problem solving about density

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math tutorials for students in introductory geoscience

• First Publication: July 19, 2011
• Revision: September 19, 2024 -- Updated website to improve accessibility of images math equations and chrome.

Density and Specific Gravity - Practice Problems

Jump to: Rock and Mineral density | Rock and mineral specific gravity You can download the questions (Acrobat (PDF) 25kB Jul24 09) if you would like to work them on a separate sheet of paper.

Calculating Densities of Rocks and Minerals

`Volume\times Density=Mass` You can then divide both sides by density to get volume alone: `Volume=\frac{Mass}{Density}` By substituting in the values listed above, `Volume=\frac{2000\ kg}{3200\ \frac{kg}{m^{3}}}`

so the volume will be 0.625 m 3 Note that the above problem shows that densities can be in units other than grams and cubic centimeters. To avoid the potential problems of different units, many geologists use specific gravity (SG), explored in problems 8 and 9, below.

`text{volume}=text{length}\times text{width}\times text{height}` .

The volume of the cube is

`2cm\times2cm\times2cm=8cm^{3}` . The density then is the mass divided by the volume: `Density=\frac{Mass}{Volume}` `Density=\frac{40g}{8cm^{3}}=5.0\frac{g}{cm^{3}}` Thus the cube is NOT gold , since the density (5.0 g/cm 3 ) is not the same as gold (19.3 g/cm 3 ). You tell the seller to take a hike . You might even notice that the density of pyrite (a.k.a. fool's gold) is 5.0 g/cm 3 . Luckily you are no fool and know about density!  

Calculating Specific Gravity of Rocks and Minerals

Take the quiz  .

If this is not how you feel, you can go back to the explanations .

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Density – problems and solutions

Wanted : density (ρ)

Density (ρ) = 250 g/cm 3

m = ρ V = (250 g/cm 3 )(5 cm 3 ) = 1250 gram

4. Mass of an metal is 120 gram and volume of an metal is 60 cm 3 . What is the density of the metal?

ρ = density, m = mass, V = volume

Wanted : density

Volume (V) = volume of spilled water = 40 ml

0.04 liters = (0.04)(0.001) m 3 = 0.00004 m 3

Mass (m) = 100 gram + 20 gram = 120 gram = 120 / 1000 kilogram = 0.120 kilogram

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Density Worked Example Problem

Calculating Density of a Substance

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Density is a measure of how much matter is in a space. It is expressed in units of mass per volume, such as g/cm 3 or kg/L. This is a worked example of how to calculate the density when given the volume and mass of a substance.

Sample Density Problem

A brick of salt measuring 10.0 cm x 10.0 cm x 2.0 cm weighs 433 grams. What is its density? Solution: Density is the amount of mass per unit volume, or: D = M/V Density = Mass/Volume Step 1: Calculate Volume In this example, you are given the dimensions of the object, so you have to calculate the volume. The formula for volume depends on the shape of the object, but it's a simple calculation for a box:

Volume = length x width x thickness Volume = 10.0 cm x 10.0 cm x 2.0 cm Volume = 200.0 cm 3 Step 2: Determine Density

Now you have the mass and the volume, which is all the information you need to calculate density. Density = Mass/Volume Density = 433 g/200.0 cm 3 Density = 2.165 g/cm 3 Answer: The density of the salt brick is 2.165 g/cm 3 .

A Note About Significant Figures

In this example, the length and mass measurements all had 3 significant figures . So, the answer for density should also be reported using this number of significant figures. You'll have to decide whether to truncate the value to read 2.16 or whether to round it up to 2.17.

• How to Calculate Density - Worked Example Problem
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Chemistry Steps

General Chemistry

Chemistry and math.

In these practice problems, we will work on determining the density, volume, and the mass of different objects. First, density is calculated by the ratio of the mass and the volume of the object:

For example , what is the density of a metal if its 2.35 g sample has a volume of 0.654 g/mL?

\[{\rm{d}}\;{\rm{ = }}\,\,\frac{{\rm{m}}}{{\rm{v}}}\,{\rm{ = }}\frac{{{\rm{2}}{\rm{.35}}\,{\rm{g}}}}{{{\rm{0}}{\rm{.654}}\,{\rm{mL}}}}\,{\rm{ = }}\;{\rm{3}}{\rm{.59}}\,{\rm{g/mL}}\]

Sometimes the volume may not be given and there are two main scenarios here depending on if the object has a regular or irregular shape . So, let’s discuss them one by one.

