Investigation Four:

Densities of Different Solids

In Investigation Three, students calculated the densities of several different liquids, confirming that the density of one type of matter is a constant and that the densities of different types of matter are different. In Investigation Four, students will use several different examples of matter in its solid phase to confirm these two findings.

As with liquids and gases, calculating the density of a solid requires that the volume and the mass are determined. The mass of a solid is easily determined using an appropriate balance but compared to measuring the volume of a liquid or a gas, measuring the volume of a solid can be problematic. Since liquids assume the shape of their container, using a graduated cylinder or other calibrated device is all that is required for measuring the volume of a liquid. Using the appropriate formula, the volume of a gas can be calculated relatively easily if the amount of gas, its pressure, and its temperature are known.

Regularly-Shaped Solid Objects

If a solid is regularly shaped such as a cube or a cylinder, its volume is easily calculated using a formula borrowed from geometry. In this Investigation, students will calculate the volumes of two regular shapes, a woodblock, which is an example of a rectangular prism, and a metal cube. The volume of both a rectangular prism and a cube can be calculated by measuring its three dimensions, height, width, and length, and multiplying them. When the dimensions are measured using centimeters, the resulting volume is expressed in cubic centimeters or cm³. In the metric system, one cubic centimeter is equivalent to one milliliter.

Irregularly-Shaped Solid Objects

The volume of a solid can also be determined using a method termed volume displacement. In this method, the volume of one substance is displaced by the volume of another substance since no two types of matter can occupy the same physical space at the same time. In this Investigation, students will use a graduated cylinder filled with a predetermined volume of water. A solid is placed in the water and the volume of water that is displaced by the solid will correspond exactly to the volume of the solid.

The volumes of the woodblock and the metal cube will be determined using the volume displacement method in addition to measuring the dimensions using the geometric formula. The volumes of the metal cube determined by both methods will correspond closely. This illustrates that one cubic centimeter is equivalent to one milliliter since the volume determined using the measurement method will be in units of cm³ and the volume determined by the volume displacement method will be in units of ml. However, the volume of the woodblock determined using the volume displacement method will be less than the volume determined using the measurement method. This results from the fact that the woodblock will float in the water and not completely displace all the water corresponding to its volume. This discrepancy will illustrate that despite the appearance of the woodblock, it is not composed entirely of one type of matter, wood, but is also composed of a second type of matter, air trapped in the wood.

The volume displacement method is well suited for determining the volume of a solid that is irregularly shaped or whose dimensions are difficult to measure. In this Investigation, students will measure the volume and mass and calculate the density of spheres of modeling clay, a third solid. Measuring the exact radius or diameter of a sphere is difficult. Additionally, the formula for the volume of a sphere is conceptually difficult since it requires using and understanding the term π and requires students to cube the radius. For these reasons, students will use a simplified formula for determining the volume of a sphere, volume = 4.2 × radius × radius × radius. Students will also use the volume displacement method to determine the volumes of spheres of modeling clay. Using the volumes and the masses of the two spheres, students will calculate the density of the modeling clay.

Calculating the density of two differently sized spheres of modeling clay clearly illustrates that the density of one type of matter is a constant that is dependent on both the mass and the volume. As discussed previously, a single type of matter possesses the same atoms, in the same proportion bonded together in the same structure to form the same molecules. This arrangement of atoms is a constant regardless of the amount of matter present. Increasing the mass of matter simply increases the number of molecules present while not affecting the types of atoms or molecules or their structure. More matter does not change the identity of the matter.

In the context of the Investigation, a larger ball of modeling clay is still the same substance as a smaller ball of modeling clay. As the mass of the clay is increased, the volume occupied increases by the same incremental amount. The same incremental increase in both mass and volume results in a constant ratio of mass divided by volume.