Problem 103 - Chemistry The Central Science In Si Units

Question

Nickel carbonyl, \(\mathrm{Ni}(\mathrm{CO})_{4},\) is one of the most toxic substances known. The present maximum allowable concentration in laboratory air during an 8 -hr workday is \(1 \mathrm{ppb}\) (parts per billion) by volume, which means that there is one mole of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for every \(10^{9}\) moles of gas. Assume \(24^{\circ} \mathrm{C}\) and 101.3 kPa pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m}\) ?

Step-by-Step Solution

Answer

The mass of Ni(CO)4 allowable is approximately 3.682 x 10^-4 grams.

Step 1: Calculate the Volume of the Room

First, calculate the volume of the laboratory, which is given by the formula for the volume of a rectangular box: \( V = \text{length} \times \text{width} \times \text{height} \). Thus, the volume is \( V = 3.5 \times 6.0 \times 2.5 = 52.5 \text{ m}^3 \).

Step 2: Convert Volume to Liters

Convert the room's volume from cubic meters to liters, knowing that 1 cubic meter is equivalent to 1000 liters. So, \( V_{\text{liters}} = 52.5 \times 1000 = 52500 \, \text{liters} \).

Step 3: Use Ideal Gas Law to Find Total Moles of Air

Use the ideal gas equation \( PV = nRT \) to find the number of moles. Where \( P = 101.3 \text{ kPa} \), \( V = 52500 \, \text{liters} \), \( R = 8.314 \, \text{J/mol} \cdot \text{K} \), and \( T = 273 + 24 = 297 \text{ K} \). Convert pressure to Pa \( P = 101300 \text{ Pa} \) and volume to \( m^3 \) \( V = 52.5 \text{ m}^3 \). Then, \[ n = \frac{PV}{RT} = \frac{101300 \times 52.5}{8.314 \times 297} \approx 2156.5 \text{ moles} \].

Step 4: Find Moles of Ni(CO)4

Given that the concentration of \( \text{Ni(CO)}_4 \) is 1 part per billion (ppb), we can find the moles of \( \text{Ni(CO)}_4 \) by using \( \frac{1}{10^{9}} \) times the total moles of air. Therefore, moles of \( \text{Ni(CO)}_4 = \frac{2156.5}{10^9} = 2.1565 \times 10^{-6} \text{ moles} \).

Step 5: Calculate Mass of Ni(CO)4

Find the molar mass of \( \text{Ni(CO)}_4 \). This is the sum of the atomic masses: Ni (58.69 g/mol) and four CO groups (4 \times (12.01 + 16.00) g/mol = 112.04 g/mol). The molar mass is \( 58.69 + 112.04 = 170.73 \text{ g/mol} \). Thus, the mass of \( \text{Ni(CO)}_4 = 2.1565 \times 10^{-6} \text{ moles} \times 170.73 \text{ g/mol} = 3.682 \times 10^{-4} \text{ grams} \).

Key Concepts

Nickel Carbonyl Toxicity

Nickel carbonyl, with the chemical formula \(\mathrm{Ni(CO)}_4\), is exceptionally toxic. It is known to be one of the most hazardous substances, and exposure to even small amounts can be lethal.

• It can cause a range of health issues, primarily affecting the lungs, heart, and liver.
• Symptoms of exposure can include headaches, dizziness, and pulmonary issues.
• Due to its high volatility, it can easily become airborne, increasing the risk of inhalation in a confined space.
• Because of these hazards, stringent regulations are in place to limit exposure to very low levels, such as 1 part per billion (ppb) in laboratory environments.

This makes it crucial for labs to continually monitor air quality to ensure that nickel carbonyl levels remain within safe boundaries.

Molar Mass Calculation

The molar mass of a compound is a crucial concept in chemistry and is essential for converting between moles and grams.
To find the molar mass of nickel carbonyl, you need to sum the atomic masses of all the atoms in the compound:

• The atomic mass of nickel (Ni) is approximately 58.69 g/mol.
• Each carbon monoxide (CO) unit has a carbon atom (12.01 g/mol) and an oxygen atom (16.00 g/mol), leading to a CO mass of \(12.01 + 16.00 = 28.01\) g/mol.

Nickel carbonyl contains one nickel atom and four CO units, so its molar mass is:
(58.69 g/mol) + 4 \times (28.01 g/mol) = 58.69 + 112.04 = 170.73 \text{ g/mol}.
Knowing the molar mass allows you to determine the mass of any amount of nickel carbonyl given in moles. This is vital for understanding and predicting the material's behavior in chemical processes.

Volume Conversion

Volume conversion is an essential skill for solving problems in chemistry, especially when dealing with gases and the ideal gas law.
This process often requires converting between different volume units, such as cubic meters (m³) and liters.

• 1 cubic meter is equivalent to 1,000 liters. So, if a room's volume is given in cubic meters, multiply by 1,000 to convert it to liters.
• For example, a room that is 3.5 m \(\times\) 6.0 m \(\times\) 2.5 m has a volume of 52.5 m³.
• Converting this to liters gives: \(52.5 \text{ m}^3 \times 1000 = 52500 \text{ liters}\).

This conversion is crucial because the ideal gas law, often used in chemical calculations, generally employs volume in liters.

Gas Concentration Measurement

Measuring gas concentration accurately is critical for ensuring safety and compliance with health standards. In this context, "ppb" or parts per billion is a common unit of measurement for evaluating minuscule amounts of gaseous substances in air.

• Gas concentration measurement often involves calculations using the total number of moles of air and the mole fraction of the gas in question.
• The mole fraction is determined by dividing the number of moles of the gas by the total number of moles of air. For nickel carbonyl at 1 ppb, it means there’s one mole of \(\text{Ni(CO)}_4\) for every \(10^9\) moles of air.
• In a room, knowing the total number of moles of air helps calculate the safe limit of a toxic gas, ensuring that its concentration does not exceed permissible levels.

Monitoring and maintaining proper gas concentrations is vital for safe laboratory practices.