The density of Objects with Regular Shapes

The first example here would be the density of an object with a cubic shape. For example , a wood block with a length of 11 in, 7.5 in width and 1.95 in thickness weighs 4.93 lb. Calculate density of the block in lb/in 3 .

The mass is given and therefore, the only thing missing is the volume of the block which we find by multiplying all the sides:

And now, we can determine the density:

\[{\rm{d}}\;{\rm{ = }}\,\,\frac{{\rm{m}}}{{\rm{v}}}\,{\rm{ = }}\frac{{{\rm{4}}{\rm{.93}}\,{\rm{lb}}}}{{{\rm{160}}{\rm{.875}}\,{\rm{i}}{{\rm{n}}^{\rm{3}}}}}\,{\rm{ = }}\;{\rm{0}}{\rm{.031}}\,{\rm{lb/i}}{{\rm{n}}^{\rm{3}}}\]

The density of Objects with Irregular Shapes

The most common example here is the one where the density of metal pellets needs to be determined. For example , a sample containing 15.4 g of metal pellets is poured into a graduated cylinder initially containing 12.0 mL of water, causing the water level in the cylinder to rise to 16.2 mL. Calculate the density of the metal.

What you need to visualize in these problems, is that the volume of pellets or anything else that was added to water, is equal to the volume of the water displaced :

So, in this case, the volume of the pellets would be:

16.2 – 12.0 = 4.20 mL

Therefore, the density is the ratio of the mass and this difference in initial and final volumes:

\[{\rm{d}}\;{\rm{ = }}\,\,\frac{{\rm{m}}}{{\rm{v}}}\,{\rm{ = }}\frac{{{\rm{15}}{\rm{.4}}\,{\rm{g}}}}{{{\rm{4}}{\rm{.20}}\,{\rm{mL}}}}\,{\rm{ = }}\;{\rm{3}}{\rm{.67}}\,{\rm{g/mL}}\]

Density When the Units are Different

Another type of problem is when the initial units are different than what they are asked to be in the answer. For example , determine the density of a plastic in g/cm 3 if a 1.39-lb piece occupies 6.48 in 3 volume.

First, we can calculate the density in lb/in 3 and then convert the units to g/cm 3 .

\[{\rm{d}}\;{\rm{ = }}\,\,\frac{{\rm{m}}}{{\rm{v}}}\,{\rm{ = }}\frac{{{\rm{1}}{\rm{.39}}\,{\rm{lb}}}}{{{\rm{6}}{\rm{.48}}\,{\rm{i}}{{\rm{n}}^{\rm{3}}}}}\,{\rm{ = }}\;{\rm{0}}{\rm{.2145}}\,{\rm{lb/i}}{{\rm{n}}^{\rm{3}}}\]

Now, remember, for converting two units, we treat them like separate units and do the conversions one by one. Check the “Multi-Step Unit Conversion” section here for more details.

\[{\rm{0}}{\rm{.2145}}\;\frac{{\cancel{{{\rm{lb}}}}}}{{\cancel{{{\rm{i}}{{\rm{n}}^{\rm{3}}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{453}}{\rm{.6}}\,{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{lb}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{(1}}\,\cancel{{{\rm{in}}{{\rm{)}}^{\rm{3}}}}}}}{{{{{\rm{(2}}{\rm{.54}}\,{\rm{cm)}}}^{\rm{3}}}}}\,{\rm{ = }}\;{\rm{5}}{\rm{.94}}\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}\,\,\]

So, in the first part, we converted pounds to grams, and the second multiplication was to convert in 3 to cm 3 . Remember, you need to apply the exponent to both the number and the unit when converting units raised to a power ( Converting Units Raised to Power ).

The density of a Cylinder

Another common question is determining the density of a cylinder. What you need to remember here is the formula that may not be given:

If you forget it, try to remember the formula for the surface of circle:

Now you can visualize the cylinder as a stack of multiple circles and therefore, its volume is the product of the circle’s surface and the height of the cylinder.

For example , a plastic cylinder has a length of 8.52 in, a radius of 2.34 in, and a mass of 5.60 lb. What is the density of the plastic in lb/in 3 ?

