Twenty Examples

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There are physical characteristics of a substance that help identify the substance. One of these characteristics is density. Density (whose most common symbol is the lowercase letter d) is defined as mass per unit volume. Density is calculated by dividing the mass of an object by its volume. This is shown in equation form, as follows:

Density = mass ÷ volume

By the way, the lower-case Greek letter rho, ρ, is also used to symbolize density. Oftentimes, the rho shape that a textbook would use looks more like the lower-case letter p. However, a lower-case d is more often used in intoductory settings like the one you are currently reading.

We can calculate the density of a solid, liquid, or gas. The density of a gas will be dealt with in a later unit, because its density is very sensitive to temperature and pressure. Although the density of liquids and solids do change with temperature and pressure changes, the amount is fairly small. We will ignore these small amounts and act as if all our density problems are at the same temperature and pressure. Note the difference in units in the formulas of the density of a solid and liquid. The unit for cubic centimeters is cm^{3} and for milliliters is mL.

solids: d = grams ÷ cubic centimeters (cm^{3}is the symbol for cubic centimeters)liquids: d = grams ÷ milliliters

Since one mL equals one cm^{3}, there is no functional difference between g/cm^{3} and g/mL. This means that sometimes the density of a liquid is described using cm^{3} and the density of a solid is described using mL.

This reminder might help you in figuring out how to solve density problems:

M –––– d V

Simply cover up whichever value you need to calculate and the other two are shown in their proper placement, be it to multiply or to divide.

For example, cover up the M. This leave you with dV (ignore the fact that it is in the denominator). Density times volume will give you mass. You can also check it out by way of the units: (g / cm^{3}) x cm^{3} cancels out the volume unit leaving grams, the desired unit for mass.

**Example #1:** A block of aluminum occupies a volume of 15.0 mL and weighs 40.5 g. What is its density?

**Solution:**

d = 40.5 g / 15.0 mLd = 2.70 g/mL

**Example #2:** Mercury metal is poured into a graduated cylinder that holds exactly 22.5 mL. The mercury used to fill the cylinder weighs 306.0 g. From this information, calculate the density of mercury.

**Solution:**

d = 306.0 g / 22.5 mLd = 13.6 g/mL

**Example #3:** What is the mass of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL.

**Solution:**

d = g / mLg = (d) (mL)

g = (0.789 g/mL) (200.0 mL) = 158 g

Note that the unit of mL cancels out.

**Example #4:** A rectangular block of copper metal weighs 1896 g. The dimensions of the block are 8.4 cm by 5.5 cm by 4.6 cm. From this data, what is the density of copper?

**Solution:**

1) Calculate the volume of the block:

(8.4 cm) (5.5 cm) (4.6 cm) = 212.52 cm^{3}

2) Calculate the density:

d = 1896 g / 212.52 cm^{3}= 8.9 g/cm^{3}Significant figures in the answer are dictated by the length measurements of two sig figs.

**Example #5:** A flask that weighs 345.8 g is filled with 225 mL of carbon tetrachloride. The weight of the flask and carbon tetrachloride is found to be 703.55 g. From this information, calculate the density of carbon tetrachloride.

**Solution:**

1) Find mass of CCl_{4} by subtraction:

703.55 g − 345.8 g = 357.75 g

2) Calculate the density:

d = 357.75 g / 225 mL = 1.59 g/mL

**Example #6:** Calculate the density of sulfuric acid if 35.4 mL of the acid weighs 65.14 g.

**Solution:**

d = 65.14 g / 35.4 mLd = 1.84 g/mL

**Example #7:** Find the mass of 250.0 mL of benzene. The density of benzene is 0.8786 g/mL.

**Solution:**

density = mass ÷ volumetherefore mass = (density) (volume)

mass = (0.8786 g/mL) (250.0 mL)

mass = 219.6 g (rounded to 4 sig fig)

**Example #8:** A block of lead has dimensions of 4.50 cm by 5.20 cm by 6.00 cm. The block weighs 1591 g. From this information, calculate the density of lead.

