Solving Density Problems - Online Tutor, Practice Problems & Exam Prep

Density, represented by the Greek letter rho (ρ), is defined as mass (m) divided by volume (V). The formula is $\rho =\frac{m}{V}$. To find mass, rearrange the equation to $m=\rho V$. For a sphere, volume is calculated using $V=4/3\pi r^3$. Understanding these concepts is essential for solving density-related problems across various geometric shapes.

Problems with Mass, Volume, & Density

Hey guys. So you may run into some problems involving density. In this video, I'm just going to give you a quick refresher on how we solve density problems just in case it's been some time since you've seen this stuff. So guys, remember that density is defined as mass, which is the amount of stuff that something has, divided by the volume, which is the amount of space that it takes up. The symbol for density is this Greek letter rho, which kinda looks like a p but it's a little curved. And it's defined as mass over volume \( m \) over \( v \). Now in the SI system, the units are kilograms per cubic meter. And really guys, all these problems are just going to relate these three variables inside of this equation, density, mass, and volume, for just different geometric shapes like rectangular prisms, cubes, spheres, cylinders, all that stuff. And then eventually, you may also have to convert some units. So that's really all there is to it. Let's just jump straight into an example.

So we've got the average density of the Earth is 55100. We're going to assume that's the sphere, which might be news to some of you. Contrary to what you see on the Internet, the Earth is actually round. The radius is 3960 and we're going to calculate what the mass is. So first things first, I'm just going to draw a quick little sketch of the Earth. It's a sphere so I'm going to just draw it like this. So here's a sphere and the radius, which I'm going to call capital R, we already know what that is. It's 3960. So I know the density of the Earth is 55100 and I know the radius of the Earth is 3960, but it's in miles. And the density is in kilograms per cubic meter. I'm going to use this information to find out what the mass of the Earth is. This is my target variable. So here's what we're going to do. And these problems involving density, you're always just going to start off just with the by writing the density equation. So remember that the density equation, rho, is just equal to mass over volume. So if I'm trying to figure out what the mass is, basically all I need to do is just rearrange the equation. So I can move this volume term up here and just say that rho times the volume is equal to the mass. Now I already know what the density is and it's in the right units, kilograms per cubic meter. So I can use this to basically figure out what the mass is. All I have to do is just figure out what the volume of the Earth is. And so notice all we have is the radius. We don't actually have the volume. So let's just go over here and figure out what that volume is. So if I can figure out the volume of the Earth, which is just modeled after a sphere, so we're going to use this volume equation up here, which is \( \frac{4}{3} \pi \) times the radius of the Earth cubed. So we can say that this is the radius of the Earth. If I can figure out this volume here, I can plug it back into my equation for density and figure out the mass. Unfortunately, what happens is that if I do this and if I use the unit that I have for radius, this is given to me in miles, which unfortunately is not the SI unit. So first, I actually have to do a unit conversion. So I'm told here that one mile is equal to approximately 1609 meters. So what that means is I can set up a little unit conversion here. I can say that 3960 times a conversion factor will give me meters. So to do that, I want to set up my conversion factor so that the miles are on the bottom, so that it'll cancel out. And so 1 mile, I'm told here, is equal to 1609 meters. And so if I work this out, the miles cancel, and I'm just left with meters, which is the SI unit that I want. So if you work this out in your calculators, you're going to get 637 or \( 6.37 \times 10^6 \), and that's in meters. You could also just display it as a full number. That's totally fine. I just chose to do it in scientific notation because it's a little bit easier that way. So here are my units in meters. So now I've got the radius. I'm just going to plug it back into this equation and figure out the volume.

So the volume of the Earth is just equal to \( \frac{4}{3} \times \pi \times (6.37 \times 10^6)^3 \), and you have to cube that. Make sure that you plug this into your calculator. Do this first and then multiply by this raised to the 3rd power. And your calculator should be handling the rest. And when you do that, you're going to get the volume of the earth is equal to \( 1.08 \times 10^{21} \) in cubic meters. So now we just plug it back into this equation. So now we have the density, which is 55100, and we have the volume which is \( 1.08 \times 10^{21} \), and you're going to get the mass of the Earth is \( 5.94 \times 10^{24} \) kilograms. So that's what we get for the mass of the Earth. It's actually pretty close. It might be just slightly off from the real value just in case you go look it up, and that's because we rounded a bit. But that's basically the right answer for the mass of the Earth. That's it for this one, guys. Let me know if you have any questions.

A wooden cylinder has a radius of 3.5 cm and a height of 6 cm. If the mass is 161 g, what is the density of the wooden cylinder?

