Metric-English Unit Conversion Examples

Metric-English Conversion Problems #1-10     Metric-English Conversion Problems #26-60
Metric-English Conversion Problems #11-25     Return to Metric Menu

The conversions between Metric and English units are not all exact values (which means they are defined relationships, like, say, 1 meter = 100 cm.). Consequently, a question that often arises is how to determine significant figures in the answer. The solution to this is to use as many digits as possible in any Metric-English conversion unit. For example, I tend to use 453.6 g/lb when converting between grams and pounds, but you could use 453.592 g/lb, if it were needed.

Put another way, what you do is look to use more significant figures in your conversions than are present in your starting value. Then, make sure to carry two or three guard digits through each calculation (if you do a several-step calculation) and round off to the final number of significant figures only when you arrive at the final answer.

There are teachers that advocate a one-line set up method called dimensional analysis (DA). Using this technique, one would not multiply anything until the entire problem was set up, so there would be no intermediate answers that got rounded off. In the problems I explain, I have tended towards the intermediate answer way, so as to break the problem's solution into steps that can be explained. However, I have included DA in a variety of the problems. Teachers that advocate DA tend to do so very passionately, insisting on an exclusive use of DA in their classes. The problem is that the teacher will then tend to under-explain the technique, doing one ot two examples and thinking that is enough. They are wrong to think so.

You have been warned!

You may want to search the Internet for information on Metric-English conversions. I will include discussion of a variety of conversions in the solutions provided below, but it i not an exhaustive discussion of all possible Metric-English conversions. Make sure you look at the end of Example #1 for a mention about the consequence of one inch equalling 2.54 cm by definition.

In the examples to follow, I will freely mix one-unit and two-unit conversions as well as conversions involving square and cube units.


Example #1: Convert 1.000 km to inches.

Solution:

1) A conversion that you should memorize is this:

1 inch = 2.54 cm
2) Based on that, I propose to first change km to cm (which is a common Metric-only conversion):
1.000 km times (105 cm / km) = 1.00 x 105 cm
3) Now, the conversion to inches:
1.00 x 105 cm times (1 inch / 2.54 cm) = 39370 inch

Since 1 inch = 2.54 cm is by definition, I propose to keep four sig figs in the answer. In other words, 2.54 is NOT three sig figs. Since it is a defined value, it plays no role in determining the number of significant figures. Some teachers will state that the defined value has an infinite number of significant figures. The ChemTeam prefers to state that defined values play no role.


Example #2: Convert 4.04 x 105 feet to centimeters

Solution:

1) Convert feet to inches:

4.04 x 105 feet x (12 inches / ft) = 4.848 x 106 in

2) Convert inches to centimeters:

4.848 x 106 in x (2.54 cm / in) = 1.23 x 107 cm (to three sig figs)

3) Here is the problem, done in one line:

This one-line is what is commonly meant by saying "dimensional analysis." All the conversions are strung together in one line, as opposed to be set out as a multi-step solution, as was done in Example #1 and in steps 1 and 2 to this example.


Example #3: Convert 1.000 yards to nanometers.

Solution:

Here is the solution, set up in dimensional analysis style:

  3 ft   12 in   2.54 cm   107 nm  
1.000 yd x ––––– x –––––– x –––––––– x ––––––– = 9.144 x 108 nm
  yd   ft   in   cm  

I omitted the yard to feet conversion in the hand-written solution.


Example #4: Convert 0.02515 ft3 to cm3

Solution #1:

1) Convert 1 foot to inches, then centimeters:

1 foot = 12 inches (by definition)

12 inches x (2.54 cm / 1 inch) = 30.48 cm

2) Calculate what one cubic foot would be in cubic centimeters:

30.48 cm x 30.48 cm x 30.48 cm = 28317 cm3

3) Determine answer to problem:

0.02515 ft3 x (28317 cm3 / ft3) = 712.2 cm3

Now, go to Google and type this in the search box:

convert 0.02515 cubic foot to cubic centimeters

and then press "Return."

