Buoyancy: Videos & Practice Problems

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

Buoyancy is a crucial concept in analytical chemistry, particularly when striving for precision and accuracy in measurements. It refers to the upward force exerted on an object submerged in a fluid, which in this context is air. When weighing an object on an analytical balance, two primary forces are at play: the downward gravitational force and the upward buoyant force caused by air. The gravitational acceleration, denoted as \( g \), has a standard value of \( 9.8 \, \text{m/s}^2 \).

Even in a controlled environment, such as a closed analytical balance, the presence of airflow can introduce uncertainty in the measured mass. This is where the concept of buoyancy becomes significant, as it affects the apparent mass recorded by the balance. To accurately determine the true mass of an object as if it were in a vacuum—where no airflow exists—one can use a specific buoyancy equation.

The equation can be expressed as follows:

\[m = m_a + (D_a - D_w) \cdot V\]

In this equation:

• m represents the true mass of the object.
• m_a is the apparent mass, or the mass read from the analytical balance.
• D_a is the density of air, approximately \( 0.00122 \, \text{g/mL} \) at 1 bar pressure and 25 degrees Celsius.
• D_w is the density of the calibration weights, typically \( 8 \, \text{g/mL} \) for standard weights, but this can vary based on the material used.
• V is the volume of the object being weighed.

It is essential to note that different materials may have varying densities, which should be accounted for when using non-standard calibration weights. By applying this equation, one can effectively eliminate the influence of buoyancy and obtain a more accurate measurement of mass. Understanding and applying this concept is vital for achieving high levels of precision in analytical measurements.

Buoyancy in Air

Pipettes used in analytical labs are typically Class A, but they require calibration to account for measurement uncertainty. A convenient method for calibrating pipettes involves weighing the water they deliver. By utilizing the density of water at a specific temperature, one can determine the volume delivered with greater accuracy.

For instance, consider a 50 milliliter pipette that needs calibration. An empty flask weighs 49.563 grams, and when water delivered from the pipette is added, the total mass is recorded as 69.618 grams. To find the mass of water delivered, we first calculate the apparent mass of the water:

Apparent Mass (m) = Total Mass with Water - Mass of Empty Flask

m = 69.618 g - 49.563 g = 20.055 g

Next, we need to account for the density of the calibration weights, which is given as 8.40 grams per milliliter. The formula to find the true mass of the water delivered, considering the density of air (0.0012 grams per milliliter) and the density of water (assumed to be 1 gram per milliliter), is:

True Mass = m × \(\frac{m'}{1 - \frac{d_A}{d_W}}\)

Where:

• m' = density of calibration weight (8.40 g/mL)
• d_A = density of air (0.0012 g/mL)
• d_W = density of water (1 g/mL)

Substituting the values, we calculate:

True Mass = 20.055 g × \(\frac{8.40}{1 - \frac{0.0012}{1}}\)

After performing the calculations, we find:

True Mass ≈ 20.079 g

This demonstrates that the apparent mass measured by the analytical balance differs from the true mass when accounting for the effects of air density. This method not only helps in determining the true mass of the water but also serves as an effective way to calibrate pipettes used in analytical labs.

Understanding buoyancy in fluids is also essential, as it relates to how objects behave when submerged, which can further influence measurements in laboratory settings.

Buoyancy in Fluid

Buoyancy in fluids is fundamentally explained by Archimedes' principle, which states that the buoyant force acting on an object submerged in a fluid is equal to the weight of the fluid displaced by that object. This relationship can be expressed as:

$$ F_b = W_d $$

where \( F_b \) is the buoyant force and \( W_d \) is the weight of the displaced fluid. Understanding buoyancy involves considering three scenarios for an object placed in water: it can float, be neutrally buoyant, or sink. These scenarios depend on the relationship between the object's density (\( \rho_A \)) and the density of the fluid, typically water (\( \rho_W \)).

1. **Floating**: An object floats when its density is less than that of the fluid (\( \rho_A < \rho_W \)). In this case, the buoyant force equals the gravitational force acting on the object.

2. **Neutral Buoyancy**: An object is neutrally buoyant when its density is equal to that of the fluid (\( \rho_A = \rho_W \)). Here, the object remains suspended in the fluid without sinking or rising.

3. **Sinking**: An object sinks when its density is greater than that of the fluid (\( \rho_A > \rho_W \)). In this scenario, the gravitational force exceeds the buoyant force, causing the object to descend.

The buoyant force can also be calculated using the formula:

$$ F_b = \rho_L \cdot V_d \cdot g $$

where \( \rho_L \) is the density of the liquid, \( V_d \) is the volume of the liquid displaced, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)). This equation highlights the interplay between the forces acting on the object: the upward buoyant force and the downward gravitational force.

In summary, the behavior of an object in a fluid is determined by its density relative to the fluid's density and the balance of forces acting upon it. Understanding these principles is crucial for solving problems related to buoyancy and fluid dynamics.

