problem solving in density with answer

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

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Density Calculations – Worked Example Problem

Density is the measurement of the amount of mass per unit volume. Density calculations are done using the formula:

Example Problems: 1. Calculate the density in g/mL of 30 mL of solution that weighs 120 grams. 2. Calculate the density in g/mL of 0.4 L of solution weighing 150 grams. 3. Calculate the density in g/mL of 3000 mL of solution weighing 6 kg.

Example 2 We want to know the density in g/mL, but our volume is in liters. First, convert the volume to mL.

Example 3 Again, we want g/mL, and our mass is in kg. Convert the mass to grams.

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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 ?

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

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How to Calculate Density - Worked Example Problem

Finding the Ratio Between Mass and Volume

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Density is the measurement of the amount of mass per unit of volume . In order to calculate density , you need to know the mass and volume of the item. The formula for density is:

density = mass/volume

The mass is usually the easy part while finding volume can be tricky. Simple shaped objects are usually given in homework problems such as using a cube, brick or sphere . For a simple shape, use a formula to find volume. For irregular shapes, the easiest solution is to measure volume displaced by placing the object in a liquid.

This example problem shows the steps needed to calculate the density of an object and a liquid when given the mass and volume.

Key Takeaways: How to Calculate Density

• Density is how much matter is contained within a volume. A dense object weighs more than a less dense object that is the same size. An object less dense than water will float on it; one with greater density will sink.
• The density equation is density equals mass per unit volume or D = M / V.
• The key to solving for density is to report the proper mass and volume units. If you are asked to give density in different units from the mass and volume, you will need to convert them.

Question 1: What is the density of a cube of sugar weighing 11.2 grams measuring 2 cm on a side?

Step 1: Find the mass and volume of the sugar cube.

Mass = 11.2 grams Volume = cube with 2 cm sides.

Volume of a cube = (length of side) 3 Volume = (2 cm) 3 Volume = 8 cm 3

Step 2: Plug your variables into the density formula.

density = mass/volume density = 11.2 grams/8 cm 3 density = 1.4 grams/cm 3

Answer 1: The sugar cube has a density of 1.4 grams/cm 3 .

Question 2: A solution of water and salt contains 25 grams of salt in 250 mL of water. What is the density of the salt water? (Use density of water = 1 g/mL)

Step 1: Find the mass and volume of the salt water.

This time, there are two masses. The mass of the salt and the mass of the water are both needed to find the mass of the salt water. The mass of the salt is given, but the only the volume of water is given. We've also been given the density of water , so we can calculate the mass of the water.

density water = mass water /volume water

solve for mass water ,

mass water = density water ·volume water mass water = 1 g/mL · 250 mL mass water = 250 grams

Now we have enough to find the mass of the salt water.

mass total = mass salt + mass water mass total = 25 g + 250 g mass total = 275 g

Volume of the salt water is 250 mL.

Step 2: Plug your values into the density formula.

density = mass/volume density = 275 g/250 mL density = 1.1 g/mL

Answer 2: The salt water has a density of 1.1 grams/mL.

Finding Volume by Displacement

If you're given a regular solid object, you can measure its dimensions and calculate its volume. Unfortunately, very few objects in the real world can have their volume measured this easily! The rest of the time, you need to calculate volume by displacement.

How do you measure displacement? Say you have a metal toy soldier. You can tell it is heavy enough to sink in water, but you can't use a ruler to measure its dimensions. To measure the toy's volume, fill a graduated cylinder about half way with water. Record the volume. Add the toy. Make sure to displace any air bubbles that may stick to it. Record the new volume measurement. The volume of the toy soldier is the final volume minus the initial volume. You can measure the mass of the (dry) toy and then calculate density.

Tips for Density Calculations

In some cases, the mass will be given to you. If not, you'll need to obtain it yourself by weighing the object. When obtaining mass, be aware of how accurate and precise the measurement will be. The same goes for measuring volume. Obviously, you'll get a more precise measurement using a graduated cylinder than using a beaker, however, you may not need such a close measurement. The significant figures reported in the density calculation are those of your least precise measurement . So, if your mass is 22 kg, reporting a volume measurement to the nearest microliter is unnecessary.

Another important concept to keep in mind is whether your answer makes sense. If an object seems heavy for its size, it should have a high density value. How high? Keep in mind the density of water is about 1 g/cm³. Objects less dense than this float in water, while those that are more dense sink in water. If an object sinks in water, your density value better be greater than 1!

