Question

At \(110^{\circ}\) and \(454 \mathrm{~mm}\) pressure, \(0.11 \mathrm{~g}\) acetic acid vapor occupies \(63.7 \mathrm{cc}\); at \(156^{\circ}\) and \(458 \mathrm{~mm}, 0.081\) g occupies \(66.4 \mathrm{cc}\). Calculate the molecular weight of acetic acid in the vapor phase at each temperature. How do you interpret these results?

Short answer

The molecular weight of acetic acid in the vapor phase at \(110^{\circ}C\) is approximately \(74.32 \, g/mol\), and at \(156^{\circ}C\) it is approximately \(74.03 \, g/mol\). These results confirm that the molecular weight does not change significantly in the vapor phase at the temperatures measured, and the Ideal Gas Law approximation is relatively accurate for describing the behavior of acetic acid vapor at these temperatures.

Step-by-step solution

Convert Temperatures to Kelvin

We are given temperatures in Celsius. We need to convert them to Kelvin using the formula \[K = °C + 273.15\]. First temperature: \(T_1 = 110^{\circ}C + 273.15 = 383.15K\) Second temperature: \(T_2 = 156^{\circ}C + 273.15 = 429.15K\)

Solve for the number of moles (n)

We will use the Ideal Gas Law to find the number of moles of acetic acid at each temperature. \[PV=nRT\] Using this formula, we can solve for n: \[n = \frac{PV}{RT}\], where the gas constant R = 0.0821 L atm / (mol K). For the first temperature: P_1 = 454 mm = 0.597 atm (1 atm = 760 mm) V_1 = 63.7 cc = 0.0637 L (1 L = 1000 cc) \[n_1 = \frac{0.597 \times 0.0637}{0.0821 \times 383.15} = 1.48\times10^{-3} \, mol\] For the second temperature: P_2 = 458 mm = 0.602 atm (1 atm = 760 mm) V_2 = 66.4 cc = 0.0664 L (1 L = 1000 cc) \[n_2 = \frac{0.602 \times 0.0664}{0.0821 \times 429.15} = 1.094\times10^{-3} \, mol\]

Calculate Molecular Weight

Now, we can calculate the molecular weight of acetic acid at each temperature. Molecular weight (M) is defined as the mass (m) divided by the number of moles (n): \[M = \frac{m}{n}\] For the first temperature: m_1 = 0.11 g \[M_1 = \frac{0.11}{1.48 \times 10^{-3}} = 74.32 \, g/mol\] For the second temperature: m_2 = 0.081 g \[M_2 = \frac{0.081}{1.094 \times 10^{-3}} = 74.03 \, g/mol\]

Interpretation of Results

Both molecular weights are fairly close to each other (74.32 g/mol and 74.03 g/mol), which suggests that the Ideal Gas Law approximation is relatively accurate for describing the behavior of acetic acid vapor at these temperatures. The molecular weight of acetic acid should be constant regardless of temperature, and these results confirm that the molecular weight does not change significantly in the vapor phase at the temperatures measured.

Key concepts

Understanding the Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that describes the behavior of ideal gases. It is expressed as:
\[PV = nRT\]
where P is the pressure of the gas, V is the volume it occupies, R is the universal gas constant, and T is the temperature in Kelvin. This law combines several previous gas laws and provides a basis for understanding how gases respond to changes in temperature, pressure, and volume.

When applying the Ideal Gas Law to real-life problems, such as calculating molecular weight from experimental data, understanding each variable is crucial. To conduct these calculations, accurate conversions of temperatures to Kelvin and pressures to atmospheres are necessary, as the gas constant R is typically given in liters atmospheres per mole Kelvin (L atm / (mol K)).

By rearranging the Ideal Gas Law, we can solve for the number of moles (n) of a gas when the pressure, volume, and temperature are known. The equation becomes:
\[n = \frac{PV}{RT}\]
This formula is especially useful in our exercise where we calculate the molecular weight of acetic acid vapor by first determining the moles of gas present.

Moles Calculation in Chemistry

The concept of moles is central to chemistry because it allows scientists to count exact quantities of atoms and molecules. One mole is defined as the quantity of any substance that has as many constituent particles as there are atoms in 12 grams of pure carbon-12. This number is known as Avogadro's number and consists of approximately \(6.022 \times 10^{23}\) entities.

Mole calculations come into play when we use the Ideal Gas Law. After determining the moles of a gas using the rearranged law, we facilitate various vital computations, such as finding molecular weights. The mole calculation in our exercise helps us connect the mass of acetic acid to the number of moles using the known pressure and temperature conditions.

A precise understanding of moles is critical because it bridges the gap between the physical world we can measure and the microscopic world of atoms and molecules. Without it, quantifying substances in chemical reactions would be nearly impossible.

Vapor Phase Chemistry and Molecular Weight

In vapor phase chemistry, understanding the behavior of substances in a gaseous state is essential. Gases behave differently from solids and liquids in that their particles are much further apart and move more freely. The molecular weight (often referred to as the molar mass) of a substance in the vapor phase is calculated by dividing the mass of vapor by the number of moles present.

In our example, we're examining acetic acid in its vapor form. Determining the molecular weight involves the careful moles calculation from the Ideal Gas Law and then using the formula:
\[M = \frac{m}{n}\]
This equation is powerful because it shows that molecular weight, a fundamental property of a substance, is consistent regardless of changes in physical conditions like temperature and pressure.

The results from our molecular weight calculations of acetic acid confirm this principle, as the weights computed at two different temperatures are nearly identical. This agreement also implies that acetic acid behaves nearly as an ideal gas under these conditions, showing minimal deviation due to intermolecular forces, which is a crucial observation in vapor phase chemistry.