Buoyancy in Fluid

To determine the mass of a wooden block floating on a lake, we start by recognizing that the block is subject to buoyancy, which is the upward force exerted by the fluid. According to the principles of buoyancy, when an object is floating, the buoyant force is equal to the weight of the object. This relationship can be expressed mathematically as:

$$ F_b = m \cdot g $$

where \( F_b \) is the buoyant force, \( m \) is the mass of the object, and \( g \) is the gravitational field strength.

Since the block is floating, we can also relate the buoyant force to the volume of water displaced, which is given by:

$$ F_b = \rho \cdot V \cdot g $$

Here, \( \rho \) represents the density of the fluid (water in this case), and \( V \) is the volume of the water displaced. By equating the two expressions for buoyant force, we can eliminate \( g \) from both sides:

$$ m \cdot g = \rho \cdot V \cdot g $$

Thus, we simplify to:

$$ m = \rho \cdot V $$

Next, we calculate the volume of the wooden block using its dimensions:

$$ V = 0.15 \, \text{m} \times 0.44 \, \text{m} \times 0.56 \, \text{m} = 0.03696 \, \text{m}^3 $$

To find the mass, we need the density of water, which is typically \( 1 \, \text{g/cm}^3 \). However, since our volume is in cubic meters, we convert it to cubic centimeters (noting that \( 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \)):

$$ V = 0.03696 \, \text{m}^3 \times 10^6 \, \text{cm}^3/\text{m}^3 = 36960 \, \text{cm}^3 $$

Now, substituting the values into the mass equation:

$$ m = \rho \cdot V = 1 \, \text{g/cm}^3 \times 36960 \, \text{cm}^3 = 36960 \, \text{g} $$

Thus, the mass of the wooden block is 36,960 grams. This example illustrates the application of Archimedes' principle in determining the mass of an object based on its buoyancy in a fluid. Understanding whether an object is floating, sinking, or at neutral buoyancy is crucial for applying the correct principles of buoyancy effectively.

Buoyancy Calculations

To determine the density of a sucrose crystal, we start with the mass and dimensions provided. The mass of the crystal is given as 5.345 milligrams, which we convert to grams for our calculations. Since there are 1000 milligrams in a gram, we can express this as:

Mass (g) = 5.345 \times 10^{-3} \text{ g}

Next, we need to calculate the volume of the crystal. The dimensions of the crystal are 2.20 millimeters, 1.36 millimeters, and 1.12 millimeters. To find the volume in cubic millimeters, we multiply these dimensions together:

Volume (mm³) = 2.20 \times 1.36 \times 1.12 = 3.35104 \text{ mm³}

To convert cubic millimeters to cubic centimeters, we use the conversion factor where 1 centimeter equals 10 millimeters. Therefore, we cube the conversion factor:

1 cm = 10 mm ⇒ 1 cm³ = 10³ mm³ = 1000 mm³

Now, we convert the volume:

Volume (cm³) = \frac{3.35104 \text{ mm³}}{1000} = 3.35104 \times 10^{-3} \text{ cm³}

Since 1 cm³ is equivalent to 1 mL, we can express the volume in milliliters as:

Volume (mL) = 3.35104 \times 10^{-3} \text{ mL}

Now that we have both mass and volume, we can calculate the density using the formula:

Density (g/mL) = \frac{\text{Mass (g)}}{\text{Volume (mL)}}

Substituting the values we have:

Density = \frac{5.345 \times 10^{-3} \text{ g}}{3.35104 \times 10^{-3} \text{ mL}} \approx 1.595 \text{ g/mL}

Rounding to three significant figures, the density of the sucrose crystal is:

Density = 1.60 g/mL

This calculation illustrates the process of converting units and applying the density formula effectively. You can apply this method to similar problems to reinforce your understanding of density calculations.

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

Answer: (C)

Buoyancy Calculations 2

To determine the density of water based on the given scenario, we start by understanding the relationship between weight, buoyant force, and density. The concrete weighs 120 newtons in air, but when submerged, its apparent weight is 97 newtons. The difference between these two weights gives us the buoyant force:

Buoyant Force (F) = Actual Weight - Apparent Weight = 120 N - 97 N = 23 N.

According to Archimedes' principle, the buoyant force is equal to the weight of the fluid displaced, which can be expressed as:

F = ρ × V × g,

where:

• ρ is the density of the fluid (water in this case),
• V is the volume of the fluid displaced,
• g is the acceleration due to gravity (approximately 9.8 m/s²).

Next, we need to convert the volume of water displaced from cubic centimeters to cubic meters. The volume given is 4200 cm³. To convert this, we use the conversion factor:

1 cm³ = 1 × 10^-6 m³.

Thus, 4200 cm³ is converted as follows:

V = 4200 cm³ × 1 × 10^-6 m³/cm³ = 4.20 × 10^-3 m³.

Now, substituting the known values into the buoyant force equation:

23 N = ρ × (4.20 × 10^-3 m³) × (9.8 m/s²).

To find the density (ρ), we rearrange the equation:

ρ = 23 N / (4.20 × 10^-3 m³ × 9.8 m/s²).

Calculating this gives:

ρ ≈ 5.58 × 10² kg/m³.

