Buoyancy - Online Tutor, Practice Problems & Exam Prep

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. Now, one of the factors that we have to take into effect is buoyancy. Buoyancy is the upward force exerted on an object in a liquid or gas. Here since we're looking at buoyancy in air, we're looking at how buoyancy works within a gas. Now, 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 9.8ms2. 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.

Here the equation below 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. If we take a look here at this buoyancy equation we're going to say m here 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 0.0012gml when the pressure is 1 bar and the temperature is 25 degrees Celsius. D_w represents the density of our calibration weights. And here if it's a standard calibration weight its density will be 8gml, 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. They each would have their own different density. So 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 you 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 with greater accuracy. 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, it's noted that 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. 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 use the equation: m = m ′ − d d w − d a / d / 1 − d a / d

To find the apparent mass, 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. Subtracting them gives us the mass of just the water as 20.055 grams, which is our apparent mass. Given d a = 0.0012 g / m L , the density of the calibration weight is 8.40 grams per ml, and using 1 gram per milliliter for the density of water (since the exact temperature is not specified), we just have to solve for our true mass. This calculation results in a true mass of 20.079 grams, showing the slight difference from the apparent mass due to air buoyancy.

This is a great way to find the true mass of an object and also a good method to calibrate the pipettes that you'll be using extensively within your analytical labs. 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, the buoyancy force equals the weight of the liquid 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 nearly, completely floats on water. 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. So in the first object, we can say here that P_w represents the density of water and P_a represents the object's density. Here, the object is floating on the water. That's because the density of 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, the densities must be equal to each other. The density of water equals the density of the object. We would coin this neutral buoyancy. 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=mg=ρVg, where F is the force of the object or actually the buoyancy force, the upward force, sorry, equals mass times gravitational field strength, so that's just acceleration due to gravity, 9.8ms2. That equals the density of the liquid times the volume of the liquid displaced. And 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 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, acting upon it, 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. Then we can take into account the forces at work. Remember, the buoyancy force is always the force that's pushing upward, trying to cause the object to float, and it fights against gravity which pushes downward. 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. So 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 is telling us that it is afloat. It's floating on the lake. That tells me that the buoyancy force is equal to mass times the gravitational field strength. If they had said that it was at neutral buoyancy, then we'd say that the buoyancy force would be greater than mass times gravity. And if it had sunk to the bottom, then that would mean that the buoyancy force would be less than mass times gravity. Again, because it's floating, we're going to say they're equal to each other. Remember here that force would then equal 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 again 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 of them have the same gravity involved, so you can just eliminate that. The equation I'm going to use is going to simply become mass equals density times volume here. All right. We have length times length times length. Length³ will give me volume. So volume equals 0.15 meters times 0.44 meters times 0.56 meters. So, when I multiply all those together, I get 0.03696 meters³. Here the density of the liquid, which is water, is just 1 gram per milliliter. But here I have meters³. I'm going to have to convert meters³ into centimeters³ because, remember, a centimeter³ is equal to a milliliter. So, we're going to put meters here on the bottom, centimeters here on top. 1 centi is 10 to the negative 2. We have to cube this entire thing so that the meters³ can cancel out and and I'll have centimeters³ at the end. So when I work that out, it really means that I have 0.03696 meters³ times 1106-1 centimeters³. So that's going to give me 36960 centimeters³. Take that and plug it in here. 36960 centimeters³. And what happens here is milliliters and centimeters³ cancel out. So, at the end, I'll have the mass as 36960 grams. So that would be our answer in terms of this question. Again, you have to determine if the object is sinking, floating, or if it is in neutral buoyancy to determine which version of force equals greater than or less than mass times gravity. From there, we're able to isolate the one variable that we needed at the end which is mass. So continue on where we're looking at examples where we're dealing with buoyancy whether it be in air or whether it be within fluids. When it's dealing with air, we have to take into account airflow. That's when we use the buoyancy equation to find the true buoyancy in fluids. In that way, we'd have to use Archimedes' principle. Remember the distinction between the two forms of buoyancy, whether it's in air or whether it's in a fluid.

Buoyancy Calculations

Here we are told that we have a small crystal of sucrose and it has a mass of 5.345 mg. We are 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? We want our density in grams per milliliter. So 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 milligram, we have 10 to the negative 3 grams. So that's 5.345×10-3g. Now we need to figure out our volume. We need milliliters. Well, realize here that they give us, millimeters here on the bottom and if we multiply all 3 together, that's going to give me millimeters 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 altogether to get millimeters cubed, convert that to centimeters cubed, and that will represent the milliliters that we need at the end for our density.

We have 2.20 millimeters times 1.36 millimeters times 1.12 millimeters. Together, that gives me 3.35104 mm3. 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 cancel out with the cubed, so I'm going to cube the entire thing. So that's going to give me 3.35104 mm3÷1000=3.35104×10-3 mL. So when we punch that in, we get 1.595 g/mL, which if we want 3 significant figures, we get 1.60 g/mL for 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 are 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. So remember here that our buoyant force, F, is equal to the weight of the water displaced. Therefore, our buoyant force is equal to the density of the liquid, the volume of the water that's displaced, and g, which is our gravitational field strength.

Now, to figure out our buoyant force, we will say the 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. We're looking for the density of the liquid. Here we have the volume of the liquid. We're going to have to convert it because of the units of our gravitational field strength, which is 9.8 newtons per kilogram. We're going to convert our volume from centimeters cubed to meters cubed. So, we will have 4,200 centimeters cubed. We need to cube the conversion factor to get centimeters cubed to cancel out, which really results in 4200 centimeters cubed times 10 to the negative 6 meters cubed on top and 1 centimeter cubed on the bottom. This is equal to 4.20 times 10 to the negative 3 meters cubed.

Next, we divide both sides by 4.20 times 10 to the negative 3 meters cubed and also by 9.8 newtons per kilogram. So at the end, we'll get our density of this liquid as 558.8 kilograms per meter cubed. If we use significant figures, assuming there's a decimal point in each value, we would keep 2 significant figures here. So our final reported density would be 5.6 times 10 to the 2 kilograms per meter cubed.

Thus, the density of the water, adopting Archimedes' principle for dealing with a solid object in a liquid, is calculated to be 560 kilograms per meter cubed. If we were dealing with an object being weighed typically, only airflow would be a concern, and we would use the buoyancy equation. Now that you've seen this example, look to see if you can tackle example 2. If you're stuck, come back to see how I approached 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're 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. 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 \(1 - \frac{0.0012 \text{ grams per milliliter}}{8.0 \text{ grams per milliliter}}\). Then on the bottom, we're going to divide by \(1 - \frac{\text{density of air}}{\text{density of the propane}}\). Here that would be 0.922 grams per milliliter, like what we're told right here. So, 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 \(1 - \frac{0.0012}{0.922}\)

This calculation 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 to 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