**Solution:**

1) Determine the volume of the lead block:

(4.50 cm) (5.20 cm) (6.00 cm) = 140.4 cm^{3}(ignore sig figs for the moment)

2) Calculate density:

1591 g / 140.4 cm^{3}= 11.3 g/cm^{3}(3 sig figs because of the length measurements on the lead block)

**Example #9:** 28.5 g of iron shot is added to a graduated cylinder containing 45.5 mL of water. The water level rises to the 49.1 mL mark, From this information, calculate the density of iron.

**Solution:**

1) Calculate the volume of the iron shot:

49.1 mL − 45.5 mL = 3.6 mL

2) Calculate the density:

d = 28.5 g / 3.6 mLd = 7.9 g/mL

**Example #10:** What volume of silver metal will weigh exactly 2500.0 g. The density of silver is 10.5 g/cm^{3}.

**Solution:**

d = g / cm^{3}cm

^{3}= g / dcm

^{3}= 2500.0 g / (10.5 g/cm^{3}) = 238 cm^{3}Note that, in the g/cm

^{3}unit, that the cm^{3}is in the denominator of the denominator. It moves to the numerator, giving us the volume unit we desire for the answer.

**Example #11:** What is a Dord?

It's a ghost word, an accidental creation that came to be reproduced in dictionaries.Webster's Second International Dictionary,1934 showed Dord as a synonym for density. The word arose when a proofreader's question about the symbol for density, "D or d?" was taken literally. Source:The Oxford Companion to the English Language,Ed.: Tom McArthur, Oxford University Press, 1992, page 440.

**Example #12:** A piece of copper foil has a mass of 4.924 g, a length of 3.62 cm, and a width of 3.02 cm. Calculate the thickness in mm, assuming the foil has uniform thickness.

**Solution:**

1) We look up the density of copper on the Intertubes and find:

8.96 g/cm^{3}

2) Calculate the volume of copper foil:

4.924 g / 8.96 g/cm^{3}= 0.54955 cm^{3}

3) Calculate the missing dimension:

(3.62 cm) (3.02 cm) (x) = 0.54955 cm^{3}x = 0.050268 cm

4) Convert cm to mm:

(0.050268 cm) (10 mm / 1 cm) = 0.503 mm (to three sig figs)

**Example #13:** A golden-colored cube is handed to you. The person wants you to buy it for $100, saying that is a gold nugget. You pull out your chemistry text and look up gold's density and read that its density is 19.32 g/cm^{3}. You measure the cube and find that it is 2.00 cm on each side, and weighs 40.0 g. What is its density? Is it gold? Should you buy it?

**Solution:**

1) Determine volume of the cube:

(2.00 cm) (2.00 cm) (2.00 cm) = 8.00 cm^{3}

2) Determine the density of the cube:

40.0 g / 8.00 cm^{3}= 5.00 g/cm^{3}

3) Conclusion:

Not gold. And, no, you should not buy it! Hint: look up the density of fool's gold.

**Example #14:** Sapphire has a density of 3.98 g/cm^{3}. The mass of gemstones is often measured in “carats,” where 1 carat equals 0.200 g. What is the volume (in cubic centimeters) of the 563.35 carat Star of India sapphire?

**Solution:**

(563.35 carat) (0.200 g / carat) = 112.67 g112.67 g / 3.98 g/cm

^{3}= 28.3 cm^{3}(to three sig figs)

**Example #15:** A titanium cube contains 3.10 x 10^{23} atoms. The density of titanium is 4.50 g/cm^{3} What is the edge length of the cube?

**Solution:**

1) Solve for the mass of Ti in the cube:

3.10 x 10^{23}atoms / 6.022 x 10^{23}atoms/mole = 0.514779 mole(0.514779 mol) (47.867 g/mole) = 24.6409 g

2) Solve for the volume of the cube:

24.6409 g / 4.50 g/cm^{3}= 5.475756 cm^{3}

3) The formula for volume of a cube is side x side x side. Solve for the edge length:

s^{3}= 5.475756 cm^{3}s = 1.76 cm (to three sig figs)

**Example #16:** The density of aluminum foil is 2.70 g/cm^{3}. A square of aluminum foil measuring 3.00 inches on each side weighs 384 mg. Find the thickness in micrometers (μm) of the aluminum foil.

