Buoyancy - Online Tutor, Practice Problems & Exam Prep

Buoyancy in air affects the precision of measurements in analytical chemistry. When weighing an object, the downward force of gravity (9.8 m/s²) and the upward buoyant force from air must be considered. The true mass can be calculated using the equation: $m_t=m_a+(d_a-d_w)\times d_o)$. Here, m_a is the apparent mass, d_a is the density of air, d_w is the density of calibration weights, and d_o is the density of the object. Understanding this concept is crucial for achieving accuracy in measurements.

Buoyancy in Air

Guys, in this video, we're going to take a look at buoyancy within air. Now, in analytical chemistry, we're trying to minimize as many factors as possible in order to ensure that we get the highest level of precision and accuracy within our measurements. One of the factors that we have to take into account is buoyancy. Buoyancy is the upward force exerted on an object in a liquid or gas. Since we're looking at buoyancy in air, we're examining how buoyancy works within a gas.

Even something as simple as weighing an object on an analytical balance has some level of uncertainty associated with it. Remember, there are two forces at work. There is the downward force upon the object as a result of gravity. We can use the variables \(a\) or \(g\) to represent it, with a value of \(9.8 \text{ meters per second squared}\). But here, when we put an object on an analytical balance, we have to take into account that there is airflow.

This airflow will distort our true mass of the object. This airflow is buoyancy; it's the upward force upon the object. So, even when you put it in a closed analytical balance, there is some level of uncertainty associated with the mass that you're recording. Below, the equation will allow us to calculate the true mass of the object as though it exists within a vacuum.

If it exists within a vacuum, this eliminates airflow entirely, so we'll get the actual mass of the object. Taking a look at this buoyancy equation, we're going to say m represents our true mass. This m here is our apparent mass, the mass that you read off of the analytical balance once you weighed your analyte. D_a represents the density of air, which has a density of \(0.00122\) grams per milliliter when the pressure is 1 bar and the temperature is 25 degrees Celsius. D_w represents the density of our calibration weights.

If it's a standard calibration weight, its density will be \(8\) grams per milliliter. But you have to pay very close attention to the question because different types of metals and alloys can be used in place of a standard calibration weight. Each would have its own different density. If it was using something that was not a standard calibration weight, they would give you a new density to input into the formula. Finally, d just represents the density of the weighed object.

So remember, anytime you weigh anything, there is a level of uncertainty associated with airflow. There's that upward force, that buoyancy force that'll distort our true mass. This equation allows us to weigh the mass as though it exists within a vacuum, which eliminates the whole idea of airflow. Now that we've seen this equation, we'll apply it to the example that we see below, so click on the next video and see how we approach that problem.

Buoyancy in Air

The pipettes that you'll be using in your analytical labs come from the manufacturer as usually Class A pipettes. But once you've obtained them, there is some level of uncertainty associated with them. You have to first calibrate them before we can use them in experiments. Here, we're going to say a convenient method in calibrating pipettes is to weigh the water delivered from them. By using the density of water at a given temperature, we can determine the volume of water with greater accuracy.

So here it says assume a 50 milliliter pipette is in need of calibration. An empty flask weighs 49.563 grams. When water delivered from the pipette is added to the empty flask, the new mass is recorded as 69.618 grams. We're asked, What is the mass of water delivered? Now, here they're saying the density of the standard weights is determined to be 8.40 grams per milliliter.

Remember, in the equation above, we said that the standard weight is 8.0 grams per milliliter. In this question, we must be using a different type of metal or alloy for those calibration weights which is why it's a new number. Again, always be careful if they're giving you a new density for the calibration weights if they do use that one.

We're going to say here \( m = m' \times \left(1 - \frac{d_a}{d_w}\right) \), divided by \( 1 - \frac{d_a}{d} \). We need to first find the apparent mass.

Well, when the flask is filled with water, the weight of the water and the flask together is 69.618 grams. When the flask is empty, it weighs that much. When we subtract them, that'll give me the mass of just the water. So that comes out to 20.055 grams. So that's our apparent mass.

So times \( \frac{1}{1 - \frac{d_a}{d}} \), So \( d_a \) is the density of air, which is 0.0012 grams per milliliter. The density of the calibration weight, we're told, is 8.40 grams per milliliter divided by \( 1 - \frac{d_a}{d_w} \). Here, they don't tell us the exact temperature. The density of water can change with changing temperatures, but most of them are around 1 gram per milliliter. We're going to use 1 gram per milliliter for the density of water.

