Chapter 1: Problem 89
Calculate these masses. (a) What is the mass of \(6.00 \mathrm{cm}^{3}\) of mercury, density \(=13.5939 \mathrm{g} / \mathrm{cm}^{3} ?\) (b) What is the mass of \(25.0 \mathrm{mL}\) octane, density \(=0.702 \mathrm{g} / \mathrm{cm}^{3} ?\)
Short Answer
Expert verified
The mass of mercury is 81.5634 g and the mass of octane is 17.55 g.
Step by step solution
01
Understanding the concept of density
Density is defined as the mass per unit volume of a substance. The formula to calculate mass based on the density and volume is given by: \( mass = density \times volume \). We will use this formula to calculate the mass of mercury and octane in the following steps.
02
Calculate the mass of mercury
Given the volume of mercury is \(6.00 \mathrm{cm}^3\) and the density is \(13.5939 \mathrm{g}/\mathrm{cm}^3\), we can calculate the mass as follows: \[ mass_{mercury} = density_{mercury} \times volume_{mercury} = 13.5939 \mathrm{g}/\mathrm{cm}^3 \times 6.00 \mathrm{cm}^3 \]. Now we compute the value to find the mass of mercury.
03
Calculate the mass of octane
Given the volume of octane is \(25.0 \mathrm{mL}\), and noting that \(1 \mathrm{mL} = 1 \mathrm{cm}^3\), the volume is also \(25.0 \mathrm{cm}^3\). The density of octane is \(0.702 \mathrm{g}/\mathrm{cm}^3\). Thus, the mass can be calculated as: \[ mass_{octane} = density_{octane} \times volume_{octane} = 0.702 \mathrm{g}/\mathrm{cm}^3 \times 25.0 \mathrm{cm}^3 \]. Compute this to find the mass of octane.
04
Performing the calculations
For mercury: \[ mass_{mercury} = 13.5939 \times 6.00 = 81.5634 \mathrm{g} \]. For octane: \[ mass_{octane} = 0.702 \times 25.0 = 17.55 \mathrm{g} \]. These are the masses of mercury and octane given the respective volumes and densities.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Mass Volume Density Relationship
When you're dealing with substances, understanding the mass volume density relationship is crucial. This relationship is a way of expressing how much matter is packed into a certain space. In simpler terms, it answers the question: 'How heavy is something for its size?' Imagine having two boxes of the same size - one filled with feathers and the other with rocks. Despite being the same size, the box with rocks is heavier because rocks have a greater density than feathers.
This relationship guides us in calculating the mass of an object if we know its density and volume. Using the formula \( mass = density \times volume \) we can figure out the mass for any object as long as we have the other two values. It's an essential concept in chemistry and in many real-life applications – from designing ships that can float to packing goods for transportation.
This relationship guides us in calculating the mass of an object if we know its density and volume. Using the formula \( mass = density \times volume \) we can figure out the mass for any object as long as we have the other two values. It's an essential concept in chemistry and in many real-life applications – from designing ships that can float to packing goods for transportation.
Density Formula
The density formula is a mathematic expression that connects mass, volume, and density. It's often expressed as \( density = \frac{mass}{volume} \) and is used to describe the compactness of a material. Understanding this formula is critical, especially in chemistry, because it allows you to solve for different variables depending on the information available.
For instance, if you're given the density and volume, you can rearrange the formula to find the mass, as shown in the original exercise. This versatility in the formula makes it a potent tool for not only learning about the characteristics of different substances but also for practical problem-solving in various scientific fields.
For instance, if you're given the density and volume, you can rearrange the formula to find the mass, as shown in the original exercise. This versatility in the formula makes it a potent tool for not only learning about the characteristics of different substances but also for practical problem-solving in various scientific fields.
Converting mL to cm^3
Converting milliliters (mL) to cubic centimeters (cm^3) is a basic but essential step in many scientific calculations, particularly in chemistry. The conversion is straightforward: 1 mL is exactly equal to 1 cm^3. This simple 1:1 relationship is because both milliliters and cubic centimeters are measurements of volume, stemming from the metric system, and they're built on the foundation of a liter (L), which is exactly 1,000 milliliters or 1,000 cubic centimeters.
This straightforward conversion is vital for calculations involving liquids in chemistry. Many chemical properties are measured in mL for liquids, but when we conduct density calculations, we need the volume to be in cubic centimeters, as seen in the given problem where we convert the volume of octane from mL to cm^3 to find its mass based on its density.
This straightforward conversion is vital for calculations involving liquids in chemistry. Many chemical properties are measured in mL for liquids, but when we conduct density calculations, we need the volume to be in cubic centimeters, as seen in the given problem where we convert the volume of octane from mL to cm^3 to find its mass based on its density.
Chemistry Problem Solving
Chemistry problem solving often involves a sequential approach where you identify the known variables, understand the relationships between them, choose the right formula, and then carry out the calculations. As seen in the textbook exercise, problem solving not only requires mathematical skills but also an understanding of chemical concepts like density.
For example, to solve the given problems, you must recognize that mass, volume, and density are interconnected and use their relationship to your advantage. Breaking down complex problems into steps, as done in the exercise provided, is a powerful strategy to ensure a structured approach and clear thinking as you navigate towards the solution.
For example, to solve the given problems, you must recognize that mass, volume, and density are interconnected and use their relationship to your advantage. Breaking down complex problems into steps, as done in the exercise provided, is a powerful strategy to ensure a structured approach and clear thinking as you navigate towards the solution.