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 not 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 is 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 cm2) Based on that, I propose to first change km to cm (which is a common Metric-only conversion):

1.000 km times (103) Now, the conversion to inches:^{5}cm / km) = 1.000 x 10^{5}cm

1.000 x 10^{5}cm times (1 inch / 2.54 cm) = 39370 inchSince 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 10^{5} feet to centimeters

**Solution:**

1) Convert feet to inches:

4.04 x 10^{5}feet x (12 inches / ft) = 4.848 x 10^{6}in

2) Convert inches to centimeters:

4.848 x 10^{6}in x (2.54 cm / in) = 1.23 x 10^{7}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 10 ^{7}nm1.000 yd x ––––– x –––––– x –––––––– x ––––––– = 9.144 x 10 ^{8}nmyd ft in cm

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

**Example #4:** Convert 0.02515 ft^{3} to cm^{3}

**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 cm^{3}

3) Determine answer to problem:

0.02515 ft^{3}x (28317 cm^{3}/ ft^{3}) = 712.2 cm^{3}

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 ft^{3}= 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 in^{3}

3) Think of 43.4592 in^{3} this way:

43.4592 in^{3}= 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 cm^{3}

**Solution #3:**

Setting it up DA-Style:

12 ^{3}in^{3}2.54 ^{3}cm^{3}0.02515 ft ^{3}x ––––––– x ––––––– = 712.2 cm ^{3}ft ^{3}1 in ^{3}

Note the cube on the 12 in 12^{3} in^{3}/ 1 ft^{3} and the cube on the 2.54 in 2.54^{3} cm^{3} / 1 in^{3}. Here's another way to show the cube:

(12 in) ^{3}(2.54 cm) ^{3}0.02515 ft ^{3}x ––––––– x ––––––– = 712.2 cm ^{3}ft ^{3}1 in ^{3}

**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 cm^{3}:

(1.3 in x 3.4 in x 5.9 in) times (2.54 cm / in)^{3}427.342 cm

^{3}

2) Convert cm^{3} to mL, then to liters:

427.342 cm^{3}= 427.342 mL <--- because 1 cm^{3}= 1 mL <--- You should memorize this conversion.427.342 mL times (1 L / 1000 mL) = 0.427 L

To two sig figs, this is 4.3 x 10

^{-1}LI 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:

(in

^{3}times cm^{3}/in^{3})

**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 styles) and then explain the pieces:

(5.70 L) (1000 mL / 1 L) (1 cm^{3}/ 1 mL) (1 inch / 2.54 cm)^{3}= 345 in^{3}

1000 mL 1 cm ^{3}1 ^{3}in^{3}5.70 L x ––––––– x ––––––– x ––––––– = 345 in ^{3}1 L 1 mL 2.54 ^{3}cm^{3}

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 cm^{3}:

(5.70 L) (1000 mL / 1 L) (1 cm^{3}/ 1 mL) = 5700 cm^{3}

4) The third calculation converts cm^{3} to in^{3}:

(5.70 L) (1000 mL / 1 L) (1 cm^{3}/ 1 mL) (1 inch / 2.54 cm)^{3}= 345 in^{3}

Note that the last conversion factor (the 1 inch / 2.54 cm) is cubed. This is because a volume (the cm^{3}) 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 cm^{2} to in^{2}), 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 cmNote the unit of mil. 1 mil is one one-thousandth of an inch. It is NOT a millimeter, whose symbol is mm.

2) Determine the volume of the foil:

(60.96 cm) (15240 cm) (0.00024384 cm) = 226.535 cm^{3}

3) Determine the mass of Al present:

(226.535 cm^{3}) (2.70 g/cm^{3}) = 611.6445 gNote the use of the density, a value not provided in the problem. Up above, I converted the volume to cm

^{3}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 10^{23}atoms/mol) = 1.365 x 10^{25}atoms

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

59.48 dollars equals 5948 cents5948 cents / 1.365 x 10

^{25}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/cm^{3}

**Solution:**

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

(2.94 lb) (453.6 g/lb) (1 cm^{3}/8.94 g) = the volume in cm^{3}

453.6 g 1 cm ^{3}2.94 lb x ––––––– x ––––––– = 149.17 cm ^{3}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 = πr^{2}hh = V / (πr

^{2})h = 149.17 cm

^{3}/ [(3.14159) (0.1865 cm)^{2}]h = 149.17 cm

^{3}/ 0.10927 cm^{2}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 factors 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/ft^{3} converted to g/cm^{3}?

**Solution:**

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

3.25 lb 453.6 g 1 ft ^{3}1 in ^{3}––––––– x ––––––– x ––––––– x ––––––– = 21.45 g/cm ^{3}ft ^{3}lb 12 ^{3}in^{3}2.54 ^{3}cm^{3}

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/ft^{3}) (453.6 g / lb) = 1474.2 g/ft^{3}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 ft^{3} to in^{3}:

(1474.2 g/ft^{3}) (1 ft^{3}/ 12^{3}in^{3}) = 0.853125 g / in^{3}

4) The third conversion is in^{3} to cm^{3}:

(0.853125 g / in^{3}) (1 in^{3}/ 2.54^{3}cm^{3}) = 0.05206 g/cm^{3}To three sig figs, 0.0521 g/cm

^{3}

5) In the DA setup, you might ask why not go directly from 1 ft^{3} to cm^{3}? 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 ft^{3} will equal 30.48^{3} cm^{3} and this results in:

3.25 lb 453.6 g 1 ft ^{3}––––––– x ––––––– x ––––––––– = 0.0521 g/cm ^{3}ft ^{3}1 lb 30.48 ^{3}cm^{3}

**Bonus Problem #1:** Convert the speed of light from meters/second to furlongs/fortnight.

**Solution #1:**

1) A favorite "trick" of teachers is to use unusual units. Furlongs and fortnights are units of length and time, respectively. Let us do some looking up:

A furlong is 220 yd. It is a popular length unit in horse racing.A fortnight is 14 days.