More Homework Help

Need more examples of help with related problems?

• Worked Example Problems : Browse different types of chemistry problems.
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Chem – Density Problem Solving

What sections should I know before attempting to learn this section?

—> Introduction to Units

—> Unit Prefixes

—> Introduction to Problem Solving

How do you solve density word problems?

To help further your problem solving skills, I will introduce you to the concept of density. Density is the mass per unit volume. Another way to state that is the mass divided by the volume. See equation below:

The letter D represents Density , the letter m represents mass , and the letter V represents volume . The density equation has 3 variables. That means in a problem where you have to use density, you will be given 2 variables and asked to solve for the 3 rd . Density can have different units like (g/mL) or (g/L) or (kg/L) or (mg/mL) just to name a few. How do we use it to solve some equations? Look to the demonstrated examples below.

VIDEO Density Problem Solving Demonstrated Example 1 : What is the density for a mass that is 345g and 789mL?

Step 1:  highlight number and units

What is the density for a mass that is 345g and 789mL ?

Step 2 :  write numbers and units

mass = 345g

volume = 789mL

Step 3 :  restate the question

density = ?

Now what is a formula or concept that relates together mass, volume, and density?

Answer: the density formula (so write down the density formula below)

Answer: fill in the numbers and units from your steps 1 through 3.

Solve for density (D)

Answer: 245g / 789mL = 0.437g/mL

COMPLETE ANSWER: D = 0.437g/mL

VIDEO Density Problem Solving Demonstrated Example 2 : What is the volume of an object that is 154g and has a density of 67g/L?

Step 1:  highlight numbers and units

What is the volume of an object that is 154g and has a density of 67g/L ?

mass = 154g

density = 67g/L

Step 3 :  restate question

How do you solve for volume?

Answer: First multiply both sides by V

Cross out V on the right side

Now divide both sides by 67g/L (red)

Cross out 67g/L on the left side

Solve for V

Answer: 154 / 67 = 2.30 L

COMPLETE ANSWER: V = 2.30 L

PRACTICE PROBLEMS : Solve for the missing variable in the word problem.

What is the density of an object with a mass of 25g and a volume of 47mL?

Answer: 0.53g/ml

What is density of something that is 36L and 97g?

Answer: 2.69g/L

What is the mass of a cube that has a density of 42g/mL and a volume of 520mL?

Answer: 21840g

An object is 34L and 369g/L. What is the mass of the object?

Answer: 12546g

What is the volume of an object that has a density of 74g/mL and a mass of 58g?

Answer: 0.78ml

If an object is 23kg and 4kg/L. What is the volume?

Answer: 5.75L

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Density – problems and solutions

1. The volume of an object is 8 cm 3 and mass of an object is 84 gram. What is the density of the object?

Volume (V) = 8 cm 3

Mass (m) = 84 gram

Wanted : density (ρ)

ρ = m / V = 84 gram / 8 cm 3 = 10.5 gram/cm 3

2. Volume of an block is 5 cm 3 . If the density of the block is 250 g/cm 3 , what is the mass of the block ?

Volume (V) = 5 cm 3

Density (ρ) = 250 g/cm 3

Wanted : Mass of the block

ρ = density, V = volume, m = mass

Mass of block :

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

3. Volume of water is 35 cm 3 and mass of water is 60 gram, what is the density of the water.

Volume of water (V) = 35 cm 3

Mass of water (m) = 60 gram

Wanted: density (ρ)

ρ = m / V = 60/35 = 1.71 gram/cm 3

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

Mass (m) = 120 gram

Volume (V) = 60 cm 3

Wanted : density

ρ = 120 gram / 60 cm 3

ρ = 2 gram/cm 3

Based on the table, an object with the same density is…

Formula of density :

ρ = density, m = mass, V = volume

6. Based on the figure below, if the mass of an object is 300 gram, what is the density of the object.

Mass (m) = 300 gram

Volume (V) = volume of spilled water = 20 cm 3

ρ = m / V = 300 gram / 20 cm 3 = 15 gram/cm 3

7. Mass of object is 316 gram, placed in a container as shown in figure. What is the density of the object.

Mass (m) = 316 gram

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

Mass = 316 gram = 316/1000 kg = 0.316 kg

Volume = 40 ml = 40/1000 liters = 4/100 liters = 0.04 liters

1 liter = 1 dm 3 = 1/1000 m 3 = 0.001 m 3

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

ρ = m / V = 0.316 kg / 0.00004 m 3 = 316 kg / 0.04 m 3 = 7900 kg/m 3

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

Volume (V) = 80 ml – 60 ml = 20 ml = 20 / 1000 liters = 2/100 liters = 0.02 liters

