2.3: Density
- Page ID
- 521707
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Calculate density, mass, or volume when given 2 of these three variables.
- Identify what units are required for the density equation.
- Review metric conversions.
- Compare the densities of different chemical substances.
- Classify a substance as being a heterogeneous or homogeneous mixture if the solubility data is provided.
- Identify where a chemical would appear in water if solubility and density data are provided.
- Compare any chemical substance's density to the density of water (please memorize this value).
After trees are cut, logging companies often move these materials down a river to a sawmill where they can be shaped into building materials or other products. The logs float on the water because they are less dense than the water they are in. Knowledge of density is important in the characterization and separation of materials. Information about density allows us to make predictions about the behavior of matter.
Various medical conditions can influence the density of urine. Patients who suffer from diabetes produce an abnormally large volume of urine with a relatively low density. In another form of diabetes, called diabetes mellitus, there is excess glucose dissolved in the urine, so the density of urine is abnormally high. The density of urine may also be abnormally high due to excess protein in the urine, which can be caused by congestive heart failure or certain renal (kidney) disorders. Thus, a urine specific gravity test can provide clues to various kinds of health problems.
Density
Density is a physical property that is defined as a substance’s mass divided by its volume:
\[ \begin{align} \text{density} &= \dfrac{\text{mass}}{\text{volume}} \label{eq1} \\[4pt] d &= \dfrac{m}{V} \label{eq2} \end{align} \]
Density is usually a measured property of a substance, so its numerical value affects the significant figures in a calculation. Notice that density is defined in terms of two dissimilar units, mass and volume. That means that density overall has derived units, just like velocity. Common units for density include g/mL, g/cm3, g/L, kg/L, and even kg/m3. Densities for some common substances are listed in Table \(\PageIndex{1}\). Memorize the density of water, along with its corresponding units.
| Substance | Density (g/mL or g/cm3) |
|---|---|
| water | 1.0 |
| gold | 19.3 |
| mercury | 13.6 |
| air | 0.0012 |
| cork | 0.22–0.26 |
| aluminum | 2.7 |
| iron | 7.87 |
Because of how it is defined, density can act as a conversion factor for switching between units of mass and volume. For example, suppose you have a sample of aluminum that has a volume of 7.88 cm3. How can you determine what mass of aluminum you have without measuring it? You can use the volume to calculate it. If you multiply the given volume by the known density (Table \(\PageIndex{1}\)), the volume units will cancel and leave you with mass units, telling you the mass of the sample:
Start with Equation \ref{eq1} \[\text{density} = \dfrac{m}{V} \nonumber \]
and insert the relevant numbers
\[\dfrac{2.7g}{cm^3} = \dfrac{m}{7.88 \, cm^3} \nonumber \]
Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.
\[7.88\cancel{cm^3}\times \dfrac{2.7\,g}{\cancel{cm^3}}= 21 g \text{ of aluminum} \nonumber \]
What is the mass of 44.6 mL of mercury?
Solution
Use the density value for mercury from Table \(\PageIndex{1}\) and the definition of density (Equation \ref{eq1})
\[density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{V} \nonumber \]
\[\dfrac{13.6g}{mL}= \dfrac{m}{44.6 \, mL} \nonumber \]
Remember to cross multiply here in order to isolate variable. Then, report answer with correct units.
\[44.6\cancel{mL}\times \dfrac{13.6\,g}{\cancel{mL}}= 607\,g \nonumber \]
The mass of the mercury is 607 g.
What is the mass of 25.0 cm3 of iron?
- Answer
-
Use the density value for iron from Table \(\PageIndex{1}\)
\[density= \dfrac{mass}{volume}\Rightarrow d= \dfrac{m}{V} \nonumber \]
\[\dfrac{7.87g}{cm^3}= \dfrac{m}{25.0\, cm^3} \nonumber \]
Cross multiplying both sides (right numerator x left denominator = left numerator x right denominator), we get the following expression with answer and appropriate unit.
\[25.0\cancel{cm^3}\times \dfrac{7.87\,g}{\cancel{cm^3}}= 197 g \text{ of iron} \nonumber \]
Another way of looking at density (some students choose to perform calculations using this method)
Density can also be used as a conversion factor to convert mass to volume—but care must be taken. We have already demonstrated that the number that goes with density normally goes in the numerator when density is written as a fraction. Take the density of gold, for example:
\[d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \]
Although this was not previously pointed out, it can be assumed that there is a 1 in the denominator:
\[d=19.3\,g/mL =\dfrac{19.3\,g}{mL} \nonumber \]
That is, the density value tells us that we have 19.3 grams for every 1 milliliter of volume, and the 1 is an exact number. When we want to use density to convert from mass to volume, the numerator and denominator of density need to be switched—that is, we must take the reciprocal of the density. In so doing, we move not only the units but also the numbers:
\[\dfrac{19.3\,\cancel{g}}{mL}= \dfrac{45\,\cancel{g}}{V} \nonumber \]
Cross-multiplying denominators with numerators, we obtain the following algebraic equation.
\[19.3 V = 45.9\, mL \nonumber \]
Then you will need to isolate the variable (volume)
\[ V = \dfrac{ 45.9\,mL}{19.3} \nonumber \]
After multiplication, the answer would be
\[ V = 2.38 \,mL \nonumber \]
A cork stopper from a bottle of wine has a mass of 3.78 g. If the density of cork is 0.22 g/mL, what is the volume of the cork? Regardless of the method that is used, you should still be able to obtain the same (and correct) answer.
