Question

The radius of a neon atom is \(69 \mathrm{pm}\), and its mass is \(3.35 \times\) \(10^{-23} \mathrm{~g}\). What is the density of the atom in grams per cubic centimeter \(\left(\mathrm{g} / \mathrm{cm}^{3}\right) ?\) Assume the nucleus is a sphere with volume \(=\frac{4}{3} \pi r^{3}\).

Short answer

The density of the neon atom is approximately 0.169 \(g/cm^3\).

Step-by-step solution

Calculate the volume of the neon atom

We are given the radius of the neon atom as 69 pm (picometers). The formula for the volume of a sphere is \(\frac{4}{3} \pi r^{3}\), where \(r\) is the radius. We use this formula to calculate the volume of the neon atom: Volume = \(\frac{4}{3} \pi (69 \times 10^{-12})^3\)

Convert picometers to centimeters

We are given the radius in picometers, and we need to find the density in grams per cubic centimeter. Therefore, we must convert the volume from cubic picometers to cubic centimeters. We know that: 1 cm = \(10^{10}\) pm So, 1 \(pm^3\) = \((10^{-10})^3 cm^3 = 10^{-30} cm^3\) Volume = \(\frac{4}{3} \pi (69 \times 10^{-12})^3 \times 10^{-30}\)

Calculate the density of the neon atom

We have the mass of the neon atom as \(3.35 \times 10^{-23}\) g and the volume in cubic centimeters. To calculate the density, we use the formula: Density = \(\frac{M}{V}\) Where \(M\) is the mass and \(V\) is the volume. Density = \(\frac{3.35 \times 10^{-23}}{\frac{4}{3} \pi (69 \times 10^{-12})^3 \times 10^{-30}}\)

Simplify the density expression

Now we can simplify the expression we obtained in Step 3: Density = \(\frac{3.35 \times 10^{-23}}{\frac{4}{3} \pi (69 \times 10^{-12})^3 \times 10^{-30}}\) Plug in the values and simplify the expression: Density ≈ 0.169 grams per cubic centimeter So, the density of the neon atom is approximately 0.169 \(g/cm^3\).

Key concepts

Picometers to Centimeters Conversion

When working with atomic-size measurements, converting from picometers to centimeters is a common task, as the latter is a more standard unit for volume and density calculations. One picometer (pm) is equal to one trillionth of a meter (\(10^{-12}\text{ meters}\)), while one centimeter (cm) is one hundredth of a meter (\(10^{-2}\text{ meters}\)).

To convert picometers to centimeters, you need to divide the number of picometers by a factor of one trillion, or move the decimal point twelve places to the left. This can be expressed as:

• \(1 \text{ pm} = 10^{-12} \text{ meters} = 10^{-10} \text{ centimeters}\)
• \(1 \text{ pm}^3 = (10^{-10} \text{ cm})^3 = 10^{-30} \text{ cm}^3\)

This conversion is crucial when calculating the volume and density of atoms, as in our exercise, where the known radius of the neon atom in picometers needed to be converted into cubic centimeters to find its density in \(g/cm^3\).

Volume of a Sphere Formula

The volume of a sphere is calculated using the simple yet important mathematical formula \(V = \frac{4}{3} \pi r^3\). Here, \(V\) represents the volume, \(\pi\) is a constant approximately equal to 3.14159, and \(r\) is the radius of the sphere.

This formula is derived from integral calculus and reflects the three-dimensional space occupied by a spherical object. Such geometric considerations become significant in chemistry when envisioning atoms as spheres to approximate their volumes – as in our exercise. To effectively use this formula, it's essential to ensure the radius is in the correct unit, typically centimeters, to calculate the volume in cubic centimeters (\(cm^3\)).

Density Formula in Chemistry

Density is a fundamental concept in chemistry, representing the mass per unit volume of a substance, commonly expressed as \(\text{g/cm}^3\) for solids and liquids. It's calculated using the formula \(\text{Density} = \frac{M}{V}\), where \(M\) is the mass and \(V\) is the volume.

The density can provide insights into properties like particle arrangement and atomic structure. Our neon atom's density was found by dividing its mass by the volume calculated from the sphere's volume formula after converting the given radius from picometers to centimeters. This calculation yields the atom's density, helping us understand its physical characteristics at an atomic level. Being familiar with this formula allows chemists and students to solve various problems involving the mass and space that substances occupy.