The volume of a cylinder is calculated by the formula v = π · r 2  · l, and therefore,

v = π · r 2  · l = 3.14 x (2.34) 2  in x 8.52 in = 146.488 in 3

The density is the ratio of the mass and the calculated volume:

\[{\rm{d}}\;{\rm{ = }}\,\,\frac{{\rm{m}}}{{\rm{v}}}\,{\rm{ = }}\frac{{{\rm{5}}{\rm{.60}}\,{\rm{lb}}}}{{{\rm{146}}{\rm{.488}}\,{\rm{i}}{{\rm{n}}^{\rm{3}}}}}\,{\rm{ = }}\;{\rm{0}}{\rm{.0382}}\,{\rm{lb/i}}{{\rm{n}}^{\rm{3}}}\]

Mass and Volume from Density

The formula for the density can be rearranged to get an expression for the mass or the volume. For example , what is the mass of a metal block with a density of 9.25 g/ml if it occupies 14.6 cm 3 volume?

Rearranging the formula for density, we ger that the mass if the product of the volume and density:

\[{\rm{m}}\;{\rm{ = }}\,{\rm{d}}\,{\rm{ \times }}\,{\rm{v}}\,{\rm{ = }}\,{\rm{9}}{\rm{.25}}\,\frac{{\rm{g}}}{{\cancel{{{\rm{mL}}}}}}\;{\rm{ \times }}\,{\rm{14}}{\rm{.6}}\,\cancel{{{\rm{mL}}}}\,{\rm{ = }}\;{\rm{135}}\;{\rm{g}}\]

Notice that 1 mL = 1cm 3 that is why we replaced cm 3 by ml for the volume and canceled them with the density units.

The volume is the ratio of the mass and density . For example, what is the volume of 154 g bromine in milliliters if it has a density of 3.10 g/cm 3 ?

\[{\rm{v}}\;{\rm{ = }}\,\frac{{\rm{m}}}{{\rm{d}}}\,{\rm{ = }}\,\frac{{{\rm{154}}\,{\rm{g}}}}{{{\rm{3}}{\rm{.10}}\,{\rm{c}}{{\rm{m}}^{\rm{3}}}}}\,{\rm{ = }}\;{\rm{49}}{\rm{.7}}\,{\rm{g/c}}{{\rm{m}}^{\rm{3}}}\]

• Significant Figures
• Significant Figures in Addition, Subtraction Multiplication, and Division
• Significant Figures Practice Problems
• Converting Units With Conversion Factors Dimensional Analysis
• Conversion Factors and Dimensional Analysis Practice Problems

How many dozen eggs are in 15,652 eggs?

How many years are in 5,489 days?

How many weeks are in 2.5 centuries? (1 yr. = 52 weeks)

Would 550 m 2 of fabric be enough to upholster 200 chairs if each requires 35.4 ft 2 fabric?

A 20.0-mL sample of a liquid has a mass of 17.8 g. What is the liquid’s density in grams per milliliter?

A sample containing 31.25 g of metal pellets is poured into a graduated cylinder initially containing 11.9 mL of water, causing the water level in the cylinder to rise to 18.7 mL. Calculate the density of the metal.

What is the mass of a 2.85 L sample of a liquid that has a density of 0.954 g/mL?

What is the volume of 100. g bromine in milliliters if it has a density of 3.10 g/cm 3 ?

Complete the missing data for the density, mass, and volume in the following table:

How many kilograms of honey with a density of 1.40 kg/L are there in a gallon container?

The density of iron is 8.96 g/cm 3 . What is its density in pounds per cubic inch (lb/in 3 )?

The density of iron is 7.86 g/cm 3 . What is the volume of 5.24 lb of iron expressed in cubic inches?

A plastic cylinder has a length of 7.25 in, a radius of 1.26 in, and a mass of 841 g. What is the density of the plastic in g/cm 3 ?

What is the radius of a steel sphere that has a mass of 65.0 g and a density of  7.86 g/cm 3 ?

