Question

You have \(357 \mathrm{~mL}\) of chlorine trifluoride gas at \(699 \mathrm{mmHg}\) and \(45^{\circ} \mathrm{C}\). What is the mass (in \(\mathrm{g}\) ) of the sample?

Short answer

1.17 g

Step-by-step solution

Convert the given values to standard units

First, convert the volume from \(\text{mL}\) to \(\text{L}\) and the temperature from \(^\text{C}\) to \(K\). \[ 357 \text{ mL} = 0.357 \text{ L} \] and \[ 45^\text{C} = 45 + 273.15 = 318.15 \text{ K}\]

Identify and write down the formula

The ideal gas law equation is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature.

Convert pressure to atmospheres

Convert \(699 \text{ mmHg}\) to atmospheres using the conversion factor \(1 \text{ atm} = 760 \text{ mmHg}\). \[ 699 \text{ mmHg} \times \frac{1 \text{ atm}}{760 \text{ mmHg}} = 0.92 \text{ atm}\]

Solve for moles of gas

Use the ideal gas law to solve for \(n\). \[n = \frac{PV}{RT} = \frac{(0.92 \text{ atm})(0.357 \text{ L})}{(0.0821 \frac{\text{L} \text{ atm}}{\text{mol} \text{ K}})(318.15 \text{ K})} = 0.01269 \text{ moles} \]

Convert moles to grams

Find the molar mass of \( \text{ClF}_3 \). The molar mass of \( \text{ClF}_3 \) is \( (1 \times 35.45) + (3 \times 19.00) = 92.45 \text{ g/mol}\). Calculate the mass: \[ 0.01269 \text{ moles} \times 92.45 \frac{\text{g}}{\text{mol}} = 1.17 \text{ g} \]

Key concepts

gas laws

Gas laws are mathematical relationships that describe the behavior of gases. One of the most important gas laws is the Ideal Gas Law, represented by the equation:

\(PV = nRT\).

Here, \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature.

The Ideal Gas Law combines several simpler gas laws:

• Boyle's Law: Describes the inverse relationship between pressure and volume, keeping temperature constant \( (PV = k)\).
• Charles's Law: Describes the direct relationship between volume and temperature, keeping pressure constant \( (V \, \text{is directly proportional to} \, T)\).
• Avogadro's Law: States that equal volumes of gases at the same temperature and pressure contain the same number of moles \(V \, \text{is directly proportional to } \, n)\).

Understanding these relationships helps to predict how a gas will behave under different conditions.

mole calculation

Calculating the number of moles in a gas involves using the formula \(n = \frac{PV}{RT}\), derived from the Ideal Gas Law.

In the given exercise, we have:

• Pressure \( (P) = 0.92\, \text{atm}\)
• Volume \( (V) = 0.357\, L\)
• Ideal Gas Constant \( (R) = 0.0821\, \frac{L \, atm}{mol \, K} \)
• Temperature \( (T) = 318.15\, K \)

Plugging these values into the formula, we get:

\[ n = \frac{(0.92 \, atm)(0.357 \, L)}{(0.0821 \, \frac{L \, atm}{mol \, K})(318.15 \, K)} = 0.01269 \, \text{moles} \]

This tells us how many moles of chlorine trifluoride gas are present in the given conditions.

unit conversion

Unit conversion is crucial to ensure that all measurements are in compatible units when applying mathematical formulas.

In this exercise, we needed to convert:

• Volume from \(357\,mL\) to \(0.357\,L\). Remember, \(1\,L = 1000\,mL\).
• Temperature from \(45^{\circ}C\) to \(318.15\,K\). Use the formula \(T(K) = T(^\circ C) + 273.15\).
• Pressure from \(699\,mmHg\) to \(0.92\,atm\). Here, \(1\,atm = 760\,mmHg\), so the conversion factor is \(\frac{1 atm}{760 mmHg}\).

These conversions align all units into the standard form required by the Ideal Gas Law.

molar mass

Molar mass is the mass of one mole of a substance. It is usually expressed in grams per mole \( g/mol \).

For chlorine trifluoride \(ClF_3\), we need to calculate it based on the atomic masses of the constituent elements:

• Chlorine (Cl): \(35.45\,g/mol\)
• Fluorine (F): \(19.00\,g/mol\)

Chlorine trifluoride has 1 chlorine atom and 3 fluorine atoms, so:

\[Molar\, mass = (1 \times 35.45) + (3 \times 19.00) = 92.45 \, g/mol\]

Finally, we convert moles to grams using this molar mass. Given \(0.01269\, \text{moles}\), the mass in grams is:

\[Mass = 0.01269 \,mol \times 92.45\, \frac{g}{mol} = 1.17\,g\] This result shows the mass of the chlorine trifluoride gas sample.