We just have to solve for our true mass now. So that's 20.055 grams times when I work all this out in here, it gives me 0.99857. Divide by the bottom when I subtract this value here from 1. What does it give me? It gives me 0.998.

Then what do we get at the end? So at the end, we're going to get 20.079 grams. So notice our apparent mass when we read it from the analytical balance was that number, but the true mass when we take into account airflow gives us this new mass for our water. Again, this is a great way in order to find the true mass of an object and also a good way to calibrate the pipettes that you'll be using extensively within your analytical labs. So now that we've seen this, we'll take a look at buoyancy when it's found within fluids.

Buoyancy in Fluid

In this video, we're taking a look at buoyancy within fluids. Now when we're taking into account buoyancy within fluids, we're looking at Archimedes' principle which states that the buoyancy force, so that's the upward force acting upon an object, is equal to the weight of the fluid displaced by the object. So, buoyancy force equals the weight of the fluid displaced. Remember, when we're trying to place an object within water, there are 3 scenarios that are possible.

We could have one scenario where the object itself floats on top of the water, partially or almost completely. We have one where it is exactly in the middle and then one where it sinks all the way to the bottom. So, these are our three scenarios. Now remember, ultimately an object's buoyancy within a fluid is determined by its density. In the first object, we can say here that PW represents the density of water and PA represents the object's density.

Here, the object is floating on the water. That's because the density of the water must be greater than the density of the object. In this one, it's floating directly in the middle of the fluid. In this case, they must be equal to each other. So, density of water equals the density of the object.

We would coin this neutral buoyancy. And then finally, the object sinks all the way down to the bottom. That's because the density of water must be less than the density of the object. When we're dealing with Archimedes' principle, the equation that we're dealing with is F, which is the force of the object or actually the buoyancy force, the upward force, equals mass times the gravitational field strength, so that's just acceleration due to gravity, which is 9.8ms2 and that equals the density of the liquid times the volume of the liquid displaced.

Again, we have the gravity due to acceleration, 9.8ms2. Now, in each of these objects, we always have 2 forces at work. We have the force that's pushing up, so this is our force due to the buoyancy force. And then you have another force due to gravity.

See, those are the two forces acting upon these objects. Here, when it's floating on the surface, we'd say that the force, the buoyancy force, is equal to the gravitational force. In this one where it's exactly equal in the middle, we would say in this case that the buoyancy force must be greater than the gravitational force. Lastly, when it's sunken all the way down to the bottom, the force that's pushing it down, the gravitational force must be greater. So again, we have two scenarios happening here where we can look at the density of the object versus the density of the water, which ultimately determines the buoyancy of the object overall.

And then we can take into account the forces at work. Remember, the buoyancy force is always a force that's pushing upward, trying to cause the object to float and it fights against gravity, which pushes downward. It's important to remember the Archimedes principle which we'll be utilizing when answering questions dealing with objects suspended in fluid and then understanding the relationship between the density of the objects. As we work more and more towards Archimedes principles, we'll take a look at examples. On the one below, we're dealing with the typical Archimedes principle type of question.

Click on the following video to see how I approach that question.

Buoyancy in Fluid

Here in this question, it says a wooden block with measurements of 0.15x0.44x0.56 meters is afloat on a lake. If it is submerged by 0.031 meters, what is its mass? Here it's telling us that it is floating on the lake. This indicates that the buoyancy force is equal to the mass times the gravitational field strength.

If they had said that it was at neutral buoyancy, then we would say that the buoyancy force would be greater than the mass times gravity. And if it had sunk to the bottom, then that would mean that the buoyancy force is less than the mass times gravity. Since it's floating, we conclude that they are equal to each other. Remember here that force equals the volume of water displaced, which is why we can continue and say that this equals the density of the liquid times the volume of the liquid displaced times the gravitational field strength. This portion I'm going to bring down.

So, mass equals gravity. Mass times gravity equals density times volume times gravity. Both have the same gravity involved, so you can just eliminate that. The equation I'm going to use simply becomes mass equals density times volume.

