Video: Calculating Density

In this lesson, we will learn how to use the formula 𝜌 = 𝑚/𝑉 to calculate the densities of different materials and objects.


Video Transcript

In this video, we’re talking about calculating density. What we’ll see is that, given some object, in order to calculate its density, we’ll need to know two things about it. Beyond that, we’ll get some practice going through the steps of calculating the densities of various materials.

To start off, we can recall that density is a measure of how much material, that is, mass, fits into a given space, a volume. This point is important because it means that density isn’t only about mass and it’s not only about volume either. For example, say that we have these three different-sized cubes. The length of each of the sides of the smallest cube we can call 𝐿. The next largest cube has side length of two 𝐿. And the biggest cube of all has sides of length five 𝐿. And let’s say further that, going left to right, the masses of these three cubes are 10 kilograms, 12 kilograms, and eight kilograms, respectively.

If we were going to calculate the relative density of each of these cubes, we might guess that the most dense cube is the largest one, the one with the largest side length and the one with the most mass. To calculate these three densities, let’s rewrite this sentence as an equation. The sentence told us that density is how much mass fits in a certain volume.

Mathematically then, we can see that density is equal to mass divided by volume. And we can even go further, letting symbols represent these terms in the equation. To represent the density of a material, we’ll use the Greek letter 𝜌. Then to represent an object’s mass, we’ll use 𝑚. And to represent volume, we’ll use capital 𝑉.

We now have a mathematical expression for the density of an object that will let us calculate that density, so long as we know its mass and its volume. Knowing this, we can now calculate the densities of these three cubes. The density of the green cube — we’ll call it 𝜌 sub 𝑔 — is equal to the mass of that cube, 10 kilograms, divided by its volume. Since this cube has sides of length two 𝐿, that means the volume it takes up is two 𝐿 times two 𝐿 times two 𝐿, or written in another way, two 𝐿 quantity cubed. And when we cube this denominator, we find a result of eight 𝐿 cubed.

So finally, if we simplify this fraction, we find that the density of the green cube is five kilograms divided by four 𝐿 cubed, where 𝐿 is some distance in meters. We can then store that result off to the side and go on to calculating the density of our orange cube, what we’ll call 𝜌 sub 𝑜. For this cube, the mass is 12 kilograms. And the volume it takes up is five 𝐿 times five 𝐿 times five 𝐿. We can also write that as five 𝐿 quantity cubed. And that expands out to 125𝐿 cubed. That’s because five times five times five is 125. We can store this result and then go on to calculating the density of our blue cube. We’ll call it 𝜌 sub 𝑏. This smallest cube of all three has a mass of eight kilograms and a volume of 𝐿 times 𝐿 times 𝐿, or 𝐿 cubed.

We can now compare these results for the three different cubes. And we see that the density of the smallest cube, the blue one, is actually the greatest. The blue cube density is greatest. And then the green cube density is greater than the density of the orange cube. What we find then is that our largest cube, the orange one, has the smallest density. And then our smallest cube in terms of volume has the greatest density.

This helps us understand that just because an object is larger, like the orange cube in this case, doesn’t mean it has a larger density. And in fact, because volume appears in the denominator of the equation for calculating density, by itself having a large volume decreases overall density.

Beyond just considering relative differences in material densities, it’s helpful to talk a bit about the units of density. We see that density is calculated by dividing mass by volume. That means that the units of density, putting these square brackets around the density symbol, will be equal to the units of mass, typically kilograms, divided by the units of volume, typically meters cubed. We say “typically” because sometimes our mass is given in units of grams or milligrams or even micrograms. And sometimes our volume is given in units of cubic centimeters or cubic millimeters, and so on.

So to equip ourselves for any units that might come up in a density calculation, it’s helpful to recall the conversions between different mass and volume units. The most common mass conversion is between units of kilograms and grams. For every one kilogram, there are 1000 grams. Sometimes though, we’re given an object’s mass in units of milligrams. And there are 1000 milligrams in one single gram. So we can see if there are 1000 milligrams in one gram and 1000 grams in one kilogram, that means there are 1000 times 1000 milligrams in one kilogram, in other words 1000000 milligrams per kilogram.

Moving on to the units of volume and converting between them, this one is a bit trickier than units of mass. That’s because when we’re calculating a volume, it has dimensions of length cubed. We can look at it this way. Say we wanna convert between meters and centimeters. To do that, we can recall that one meter is equal to 100 centimeters. But remember, this one meter is a length, as is 100 centimeters, whereas for our density calculation, we want a volume, a length cubed.

So let’s say we go to our one meter equals 100 centimeters equation. And then we cube both sides of it. This way, we have a volume on either side of the equation. Looking on this left-hand side, when we apply the cube to both the meters as well as the number, one, we find a result of one cubed times meters cubed. But since one cubed is just one, then it simplifies to one meter cubed.

On the right-hand side of the expression, when we do the same thing, when we cube the unit as well as the number, we find a result of 100 cubed times centimeters cubed. And 100 cubed, which is 100 times 100 times 100, is 1000000. We see then that when we’re talking about a volume, a length cubed, we need to be careful to keep track of our conversions. They’re not as easy as when we’re working with just a length to another length.

Perhaps the best way to take care when calculating volumes is to start with a length conversion, say, for example, centimeters to millimeters. And then when calculating a volume, cube both sides of the equation and then calculate each side separately. In this instance, when we do that, we find that one cubic centimeter is equal to 1000 millimeters cubed. We now know enough to try out a few examples that will let us calculate density for ourselves.

