1.15: Density Applications

Learning Objectives

• Apply density as a conversion factor.
• Compare two densities to determine whether an object will float or sink in a given liquid.

Densities have several significant applications, two of which will be discussed in this section.  First, densities can be applied as ("hidden") conversion factors.  The density of an object will also determine whether it will float or sink in another substance. 

Density as a Conversion Factor

The units for density, "g/mL" and "g/cm³," which are read as "grams per milliliter" and "grams per cubic centimeter," respectively, each contain the indicator word "per."  As discussed in the previous section, this word means that the given information (a density, in this case) can be used as a "hidden" conversion factor.

Consider chloroform, which has a density of 1.49 grams per milliliter.  Calculate the volume of 19.7 grams of chloroform.

This problem can be solved either by using the density equation or by applying the density of chloroform as a "hidden" conversion factor.

Method 1: Using the Density Equation

The equation

\(\text{d} = \dfrac{\text{m}}{\text{V}}\)

can be applied by replacing the density and mass variables with the corresponding given information, as shown below.

\(\text{1.49 g/mL} = \dfrac{\text{19.7 g}}{\text{V}}\)

The algebra required to solve this equation is fairly complex, due to the location of the remaining variable.  The most reliable mathematical technique for solving equations that involve fractions is cross-multiplication and division.  This process eliminates the fractions by multiplying the numerator on each side of the equal sign by the denominator on the other side, and then equating those values.  The "slash" written within the density unit implies that the unit "milliliter" could be written as a denominator, and the numerical value associated with this unit is understood to be an unwritten "1."

\(\dfrac{\text{1.49 g}}{\text{mL}} = \dfrac{\text{19.7 g}}{\text{V}}\)

Cross-multiplication would result in the relationship shown below.

\( {\text{1.49 g}} \times {\text {V}} = {\text{19.7 g}} \times {\text {mL}}\)

Dividing both sides of the equation by "1.49 g" completes the process of isolating the desired variable, volume.

\( \text {V} = \dfrac{\text{19.7 g × mL}}{\text {1.49 g}}\)

The unit "grams" appears in both the numerator and the denominator on the right side of this equation and, therefore, cancels.

\( \text {V} = \dfrac{19.7 \; \cancel{\rm{g  }}\; \rm{ × }\; \rm{mL}}{1.49 \; \cancel{\rm{g}}}\)

The calculation that remains after this unit cancelation is

\( {\text {V}} = {\text {19.7 mL ÷ 1.49}} = {\text {13.2214765... mL}}\)

After the appropriate number of significant figures are applied, the final answer would be reported as 13.2 mL.

Method 2: Applying Density as a Conversion Factor

The process of solving this problem can be significantly simplified by recognizing that the "per" in the density ("1.49 grams per milliliter") indicates that this quantity is a "hidden" conversion factor.  Based on the process developed in the previous section, the remaining quantity, "19.7 grams" must be the given quantity.  Since a volume is the desired quantity, "milliliters" must be the desired unit.  In order to achieve unit cancelation, the density conversion must be written such that the unit "g" is written in the denominator, as shown below.

\( {19.7 \; \cancel{\rm{g}}} \times \dfrac{\rm{mL}}{1.49 \;\cancel{\rm{g}}} = {\text {13.2214765... mL}}\)

Finally, the correct number of significant figures must be applied to the calculated quantity.  Remember that since density is a "hidden" conversion factor, the number of significant figures present in its numerical value must be considered when determining the number of significant figures that the answer should have.  After applying the correct number of significant figures, the same volume as above, 13.2 mL, is reported as the final answer.

Example \(\PageIndex{1}\)

A student obtains 75.0 cubic centimeters of corn oil, which has a density of 0.922 grams per cubic centimeter.  What is the mass of the corn oil? 

Solution

As was the case with the word problem given above, this problem can be solved either by using the density equation or by applying the density as a "hidden" conversion factor.  While both methods are equally viable, the latter method was shown to be much more straight-forward and so will be applied here. 

The "per" in the phrase "0.922 grams per cubic centimeter" indicates that density is a "hidden" conversion factor and should be expressed as such.  The remaining quantity, "75.0 cubic centimeters" must be the given quantity.  Since a mass is the desired quantity, "grams" must be the desired unit.  In order to achieve unit cancelation, the density conversion must be written such that the unit "cm³" is written in the denominator, as shown below.

\( {75.0 \; \cancel{\rm{cm^3}}} \times \dfrac{0.922 \; \rm{g}}{\cancel{\rm{cm^3}}} = {\text {69.15 g}}\)

After the correct number of significant figures are applied, the final answer would be reported as 69.2 g.

Exercise \(\PageIndex{1}\)

What is the volume, in milliliters, of 879.85 grams of liquid water?

Answer

This problem is particularly challenging, as some information is seemingly missing.  Using the previously-established methodology for solving word problems, the given unit is "grams" and the desired unit is "milliliters."  Based on the information in this section, these quantities could be related by the density of liquid water.  Note, however, that this quantity is not given in the problem, because it was highlighted as an important quantity to memorize in the previous section.  In order to achieve unit cancelation, the density conversion must be written such that the unit "g" is written in the denominator, as shown below.

\( {879.85 \; \cancel{\rm{g}}} \times \dfrac{\rm{mL}}{1.00 \; \cancel{\rm{g}}} = {\text {879.85 mL}}\)

Since the number of significant figures in the answer must be limited to the lesser count of significant figures, the density conversion factor is the limiting quantity, and the final answer would be reported as 880. mL.  Note that this quantity must be expressed using a written decimal, in order to establish the final zero as a significant figure.

Density Comparisons

Whether an object sinks or floats depends solely on the relative densities of the object and the liquid in which it is being placed.  If the density of the object (d_object) is greater than the density of the liquid (d_liquid), the object will sink.  Conversely, if the density of the object (d_object) is less than the density of the liquid (d_liquid), the object will float.  A common misconception is that the "heaviness" of an object is what will determine whether or not something will sink. However, ships are extremely "heavy," but they still float in water!  Therefore, words like "heavier" and "lighter" should never be used to explain whether an object floats or sinks.

Exercise \(\PageIndex{2}\)

Table 1.12.1 from the previous section is reproduced below. 

Table 1.12.1: Densities of Four Common Liquids

Substance	Density at 25°C (g/cm^3)
corn oil	0.922
liquid water	1.00
honey	1.42
mercury	13.6

Which of these substances will sink in liquid water? Which will float in liquid water? Provide specific rationales for each of your conclusions.

Answer

Since dcorn oil (0.922 g/cm³) is less than dliquid water (1.00 g/cm³), corn oil will float in liquid water.  Since both dhoney (1.42 g/cm³) and dmercury (13.6 g/cm³) are greater than dliquid water (1.00 g/cm³), honey and mercury will both sink in liquid water.