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    M D  =  ––––     V     25.1 g 13.6 g/cm 3  =  ––––––     V V = 1.845588 cm 3

    M D  =  ––––     V     M 11.4 g/cm 3  =  ––––––––––––     1.845588 cm 3 V = 21.0 g

V Pb = V Hg

    M Pb V Pb  =  ––––     D Pb and     M Hg V Hg  =  ––––     D Hg

M Pb   M Hg ––––  =  –––– D Pb   D Hg

M Pb   25.1 ––––  =  –––– 11.4   13.6 M Pb = 21.0 g

Density = mass / volume rearrange to: Volume = mass / density V = 2.25 g / 2.70 g/cm 3 = 0.83 cm 3 Note: since 1 mL = 1 cm 3 , this is 0.83 mL

11.20 mL + 0.83 mL = 12.03 mL

(a) Calculate the volume (density of Al = 2.70 g cm¯ 3 (b) Calculate the thickness of the piece of Al in mm

1.50 g / 2.70 g/cm 3 = 0.555555 cm 3 (answer to a) 0.555555 cm 3 = (x) (24.0 cm) (30.0 cm) x = 0.000772 cm (to 3 sig figs) There are 10 mm in one cm, so divide the cm answer by 10 to get the thickness in mm. 0.0000772 mm (answer to b)

That means 92.276 cm 3 is Zn and 7.724 cm 3 is Cu.

(92.276 cm 3 ) (7.14 g/cm 3 ) = 658.85064 g (7.724 cm 3 ) (8.96 g/cm 3 ) = 69.20704 g

728.05768 g / 100. cm 3 = 7.28 g/cm 3 (that's the answer to three sig figs)

1.00 x 10 5 kg = 1.00 x 10 8 g 1.00 x 10 8 g / 6.62 g cm¯ 3 = 1.51 x 10 7 cm 3 1.51 x 10 7 cm 3 3 = 247 cm = 2.47 m

D = m / V V = m / D V = 71.7 g / 7.87 g/cm 3 V = 9.11 cm 3

63.0 mL + 9.11 mL = 72.1 mL (rounded off to the proper number of sig figs)

V = 71.7 g Fe x (1 cm 3 / 7.87 g Fe) = 9.11 cm 3

Let us assume 1.00 cm 3 of Al is present. That means 2.70 g of Al is present. In the rod of tin that is hollow, the mass of tin present weighs 2.70 g. We know this because the average densities of the two rods are equal. Remember, the average density of the tin rod includes the empty space. What volume of solid tin is required to provide 2.70 g? The answer is 0.370 cm 3 (from 2.70 g divided by 7.31 g/cm 3 ). The rest of the tin rod must be hollow. This is 0.670 cm 3 . Since the overall volume was 1.00 cm 3 , the fraction of the tin rod that is hollow is 0.670

Identical external dimensions means overall volume is the same for both rods. Let's call that V. It must be that mass Al = mass Sn. So, let's convert from volume Al to volume Sn with equivalent mass:   2.70 g Al   1 g Sn   1 cm 3   V x  –––––––  x  –––––––  x  –––––––  = 0.370V   1 cm 3   1 g Al   7.31 g Sn  

(V − 0.370V) / V 1 − 0.370 = 0.630

Kilograms will be changed to grams by multiplying the numerator by 1000. Liters will be changed to mL by multiplying the denominator by 1000. Those two multiplications cancel each other out. And one mL equals one cm 3 . The answer is 22.59 g/cm 3 . Here's a dimensional analysis set up showing the solution: 22.59 kg   1000 g   1 L   1 mL   –––––––  x  ––––––  x  –––––––  x  –––––  = 22.59 g/cm 3 1 L   1 kg   1000 mL   1 cm 3  

One mole of gold weighs 196.9665 g. V = mass / density = 196.9665 g / 19.32 g cm¯ 3 = 10.195 cm 3 The formula for volume of a cylinder is V = πr 2 L The length (L) of the wire is therefore: L = 10.195 cm 3 / [(3.14159) (0.05000 cm) 2 ] = 1298 cm By the way, note the 0.05 cm for the radius. The wire was 1 mm in diameter, which is 0.1 cm. And then half of that for the radius is 0.05 cm.

0.050 L = 50. mL = 50. cm 3 (11.34 g/cm 3 ) (50. cm 3 ) = 567 g 0.50 kg = 500 g The 0.050 L contains more mass.

0.50 kg = 500 g 500 g / 11.34 g/cm 3 = 44.1 cm 3 44.1 cm 3 = 44.1 mL = 0.0441 L The larger volume has the greater mass since both samples are lead. The answer to the problem is 0.050 L.