0.02 liters = 0.02 x 0.001 m 3 = 0.00002 m 3

ρ = m / V = 0.120 kg / 0.00002 m 3 = 120 kg / 0.02 m 3 = 6000 kg/m 3

• Answer : Density is a measure of how much mass is contained in a given volume. It is calculated by dividing the mass of an object by its volume: Density=MassVolume Density = Volume Mass ​ .
• Answer : Objects with a density greater than the fluid will sink, while objects with a density less than the fluid will float.
• Answer : It’s all about displaced volume. A ship, while heavy, has a shape that displaces a large volume of water. If the weight of the water displaced is greater than or equal to the weight of the ship, the ship will float.
• Answer : Generally, as temperature increases, the density of a substance decreases because the molecules move more and tend to occupy a larger volume. There are exceptions, such as the anomalous expansion of water below 4°C.
• Answer : The density value of 1 g/cm 3 for water is a convenient benchmark. Substances with densities greater than this will sink in water, while those with densities less will float.
• Answer : The two objects could be made of different materials (and thus have different atomic or molecular structures) or they could have different volumes. An object with a larger volume and the same mass as a smaller object will have a lower density.
• Answer : Compressing a gas decreases its volume. Since the mass remains constant, reducing the volume will increase the density of the gas.
• Answer : The “heaviness” you feel is actually the density. Lead has a higher density than many other metals, so a piece of lead feels heavier than another metal of the same volume.
• Answer : Surface tension can sometimes support objects denser than the fluid, especially if the object is small or has a shape that can be “cradled” by the fluid’s surface. Over time, however, if the object’s density is greater, it will likely sink.
• Can the density of a homogeneous object be different at different points inside the object?
• Answer : No. For a homogeneous object, the density is uniform throughout, meaning it has the same value at every point inside the object.

• 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 Practice Questions

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What Can Generative AI Do?

How exactly can generative AI be used to interpret data? Use cases include:

· Driving faster and better decision making through better insights: Through real-time tracking of data, decision makers can gain a better grasp of what’s happening across the business and be presented with actionable insights. And this can be achieved through natural language prompts, such as “What are our top three customer behavior trends this month?”

· Acting as a decision-making co-pilot: Thanks to generative AI’s conversational abilities, these tools can function as virtual advisors – a sounding board to help discuss and generate ideas.

· Generating summaries of data: Generative AI can sift through vast quantities of data and create executive summaries that pull out the key points, along with best-practice recommendations.

· Visualizing data: Generative AI can generate analytics reports in an easy-to-digest format – presenting insights from the data not just as text narratives but also in a visual format (graphs, charts, etc.).

· Automating data analytics: Generative AI can potentially automate the data analysis process and provide automatic notifications for, well, anything you want. Spikes in sales, trending website activity, a drop in factory machine performance, increased sick leave, you name it…

· Harnessing predictive capabilities: As well as understanding what’s going on in the business right now , generative AI can help decision makers pre-empt what might be coming down the line.

· Using synthetic data to test ideas and scenarios: By creating large amounts of synthetic data that mimic real-world data, leaders can model scenarios that may be difficult to model with real-world data (for example, because an event is a rare, but impactful occurrence, or because gathering that much data would be difficult and expensive).

· Preparing data: Generative AI can also be used to take care of data preparation tasks such as tagging, classification, segmentation, and anonymization.

· Helping to clean up data for better analysis results: Because generative AI is so good at spotting patterns, it can be used to detect anomalies and inconsistencies in your data – things that could potentially skew results.

Another advantage is that generative AI can, in theory, work with all sorts of messy, unstructured data, including photos and video data, customer feedback comments, and social media posts – meaning, it isn’t just limited to neatly structured data in databases.

Best of all, these incredible capabilities make data much more actionable for decision-makers across the organization – regardless of their data expertise . So, you don't need to be a data expert to harness data in your everyday work. Decision paralysis, begone!

