Question

An unknown compound has the formula \(\mathrm{C}_{x} \mathrm{H}, \mathrm{O}_{z} .\) 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?

Short answer

The empirical and molecular formula is \( C_4H_8O \).

Step-by-step solution

Calculate Moles of Carbon in CO2

To find the moles of carbon, use the formula: \[ \text{moles of } C = \frac{\text{mass of } CO_2}{\text{molar mass of } CO_2} \times \text{moles of C in one mole of } CO_2 \]. The molar mass of \( CO_2 \) is 44 g/mol. Thus, the moles of carbon is \( \frac{0.3718}{44} = 0.00845 \text{ moles of C} \).

Calculate Moles of Hydrogen in H2O

To find the moles of hydrogen, use the formula: \[ \text{moles of } H = \frac{\text{mass of } H_2O}{\text{molar mass of } H_2O} \times \text{moles of H in one mole of } H_2O \]. The molar mass of \( H_2O \) is 18 g/mol. Thus, the moles of hydrogen is \( \frac{0.1522}{18} \times 2 = 0.0169 \text{ moles of H} \).

Calculate Mass of Oxygen in Compound

The mass of oxygen in the compound can be calculated by subtracting the mass of carbon and hydrogen from the total mass: \[ \text{mass of } C = 0.00845 \times 12 = 0.1014 \text{ g} \] \[ \text{mass of } H = 0.0169 \times 1.01 = 0.0171 \text{ g} \] \[ \text{mass of } O = 0.1523 - 0.1014 - 0.0171 = 0.0338 \text{ g} \].

Calculate Moles of Oxygen in Compound

To find the moles of oxygen, use: \[ \text{moles of } O = \frac{\text{mass of } O}{\text{molar mass of } O} \]. The molar mass of oxygen is 16 g/mol. Therefore, the moles of oxygen is \( \frac{0.0338}{16} = 0.00211 \text{ moles of O} \).

Determine Empirical Formula

Compare the ratios of the number of moles for each element. Divide each by the smallest number of moles calculated: \[ C: \frac{0.00845}{0.00211} = 4 \] \[ H: \frac{0.0169}{0.00211} = 8 \] \[ O: \frac{0.00211}{0.00211} = 1 \]. The empirical formula is \( C_4H_8O_1 \) or simply \( C_4H_8O \).

Calculate Empirical Formula Mass

The empirical formula mass of \( C_4H_8O \) is calculated as: \[ 4(12) + 8(1) + 16 = 72 \].

Determine Molecular Formula

Compare the molar mass with the empirical formula mass: \( \frac{72.1}{72} = 1 \). Since the ratio is 1, the empirical formula is also the molecular formula. Thus, the molecular formula is \( C_4H_8O \).

Key concepts

Molecular Formula

The molecular formula tells us the exact number of each type of atom present in a molecule of a substance. It is essentially the chemical makeup of the compound.
For the compound in the exercise with a given molar mass of 72.1 g/mol, the molecular formula can be determined once the empirical formula is found.
The molecular formula can be the same as the empirical formula or a whole number multiple of it.
This depends on whether the molar mass is equivalent to the molar mass of the empirical formula or a simple multiple.
In this case, the molar mass of the empirical formula, "C₄H₈O", equaled the compound's molar mass of 72.1 g/mole, resulting in both formulas being identical.
The molecular formula offers insight into the molecule's structure and composition, which aids in understanding its behavior and reactivity.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the molar masses of all the atoms in the empirical formula.
For example, carbon (C) has a molar mass of 12 g/mol, hydrogen (H) is about 1 g/mol, and oxygen (O) is 16 g/mol.
In the exercise, the molar mass of the empirical formula "C₄H₈O" is derived as follows:

• 4 carbon atoms: \(4 \times 12 = 48\) g/mol
• 8 hydrogen atoms: \(8 \times 1 = 8\) g/mol
• 1 oxygen atom: 16 g/mol

Overall, the molar mass of the empirical formula "C₄H₈O" is \(48 + 8 + 16 = 72\) g/mol.
This is a crucial calculation, as it informs the determination of the molecular formula as well.

Chemical Reactions

Chemical reactions involve the transformation of substances through breaking and forming of chemical bonds.
In our exercise, a combustion reaction took place, where an unknown compound burns, producing carbon dioxide and water.
This reaction is used to determine the amounts of carbon and hydrogen in the compound.
During the combustion of a hydrocarbon:

• Carbon from the compound forms carbon dioxide (CO₂)
• Hydrogen forms water (H₂O)

By isolating and measuring the masses of CO₂ and H₂O formed, one can backtrack to find how much carbon and hydrogen were in the original compound.
This is a foundational method in chemistry for analyzing unknown substances.
Knowing the masses and molar masses of these products helps us find the moles of each constituent element present in the original compound.

Empirical Formula Calculation

The empirical formula represents the simplest whole-number ratio of elements in a compound. Calculating it involves:

• Determining the moles of each element present
• Finding the ratio of these moles
• Simplifying to the smallest whole numbers

In our exercise:

• The moles of carbon in the produced CO₂ and hydrogen in H₂O were calculated using their respective masses and molar masses.
• The difference in mass accounted for the oxygen in the compound.
• The mole ratios were found by dividing by the smallest number of moles obtained.

This confirmed the empirical formula "C₄H₈O".
This process is key for converting measured mass data into a tangible chemical formula, offering insights into the composition and properties of the compound.