Solution #2:

1) Think of the volume in this manner:

0.02515 ft3 = 0.02515 ft x 1 ft x 1 ft

2) Now, convert it to cubic inches:

(0.02515 ft x 12 in/ft) x 12 in x 12 in = 43.4592 in3

3) Think of 43.4592 in3 this way:

43.4592 in3 = 43.4592 in x 1 in x 1 in

4) Convert to cm:

(43.4592 in x 2.54 cm/in) x 2.54 cm x 2.54 cm = 712.2 cm3

Solution #3:

Setting it up DA-Style:

    123 in3   2.543 cm3  
0.02515 ft3 x ––––––– x ––––––– = 712.2 cm3
    ft3   1 in3  

Note the cube on the 12 in 123 in3/ 1 ft3 and the cube on the 2.54 in 2.543 cm3 / 1 in3


Example #5: The three dimensions of a box are measured to be 1.3 in, 3.4 in, and 5.9 in. What is the volume of the box in liters?

Solution:

1) Calculate volume of the box in cm3:

(1.3 in x 3.4 in x 5.9 in) times (2.54 cm / in)3

427.342 cm3

2) Convert cm3 to mL, then to liters:

427.342 cm3 = 427.342 mL <--- because 1 cm3 = 1 mL

427.342 mL times (1 L / 1000 mL) = 0.427 L

To two sig figs, this is 4.3 x 10-1 L

I had to use scientific notation for the answer because expressing the volume as 0.430 L would have been three sig figs.

Comment: for step one, I could have written this:

(1.3 in x 2.54 cm / in) x (3.4 in x 2.54 cm / in) x (5.9 in x 2.54 cm / in)

The above simply converts each inch value separately. In step one above, the units would have been written this way:

(in3 times cm3/in3)


Example #6: Calculate the displacement (in cubic inches) of a 5.70 L engine.

Solution:

1) I'll set it up in dimensional analysis (in two different presentation stypes) and then explain the pieces:

(5.70 L) (1000 mL / 1 L) (1 cm3 / 1 mL) (1 inch / 2.54 cm)3 = 345 in3
  1000 mL   1 cm3   13 in3  
5.70 L x ––––––– x ––––––– x ––––––– = 345 in3
  1 L   1 mL   2.543 cm3  

For the individual pieces, I'll use the first style just above.

2) The first calculation converts L to mL:

(5.70 L) (1000 mL / 1 L) = 5700 mL

3) The second calculation converts mL to cm3:

(5.70 L) (1000 mL / 1 L) (1 cm3 / 1 mL) = 5700 cm3

4) The third calculation converts cm3 to in3:

(5.70 L) (1000 mL / 1 L) (1 cm3 / 1 mL) (1 inch / 2.54 cm)3 = 345 in3

Note that the last conversion factor (the 1 inch / 2.54 cm) is cubed. This is because a volume (the cm3) has three dimensions, each of which must be converted from cm to inch. Consequently, the conversion is cubed because it gets used three times, once for each of the three dimensions.

Comment: If you were converting an area (like cm2 to in1), the conversion factor (the 1 inch / 2.54 cm) would be squared.


Example #7: A roll of aluminum foil was recently purchased. The roll was 24.0 in wide, 500.0 ft long and 0.096 mil thick. The cost was $59.48. Determine the cost (in pennies) per aluminum atom.

Solution:

1) Convert 24.0 in, 500.0 ft, and 0.096 mil to centimeters:

(24.0 inch) (2.54 cm / inch) = 60.96 cm
(500.0 feet) (12 inch / ft) (2.54 cm / inch) = 15240 cm
(0.096 mil) (1 inch / 1000 mil) (2.54 cm / inch) = 0.00024384 cm

Note the unit of mil. 1 mil is one one-thousandth of an inch. It is NOT a millimeter, whose unit is mm.

2) Determine the volume of the foil:

(60.96 cm) (15240 cm) (0.00024384 cm) = 226.535 cm3

3) Determine the mass of Al present:

(226.535 cm3) (2.70 g/cm3) = 611.6445 g

Note the use of the density, a value not provided in the problem. Up above, I converted the volume to cm3 because I knew I would use the density.

4) Convert the mass to moles. This uses another "missing" value, the atomic weight of aluminum.

611.6445 g / 26.98 g/mol = 22.67 mol

5) Use Avogadro's Number (also not mentioned in the problem) and the moles to get the number of Al atoms.

(22.67 mol) (6.022 x 1023 atoms/mol) = 1.365 x 1025 atoms

6) Divide the money by the number of atoms to get the per atom cost.

