Question

A "canned heat" product used to warm chafing dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin that has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{5}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin in formulating the mixture if the vapor pressure of ethanol at \(35^{\circ} \mathrm{C}\) over the mixture is to be 8 torr? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

Short answer

Approximately 7.38 kg of ethanol should be added to 620 kg of paraffin to achieve a vapor pressure of 8 torr at 35°C.

Step-by-step solution

Write down the Raoult's Law formula.

Raoult's Law states that the vapor pressure of a component in a solution is equal to the mole fraction of that component multiplied by its vapor pressure in the pure state. Mathematically, it is given as: \(P_A = X_A P_A^*\) Where, \(P_A\) = vapor pressure of component A in the mixture \(X_A\) = mole fraction of component A in the mixture \(P_A^*\) = vapor pressure of pure component A

Calculate the moles of paraffin.

We are given that we have 620 kg of paraffin (C24H50) in the mixture. To find the moles of paraffin, we will first need to find its molar mass. Molar mass of paraffin = \(24 * molar~mass~of~C + 50 * molar~mass~of~H\) Molar mass of paraffin = \(24 * 12.01 + 50 * 1.01 = 288.24 + 50.5 = 338.74~g/mol\) Now, we can find the moles of paraffin in the mixture: Moles of paraffin = \(\frac{mass~of~paraffin}{molar~mass~of~paraffin}\) Moles of paraffin = \(\frac{620,000~g}{338.74~g/mol} = 1830.49~mol\)

Calculate the mole fraction of ethanol in the mixture.

We are given the vapor pressure of ethanol over the mixture at 35°C, which is 8 torr. We are also given the vapor pressure of pure ethanol at this temperature, which is 100 torr. From Raoult's Law, we can calculate the mole fraction of ethanol in the mixture: \(P_A = X_A P_A^*\) \(X_{C_2H_5OH} = \frac{P_{C_2H_5OH}}{P_{C_2H_5OH}^*}\) \(X_{C_2H_5OH} = \frac{8~torr}{100~torr} = 0.08\)

Calculate the moles of ethanol in the mixture.

Now we need to find the moles of ethanol in the mixture. Since the mole fraction of ethanol is 0.08, the mole fraction of paraffin is 0.92. We can write this relationship as: \(X_{C_2H_5OH} = \frac{moles~of~ethanol}{moles~of~ethanol + moles~of~paraffin}\) \(0.08 = \frac{moles~of~ethanol}{moles~of~ethanol + 1830.49}\) Now, we can solve for the moles of ethanol: moles of ethanol = 0.08 * (moles of ethanol + 1830.49) Upon solving, we get: moles of ethanol = 160.05

Calculate the mass of ethanol to be added to the paraffin.

Now, we can calculate the mass of ethanol that needs to be added to the 620 kg of paraffin to achieve the desired vapor pressure. We will first need to find the molar mass of ethanol. Molar mass of ethanol = \(2 * molar~mass~of~C + 6 * molar~mass~of~H + 1 * molar~mass~of~O\) Molar mass of ethanol = \(2 * 12.01 + 6 * 1.01 + 1 * 16.00 = 24.02 + 6.06 + 16 = 46.08~g/mol\) Now, we can find the mass of ethanol: Mass of ethanol = moles of ethanol * molar mass of ethanol Mass of ethanol = \(160.05~mol * 46.08~g/mol = 7376.06~g\) So, 7376.06 grams (approximately 7.38 kg) of ethanol should be added to 620 kg of paraffin to achieve a vapor pressure of 8 torr at 35°C.

Key concepts

Vapor Pressure

Vapor pressure is a key concept in understanding how substances behave when mixed. It refers to the pressure exerted by a vapor in equilibrium with its liquid or solid form. When a liquid evaporates, its particles escape into the air, creating a vapor. The vapor pressure increases until it's balanced with the liquid's evaporation rate.
This balance is affected by temperature: higher temperatures generally increase vapor pressure.

• For example, ethanol, at 35°C, has a vapor pressure of 100 torr.
• In mixtures, the vapor pressure depends on the presence and concentration of other substances, according to Raoult's Law.

Understanding vapor pressure helps in designing mixtures with specific properties, like ethanol in a paraffin mixture to achieve a particular vapor pressure.

Mole Fraction

The mole fraction is a way to express the concentration of a component in a mixture.
It's calculated by dividing the moles of the component by the total moles of all components in the mixture. This gives us a dimensionless number that indicates how much of a particular component is present.

• The mole fraction of ethanol in our solution is found using Raoult's Law.
• Here, the desired vapor pressure of ethanol is 8 torr, making its mole fraction 0.08 as calculated by dividing 8 torr by the pure vapor pressure of ethanol (100 torr).

Using mole fractions helps determine the correct proportions of each substance to achieve the desired physical and chemical characteristics.

Ethanol

Ethanol is an organic compound with the formula \( ext{C}_2 ext{H}_5 ext{OH}\). It's commonly known for its use in alcoholic beverages, but its properties make it valuable in various industrial applications. Ethanol evaporates easily, and at 35°C, its pure vapor pressure is 100 torr.
In mixtures, ethanol's behavior is described by Raoult's Law, important to know how much should be added to another substance for targeted results, like vapor pressures.

• To reach a vapor pressure of 8 torr in our mixture, we calculated that we need approximately 7.38 kg of ethanol.
• This calculation involved understanding ethanol's interactions and contributions to vapor pressure in the mixture.

Recognizing ethanol's role across different contexts helps grasp its diverse applications.

Paraffin

Paraffin is a waxy, hydrocarbon substance with a typical formula of \( ext{C}_{24} ext{H}_{50}\). It's often used in candles, waterproofing, and creating fuel mixtures, such as canned heat.
In this particular setup, paraffin acts as a base in which ethanol is mixed to manipulate the total vapor pressure.

• Given 620 kg of paraffin, we calculated its moles based on its molar mass (338.74 g/mol).
• Knowing the amount of paraffin helps determine how much ethanol is required to adjust the vapor pressure to the desired level, using Raoult's Law and mole fraction calculations.

Understanding the properties and role of paraffin in mixtures is crucial for industries that rely on its stable characteristics.