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Metric conversion where only one unit is converted

Go to 10 two-unit metric problems

Doing this type of problem is simply a succession of conversions from one unit to another. You first convert one side of the fraction, say, the numerator, then you do the denominator.

Often, a teacher will present these solutions as one long string of conversions. You also see this type of presentation in textbooks. It can be quite confusing when you first see it.

This technique is called "dimensional analysis" (the older term), with "factor label method" being the newer term. DA can also called "unitary conversions," or "unitary rates." The word unitary comes from the idea that the numerator and the denominator in a conversion fact both describe the same quantity.

As an example, take this conversion factor:

1000 mL / 1 L

Both 1000 mL and 1 L describe the same-sized volume, so 1000 mL / 1 L is referred to as a unitary rate. Since 1000 mL and 1 L describe the same volume, we can think of 1000 mL / 1 L as being like multiplying some number by 1. The description of the volume changes units, but it still describes the same sized volume.

The ChemTeam tends to present two-unit conversion problems as a sequence of one-step calculations. However, I will also reference the one-line type presentation that is often used. On a professional basis, I do not believe the one-line approach is the proper tool to use when teaching these types of problems. There are those that disagree with me.

Doing DA problems, to me, are like balancing equations or predicting products of a reaction. There are LOTS of little bits that you have to remember and, when that is the case, experience is really, really important. The problem you would face is to be able to read a one-line solution and back-track to the logic the writer used. That can be difficult, especially for a rookie.

Conclusion: lots of examples for you to study!

**Example #1:** Convert the speed of light (3.00 x 10^{8} m/sec) to km/year.

**Solution:**

We'll start with the numerator, since that's an easy, one step conversion.

This gives an answer of 3.00 x 10^{5} km/sec.

Now, we have to focus on converting seconds to years. This is done in a step-by-step manner. For example, I happen to have memorized that there are 3600 seconds in one hour. So, we do that conversion.

Continuing the calculations, we move step-by-step to days and then to years (we can skip months, since we know how many days there are in a year.

Converting to scientific notation and rounding to three significant figures, we get 9.46 x 10^{12} km/yr as the answer.

If I were to present it as a one-line type calculation (the usual presentation form in dimensional analysis), it would be this:

3.00 x 10^{8}m/s x (1 km / 1000 m) x (3600 s / hr) x (24 hr / day) x (365 day / yr)

One advantage to the above presentation is that you simply carry out the steps in sequence (divide by 1000, multiply by 3600, mult by 24 and mult by 365) on your calculator and then round off.

Doing it step-by-step results in intermediate answers along the way, but I think it's better to teach the steps rather than confront a student with the one-line setup right from the start of instruction.

Notes on variations of the above problem:

1) Notice that I used 365 days rather than 365.25. Using the latter figure results in an answer of 9.46728 x 10

^{12}km/yr, which rounds off to 9.47 x 10^{12}km/yr.2) This problem can start with cm/s rather than m/s. The speed of light in cm/s is 3.00 x 10

^{10}cm/s.3) Often, this problem ends in km/hr. Another common question asks for the conversion from cm/s to km/hr.

**Example #2:** Light travels at a speed of 3.00 x 10^{10} cm/s. What is the speed of light in kilometers/hours?

**Solution:**

1) Convert cm/s to km/s:

3.00 x 10^{10}cm/s times (1 m / 100 cm) times (1 km / 1000 m) = 3.00 x 10^{5}km/s

2) Convert seconds to hours:

3.00 x 10^{5}km/s times (60 s / 1 min) times (60 min / 1 hr) = 1.08 x 10^{9}km/hr

3) A slightly more compact version:

3.00 x 10^{10}cm/s times (1 km / 10^{5}cm) = 3.00 x 10^{5}km/s3.00 x 10

^{5}km/s times (3600 s / 1 hr) = 1.08 x 10^{9}km/hr

**Example #3:** Convert 64.3 g/mL to its equivalent in kg/L.

**Solution:**

1) Convert grams to kilograms:

64.3 g/mL x (1 kg/1000 g) = 0.0643 kg/mL

2) Convert mL to L:

0.0643 kg/mL x (1000 mL/L) = 64.3 kg/L

Comment: teachers like to ask this type of question on tests. The ChemTeam did!

**Example #4:** A cylindrical piece of metal is 4.50 dm in height with radius of 5.50 x 10¯^{5} km.

a) Calculate the volume in milliliters to the correct significant figures givenV = π rfor a cylinder.^{2}h

b) Calculate the volume in mm^{3}

c) Calculate the density in units g/L to the correct number of significant figures given it has a mass of 6.54 x 10^{5}grams

The key to solving part a is to remember that cm^{3} and mL are the same volume, so 1 cm^{3} = 1 mL. So, if we convert both measurements of the cylinder to cm, like this:

then, all we need to do is plug our numbers into the volume formula provided to get cm^{3}. We find that the height converts to 45.0 cm and the radius to 5.50 cm. Solving for the volume gives 4276.5 cm^{3}. Converting to proper sig figs and mL gives 4280 mL.