Look Out For Generative AI-Powered Tools

Providers of analytics software and platforms are beginning to build generative AI functions into their tools to enable more intelligent data analytics. For example, tools such as Microsoft Power BI, Teradata VantageCloud, Tableau AI, and Qlik Cloud now incorporate generative AI capabilities. This generally allows for natural language querying of data, easy summaries, tailored reports, and more.

What we’re seeing, then, is a democratization of generative AI and data. This will help to level the playing field between large corporations and smaller enterprises because you no longer need an army of data scientists to gain a competitive advantage.

We urgently need people to become more confident and competent at working with data. I believe generative AI will help to achieve this vision and solve the data overwhelm problem – by giving anyone the ability to analyze vast amounts of data in a more intuitive way. In other words, all you need to do is ask the right questions!

Read more about generative AI and its impact in my new book, Generative AI in Practice, 100+ Amazing Ways Generative Artificial Intelligence Is Changing Business And Society.

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1.5: Density and Percent Composition - Their Use in Problem Solving

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Learning Objectives

• To be introduced to the concepts of Density and Percent Composition as important properties of matter.

Density and percent composition are important properties in chemistry. Each have basic components as well as broad applications. Components of density are: mass and volume, both of which can be more confusing than at first glance. An application of the concept of density is determining the volume of an irregular shape using a known mass and density. Determining Percent Composition requires knowing the mass of entire object or molecule and the mass of its components.

Introduction

Which one weighs more, a kilogram of feathers or a kilogram of bricks? Though many people will say that a kilogram of bricks is heavier, they actually weigh the same! However, many people are caught up by the concept of density \(\rho\), which causes them to answer the question incorrectly. A kilogram of feathers clearly takes up more space, but this is because it is less "dense." But what is density, and how can we determine it?

In the laboratory, density can be used to identify an element, while percent composition is used to determine the amount, by mass, of each element present in a chemical compound. In daily life, density explains everything from why boats float to why air bubbles will try to escape from soda. It even affects your health because bone density is very important. Similarly, percent composition is commonly used to make animal feed and compounds such as the baking soda found in your kitchen.

Density is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows:

\[ \text{Density} = \dfrac{\text{Mass}}{\text{Volume}} \label{1.5.4}\]

\[\rho = \dfrac{m}{v} \label{1.5.5}\]

Based on this equation, it's clear that density can, and does, vary from element to element and substance to substance due to differences in the relation of mass and volume. But let us break it down one step further. What are mass and volume? We cannot understand density until we know its parts: mass and volume. The following two sections will teach you all the information you need to know about volume and mass to properly solve and manipulate the density equation.

Mass concerns the quantity of matter in an object. The SI unit for mass is the kilogram (kg), although grams (g) are commonly used in the laboratory to measure smaller quantities. Often, people mistake weight for mass. Weight concerns the force exerted on an object as a function of mass and gravity. This can be written as

\[\text{Weight} = \text{mass} \times \text{gravity} \label{1.5.6}\]

\(Weight = {m}{g} \label{1.5.7}\)

Since on the earth, the \(g\) in the equation is equal to one, weight and mass are considered equal on earth. It is important to note that although \(g\) is equal to one in basic equations, it actually differs throughout earth by a small fraction depending on location; gravity at the equator is less than at the poles. On other astronomical objects, gravity and hence weight, highly differs. This is because weight changes due to variations in gravity and acceleration. The mass, however, of a 1 kg cube will continue to be 1 kg whether it is on the top of a mountain, the bottom of the sea, or on the moon.

Another important difference between mass and weight is how they are measured. Weight is measured with a scale, while mass must be measured with a balance. Just as people confuse mass and weight, they also confuse scales and balances. A balance counteracts the effects of gravity while a scale incorporates it. There are two types of balances found in the laboratory: electronic and manual. With a manual balance, you find the unknown mass of an object by adjusting or comparing known masses until equilibrium is reached. With an electronic balance, which is what you will work with in the UC Davis laboratory, the mass is found by electronic counterbalancing with little effort from the user. An electronic balance can be far more accurate than any other balance and is easier to use, but they are expensive. Also, keep in mind that all instruments used in the laboratory have systematic errors .

