### Metric Cubic Unit ConversionProblems #1-10

Problem #1: Convert 0.500 cubic feet to liters

Solution:

1) Let's set up the dimensional analysis first and then comment on it:

 (12 inch)3 (2.54 cm)3 1 mL 1 L 0.5 ft3 x ––––––– x ––––––– x ––––– x ––––––– =  14.2 L 1 ft3 (1 in)3 1 cm3 1000 mL

2) The conversions:

first ---> convert cubic feet to cubic inches
second ---> convert in3 to cubic centimeters
third ---> convert cm3 to mL
fourth ---> convert mL to L

3) Comment:

Note the (12 inch)3. You must remember to "distribute" the cube. While it is true that 12 inches equals 1 foot, you have to remember that 12 in3 DOES NOT equal 1 ft3. You have to cube the 12, as in:
1 ft3 = (12 in) (12 in) (12 in) = 1728 in3

The number 12 AND the unit inch both get cubed.

Problem #2: Convert 0.500 cubic feet to dm3

Solution:

1) The dimensional analysis set up:

 (12 inch)3 (2.54 cm)3 (1 dm)3 0.5 ft3 x ––––––– x ––––––– x ––––––– =  14.2 dm3 1 ft3 (1 in)3 (10 cm)3

2) The conversions:

first ---> convert cubic feet to cubic inches
second ---> converts in3 to cubic centimeters
third ---> converts cm3 to dm3

3) Comment:

There are 10 cm in 1 dm. Remember, deci- means 0.1 while centi- means 0.01. There are 1000 cm3 in 1 dm3, just like there are 1000 mL in 1 L.

Problem #3: Convert the density of gold, 19.3 g/cm3, to kg/L.

Solution:

1) Convert grams to kg:

19.3 g/cm3 times (1 kg / 1000 g) = 0.0193 kg/cm3

2) Convert cm3 to mL:

0.0193 kg/cm3 times (1 mL / 1 cm3) = 0.0193 kg/mL

3) Convert mL to L:

0.0193 kg/mL times (1000 mL / L) = 19.3 kg/L

4) Often, a teacher will want you to write out the conversions in one long line:

19.3 g/cm3 x (1 kg / 1000 g) x (1 mL / 1 cm3) x (1000 mL / L) = 19.3 kg/L

Problem #4: Change 1.0 kg/cm3 to g/mm3

Solution:

(1.0 kg / cm3) x (1000 g / kg) = 1000 g/cm3

(1000 g / cm3) x (1 cm3 / 1000 mm3) = 1.0 g / mm3

Remember, 10 mm in one cm, so 10 mm x 10 mm x 10 mm equals 1000 mm3 in one cm3.

Teachers like these conversions where you start with 1 and end with 1. Students don't expect it and think they have done something wrong. The video has another example.

### Convert kg/m3 to g/L

Problem #5: Convert 25.0 mL to mm3 using dimensional analysis.

Solution via cm3:

 1 cm3 103 mm3 25.0 mL x ––––– x ––––––– = 2.50 x 104 mm3 1 mL 1 cm3

The 1 mL equals 1 cm3 conversion is very handy.

There are 10 mm in every cm, so 10 cubed is 103.

Solution via dm3:

 1 L 1 dm3 106 mm3 25.0 mL x ––––– x ––––– x ––––––– = 2.50 x 104 mm3 103 mL 1 L 1 dm3

The 1 L equals 1 dm3 conversion is very handy.

There are 100 mm in every dm, so 100 cubed is 106

Problem #6: Convert 5.51 g/cm3 to lb/ft3.

Solution:

1) Do the non-cubic conversion first:

 1 lb 5.51 g/cm3 x –––––– = the answer 453.6 g

I'm not going to work it out yet, I simply wanted to show the above conversion. I'll leave the working out until the cm3 to ft3 is added in.

2) We can look up a direct conversion from cm3 to ft3 (you may, I will not bother). The usual technique (which is what I follow) is to go via common conversions, ones that teachers tend to have their students memorize. In this specific case, we would go from cm3 to cubic inch (since 1 inch exactly equals 2.54 cm) and then from in3 to cubic foot (because 1 foot exactly equals 12 inches).

 1 lb (2.54 cm)3 (12 in)3 5.51 g/cm3 x –––––– x –––––––– x –––––––– = 344 lb/ft3 453.6 g (1 in)3 (1 ft)3

Notice the style I used for the cubic units: (1 ft)3, (2.54 cm)3, and so on. Sometimes, you see it done without the parentheses: 1 ft3 or 2.543 cm3 might be what you woud see. Notice how the cube on the numeral 1 is eliminated.

5.51 g/cm3 is the density of the Earth.

Problem #7: How many 1 cm cubes would it take to construct a cube that is 4 cm on edge?

Solution:

1) The formula for volume of a cube is:

V = l*w*h

2) Insert 4 cm for each of the three dimensions:

V = 4 cm times 4 cm times 4 cm = 64 cm3

Sixty-four 1 cm cubes would be required.

