of a substance accurately.

Stoichiometry

Stoichiometry is all about the calculations of reactants and products in chemical reactions based on balanced chemical equations. It tells us how much of one substance we need to completely react with another.
For unknown compounds such as in combustion analysis, stoichiometry helps us establish the relationships between different substances as they react.

• It provides ratios, allowing us to predict the amount of products formed from given reactants.
• This is useful when determining empirical formulas, as we learn the relative amounts of each element present in compounds.

For our exercise, once we know the moles of each element, stoichiometry permits us to calculate the empirical formula. By dividing the mole quantities by their smallest value, we obtain the simplest whole-number ratio, which represents the empirical formula. Understanding stoichiometry not only helps determine formulas but also predicts the quantities needed in reactions.

Hydrocarbons

Hydrocarbons are organic compounds composed of solely carbon and hydrogen atoms. They are the fundamental building blocks of many types of organic molecules and
In chemical analysis, identifying the hydrocarbon-type structure or pattern is vital, especially as they react with oxygen to form carbon dioxide and water.

• Hydrocarbons can vary widely in structure, from simple linear forms to complex rings.
• Such compounds are often examined through combustion to deduce their composition.

Within our example, recognizing the hydrocarbon aspect simplifies understanding the empirical formula calculation. Despite having an oxygen atom in its empirical formula, the compound behaves like a hydrocarbon when combusted, simplifying analysis and helping students predict reactions and compositions of molecules
Understanding hydrocarbons' behavior can provide insights into both basic chemistry and complex organic reactions.