59.48 dollars equals 5948 cents

5948 cents / 1.365 x 1025 atoms = 4.36 x 10¯22 cents per atom


Example #8: A copper ingot has a mass of 2.94 lb. If the copper is drawn into wire whose diameter is 3.73 mm, how many inches of copper wire can be obtained from the ingot? The density of copper is 8.94 g/cm3

Solution:

1) We need to know the volume of the copper:

(2.94 lb) (453.6 g/lb) (1 cm3/8.94 g) = the volume in cm3

    453.6 g   1 cm3  
2.94 lb x ––––––– x ––––––– = 149.17 cm3
    1 lb   8.94 g  

Notice how the density is set up. We are dividing the mass of copper (in grams) by the density in order to obtain the volume.

This is the volume of the wire with the 3.73 mm diameter.

2) The wire can be thought of as a cylinder. We want 'h,' the height of the cylinder:

V = πr2h

h = V / (πr2)

h = 149.17 cm3 / [(3.14159) (0.1865 cm)2]

h = 149.17 cm3 / 0.10927 cm2

h = 1365.15 cm

The 0.1865 cm is the 3.73 mm converted to cm and divided by 2 so as to have the radius.

3) Convert cm to inches:

(1365.15 cm) (1 in./2.54 cm) = 537 in (to three sig figs)

Example #9: Zippy the snail traveled 3.20 feet in 8.00 hours. What is his speed in cm per hour?

Solution:

(3.20 ft / 8.00 hr) <--- given in the problem

(3.20 ft / 8.00 hr) (12.0 inch / 1 ft) <--- this changes feet to inches because I have memorized the relationship between cm and inch. You could combine the two factos below into a cm/ft conversion if you so desired.

(3.20 ft / 8.00 hr) (12.0 inch / 1 ft) (2.54 cm / 1 inch) <--- notice how feet cancels in the first two factors and inch cancels between the second and third factors

Written in dimensional analysis style:

3.20 ft   12 in   2.54 cm  
––––––– x ––––––– x ––––––– = 12.192 cm/hr
8.00 hr   ft   in  

We don't have to bother with converting hours. If the problem was asking for cm/second, then we'd have to include more conversions.

The answer is 12.2 cm/hr, when rounded to three significant digits.


Example #10: What is 3.25 lb/ft3 converted to g/cm3?

Solution:

1) Here is the entire solution, set up in dimensional analysis style:

3.25 lb   453.6 g   1 ft3   1 in3  
––––––– x ––––––– x ––––––– x ––––––– = 21.45 g/cm3
ft3   lb   123 in3   2.543 cm3  

Notice that I tend to not use a 1 with units in the denominator, but I do use a 1 with units in the numerator. In other words, a 1 is assumed to be present in front of a unit with no numerical value shown.

What follows is a brief description of each conversion is doing.

2) The first conversion is pounds to grams:

(3.25 lb/ft3) (453.6 g / lb) = 1474.2 g/ft3

Note how the intermediate values do not appear in the DA set up. Make sure, if you write out a setp-by=step solution, to include some guard digits. Do not round off to the correct number of significant figures until the end.

3) The second conversion is ft3 to in3:

(1474.2 g/ft3) (1 ft3 / 123 in3) = 0.853125 g / in3

4) The third conversion is in3 to cm3:

(0.853125 g / in3) (1 in3 / 2.543 cm3) = 0.05206 g/cm3

To three sig figs, 0.0521 g/cm3

5) In the DA setup, you might ask why not go directly from 1 ft3 to cm3? You can do that if so desired. I happen to have memorized that there are 12 inches in one foot and 2.54 cm in one inch, so I used what I had memorized. Since 1 foot = 30.48 cm, one ft3 will equal 30.483 cm3 and this results:

3.25 lb   453.6 g   1 ft3  
––––––– x ––––––– x ––––––––– = 0.0521 g/cm3
ft3   1 lb   30.483 cm3  

Bonus Problem: How many slugs are there in 54.29 kg?

Hint: use Google to convert slugs to kilograms

A slug is a very old-school unit of mass. The ChemTeam has heard of it (as well as dyne and erg), but has never used it in a computation, either as a student or as a teacher.

The ChemTeam would shout "Erg!" when putting the shot in high school and college. He thought this was quite funny, knowing that the erg is a unit of energy.

Sometimes, the ChemTeam just cracks himself up.


Metric-English Conversion Problems #1-10     Metric-English Conversion Problems #26-60
Metric-English Conversion Problems #11-25     Return to Metric Menu