• A. 222 kg/m³
• B. 3.767×10³ kg/m³
• C. 697 kg/m³
• D. 2440 kg/m³

Finding the Side Length of a Cube

Hey, guys. Hopefully, you got a chance to check out this example problem. We've got an iron cube with a mass. We're told what the density of this iron cube is, and we're going to figure out the length of the sides of the cube. So, the first thing that I like to do is just draw a quick little sketch or diagram. We're told we have an iron cube, so I'm going to just draw a little cube here. A cube is basically like a special case of a rectangular prism in which all the lengths of the sides are the same. So I'm going to call this side s here. You don't have to say l, length, width, and height because they're all the same exact number.

What else do I know? I'm told the mass, the mass is just equal to 0.515. It's in kilograms, so it's the right unit. I'm also told the density of iron, 7.87×103 kilograms per cubic meters. And so I'm going to use this information to figure out what the side lengths of the cube are. That's just the letter s. So how do I use this to figure out what s is? That's my target variable.

Well, like we said, every time we have a density, or like densities, masses, and volumes involved, just start off with a density equation. So I've got my ρ is equal to mass divided by volume. So how does this get me any closer to figuring out what the side lengths are? Well, I already know what the density is, and I also already know what the mass is. So it's going to have to do something with this volume term. So the volume of a rectangular prism, remember, is just given by this equation, length times width times height. But in a special case of a cube, where all these letters are the same, length, width, and height, it's actually an even simpler equation. It's just the side length cubed. So this is my target variable. So this is actually what I have to figure out. So I can relate this back to the volume, and the volume I can get from using the density equation. So that's what I'm going to do.

So I'm going to use this density equation to figure out my volume and then basically just pass it back into this equation and then figure out the side length. So let's go ahead and do that. If I'm looking for the volume, I just need to move it over to one side. So what I can do is I can trade places with this ρ, with this density, and I can say that volume is equal to mass divided by the density. So my volume is just my mass, which is 0.515 divided by the density, 7.87×103. And notice how I've already checked the units and I don't have to do any unit conversions because they're all compatible with each other. So I just go ahead and plug this in and what I get is 6.54×10-5m³.

So now, basically, exactly like what we said, we can pass this number back into this equation and then figure out the side length. So that means my volume now, which is 6.54×10-5, is equal to the side length cubed. So, how do I get rid of this cubed, and how do I just get s? Well, I have to take the cube root of both sides. So I have to take the cube root here. So what happens is the cube root and the cube will cancel out, leaving me just the side length. And so if you plug in the cube root of this number, 6.54×10-5, you're just going to get 0.04 meters or, that would just be also 4 centimeters. So either one of these is correct. It just depends on which unit you would have to express it in. But that's really it. So let me know if you guys have any questions. That's it for this.

Copper has a density of 8.96 g/cm³. If a single copper atom as a mass of 1.055×10^-25 kg, what is the volume of a copper atom?

• A. 1.18×10^-26 m³
• B. 9.45×10^-25 m³
• C. 1.18×10^-29 m³
• D. $1.18\times10^{-26}\operatorname{\mathrm{cm}}^3$

Here’s what students ask on this topic:

How do you calculate the density of an object?

To calculate the density of an object, use the formula:

$\rho =\frac{m}{V}$

where $\rho$ is the density, $m$ is the mass, and $V$ is the volume. Ensure that the mass is in kilograms (kg) and the volume is in cubic meters (m³) for SI units. For example, if an object has a mass of 10 kg and a volume of 2 m³, its density would be:

$\rho =\frac{10}{2}=5 \text{kg/m}{3}^{}$

What is the formula for the volume of a sphere?

The formula for the volume of a sphere is:

$V=\frac{4}{3}\pi r^3$

where $V$ is the volume and $r$ is the radius of the sphere. For example, if the radius of a sphere is 3 meters, the volume would be:

$V=\frac{4}{3}\pi \times 3^3=113.1 \text{m}{3}^{}$

How do you convert units from miles to meters?

To convert units from miles to meters, use the conversion factor:

$1 \text{mile}=1609 \text{meters}$

For example, to convert 5 miles to meters, multiply by the conversion factor:

$5 \text{miles}\times 1609 \text{meters/mile}=8045 \text{meters}$

How do you find the mass of an object if you know its density and volume?

To find the mass of an object if you know its density and volume, use the formula:

$m=\rho \times V$

where $m$ is the mass, $\rho$ is the density, and $V$ is the volume. For example, if an object has a density of 500 kg/m³ and a volume of 2 m³, its mass would be:

$m=500\times 2=1000 \text{kg}$

What are the SI units for density?

The SI units for density are kilograms per cubic meter (kg/m³). This unit is derived from the formula for density:

$\rho =\frac{m}{V}$

where $m$ is the mass in kilograms (kg) and $V$ is the volume in cubic meters (m³). For example, if an object has a mass of 10 kg and a volume of 2 m³, its density would be:

$\rho =\frac{10}{2}=5 \text{kg/m}{3}^{}$