Problem 65 - General Chemistry: Principles and Modern Applications

Question

Without doing detailed calculations, determine which of the following samples has the greatest number of atoms: (a) a cube of iron with a length of \(10.0 \mathrm{cm}\) \(\left(d=7.86 \mathrm{g} / \mathrm{cm}^{3}\right)\) (b) \(1.00 \mathrm{kg}\) of hydrogen contained in a \(10,000 \mathrm{L}\) balloon (c) a mound of sulfur weighing \(20.0 \mathrm{kg}\) (d) a 76 lb sample of liquid mercury \((d=13.5 \mathrm{g} / \mathrm{mL})\)

Step-by-Step Solution

Answer

The sample of hydrogen has the greatest number of atoms.

Step 1: Cube of Iron

Calculate the mass of the cube of iron using its volume and density. Given that the length, height, and width of the cube are each equal to \(10.0 cm\), its volume is \(1000 cm^3\). Then, given that density is defined as mass divided by volume, the iron's mass is its volume times its density \(d=7.86 g/cm^3\). Therefore, the iron cube's mass is \(7860 g\). Using the molar mass of iron (\(55.85 g/mol\)), we can then calculate the number of moles in the cube, and then the number of atoms using Avogadro's constant (\(6.022 \times 10^{23} atoms/mol\)).

Step 2: Hydrogen Balloon

The mass of hydrogen is given directly as \(1.00 kg\), which is equivalent to \(1000 g\). We can then use this mass and the molar mass of hydrogen (\(1.01 g/mol\)) to calculate the number of moles and then the number of atoms, as in Step 1.

Step 3: Sulfur Mound

The mass of the mound of sulfur is given directly as \(20.0 kg\), which is equivalent to \(20000 g\). We can then use this mass and the molar mass of sulfur (\(32.07 g/mol\)) to calculate the number of moles, and then the number of atoms as in the previous steps.

Step 4: Mercury Sample

First, convert the weight of the mercury sample from pounds to grams (\(1 lb = 454 g\)), obtaining roughly \(34504 g\). Then, using the molar mass of mercury (\(200.59 g/mol\)), calculate the number of moles, and then the number of atoms as in previous steps.

Step 5: Comparison

Now compare the four calculated numbers of atoms. The sample with the greatest number has the greatest quantity of atoms.

Key Concepts

Moles and Molar Mass

To understand chemical calculations, moles and molar mass are key concepts. A mole is a unit that measures the amount of a substance. It's similar to how we use dozens for counting eggs. The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol). Knowing the molar mass allows you to convert between the mass of a substance and the number of moles. For example, in the problem above, we use the molar mass to calculate how many moles of the substance are present. This is essential for finding out how many atoms or molecules are in the sample. Let’s take iron as an example: its molar mass is 55.85 g/mol. So, if you have 7860 g of iron, you can find the number of moles by dividing the mass by the molar mass. That is: \[moles = \frac{mass}{molar\, mass} = \frac{7860\, g}{55.85\, g/mol}\approx 140.74\, moles.\]

Avogadro's Constant

Avogadro's constant is a crucial number in chemistry. It is \(6.022 \times 10^{23}\) atoms/mol, representing the number of atoms, ions, or molecules in one mole of a substance. This constant allows us to translate moles into something more tangible, like the number of atoms. For each of the elements in the exercise, after calculating the moles, we use Avogadro’s constant to find the total number of atoms. For instance, with the cube of iron, once you've calculated the moles (around 140.74 moles), you multiply by Avogadro’s constant to find the total number of atoms: \[Atoms = moles \times Avogadro’s\, constant = 140.74\, moles \times 6.022 \times 10^{23}\, atoms/mol = \approx 8.47 \times 10^{25} \text{ atoms}.\] This shows how important Avogadro's number is for bridging the gap between the microscopic scale and the quantities we can measure.

Density and Volume

Density and volume are important concepts that help determine how much material you have in a given space. Density is defined as mass per unit volume, and is typically expressed in grams per cubic centimeter (g/cm³) or grams per milliliter (g/mL).

• By knowing the density, you can find the mass if you know the volume, using the formula: \[mass = density \times volume.\]
• Conversely, if you know the mass, you can find the volume by rearranging to: \[volume = \frac{mass}{density}.\]

For instance, in the problem above, the iron cube's volume is given as 1000 cm³. With iron’s density at 7.86 g/cm³, you can calculate its mass as: \[mass = 7.86\, g/cm³ \times 1000\, cm³ = 7860\, g.\] This is fundamental in solving the exercise because knowing the mass enables further calculations for moles and atoms.

Mass Conversion

Mass conversion is a practical skill, especially in chemistry where you often switch between different units.

• Understanding conversions is crucial for calculations involving different systems, like converting pounds to grams or kilograms to grams.
• For example, in the given exercise, the mercury sample's weight is converted from pounds to grams using \(1\, lb = 454\, g\):\[76\, lbs \times 454\, g/lb = 34504\, g.\]

This conversion is needed to calculate how many moles, and subsequently, how many individual atoms are present in the mercury sample. Converting between mass units is often the first step in chemical calculations, allowing you to work seamlessly with the rest of the data.