Volume describes the quantity of three dimensional space than an object occupies. The SI unit for volume is meters cubed (m 3 ), but milliliters (mL), centimeters cubed (cm 3 ), and liters (L) are more common in the laboratory. There are many equations to find volume. Here are just a few of the easy ones:

Volume = (length) 3

Volume = (length)(width)(height)

Volume = (base area)(height)

Density: A Further Investigation

We know all of density's components, so let's take a closer look at density itself. The unit most widely used to express density is g/cm 3 or g/mL, though the SI unit for density is technically kg/m 3 . Grams per centimeter cubed is equivalent to grams per milliliter (g/cm 3 = g/mL). To solve for density, simply follow the equation 1.5.1. For example, if you had a metal cube with mass 7.0 g and volume 5.0 cm 3 , the density would be \(\rho = \frac{7\; g}{5\;cm^3}= 1.4\; g/cm^3\). Sometimes, you have to convert units to get the correct units for density, such as mg to g or in 3 to cm 3 .

Density in the Periodic Table

Density can be used to help identify an unknown element. Of course, you have to know the density of an element with respect to other elements. Below is a table listing the density of a few elements from the Periodic Table at standard conditions for temperature and pressure, or STP. STP corresponds to a temperature of 273 K (0° Celsius) and 1 atmosphere of pressure.

As can be seen from the table, the most dense element is Osmium (Os) with a density of 22.6 g/cm 3 . The least dense element is Hydrogen (H) with a density of 0.09 g/cm 3 .

Density and Temperature

Density generally decreases with increasing temperature and likewise increases with decreasing temperatures. This is because volume differs according to temperature. Volume increases with increasing temperature. Below is a table showing the density of pure water with differing temperatures.

As can be seen from Table \(\PageIndex{2}\), the density of water decreases with increasing temperature. Liquid water also shows an exception to this rule from 0 degrees Celsius to 4 degrees Celsius, where it increases in density instead of decreasing as expected. Looking at the table, you can also see that ice is less dense than water. This is unusual as solids are generally denser than their liquid counterparts. Ice is less dense than water due to hydrogen bonding. In the water molecule, the hydrogen bonds are strong and compact. As the water freezes into the hexagonal crystals of ice, these hydrogen bonds are forced farther apart and the volume increases. With this volume increase comes a decrease in density. This explains why ice floats to the top of a cup of water: the ice is less dense.

Even though the rule of density and temperature has its exceptions, it is still useful. For example, it explains how hot air balloons work.

Density and Pressure

Density increases with increasing pressure because volume decreases as pressure increases. And since density=mass/volume , the lower the volume, the higher the density. This is why all density values in the Periodic Table are recorded at STP, as mentioned in the section "Density and the Periodic Table." The decrease in volume as related to pressure is explained in Boyle's Law: \(P_1V_1 = P_2V_2\) where P = pressure and V = volume. This idea is explained in the figure below.

Archimedes' Principle

The Greek scientist Archimedes made a significant discovery in 212 B.C. The story goes that Archimedes was asked to find out for the King if his goldsmith was cheating him by replacing his gold for the crown with silver, a cheaper metal. Archimedes did not know how to find the volume of an irregularly shaped object such as the crown, even though he knew he could distinguish between elements by their density. While meditating on this puzzle in a bath, Archimedes recognized that when he entered the bath, the water rose. He then realized that he could use a similar process to determine the density of the crown! He then supposedly ran through the streets naked shouting "Eureka," which means "I found it!" in Latin.

Archimedes then tested the king's crown by taking a genuine gold crown of equal mass and comparing the densities of the two. The king's crown displaced more water than the gold crown of the same mass, meaning that the king's crown had a greater volume and thus had a smaller density than the real gold crown. The king's "gold" crown, therefore, was not made of pure gold. Of course, this tale is disputed today because Archimedes was not precise in all his measurements, which would make it hard to determine accurately the differences between the two crowns.

Archimedes' Principle states that if an object has a greater density than the liquid that it is placed into, it will sink and displace a volume of liquid equal to its own. If it has a smaller density, it will float and displace a mass of liquid equal to its own. If the density is equal, it will not sink or float (Figure \(\PageIndex{2}\)). The principle explains why balloons filled with helium float. Balloons, as we learned in the section concerning density and temperature, float because they are less dense than the surrounding air. Helium is less dense than the atmospheric air, so it rises. Archimedes' Principle can also be used to explain why boats float. Boats, including all the air space, within their hulls, are far less dense than water. Boats made of steel can float because they displace their mass in water without submerging all the way. The table below gives the densities of a few liquids to put things into perspective.

