Copper(II) hydrogen arsenite (CuHAsO \(_{3}\) ) is a green pigment once used in wallpaper. In damp conditions, mold metabolizes this compound to trimethylarsine \(\left[\left(\mathrm{CH}_{3}\right)_{3} \mathrm{As}\right],\) a highly toxic gas. (a) Calculate the mass percent of As in each compound. (b) How much CuHAsO must react to reach a toxic level in a room that measures \(12.35 \mathrm{~m} \times 7.52 \mathrm{~m} \times 2.98 \mathrm{~m}\) (arsenic is toxic at \(\left.0.50 \mathrm{mg} / \mathrm{m}^{3}\right) ?\)

Short Answer

Expert verified
Mass percent of As in CuHAsO _{3}: 39.95% and in (CH_{3})3As: 71.94%. For 138.30 mg of arsenic to be toxic, 346.32 mg (0.346 g) of CuHAsO _{3} is required.

Step by step solution

01

Determine Molecular Mass of CuHAsO _{3}

Calculate the molecular mass of Copper(II) hydrogen arsenite (CuHAsO _{3}). Use the atomic masses: Cu (63.55 g/mol), H (1.01 g/mol), As (74.92 g/mol), and O (16.00 g/mol).\[ \text{Molecular Mass of } \text{CuHAsO}_{3} = 63.55 + 1.01 + 74.92 + 3 \times 16.00 = 187.48 \text{ g/mol} \]
02

Calculate Mass Percent of As in CuHAsO _{3}

Determine the mass percent of As in CuHAsO _{3}. Mass \text{percent} = \[ \left(\frac{\text{Mass of As}}{\text{Molecular mass of CuHAsO}_{3}}\right) \times 100 \] \[ \frac{74.92}{187.48} \times 100 = 39.95\text{%} \]
03

Determine Molecular Mass of \text{Trimethylarsine}

Calculate the molecular mass of Trimethylarsine \boldmath{((\text{CH}_{3})_{3}\text{As})}. Use the atomic masses: As (74.92 g/mol), C (12.01 g/mol) and H (1.01 g/mol). \[ \text{Molecular Mass of } (\text{CH}_{3})_{3}\text{As} = 74.92 + 3 \times 12.01 + 9 \times 1.01 = 104.15 \text{ g/mol}\]
04

Calculate Mass Percent of As in \text{Trimethylarsine}

Determine the mass percent of As in Trimethylarsine \[ (\text{CH}_{3})_{3}\text{As} \]. \[ \text{Mass percent As} = \left(\frac{74.92}{104.15}\text{\textbf{}} \right) \times 100 = 71.94\text{%} \]
05

Calculate Toxic Level of As

Calculate the toxic level of arsenic in the room by finding the volume of the room in cubic meters. \[\text{Volume} = 12.35 \text{ m} \times 7.52 \text{ m} \times 2.98 \text{ m} = 276.59 \text{ m}^3\] Next, multiply by the toxic concentration level of arsenic. The toxic concentration of arsenic is 0.50 \text{ mg/m}^3. \[\text{Total Arsenic Concerntration} = 276.59 \text{ m}^3 \times 0.50 \text{ mg/m}^3 = 138.30 \text{ mg}\]
06

Convert the Mass of As to Mass of \text{Copper(II) hydrogen arsenite}

Use the mass percent of As in CuHAsO_{3} to convert the total mass of arsenic to the corresponding mass of CuHAsO_{3}.\[ \text{Mass of CuHAsO}_{3} = \left(\frac{\text{138.30 mg arsenic}}{\frac{39.95}{100}\right)} = 346.32 \text{ mg} \~or~ 0.346 \text{g}\]

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Key Concepts

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

Mass Percent Composition
To calculate the mass percent composition of an element in a compound, you need the molecular mass of the compound and the mass of the specific element in it. The formula to determine mass percent is given by:
\text{Mass percent} = \left( \frac{\text{Mass of element}}{\text{Molecular mass of compound}} \right) \times 100
For example, in Copper(II) hydrogen arsenite (CuHAsO \( _{3} \)), we calculated the mass percent composition of arsenic (As). Using the atomic masses: Cu (63.55 g/mol), H (1.01 g/mol), As (74.92 g/mol), and O (16.00 g/mol), the molecular mass of CuHAsO \( _{3} \) was found to be 187.48 g/mol. Since arsenic's mass is 74.92 g/mol, we calculated:
\( \text{Mass percent As} = \left( \frac{74.92}{187.48} \right) \times 100 = 39.95 \% \)
This calculation shows the fraction of the compound's total mass that is due to arsenic, expressed as a percentage.
Chemical Toxicity
Understanding chemical toxicity is crucial, especially when dealing with compounds like Copper(II) hydrogen arsenite (CuHAsO \( _{3} \)), which can be metabolized into trimethylarsine, a highly toxic gas. Toxicity is typically measured in terms of concentration, such as mg/m\( ^{3} \), indicating how much of a substance is present in a specific volume of air or water.
To determine how much CuHAsO \( _{3} \) is needed to reach a toxic level, identify the toxic concentration threshold of arsenic. In our example, arsenic is toxic at 0.50 mg/m\( ^{3} \). By calculating the room volume (length \( \times \) width \( \times \) height) and multiplying it by this threshold, we found the total toxic concentration:
\( \text{Volume} = 12.35 \text{ m} \times 7.52 \text{ m} \times 2.98 \text{ m} = 276.59 \text{ m}^{3} \)
\( \text{Total Arsenic Concentration} = 276.59 \text{ m}^{3} \times 0.50 \text{ mg/m}^{3} = 138.30 \text{ mg} \)
Therefore, the toxic concentration translates to the total mass of arsenic that can lead to toxicity in that defined space.
Volume Calculation
Volume calculation is a fundamental concept in both chemistry and physics. It involves determining the amount of space an object or substance occupies. For this exercise, we were given room dimensions and needed to find the room's volume to then calculate the toxic concentration of arsenic.
The formula for volume, given a rectangular room, is:
Volume = length \( \times \) width \( \times \) height.
Applying this to the problem:
\( \text{Volume} = 12.35 \text{ m} \times 7.52 \text{ m} \times 2.98 \text{ m} = 276.59 \text{ m}^{3} \).
Once you have the volume, you can calculate other values, such as the total concentration of a toxic substance. For the given room, knowing the toxic level is 0.50 mg/m\( ^{3} \), we can find the total arsenic concentration:
\( \text{Total Arsenic Concentration} = 276.59 \text{ m}^{3} \times 0.50 \text{ mg/m}^{3} = 138.30 \text{ mg}\).
This helps us understand how much of a compound might be hazardous in a given space.

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Most popular questions from this chapter

Zeolite \(\mathrm{A}, \mathrm{Na}_{12}\left[\left(\mathrm{AlO}_{2}\right)_{12}\left(\mathrm{SiO}_{2}\right)_{12}\right] \cdot 27 \mathrm{H}_{2} \mathrm{O},\) is used to soften water because it replaces \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) dissolved in the water with \(\mathrm{Na}^{+}\). Hard water from a certain source is \(4.5 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ca}^{2+}\) and \(9.2 \times 10^{-4} \mathrm{M} \mathrm{Mg}^{2+},\) and a pipe delivers \(25,000 \mathrm{~L}\) of this hard water per day. What mass (in kg) of zeolite A is needed to soften a week's supply of the water? (Assume zeolite A loses its capacity to exchange ions when \(85 \mathrm{~mol} \%\) of its \(\mathrm{Na}^{+}\) has been lost.)

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