2) Some pre-planning:

(a) It seems pretty obvious that we should go from seconds to minutes to days to fortnights.

(b) For the meters, I plan to go first to centimeters because I can then use one inch = 2.54 cm to move from metric measurements to English measurements. I'll then move to yards and thence to furlongs.

3) First, I will do just the time unit:

3.00 x 10 ^{8}m60 s 60 min 24 hr 14 day –––––––––– x ––––– x –––––– x ––––– x –––––––– = 3.6288 x 10 ^{14}m/fortnight <--- note the meters is still there1 s 1 min 1 hr 1 day 1 fortnight

4) Now, the length unit:

3.6288 x 10 ^{14}m100 cm 1 inch 1 foot 1 yard 1 furlong ––––––––––––– x ––––– x –––––– x ––––– x –––––––– x –––––––– = 1.80 x 10 ^{12}furlong/fortnight1 fortnight 1 m 2.54 cm 12 inch 3 feet 220 yard

**Solution #2:**

1) In this solution, I'm going to first convert 1 m/s to its equivalent in furlongs/fortnights:

1 m 86400 s 14 day 1 inch 1 foot 1 yard 1 furlong –––– x ––––––– x –––––––– x –––––––– x ––––– x ––––– x ––––––– = 6012.885 furlongs/fortnight 1 s 1 day 1 fortnight 0.0254 m 12 inch 3 feet 220 yard Note the use of 1 inch = 0.0254 m rather than the more common 1 inch = 2.54 cm.

2) The next step is to multiply both sides by 3.00 x 10^{8}. This is because 1 m/s times 3.00 x 10^{8} equals the speed of light:

1 m/s = 6012.885 furlongs/fortnight(1 m/s) (3.00 x 10

^{8}) = (6012.885 furlongs/fortnight) (3.00 x 10^{8})3.00 x 10

^{8}m/s = 1.80 x 10^{12}furlongs/fortnight

**Bonus Problem #2:** Convert 35.00 cm^{2}/s^{2} to mi^{2}/day^{2}

**Solution using dimensional analysis:**

1) The numerator:

(1 inch) ^{2}(1 foot) ^{2}(1 mi) ^{2}35.00 cm ^{2}x––––––– x ––––––– x ––––––– = 1.351358 x 10¯ ^{9}mi^{2}(2.54 cm) ^{2}(12 inch) ^{2}(5280 ft) ^{2}

2) The denominator:

(1 min) ^{2}(1 hr) ^{2}(1 day) ^{2}s ^{2}x––––––– x ––––––– x ––––––– = 1.33959 x 10¯ ^{10}day^{2}(60 s) ^{2}(60 min) ^{2}(24 hr) ^{2}Notice that I convert s

^{2}as if it is a numerator. The day^{2}value will be put into the denominator in the next step.

3) Divide the numerator by the denominator:

1.351358 x 10¯ ^{9}mi^{2}––––––––––––––––– = 10.09 mi ^{2}/day^{2}1.33959 x 10¯ ^{10}day^{2}

4) Three comments:

(a) The answer can be obtained using Google Convert.(b) Here's the problem answered on an "answers" website:

You just have to square everything, for example 1 sq ft is (12 x 12 or 144) sq in.numerator = [36 cm

^{2}] [(1 in / 2.54 cm)^{2}] [(1 ft / 12 in)^{2}] [1 mi / 5280 ft)^{2}] = N mi^{2}denominator = [s

^{2}] [(1 hr / 3600 s)^{2}] [(1 day / 24 hr)^{2}] = D day^{2}Every unit in the numerator cancels out but mi

^{2}, everything unit in the denominator cancels out except for day^{2}.1.4 x 10¯

^{9}/ 1.34 x 10¯^{10}which is about 10.4 mi^{2}/ day^{2}(c) The units of cm

^{2}/s^{2}and mi^{2}/day^{2}have no use in science (an area accelerating?). They are simply contrived units for the purpose of giving you a more complex problem to solve.

5) Here's the numerator and denominator calculation above, but done as one whole unit:

35.00 cm ^{2}(1 inch) ^{2}(1 foot) ^{2}(1 mi) ^{2}(60 s) ^{2}(60 min) ^{2}(24 hr) ^{2}–––––––– x ––––––– x ––––––– x ––––––– x –––––– x ––––––– x ––––––– = the answer s ^{2}(2.54 cm) ^{2}(12 inch) ^{2}(5280 ft) ^{2}(1 min) ^{2}(1 hr) ^{2}(1 day) ^{2}In this combined calculation, notice how the s

^{2}starts out in the denominator, so all the conversions utilized in step 2 above get flipped.

Metric-English Conversion Problems #1-10 | Metric-English Conversion Problems #26-60 |

Metric-English Conversion Problems #11-25 | Return to Metric Menu |