Problem 43 - Introductory Chemistry Atoms First

Question

The density of air is \(0.00130 \mathrm{~g} / \mathrm{mL}\). What is the mass in grams of \(500.0 \mathrm{~L}\) of air? What is this mass in kilograms?

Step-by-Step Solution

Answer

The mass of 500.0 L of air is approximately 650 grams or 0.65 kilograms.

Step 1: Identifying given information and formula

We are given the following information: Density of air: 0.00130 g/mL Volume of air: 500.0 L We need to find the mass of the air and then convert it to kilograms. First, we'll use the density formula: Density = Mass / Volume We'll rearrange the formula to find the mass: Mass = Density × Volume

Step 2: Convert liters to milliliters

Before we proceed with the mass calculation, we need to convert the given volume to milliliters, as the density is given in g/mL: 1 L = 1000 mL Therefore, 500.0 L = 500.0 × 1000 = 500000 mL

Step 3: Calculate the mass of air in grams

Now, we can use the mass formula with the given density and converted volume: Mass = 0.00130 g/mL × 500000 mL Mass ≈ 650 g So, the mass of 500.0 L of air is approximately 650 grams.

Step 4: Convert mass to kilograms

To convert the mass to kilograms, we'll use the following conversion factor: 1 kg = 1000 g So, the mass in kilograms is: Mass = 650 g × (1 kg / 1000 g) Mass ≈ 0.65 kg Therefore, the mass of 500.0 L of air is approximately 0.65 kilograms.

Key Concepts

Mass-Volume Relationship

Understanding the mass-volume relationship is crucial when studying the properties of different substances in fields like chemistry and physics. This relationship is fundamental to determining density, which is a measure of how much mass is contained in a given volume. Imagine filling two identical boxes, one with feathers and the other with rocks. Although the volume of the boxes is the same, the box filled with rocks has more mass due to the higher density of rocks compared to feathers.

To quantify this relationship, we use the basic formula for density: \( Density = \frac{Mass}{Volume} \). This equation can be rearranged to find the mass if the volume and density are known. For instance, as in our original exercise, if we want to know the mass of a certain volume of air, we can multiply the volume of the air by its density. Since the density of air is low, a large volume is needed to accumulate significant mass.

Unit Conversion

Unit conversion is an essential skill in various scientific calculations. Given that different countries and fields of study use different units of measurement, it's important to be able to convert between these units accurately. For example, in some instances, volume may be given in liters but density might be in grams per milliliter (g/mL), as seen in the air density exercise. To perform calculations, we need to express all related quantities in compatible units.

A common conversion is between liters and milliliters: \(1 L = 1000 mL\). In our exercise, to calculate the mass of air, we converted 500.0 liters into milliliters before multiplying by the density because the given density was in grams per milliliter. Similarly, for mass, the conversion between grams and kilograms is vital: \(1 kg = 1000 g\). Knowing these conversions allowed us to first find the mass of air in grams and then convert it to kilograms to match the scale often used in physics and engineering.

Density Formula

The density formula, expressed as \( Density = \frac{Mass}{Volume} \), is a fundamental physical property that represents the mass of an object per unit volume. In practical use, density helps determine the buoyancy of an object, how a substance will stack or pack, and is even used in identifying substances because each one has its unique density.

In our exercise, we used the density formula to find the mass of air. By rearranging the formula to \( Mass = Density \times Volume \), we calculated the mass based on the known density and volume of the air. This straightforward application of the density formula provides insights into the substance's characteristics and is essential for solving a wide range of real-world problems, from the design of ships to the manufacturing of aerated food products. Remembering this key formula and how to manipulate it forms the basis for understanding more complex scientific concepts.