**Solution:**

1) Determine the volume of the Al foil:

384 mg = 0.384 g0.384 g / 2.70 g/cm

^{3}= 0.142222 cm^{3}

2) Determine the thickness of the foil in cm:

1.00 inch = 2.54 cm (by definition)3.00 inch = 7.62 cm

(7.62 cm) (7.62 cm) (x) = 0.142222 cm

^{3}x = 0.00245 cm (to three sig figs)

3) Convert cm to μm:

(0.00245 cm) (10^{4}μm / 1 cm) = 24.5 μm

**Example #17:** A sheet of aluminum foil that has a thickness of 0.100 micrometers. The other two dimensions are 24.0 cm and 15.0 cm. How many milligrams does this sample weigh?

**Solution:**

1) Change 0.100 μm to cm:

(0.100 μm) (1 cm / 10^{4}μm) = 1.00 x 10¯^{5}cm

2) Determine volume of foil:

(24.0 cm) (15.0 cm) (1.00 x 10¯^{5}cm) = 0.0036 cm^{3}

3) Determine mass (the density of aluminum is well-known and can easily be looked up):

(0.0036 cm^{3}) (2.70 g/cm^{3}) = 0.00972 g

4) Change 0.00972 g to mg

(0.00972 g) (1000 mg / 1 g) = 9.72 mg

**Example #18:** I thought I'd post a link to some very nice density problems. The author starts with some standard density problems and then moves into several more complex density problems. Number 7 is a very good problem.

**Example #19:** Calculate the density of a 50. mL solution that is 5% water and 95% ethanol.

**Solution:**

1) Supposing the given percentages are by volume:

(5% of 50. mL) = 2.5 mL water(95% of 50. mL) = 47.5 mL ethanol

2) Using the densities of both:

(2.5 mL water) (1.00 g/mL) = 2.5 g water(47.5 mL ethanol) (0.7893 g/mL) = 37.49 g ethanol

3) Add 'em up and divide:

(2.5 g water + 37.49 g ethanol) / (50. mL) = 0.80 g/mL

4) There is also the assumption that the volume of water and ethanol are additive when mixed. This is not strictly correct, but the fact that 2.5 mL of water and 47.5 mL of ethanol, when mixed, do not form 50 mL of solution will be ignored.

**Example #20:** The SI unit for density is kg/m^{3}. Convert the density of ethanol (789 kg/m^{3}) to the more commonly-used unit of g/cm^{3}

**Solution:**

1) Convert kg/m^{3} to g/m^{3}:

(789 kg/m^{3}) (1000 g / 1 kg) = 789000 g/m^{3}

2) Convert g/m^{3} to g/cm^{3}

(21450000 kg/m^{3}) (1 m^{3}/ 100^{3}cm^{3}) = 0.789 g/cm^{3}Note the use of 100

^{3}. 1 m^{3}is a cube 100 cm on a side: 100 cm x 100 cm x 100 cm = 100^{3}cm^{3}.

3) Many teachers that teach dimensional analysis want the solution in one line of calculation steps:

(789 kg/m^{3}) (1000 g / 1 kg) (1 m^{3}/ 100^{3}cm^{3}) = 0.789 g/cm^{3}Note the interim values/units such as 789000 g/m

^{3}do not appear in a one-line dimensional analysis presentation. Here is the one-line set up presented a different way:

789 kg 1000 g 1 m ^{3}––––––– x ––––––– x ––––––– = 0.789 g/cm ^{3}1 m ^{3}1 kg 100 ^{3}cm^{3}

Why did I show you this problem? Because it is easy to predict a situation where a problem (test or homework) gives you the density in kg/m^{3}, but use of the g/cm^{3} value is required in the solution to the problem.

In the ChemTeam section on the metric system, I go into what I call 'two-unit conversions' and I have a problem similar to the one above in that section.

Probs #1-10 | Probs #11-25 | Probs #26-50 | All the examples & problems, no solutions | Significant Figures Menu |