Problem 106 - Chemistry The Central Science

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 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 1.00 atm pressure. What mass of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is allowable in a laboratory room that is \(12 \mathrm{ft} \times 20 \mathrm{ft} \times 9 \mathrm{ft}\) ?

Step-by-Step Solution

Answer

The allowable mass of Ni(CO)₄ in the laboratory room under the given conditions is approximately 0.42446 grams.

Step 1: Convert room dimensions to meters

To convert the room dimensions from feet to meters, we can use the conversion factor: 1 ft = 0.3048 meters. Length = 12 ft × 0.3048 m/ft = 3.6576 m Width = 20 ft × 0.3048 m/ft = 6.0960 m Height = 9 ft × 0.3048 m/ft = 2.7432 m

Step 2: Calculate the volume of the room

Multiply the length, width, and height to find the volume of the room in cubic meters. Volume = Length × Width × Height = 3.6576 m × 6.0960 m × 2.7432 m = 61.3084 m³

Step 3: Find the total number of moles of gas in the room

For this step, we'll be using the Ideal Gas Law: PV = nRT. P = pressure = 1.00 atm V = volume of the gas = 61.3084 m³ (from Step 2) T = temperature = (24 + 273.15) K = 297.15 K R = gas constant = 0.0821 L atm/mol K First, we need to convert the volume from cubic meters to liters: 61.3084 m³ × (1000 L/m³) = 61308.4 L. Now, we can rearrange the Ideal Gas Law to solve for 'n': n = PV / RT n = (1.00 atm × 61308.4 L) / (0.0821 L atm/mol K × 297.15 K) n = 2492.35 moles (total moles of gas in the room)

Step 4: Determine the number of moles of Ni(CO)₄

The allowable concentration of Ni(CO)₄ is 1 ppb, which means 1 mole of Ni(CO)₄ for every \(10^9\) moles of gas. To find the number of moles of Ni(CO)₄ in the room, use the following proportion: 1 mole Ni(CO)₄ / \(10^9\) moles of gas = x moles Ni(CO)₄ / 2492.35 moles of gas Solving for 'x': x = (1 mole Ni(CO)₄ × 2492.35 moles of gas) / \(10^9\) moles of gas x = 2.49235 × \(10^{-6}\) moles Ni(CO)₄

Step 5: Convert moles of Ni(CO)₄ to mass

To find the mass of Ni(CO)₄, use the molar mass of Ni(CO)₄: Molar mass of Ni(CO)₄ = (Ni = 58.69 g/mol) + (4 × (C = 12.01 g/mol + O = 16.00 g/mol)) Molar mass of Ni(CO)₄ = 58.69 + 4 × (12.01 + 16.00) = 170.33 g/mol Now multiply the number of moles of Ni(CO)₄ by its molar mass to find the mass: Mass = 2.49235 × \(10^{-6}\) moles × 170.33 g/mol = 0.42446 g The allowable mass of Ni(CO)₄ in the laboratory room under the given conditions is approximately 0.42446 grams.

Key Concepts

Nickel Carbonyl

Nickel carbonyl, denoted as \( \text{Ni(CO)}_4 \), is a highly toxic, volatile compound composed of nickel and carbon monoxide. It is a coordination complex where the nickel center is bonded to four carbon monoxide ligands. The compound is notable for its ability to form even under mild conditions, releasing carbon monoxide upon decomposition. It is used in the nickel refining process and as a catalyst in some chemical reactions, but extreme care must be taken due to its toxicity. The permissible exposure limit to nickel carbonyl in the air during an 8-hour workday is just 1 part per billion (ppb) by volume, indicating its hazardous potential in laboratory and industrial environments.

Molar Mass Calculation

The molar mass of a compound is central to converting moles into grams. For nickel carbonyl, the calculation of its molar mass starts with determining the atomic masses of its constituent elements: nickel (Ni) and carbon monoxide (CO). Nickel has an atomic weight of approximately 58.69 g/mol. Each carbon atom contributes 12.01 g/mol, and each oxygen atom contributes 16.00 g/mol. Since \( \text{Ni(CO)}_4 \) has four CO groups, the molar mass is calculated as follows:

• Molar mass of Ni = 58.69 g/mol
• Molar mass of CO = 12.01 g/mol + 16.00 g/mol = 28.01 g/mol
• Total molar mass of \( \text{Ni(CO)}_4 \) = 58.69 + 4 × 28.01 = 170.33 g/mol

This value is essential for converting moles of \( \text{Ni(CO)}_4 \) into grams, allowing us to determine the mass of the compound in a given volume of air.

Conversion of Units

In scientific calculations, ensuring consistent units is crucial. For this exercise, it was necessary to convert room dimensions from feet to meters. The standard conversion factor is 1 foot equals 0.3048 meters. Thus, to obtain the room's volume in cubic meters:

• Convert each dimension:
  • 12 ft to meters: \( 12 \times 0.3048 = 3.6576 \) m
  • 20 ft to meters: \( 20 \times 0.3048 = 6.096 \) m
  • 9 ft to meters: \( 9 \times 0.3048 = 2.7432 \) m
• Calculate the volume: \( 3.6576 \times 6.096 \times 2.7432 = 61.3084 \) m³

Converting all dimensions to meters and subsequently the total volume to liters (1 m³ = 1000 L) helps in using the Ideal Gas Law for further calculations.

Gas Concentration

Gas concentration, often expressed in parts per billion (ppb), provides a measure of how much of a particular gas is present in a mixture, relative to one billion moles of all gases. For \( \text{Ni(CO)}_4 \), a concentration of 1 ppb implies there is 1 mole of nickel carbonyl for every \( 10^9 \) moles of air. Understanding and calculating gas concentration are crucial for assessing exposure risks in environments like laboratories.To find the number of moles of \( \text{Ni(CO)}_4 \) permissible in the air, based on the 1 ppb limit, one needs to use the total moles of air calculated from the room volume using the Ideal Gas Law. For a room containing 2492.35 moles of gas:

• The moles of \( \text{Ni(CO)}_4 \) calculated are: \( x = \frac{2492.35}{10^9} \approx 2.49235 \times 10^{-6} \) moles

This figure is then used with the molar mass to find the mass, aiding in compliance with safety standards.