Percent Composition

Percent composition is very simple. Percent composition tells you by mass what percent of each element is present in a compound. A chemical compound is the combination of two or more elements. If you are studying a chemical compound, you may want to find the percent composition of a certain element within that chemical compound. The equation for percent composition is

\[\text{Percent Composition} = \dfrac{\text{Total mass of element present}}{\text{Molecular mass}} \times 100\% \label{1.5.3}\]

If you want to know the percent composition of the elements in an compound, follow these steps:

Steps to Solve:

• Find the molar mass of all the elements in the compound in grams per mole.
• Find the molecular mass of the entire compound.
• Divide the component's molar mass by the entire molecular mass.
• You will now have a number between 0 and 1. Multiply it by 100 to get percent composition!

Tips for solving:

• The compounds will always add up to 100%, so in a binary compound, you can find the % of the first element, then do 100%-(% first element) to get (% second element)
• If using a calculator, you can store the overall molar mass to a variable such as "A". This will speed up calculations, and reduce typographical errors.

These steps are outlined in the figure below.

For another example, if you wanted to know the percent composition of hydrochloric acid (HCl), first find the molar mass of Hydrogen. H = 1.00794g . Now find the molecular mass of HCl: 1.00794g + 35.4527g = 36.46064g . Follow steps 3 and 4: (1.00794g/36.46064g) x 100 = 2.76% Now just subtract to find the percent by mass of Chlorine in the compound: 100%-2.76% = 97.24% Therefore, HCl is 2.76% Hydrogen and 97.24% Chlorine by mass.

Percent Composition in Everyday Life

Percent composition plays an important role in everyday life. It is more than just the amount of chlorine in your swimming pool because it concerns everything from the money in your pocket to your health and how you live. The next two sections describe percent composition as it relates to you.

Example \(\PageIndex{1}\): Nutritional Labels

The nutrition label found on the container of every bit of processed food sold by the local grocery store employs the idea of percent composition. On all nutrition labels, a known serving size is broken down in five categories: Total Fat, Cholesterol, Sodium, Total Carbohydrate, and Protein. These categories are broken down into further subcategories, including Saturated Fat and Dietary Fiber. The mass for each category, except Protein, is then converted to percent of Daily Value. Only two subcategories, Saturated Fat and Dietary Fiber are converted to percent of Daily Value. The Daily Value is based on a the mass of each category recommended per day per person for a 2000 calorie diet. The mass of protein is not converted to percent because their is no recommended daily value for protein. Following is a picture outlining these ideas.

For example, if you wanted to know the percent by mass of the daily value for sodium you are eating when you eat one serving of the food with this nutrition label, then go to the category marked sodium. Look across the same row and read the percent written. If you eat one serving of this food, then you will have consumed about 9% of your daily recommended value for sodium. To find the percent mass of fat in the whole food, you could divide 3.5 grams by 15 grams, and see that this snack is 23.33% fat.

Example \(\PageIndex{2}\): The Lucky Penny

The penny should be called "the lucky copper coated coin." The penny has not been made of solid copper since part of 1857. After 1857, the US government started adding other cheaper metals to the mix. The penny, being only one cent, is literally not worth its weight in copper. People could melt copper pennies and sell the copper for more than the pennies were worth. After 1857, nickel was mixed with the more expensive copper. After 1864, the penny was made of bronze. Bronze is 95% copper and 5% zinc and tin. For one year, 1943, the penny had no copper in it due to the expense of the World War II. It was just zinc coated steel. After 1943 until 1982, the penny went through periods where it was brass or bronze.

The percent composition of a penny may actually affect health, particularly the health of small children and pets. Since the newer pennies are made mainly of zinc instead of copper, they are a danger to a child's health if ingested. Zinc is very susceptible to acid. If the thin copper coating is scratched and the hydrochloric acid present in the stomach comes into contact with the zinc core it could cause ulcers, anemia, kidney and liver damage, or even death in severe cases. Three important factors in penny ingestion are time, pH of the stomach, and amount of pennies ingested. Of course, the more pennies swallowed, the more danger of an overdose of zinc. The more acidic the environment, the more zinc will be released in less time. This zinc is then absorbed and sent to the liver where it begins to cause damage. In this kind of situation, time is of the essence. The faster the penny is removed, the less zinc is absorbed. If the penny or pennies are not removed, organ failure and death can occur.

