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Chapter 10: Problem 103

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 \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m}\) ?

Short Answer

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

Step by step solution

01

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 \( V = 3.5 \times 6.0 \times 2.5 = 52.5 \text{ m}^3 \).
02

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, \( V_{\text{liters}} = 52.5 \times 1000 = 52500 \, \text{liters} \).
03

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, \[ n = \frac{PV}{RT} = \frac{101300 \times 52.5}{8.314 \times 297} \approx 2156.5 \text{ moles} \].
04

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 \( \text{Ni(CO)}_4 = \frac{2156.5}{10^9} = 2.1565 \times 10^{-6} \text{ moles} \).
05

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 \( \text{Ni(CO)}_4 = 2.1565 \times 10^{-6} \text{ moles} \times 170.73 \text{ g/mol} = 3.682 \times 10^{-4} \text{ grams} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (m鲁) 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 m鲁.
  • 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鈥檚 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.

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Most popular questions from this chapter

Calculate each of the following quantities for an ideal gas: (a) the volume of the gas, in liters, if \(1.50 \mathrm{~mol}\) has a pressure of \(126.7 \mathrm{kPa}\) at a temperature of \(-6^{\circ} \mathrm{C} ;(\mathbf{b})\) the absolute temperature of the gas at which \(3.33 \times 10^{-3}\) mol occupies \(478 \mathrm{~mL}\) at \(99.99 \mathrm{kPa} ;(\mathbf{c})\) the pressure, in pascals, if \(0.00245 \mathrm{~mol}\) occupies \(413 \mathrm{~mL}\) at \(138^{\circ} \mathrm{C} ;\) (d) the quantity of gas, in moles, if \(126.5 \mathrm{~L}\) at \(54^{\circ} \mathrm{C}\) has a pressure of \(11.25 \mathrm{kPa} .\)

(a) If the pressure exerted by ozone, \(\mathrm{O}_{3}\), in the stratosphere is \(304 \mathrm{~Pa}\) and the temperature is \(250 \mathrm{~K}\), how many ozone molecules are in a liter? (b) Carbon dioxide makes up approximately \(0.04 \%\) of Earth's atmosphere. If you collect a \(2.0-\mathrm{L}\) sample from the atmosphere at sea level ( \(101.33 \mathrm{kPa}\) ) on a warm day \(\left(27^{\circ} \mathrm{C}\right),\) how many \(\mathrm{CO}_{2}\) molecules are in your sample?

Which one or more of the following statements are true? (a) \(\mathrm{O}_{2}\) will effuse faster than \(\mathrm{Cl}_{2}\). (b) Effusion and diffusion are different names for the same process. (c) Perfume molecules travel to your nose by the process of effusion. (d) The higher the density of a gas, the shorter the mean free path.

A glass vessel fitted with a stopcock valve has a mass of \(337.428 \mathrm{~g}\) when evacuated. When filled with \(\mathrm{Ar}\), it has a mass of \(339.854 \mathrm{~g}\). When evacuated and refilled with a mixture of Ne and Ar, under the same conditions of temperature and pressure, it has a mass of \(339.076 \mathrm{~g}\). What is the mole percent of Ne in the gas mixture?

A 500 mL incandescent light bulb is filled with \(1.5 \times 10^{-5}\) mol of xenon to minimize the rate of evaporation of the tungsten filament. What is the pressure of xenon in the light bulb at \(25^{\circ} \mathrm{C} ?\)
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