So we have length times width times height. Length cubed will give me volume. So, volume equals 0.15 meters times 0.44 meters times 0.56 meters. When I multiply all those together, I get 0.03696 meters cubed. Here the density of the liquid, which is water, is just 1 gram per milliliter.

But here I have meters cubed. I'm going to have to convert meters cubed into centimeters cubed because remember, a centimeter cubed is equal to a milliliter. So we're going to put meters here on the bottom, centimeters here on top. 1 centimeter is \(10^{-2}\). We have to cube this entire thing so that the meters cubed cancel out, and I'll have centimeters cubed at the end.

So when I work that out, it really means that I have \(0.03696 \text{ meters cubed} \times \frac{1 \text{ centimeter cubed}}{10^{-6} \text{ meters cubed}}\). So that's going to give me 36960 centimeters cubed. Plugging it in here: 36960 centimeters cubed. The milliliters and centimeters cubed cancel out.

So at the end, I'll have the mass as 36,960 grams. This answers the question in terms of this situation. Continue to determine if the object is sinking, floating, or at neutral buoyancy to determine which version of force is equal to, greater than, or less than mass times gravity. From there, we isolate the one variable we needed at the end, which is mass. Continue onwards looking at examples where we're dealing with buoyancy, whether it be in air or within fluids.

When dealing with air, we have to take into account airflow. That's when we use the buoyancy equation to find the truth in this set of questions, we're dealing with buoyancy in fluids. In that context, we would need to use the Archimedes' principle. So remember the distinction between the two forms of buoyancy, whether it's in air or in a fluid.

Buoyancy Calculations

Here we're told that we have a small crystal of sucrose and it has a mass of 5.345 milligrams. We're told that the dimensions of the box-like crystal were 2.20 millimeters by 1.36 millimeters by 1.12 millimeters. We're asked what is the density of the sucrose crystal expressed in grams per milliliter? So we want our density in grams per milliliter. What we're going to do first is we're going to convert the milligrams given to us into grams.

So that's 5.345 milligrams. We're going to say here for every 1 mg, we have 10 to the negative 3rd. So that's 5.345 times 10^-3 grams. So we're going to put that here up top. Now we need to figure out our volume.

We need milliliters. We'll realize here that they give us, millimeters here on the bottom and if we multiply all three together, that's going to give me milliliters cubed. Also, remember that if we can get centimeters cubed, that a centimeter cubed is equal to a milliliter. So what we're going to do here is we're going to multiply those dimensions all together to get millimeters cubed, convert that to centimeters cubed and that will represent the milliliters that we need at the end for our density. So we have 2.20 millimeters times 1.36 millimeters times 1.12 millimeters.

Together, that gives me 3.35104 millimeters cubed. All we have to do now is convert millimeters cubed to centimeters cubed. Remember, the conversion is 1 centimeter is equal to 10 millimeters. Those millimeters can't cancel out with the centimeters with the millimeters cubed, so I'm going to cube the entire thing. So that's going to give me 3.35104-3 milliliters.

So when we punch that in, we get 1.595 grams per milliliter, which if we want 3 significant figures, we get 1.60 grams per milliliter of this box-like crystal and its density. Take this approach and what we've learned thus far and see if you can attempt the practice question here on the bottom. Again, attempt it on your own. If you get stuck, just come back to the next video and see how I approach answering that question. Good luck, guys.

An empty container weighing 73.190 g is filled with an unknown liquid and the combined mass is recorded as 87.308 g. The container was then emptied and filled with water and recorded a new mass of 88.442 g at a temperature of 19 ^oC. (d = 1.0027 g/mL). Calculate the density of the unknown liquid. 

• A. 0.92861 g/mL
• B. 0.92565 g/mL
• C. 0.92815 g/mL
• D. 0.93000 g/mL

Buoyancy Calculations 2

Here we're told that a piece of concrete weighs 120 newtons and when it's fully submerged, its apparent weight is 97 newtons. Determine the density of the water if the volume of the water displaced is 4200 centimeters cubed. All right. So remember here that our buoyant force is f, and we're going to say here that our buoyant force is equal to the weight of the water displaced. So that's equal to the density of the liquid itself.

We have the volume of the water that's displaced and then we have g, which is our gravitational field strength. Now, we're going to say to figure out our buoyant force, we're going to say buoyant force F is equal to the actual weight minus the apparent weight. So that's going to be 120 newtons minus 97 newtons, which gives us 23 newtons. So it's going to be 23 newtons here. We're looking for the density of the liquid.