Below is a picture of a scratched penny before and after it had been submerged in lemon juice. Lemon juice has a similar pH of 1.5-2.5 when compared to the normal human stomach after food has been consumed. Time elapsed: 36 hours.

As you can see, the copper is vastly unharmed by the lemon juice. That's why pennies made before 1982 with mainly copper (except the 1943 penny) are relatively safe to swallow. Chances are they would pass through the digestive system naturally before any damage could be done. Yet, it is clear that the zinc was partially dissolved even though it was in the lemon juice for only a limited amount of time. Therefore, the percent composition of post 1982 pennies is hazardous to your health and the health of your pets if ingested.

Following are examples of different types of percent composition and density problems.

Density Problems: These problems are meant to be easy in the beginning and then gradually become more challenging. Unless otherwise stated, answers should be in g/mL or the equivalent g/cm 3 .

• If you have a 2.130 mL sample of acetic acid with mass 0.002234 kg, what is the density?
• Calculate the density of a .03020 L sample of ethyl alcohol with a mass of 23.71002 g.
• Find the density of a sample that has a volume of 36.5 L and a mass of 10.0 kg.
• Find the volume in mL of an object that has a density of 10.2 g/L and a mass of 30.0 kg.
• Calculate the mass in grams of an object with a volume of 23.5 mL and density of 10.0 g/L.
• Calculate the density of a rectangular prism made of metal. The dimensions of the prism are: 5 cm by 4 cm by 5 cm. The metal has a mass of 50 grams.
• Find the density of an unknown liquid in a beaker. The beaker's mass is 165 g when there is no liquid present. With the unknown liquid, the total mass is 309 g. The volume of the unknown is 125 mL.
• Determine the density in g/L of an unknown with the following information. A 55 gallon tub weighs 137.5lb when empty and 500.0 lb when filled with the unknown.
• A ring has a mass of 5.00g and a volume of 0.476 mL. Is it pure silver?
• What is the density of the solid in the image if the mass is 40 g? Make your answer have 3 significant figures.

• Below is a model of a pyramid found at an archeological dig made of an unknown substance. It is too large to find the volume by submerging it in water. Also, the scientists refuse to remove a piece to test because this pyramid is a part of history. Its height is 150.0m. The length of its base is 75.0m and the width is 50.0m. The mass of this pyramid is 5.50x10 5 kg. What is the density?

Percent Composition Problems: These problems will follow the same pattern of difficulty as those of density.

• Calculate the percent by mass of each element in Cesium Fluoride (CsF).
• Calculate the percent by mass of each element present in carbon tetrachloride (CCl 4 )
• A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/mL. What mass of the salt in grams is in 5.00L of this solution?
• A solution of water and HCl contains 25% HCl by mass. The density of the solution is 1.05 g/mL. If you need 1.7g of HCl for a reaction, what volume of this solution will you use?
• A solution containing 42% NaOH by mass has a density of 1.30 g/mL. What mass, in kilograms, of NaOH is in 6.00 L of this solution?

Here are the solutions to the listed practice problems.

Density Problem Solutions

• 0.7851 g/mL
• 2.94 x 10 6 mL
• 0.195 g/cm 3
• 29.3 g/cm 3

Percent Composition Problem Solutions

• CsF is 87.5% Cs and 12.5% F by mass
• CCl 4 is 92.2% Cl and 7.8% C by mass
• AUTOR , ARQUIMEDES , and Thomas Little . The Works of Archimedes . Courier Dover Publications, 2002.
• Chande, D. and T. Fisher (2003). "Have a Penny? Need a Penny? Eliminating the One-Cent Coin from Circulation." Canadian Public Policy/Analyse de Politiques 29 (4): 511-517.
• Jefferson, T. (1999). "A Thought for Your Pennies." JAMA 281 (2): 122.
• Petrucci , Ralph , William Harwood , and Geoffrey Herring . Principles and Modern Application. ninth . New Jersey : Peason Eduation , 2007.
• Rauch, F., H. Plotkin, et al. (2003). "Bone Mass, Size, and Density in Children and Adolescents with Osteogenesis Imperfecta: Effect of Intravenous Pamidronate Therapy." Journal of Bone and Mineral Research 18 : 610-614.
• Richardson, J., S. Gwaltney-Brant, et al. (2002). "Zinc Toxicosis from Penny Ingestion in Dogs." Vet Med 97 (2): 96-99.
• Tate, J. "Archimedes’ Discoveries: A Closer Look."