Considering significant figures, we can round this to 5.6 × 10² kg/m³, as the least precise measurement has two significant figures.

This example illustrates the application of Archimedes' principle in determining the density of a fluid based on the buoyant force experienced by a submerged object. Understanding these principles is crucial for solving problems related to fluid mechanics and buoyancy.

Buoyancy Calculations 2

The density of propane, an odorless hydrocarbon commonly used in cooking, is 0.922 grams per milliliter. When a sample of propane is weighed on an analytical balance, it registers a mass of 8.15 × 10⁸ nanograms. To find the true mass of propane, we need to apply the buoyancy correction formula, which accounts for the effect of air density on the measured mass.

The formula for calculating the true mass (\(m_{true}\)) is given by:

\(m_{true} = m_{measured} \times \frac{1 - \frac{\rho_{air}}{\rho_{water}}}{1 - \frac{\rho_{air}}{\rho_{object}}}\)

Where:

• \(m_{measured}\) is the mass measured by the analytical balance.
• \(\rho_{air}\) is the density of air (0.0012 grams per milliliter).
• \(\rho_{water}\) is the density of water (approximately 8.0 grams per milliliter).
• \(\rho_{object}\) is the density of the object (0.922 grams per milliliter for propane).

First, we convert the mass from nanograms to grams:

\(8.15 \times 10^{8} \text{ nanograms} = 8.15 \times 10^{-1} \text{ grams}\)

Next, we calculate the components of the buoyancy correction:

\(1 - \frac{\rho_{air}}{\rho_{water}} = 1 - \frac{0.0012}{8.0} \approx 0.99985\)

\(1 - \frac{\rho_{air}}{\rho_{object}} = 1 - \frac{0.0012}{0.922} \approx 0.998698\)

Now, substituting these values into the buoyancy correction formula gives:

\(m_{true} = 0.815 \times \frac{0.99985}{0.998698} \approx 0.816\) grams

This result indicates that the true mass of the propane, after accounting for the buoyancy effect of air, is approximately 0.816 grams. When performing similar calculations, it is essential to maintain significant figures, as the original measurement (8.15) has three significant figures. This practice ensures accuracy and precision in scientific measurements.

For further practice, consider attempting additional problems that involve buoyancy corrections and density calculations to reinforce your understanding of these concepts.

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

Answer: (B)

Here’s what students ask on this topic:

What is buoyancy in air and how does it affect measurements in analytical chemistry?

Buoyancy in air refers to the upward force exerted on an object by the air surrounding it. In analytical chemistry, this buoyant force can affect the precision and accuracy of measurements. 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 of an object can be calculated using the equation: $m_t=m_a+(d_a-d_w)\times d_o$. Here, ma is the apparent mass, da is the density of air, dw is the density of calibration weights, and do is the density of the object. Understanding this concept is crucial for achieving accurate measurements.

How do you calculate the true mass of an object considering buoyancy in air?

To calculate the true mass of an object considering buoyancy in air, you can use the following equation: $m_t=m_a+(d_a-d_w)\times d_o$. In this equation, mt represents the true mass, ma is the apparent mass (the mass read from the analytical balance), da is the density of air (0.0012 g/mL at 1 bar and 25°C), dw is the density of the calibration weights (typically 8 g/mL), and do is the density of the object being weighed. This equation helps to account for the buoyant force exerted by the air, providing a more accurate measurement of the object's true mass.

Why is it important to consider buoyancy when weighing objects in analytical chemistry?

It is important to consider buoyancy when weighing objects in analytical chemistry because buoyancy introduces an upward force that can distort the true mass of the object. This distortion can lead to inaccuracies in measurements, which is critical in analytical chemistry where precision and accuracy are paramount. By accounting for buoyancy, we can correct for this upward force and obtain a more accurate measurement of the object's true mass. This is achieved using the equation: $m_t=m_a+(d_a-d_w)\times d_o$, where the variables account for the densities of air, calibration weights, and the object itself.

What factors influence the density of air in the context of buoyancy in analytical chemistry?

The density of air, which influences buoyancy in analytical chemistry, is affected by several factors including temperature, pressure, and humidity. At standard conditions (1 bar pressure and 25°C temperature), the density of air is approximately 0.0012 g/mL. However, changes in temperature and pressure can alter this density. For example, increasing the temperature generally decreases the density of air, while increasing the pressure increases the density. Humidity also plays a role, as moist air is less dense than dry air. These variations must be considered when calculating the true mass of an object to ensure accurate measurements.

How does the density of calibration weights affect the calculation of true mass in analytical chemistry?

The density of calibration weights affects the calculation of true mass in analytical chemistry because it is a key variable in the buoyancy correction equation: $m_t=m_a+(d_a-d_w)\times d_o$. Here, dw represents the density of the calibration weights. Standard calibration weights typically have a density of 8 g/mL, but different metals or alloys used as calibration weights can have varying densities. Using the correct density value for the calibration weights ensures that the buoyancy correction is accurate, leading to a precise determination of the true mass of the object being weighed.