Here we have the volume of the liquid. Here we're going to have to convert it. And that's because of the units of our gravitational field strength, which is 9.8 newtons divided by kilograms. So what we're going to do here is we're going to convert our volume from centimeters cubed to meters cubed.

And that's because remember, a newton is equal to meters times kilograms over seconds squared. Because meters are involved, we should change our volume to meters. So we're going to have 4,200 centimeters cubed. We're going to say here for every 1 centimeter, it's 10 to the negative 2 meters. We want centimeters cubes to cancel out, so we're going to have to cube this whole thing.

But what effect does that really have? That's really saying 4200 centimeters cubed times 10 to the negative 6 meters cubed on top and 1 centimeters cubed on the bottom. Okay, so that's equal to 4.20 times 10 to the negative 3 meters cubed here. All right. So now what we're going to do is we're going to divide both sides here by 4.20 times 10 to the negative 3 meters cubed and also the 9.8 newtons over kilograms.

Okay, do that on both sides here. So this cancels out with this. Our newtons will cancel out. So at the end, we'll get our density of this liquid as being 558.8 kilograms over meters cubed.

Here if we wanted to do significant figures, this has let's just assume there's a decimal there. Let's say we have a decimal there. We'll say that has 3 significant figures, this has 2 significant figures. Let's say there's a decimal here, this has 4 significant figures. So if we wanted 2 significant figures here, we can say this is 5.56 times 10 to the 2 kilograms over meters cubed.

So that'll be the density of our liquid. And remember, we know that we're using this version of an equation because we're dealing with a solid object, and its equation because we're dealing with a solid object, and it's placed in a fluid. And when we're talking about fluids, we're going to adopt Archimedes' principle in order to figure out our answer. If we're dealing with just an analyte being weighed on a typical analytical balance, all we'd have to worry about is airflow and in that case, we'd use the buoyancy equation. Now that you've seen this example, look to see if you can tackle example 2.

Again, as always, if you're stuck, just come back and see how I approach answering that question.

Buoyancy Calculations 2

So here it says the density of propane, an odorless hydrocarbon compound used in cooking, is 0.922 grams per milliliter. We are asked when a sample of it is placed on an analytical balance, a weight of \(8.15 \times 10^8\) nanograms is obtained. Calculate the true mass of propane. All right, so we're looking for the true mass. That means we're looking for the mass of an analyte when we're weighing it on an analytical balance.

Now we're going to have to use the buoyancy equation. So that is: the true mass of my object is equal to the mass of it when it's on the analytical balance times \(1 - \frac{\text{density of air}}{\text{density of water}}\). And then on the bottom, we have \(1 - \frac{\text{density of air}}{\text{density of the object or analyte itself}}\). Now, we need to have the mass of the analyte in grams. It's given to us here in nanograms.

So, just remember we have \(8.15 \times 10^8\) nanograms. For every 1 nanogram, it's \(10^{-9}\) grams. So that means that the weight according to the analytical balance is \(8.15 \times 10^{-1}\) grams. We're going to have \(1 - \frac{0.0012}{8.0}\) grams per milliliter. Then on the bottom, we're going to divide by \(1 - \frac{0.0012}{0.922}\).

Now here, when I do \(1 - \frac{0.0012}{8.0}\), I'm going to get here, 0.99985. The density units cancel out. Then, we're going to divide by \(1 - \frac{0.0012}{0.922}\) gives me 0.998698.

Now here, we're going to want our answer to have 3 significant figures because 8.15 has 3 significant figures in it. When we do that, we get 0.81594 grams which at the end gives us 0.816 grams as the true mass of my object once I've taken into account airflow. Now finally, take a look at the final practice example that's left at the bottom of the page. Attempt to do it on your own. But again, as always, if you get stuck, don't worry.

Just come back and take a look at how I approach answering that practice question.

A 39.0 g piece of aluminum metal has a volume of 14.4 cm³. Calculate the apparent weight of the piece of metal when it is immersed in chloroform. Density of chloroform at 25 °C is 1.49 g/cm³.

• A. 0.382 N
• B. 0.210 N
• C. 0.253 N
• D. 0.115 N