Solution
To use density as a conversion factor, we need to take the reciprocal so that the mass unit of density is in the denominator. Taking the reciprocal, we find
\[\dfrac{0.22\,\cancel{g}}{mL}= \dfrac{3.78\,\cancel{g}}{V} \nonumber \]
Cross-multiplying denominators with numerators, we obtain the following algebraic equation.
\[0.22 V = 3.78\,mL \nonumber \]
then you will need to isolate the variable (volume)
\[ V = \dfrac{3.78\,mL}{0.22} \nonumber \]
so, the volume of the cork is 17.2 mL.
What is the volume of 3.78 g of gold?
- Answer
-
Before attempting this question, be sure to obtain the density of gold in the table above. If you were to need this value on a quiz or a test, then it would be provided. Once you have this value, plug it into the density equation. Next, you will need to isolate the volume variable (basic algebra). The final answer should be 0.196 cm3.
Care must be used with density as a conversion factor. Make sure the mass units are the same or the volume units are the same, before using density to convert to a different unit. Often, the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.
Density as a Conversion Factor
Conversion factors can also be constructed to convert between different units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows:
13.6 g mercury = 1 mL mercury
This relationship can be used to construct two conversion factors:
Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 16 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top so that our final answer has a unit of mass:
In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates.
Density can be used as a conversion factor between mass and volume.
An \(18.2 \: \text{g}\) sample of zinc metal has a volume of \(2.55 \: \text{cm}^3\). Calculate the density of zinc.
Solution
Step 1: List the known quantities and plan the problem.
Known
- Mass \(= 18.2 \: \text{g}\)
- Volume \(= 2.55 \: \text{cm}^3\)
Unknown
- Density \(= ? \: \text{g/cm}^3\)
Use Equation \ref{eq1} to solve the problem.
Step 2: Calculate
\[D = \frac{m}{V} = \frac{18.2 \: \text{g}}{2.55 \: \text{cm}^3} = 7.14 \: \text{g/cm}^3\]
Step 3: Think about your result.
\(1 \: \text{cm}^3\) of zinc has a mass of 7.14 grams. Metals are expected to have a density greater than that of water and zinc's density falls within the range of the other metals listed above.
Since density values are known for many substances, density can be used to determine an unknown mass or an unknown volume. Dimensional analysis will be used to ensure that units cancel appropriately.
What is the mass of \(2.49 \: \text{cm}^3\) of aluminum?
Solution
Step 1: List the known quantities and plan the problem.
Known
- Density \(= 2.70 \: \text{g/cm}^3\)
- Volume \(= 2.49 \: \text{cm}^3\)
Unknown
- Mass \(= ? \: \text{g}\)
Use the equation for density, \(D = \frac{m}{V}\), and dimensional analysis to solve each problem.
Step 2: Calculate
\[ \: \: 2.49 \: \text{cm}^3 \times \frac{2.70 \: \text{g}}{1 \: \text{cm}^3} = 6.72 \: \text{g}\]
In this problem, the mass is equal to the density multiplied by the volume.
Step 3: Think about your results.
Because a mass of \(1 \: \text{cm}^3\) of aluminum is \(2.70 \: \text{g}\), the mass of about \(2.5 \: \text{cm}^3\) should be about 2.5 times larger.
What is the volume of \(50.0 \: \text{g}\) of aluminum?
Solution
Step 1: List the known quantities and plan the problem.
Known
- Density \(= 2.70 \: \text{g/cm}^3\)
- Mass \(= 50.0 \: \text{g}\)
Unknown
- Volume \(= ? \: \text{cm}^3\)
Use the equation for density, \(D = \frac{m}{V}\), and dimensional analysis to solve each problem.
Step 2: Calculate
\[ \: \: 50.0 \: \text{g} \times \frac{1 \: \text{cm}^3}{2.70 \: \text{g}} = 18.5 \: \text{cm}^3\]
In problem 2, the volume is equal to the mass divided by the density.
Step 3: Think about your results.
The \(50 \: \text{g}\) of aluminum is substantially more than its density, so that amount should occupy a relatively large volume.
Using Density in Environmental Applications
Along with solubility, density can help determine how a contaminant could affect an aquatic system. For example, imagine mercury has been spilled in Furman Lake. Looking at this element's density value and comparing it to liquid water, one could determine the location of the insoluble (you would be given solubility information) mercury layer. The denser mercury layer would reside at the bottom of Furman Lake. If one were to take a cross-section of the lake, one could see that a heterogeneous mixture would result.
In contrast, spilling ethanol (density = 0.789g/mL) would result in the formation of a homogeneous mixture. Ethanol (grain alcohol) is soluble in water. This would make it miscible (mixable to form a solution) in water, and one would not be able to denote separate layers. According to the density, an alcohol layer would remain on top, but would ultimately dissolve.
What difficulties would arise from the separation and removal of contaminants?
- Hg in Furman Lake
- Ethanol in Furman Lake
- oil (less dense, insoluble) in Furman Lake
Watch this video and record your observations.
- What component was different in the two types of beverages (mass or volume)?
- How does the above-mentioned difference affect the density equation?
- Which beverage is denser than water?