Problem #8: Convert 6.230 x 10¯2 kg/mm3 to g/L

Solution:

1) You have to do a kg to g conversion and then take mm3 to cm3 to mL to L. Here it is:

 0.06230 kg 1000 g (10 mm)3 (1 cm)3 1000 mL ––––––––– x ––––– x ––––––– x ––––––– x ––––––– =  6.230 x 107 g/L mm3 1 kg (1 cm)3 1 mL 1 L

2) Notice that kg/mm is in the form of a density. When I formatted the above problem, I thought to explore a 'calculate the density of a black hole' problem, but I decided against it. However, here is a density conversion problem that might interest you. If you scroll up to the top of the file, you'll find some discussion about the density of a black hole.

Problem #9: If a box measures 3.00 cm wide, 50.5 mm long and 0.520 m high, what is its volume in cubic feet?

Solution:

1) One of the common pathways from metric units to English units is 2.54 cm = 1 inch. Let's take each measurement to cm and then calculate the volume in cm3:

(50.5 mm) (1 cm / 10 mm) = 5.05 cm
(0.520 m) (100 cm / 1 m) = 520. cm

volume ---> (3.00 cm) (5.05 cm) (520. cm) = 7878 cm3

2) Convert from cm3 to cubic inches:

(7878 cm3) (1 inch / 2.54 cm)3 = 480.745 in3

Notice how I cubed the entire conversion factor rather than cubing them individually, as in (1 in)3 / (2.54 cm)3

3) Convert from cubic inches to cubic feet:

(480.745 in3) (1 foot / 12 inches)3 = 0.278 ft3 (to three sig figs)

Problem #10: Convert 1 m3 to dm3 and then to L

Solution:

(1 m)3 (10 dm / 1 m)3 = 1000 dm3

Note how I wrote 1 m3

Since 1 dm3 = 1 L, 1000 dm3 = 1000 L

An alternate way to write the above conversion:

(1 m3) (103 dm3 / 13 m3) = 1000 dm3

Bonus Problem #1: Convert 2.70 g/cm3 to kg/m3 then to lb/ft3

Solution:

1) Convert grams to kilograms:

 1 kg 2.70 g / cm3 x ––––– =  0.00270 kg / cm3 1000 g

2) Convert kg/cm3 to kg/cm3:

 (100 cm)3 0.00270 kg / cm3 x –––––––– =  2.70 x 103 kg / m3 (1 m)3

3) Done in dimensional analysis style:

 2.70 g 1 kg (100 cm)3 ––––––– x ––––––– x ––––––– = 2.70 x 103 kg / m3 1 cm3 1000 g (1 m)3

4) To continue on, we need to look up the kg-to-pound conversion (1 kg = 2.2046226 lb) and the meter-to-foot conversion (1 m = 3.28084 ft). Here is the full conversion in dimensional analysis style:

 2.70 g 1 kg (100 cm)3 (1 m)3 2.2046226 lb ––––––– x ––––––– x ––––––– x –––––––––– x ––––––––––– = 168 lb / ft3 (to three sig figs) 1 cm3 1000 g (1 m)3 (3.28084 ft)3 1 kg

Bonus Problem #2: The German chemist Fritz Haber proposed paying off the reparations imposed against Germany after World War I by extracting gold from seawater. Given the following data, what was the dollar amount of the gold that could have been extracted from 3.00 m3 of seawater? (The price of gold at the time was \$0.68 per gram of gold. Gold occurs in seawater to the extent of 5.15 x 1011 atoms per gram of seawater. The density of seawater is 1.0273 g/cm3.)

Solution:

1) Convert 3.00 m3 to cm3:

 (100 cm)3 3.00 m3 x ––––––– =  3.00 x 106 cm3 1 m3

2) Determine the mass of sea water in 3.00 x 106 cm3:

(3.00 x 106 cm3) (1.0273 g/cm3) = 3.0819 x 106 g of sea water

3) Determine atoms of gold present:

(3.0819 x 106 g) (5.15 x 1011 atoms/g) = 1.58718 x 1018 atoms of gold

4) Determine moles of gold:

1.58718 x 1018 atoms of gold / 6.022 x 1023 atoms/mole = 2.635636 x 10¯6 mole

5) Determine grams of gold:

(2.635636 x 10¯6 mole) (196.96657 g/mol) = 0.000519132 g of gold

6) Determine the dollar amount:

(0.68 dollar/g) (0.000519132 g) = \$0.000353

7) Dimensional analysis, baby!!

 1003 cm3 1.0273 g 5.15 x 1011 atoms 1 mol 196.96657 g \$0.68 3.00 m3 x ––––––– x ––––––– x ––––––––––––––– x ––––––––––––––– x –––––––––– x ––––– = \$0.000353 1 m 3 1 cm3 1 g 6.022 x 1023 atoms 1 mol 1 g

8) Let's reverse the problem: how many cubic meters of sea water would be required to supply 1.00 dollar of gold? For this problem, I will provide only the dimensional analysis set up. Notice that it is the reverse of the previous problem.

 1 g 1 mol 6.022 x 1023 atoms 1 g 1 cm3 1 m3 \$1.00 x ––––– x –––––––––– x ––––––––––––––– x –––––––––––––– x ––––––– x –––––––– = 8498 m3 \$0.68 196.96657 g 1 mol 5.15 x 1011 atoms 1.0273 g 1003 cm3