Gold can be hammered into very thin sheets. If a thin sheet of gold is 30 centimeters wide and 40 centimeters long and has a mass of 9.4 grams, find the thickness of the sheet. Use a value of 19320 kilograms per cubic meter for the density of gold. Give your answer in millimeters to two significant figures.

Okay, so we have here this very thin sheet of gold. It has a width of 30 centimeters and a length of 40 centimeters. And we’ll say that it has a thickness that we’ll call 𝑡. That’s what we want to solve for. In the problem statement, we’re told what the density of gold is. But interestingly, we’re given that density in units of kilograms per cubic meter, whereas we’re given these lengths in units of centimeters. And to complicate it even more, we want to give our answer in units of millimeters.

So clearly, in answering this question, we’re going to be doing some unit conversion. There are a number of different ways we could convert units so that, in the end, we give our answer in millimeters to two significant figures. What we’re going to do is we’ll leave the density of gold, what we’ll refer to as 𝜌 sub 𝑔, in its original units of kilograms per cubic meter. Then what we’ll do is we’ll convert the given dimensions of our gold sheet, centimeters, to units of meters. And then we’ll convert the mass of the sheet given in units of grams to kilograms.

So let’s start doing that now. And as an overarching principle, if we were to write out the volume of this gold sheet in units of cubic meters and then write out the mass of this gold sheet in units of kilograms, we know what the ratio of those two values will be. It will be equal to 𝜌 sub 𝑔, the density of gold. So that’s what we’ll work on.

And to get started, let’s recall that one meter is equal to 100 centimeters. That’s the conversion between these two length units. This means that the volume of our gold sheet, what we can call 𝑉 sub 𝑠, is equal not to 30 centimeters by 40 centimeters by 𝑡, but 0.30 meters by 0.40 meters times 𝑡, where 𝑡 is now in units of meters.

Now that we have an expression for the volume of our gold sheet written in terms of the variable we want to solve for, 𝑡, we can now turn our attention to the mass of the sheet. As we saw, that mass is given in units of grams. But we’d like to express it in units of kilograms, to match the units of our given density.

We can now recall that the conversion between kilograms and grams is that one kilogram is equal to 1000 grams. Based on this, we can write the mass of our gold sheet, 𝑚 sub 𝑠, this way. We can say it’s equal to 0.0094 kilograms. Or writing this in scientific notation, that mass is 9.4 times 10 to the negative third kilograms.

We now have expressions for the mass and the volume of our gold sheet. And we can recall at this point that, in general, the density, 𝜌, of an object is equal to the mass of that object divided by its volume. This means the density of our sheet, which is equal to 𝜌 sub 𝑔 since the sheet is entirely made of gold, is equal to the mass of the sheet divided by its volume.

When we substitute in the values we’ve found for these two terms, we find that density is 9.4 times 10 to the negative third kilograms divided by 0.30 meters times 0.40 meters times 𝑡. And this whole fraction is equal to what we were given for the density of gold earlier, 19320 kilograms per cubic meter. That’s a wonderful thing because now we can solve for 𝑡 in the left-hand side of this expression.

To do it, we’ll multiply both sides of the equation by 𝑡 so that that term cancels out on the left-hand side. And then we’ll multiply both sides of the equation by the inverse of this gold density. That is, we’ll multiply by one meter cubed divided by 19320 kilograms. Looking at the right-hand side of our equation, that means that 19320 kilograms cancels in the denominator and numerator, as does the meters cubed term. This means we’re left just with 𝑡 all by itself, the variable we want to solve for.

And then on the left-hand side of the equation, notice what units we have now. On the top in the numerator, we’re multiplying by meters cubed. And in the bottom on the denominator, we’re dividing by meters squared. So this means two factors of meters cancel out from top and bottom. Along with that, both numerator and denominator have a factor of kilograms. So that unit cancels as well.

What we’re left with is a bunch of numbers, 9.4 times 10 to the negative third divided by 19320 times 0.30 times 0.40, and one single unit, a unit of meters. We can rearrange this left side this way. We can write it so that now we have one expression, which when we calculate it will just be some number, and then a unit, the unit of meters.

And remember, this left side is designed to calculate the thickness, 𝑡, of our gold sheet. And we see that we’ll get that thickness now in units of meters. But at this point, we can recall that we don’t want that answer ultimately in meters. We want it in millimeters to two significant figures.

So then let’s recall the conversion from meters to millimeters. One meter of distance is equal to 1000 millimeters of distance, all of which means if we wanna replace this meter with some amount of millimeters, that one meter will be equal to 1000 millimeters.

And then with that calculation done, we’re finished all of our setup. We now have an expression for the thickness of the gold sheet in units of millimeters. All that’s left for us to do is to calculate this expression and then round it to two significant figures. When we do that, when we calculate as well as round, we find a result of 0.0041 millimeters. That’s the thickness of this gold sheet in those units.

Let’s take a moment now to summarize what we’ve learned so far about calculating density. In this lesson, we learned that the density of an object tells us how much of its mass fits within a certain volume. Written as an equation, that’s expressed this way, that an object’s density is equal to its mass divided by its volume. And we saw that a shorthand way to write this is to use the Greek letter 𝜌 to symbolize density, 𝑚 for mass, and capital 𝑉 for volume.

Furthermore, we learned that, to calculate an object’s density, unit conversion is often needed. This is because we may wanna solve for object density in units of kilograms per cubic meter or grams per cubic centimeter or milligrams per cubic millimeter, or even some other set of units. To calculate density in any units, we’ll want to recall the conversions between kilograms and grams; grams and milligrams; and meters, centimeters, and millimeters. Those are the primary unit conversions involved in calculating density.

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