3.6: Density
- Page ID
- 283824
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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 densities of different chemical substances.
- Compare any chemical substance's density to the density of water.
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 has a compound unit; it always has multiple parts. 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 \(\PageIndex{1}\). The density units used in the table are equal to one another because 1 mL = 1 cm3. Usually mL is used for liquids and cm3 is used for solids, but this can vary based on how the volume was measured. Note: Densities change with temperature so the values in the table will vary at high or low temperatures.
| 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 |
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 a scale or balance? You can use the volume and density to calculate it. If you multiply the given volume by the known density from (Table \(\PageIndex{1}\)), the volume units will divide out and leave you with mass units, telling you the mass of the sample.
There are two ways of knowing that multiplication is needed in this case:
If using unit analysis, start with the value with a simple unit. In this case we will consider the volume unit, cm3, to be simpler than the density unit, g/mL or g/cm3.
\[7.88\:cm^3 \times \dfrac{}{\:\:cm^3}\]
To use the density as a conversion factor, split it into two parts. For aluminum the density tells us that 2.7 g = 1 mL or 2.7 g = 1 cm3. These relationships are not always true, they are specific to aluminum. Density is needed to relate mass and volume for a substance.
\[7.88\:cm^3 \times \dfrac{7.88 \: g}{1\:cm^3} = 21\:g \nonumber \]
If using algebra, 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? The density of mercury is 13.6 g/mL.
Solution
Option 1: Unit Analysis
Start with the given number. Set up a conversion factor and put the unit from the given number on the bottom. (Leave room to put a number in front of the unit.)
\(44.6\:mL \times \frac{}{\:\:mL}\)
Consider what unit will go on the top of the conversion factor. We are asked to solve for mass and the density provides a relationship between volume and mass. To use the density, which has a compound unit, we split the unit into two lines or, in other words, we make the compound unit into a conversion factor. For mercury 13.6 g = 1 mL (this is only true for mercury. You cannot use 13.6 g = 1 mL in other problems).
\(44.6\:mL \times \frac{13.6\:g}{1\:mL}\)
The diagonal division sign in 13.6 g/mL becomes the horizontal division sign in \( \frac{13.6\:g}{1\:mL}\). The number, 13.6, goes with g because 13.6 g/mL means 13.6 grams per milliliter or 13.6 grams per 1 mL.
Calculate the answer:
\(44.6\:mL \times \frac{13.6\:g}{1\:mL} = 606.56\: g\) or 607 g with three significant figures.
Option 2: Algebra
If you need to review how to isolate variables, please see the appendix of this textbook.
Use 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.
\[25.0\cancel{cm^3}\times \dfrac{7.87\,g}{\cancel{cm^3}}= 197 g \text{ of iron} \nonumber \]
Explore how different masses and volumes affect density.
- Does changing an object's mass or volume affect its density?
- How does density relate to whether an object sinks or floats in water?
Care must be used with density as a conversion factor. Make sure that either the mass units are the same or the volume units are the same before using density as a conversion factor. Sometimes the unit of the given quantity must be first converted to the appropriate unit before applying density as a conversion factor.
A cork stopper from a bottle of wine has a mass of 0.133 oz. If the density of cork is 0.22 g/mL, what is the volume of the cork in milliliters? Note: 1 lb. = 16 oz. and 1 lb. = 453.6 g. Regardless of the method that is used, you should still be able to obtain the same (and correct) answer.
Solution
If you plan to use unit analysis to solve the problem, start by identifying the number with the simple unit.
\(0.133 \:oz.\times \frac{}{\:\:oz.}\)
Consider whether the starting unit matches half of the density unit. It does not, so we need to do a conversion before using the density. Since the starting value is a mass we should convert it to the same mass unit found in the density, grams.
\(0.133 \:oz.\times \frac{1\:lb.}{16\:oz.}\times \frac{453.6\:g}{1\:lb.}\)
Now we can use the density to convert from mass to volume. Use the density as a conversion factor by splitting it into two parts.
\(0.133 \:oz.\times \frac{1\:lb.}{16\:oz.}\times \frac{453.6\:g}{1\:lb.}\times \frac{1\:mL}{0.22\:g}\)
Keep following the pattern with the units (where the unit from the top of the previous step is written on the bottom of the current step). This ensures that everything except the unit that is needed for the answer divides out. A density of 0.22 g/mL means that 1 mL of cork weighs 0.22 g. The number goes with the unit that it is next to; the other unit has a value of 1 unless something else is specified.
\(0.133 \:oz.\times \frac{1\:lb.}{16\:oz.}\times \frac{453.6\:g}{1\:lb.}\times \frac{1\:mL}{0.22\:g} \) = 17 mL
Because the density only had 2 significant figures, the answer can only have 2 significant figures. Densities are measured, not exact.
What is the volume in liters of 3780 g of gold? The density of gold is 19.3 g/cm3. Note: 1 mL = 1 cm3
- Answer
-
0.196 L
What is the volume in liters of 5.0 lbs. of iron? Iron has a density of 7.87 g/mL or g/cm3, 1 lb. = 453.6 g, and 1 cm3 = 1 mL.
- Answer
-
0.29 L
Summary
- Density is the relationship between mass and volume for a substance
- Densities are different for different substances. To convert between mass and volume for a sample its density must be known.
- Density problems can be solved using unit analysis, algebra, or cross multiplying.
Contributors and Attributions
- Hayden Cox (Furman University, Class of 2018)