0.500 lb / 0.4098 lb/in 3 = 1.22 in 3

1 inch = 2.54 cm (1.00 in) 3 = (2.54 cm) 3 1.00 in 3 = 16.387 cm 3

11.34 g   16.387 cm 3   ––––––  x  ––––––––  =  185.828 g/in 3 cm 3   in 3  

0.500 lb = 226.796 g

    1 in 3   226.796 g  x  ––––––––  =  1.22 in 3     185.828 g  

• Which is more massive on the surface of the Earth ?
• Which is more massive on the surface of the Moon ?
• Which is heavier on the surface of the Earth ?
• Which is heavier on the surface of the Moon ?
• The phrase "more massive" should be read literally as "has more mass" not "fills more space".
• The phrase "heavier" should be read as "is pulled down more strongly by gravity" not "is more dense".

• the average density of the entire Earth
• the percent of the Earth's mass located in the mantle, and
• the average density of the core.
• Write something completely different.
• Mayonnaise is essentially a mixture of vegetable oil and water with a bit of egg yolk added as an emulsifier (a substance that keeps the oil and water from separating). Traditional mayonnaise has a density of about 910 kg/m 3 while reduced fat, low calorie, or "light" mayonnaise has a density of about 1,000 kg/m 3 . Why is "light" (low calorie) mayonnaise "heavier" (more dense) than traditional mayonnaise?
• Why does "heavy cream" have a lower density than "light cream"? Explain this apparent contradiction.
• Find the mass of the air contained in a room that is 16.40 m long by 4.5 m wide by 3.26 m high.

• Compute the density of gold using only the values in the table above.
• What would be the length of a side of the Castello Cube if it was crushed into a cube that was no longer hollow?
• What would be the mass of the Castello Cube if it was entirely filled with gold?

• cubic millimeters
• cubic centimeters (milliliters)
• cubic decimeters (liters)
• cubic meters
• the density of the IPK in kg/m 3

• the volume of the IPK that is platinum
• the volume of the IPK that is iridium
• the volume of the IPK as a whole given the results of your two previous calculations
• the height and diameter of the IPK in millimeters using the results of your previous calculation and assuming that both measurements were meant to be the same, which was the intention of the designers

statistical

• Complete the table above.
• Determine the density of the moon's atmosphere in kg/m 3 .

Density, Area and Volume

Related Topics: Common Core (Geometry) Common Core for Mathematics

Examples, solutions, videos, and lessons to help High School students learn how to apply concepts of density based on area and volume in modeling situations (e.g., persons per square mile, BTUs per cubic foot).

Common Core: HSG-MG.A.2

Density, Mass and Volume How to use the Density Mass Volume formula triangle and find volume of shapes. It goes through a couple of examples of finding mass. Example:

• The density of this metal block is 5 g/cm 3 . Find the mass of the block.
• The density of this prism is 4 g/cm 3 . Find the mass of the prism.

Density Word Problems Simple problems dealing with density formula (D = m / v). Example:

• What is the density of an aluminum metal block that has a mass of 27.0 g and a volume of 10.0 cm 3 ?
• What is the density of a steel block that has a mass of 7.8 g and a volume of 1.0 cm 3 ?
• What is the density of a nail that has a mass of 12.5 g and a volume of 1.6 cm 3 ?
• On Great Skies Airlines a carry-on suitcase can be no more than 12 kg and 30,000 cm 3 . Does the following suitcase qualify as a carry-on.

Density Calculations Solving density word problems using the formula triangle. Example:

• A loaf of bread has a volume of 2,270 cm 3 and a mass of 454 g. Whas the density of the bread?
• What is the volume of a tank that can hold 18.754 g of methanol whose density is 0.788 g/cm 3 ?
• A bottle has a capacity of 1.2 liters. If the density of ether is 0.74 g/ml., what mass of ether can the bottle hold?

Density Facts and Practice Problems Learn more about density and practice calculating using word problems. Density is how much mass is contained in a given volume? Example:

• What is the density of carbon dioxide gas if 0.196 g occupies a volume of 100 mL?
• A block of wood 3 cm on each side has a mass of 27 g. What is the density of the block?
• A 10.0 cm 3 sample of copper has a mass of 89.6 g. What is the density of copper?

Density Practice Problems This video shows the equation for Density and gives 2 example problems. Example:

• What is the density of a bolt that has a mass of 14 g and a volume of 2 ml?
• What is the volume of a liquid silver sample that has a density of 0.125 g/ml and a mass of 1.2g?

What is BTU (British Thermal Unit)? A short explanation of how to calculate water temperature rise in ц, given certain BTUs per hour.

Understanding the BTU - Energy produced

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