Part b: the unit we need on the height and radius is mm, so convert 45.0 cm to 450 mm and 5.50 cm to 55.0 mm. Then plug back into the same formula as part a.

Or, you could notice that both numbers got increased by a factor of 10 and then within the volume formula, there is a total factor increase of 10^{3}.

That means the answer to part b is the answer to part a times 1000 or 4.28 x 10^{6} mm^{3}.

Part c: we have to take our mL volume to liters. You SHOULD already know how do that. 4.28 L is the answer. So we then divide 6.54 x 10^{5} by 4.28 L to get 1.53 x 10^{5} grams / L.

**Example #5:** Convert 4.09 x 10^{-6} kg/L to mg/cm^{3} using dimensional analysis.

**Solution:**

When dimensional analysis is specified in a problem, the usual answer desired is in the form of all the conversions gathered together into one line. I will build the final answer up one conversion at a time. Each comment with an arrow is about the last conversion in each line.

4.09 x 10^{-6}kg/L x (1000 g / kg) <--- converts kg to g4.09 x 10

^{-6}kg/L x (1000 g / kg) x (1000 mg / 1 g) <--- converts g to mg4.09 x 10

^{-6}kg/L x (1000 g / kg) x (1000 mg / 1 g) x (1 L / 1000 mL) <--- converts L to mL4.09 x 10

^{-6}kg/L x (1000 g / kg) x (1000 mg / 1 g) x (1 L / 1000 mL) x (1 cm^{3}/mL) <--- converts mL to cm^{3}the answer is 0.00409 mg/cm

^{3}

Notice that the above conversion converted through the base unit of grams, as in kg to g, then g to mg. You can combine those two conversions if so desired:

4.09 x 10^{-6}kg/L x (10^{6}mg / kg) x (1 L / 1000 mL) x (1 cm^{3}/mL) = 0.00409 mg/cm^{3}

The difference is stylistic only. Some teachers prefer one method over the other, others do not care. Be sure to check what your teacher desires.

**Example #6a:** Convert 303 mi/hr to feet/min.

**Solution:**

V = ( 303 mi/hr ) ( 1 hr / 60 min ) ( 5280 ft / mile ) = 26600 ft/min

303 mi 1 hr 5280 ft ––––––– x ––––––– x ––––––– = 26664 ft/min 1 hr 60 min 1 mi

Comments:

The hr/min factor converts 303 mi per hr to 5.05 mi per min.The ft/mi factor converts 5.05 mi per min to 26664 ft per min.

The factors used algebraically cancel units to give the units wanted.

The final answer is 26700 ft/s. It has been rounded off to three significant figures. Note that this example uses English units. The principles of converting are the same as with metric units.

**Example #6b:** Convert 303 mi/hr to feet/second.

**Solution:**

The dimensional analysis set-up will be presented without comment.

303 mi 1 hr 1 min 5280 ft ––––––– x ––––––– x ––––––– x ––––––– = 444 ft/s <--- rounded to 3 sig figs 1 hr 60 min 60 sec 1 mi

**Bonus Example:** The SI unit for density is kg/m^{3}. Convert the density of platinum (21450 kg/m^{3}) to the more commonly-used unit of g/cm^{3}

**Solution:**

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

(21450 kg/m^{3}) (1000 g / 1 kg) = 21450000 g/m^{3}

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

(21450000 kg/m^{3}) (1 m^{3}/ 100^{3}cm^{3}) = 21.45 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:

(21450 kg/m^{3}) (1000 g / 1 kg) (1 m^{3}/ 100^{3}cm^{3}) = 21.45 g/cm^{3}Note the interim values/units such as 21450000 g/m

^{3}do not appear in a one-line dimensional analysis presentation.

Textbooks will often present a dimensional analysis set-up in this manner:

21450 kg 1000 g 1 m ^{3}––––––– x ––––––– x ––––––– = 21.45 g/cm ^{3}1 m ^{3}1 kg 100 ^{3}cm^{3}

Your teacher may require it in that manner as well. One of the advantages to the above set-up is that it's much more obvious which units cancel. For example, you can clearly see the kg in the numerator of the first factor and in the denominator of the second factor.

Comment: it is easy to imagine a situation (test or homework) where, in the problem, you are given the density of a substance in units of kg/m^{3} but, in the problem solution, you must use the density in units of g/cm^{3}. Consequently, I recommend that the above conversion be in your "bag of tricks."

By the way, please notice that the net effect of the above conversion is to divide the kg/m^{3} value by 1000 to get the g/cm^{3} value. if you are not required to show the conversion as I did above, you can use the 'divide by 1000" step as a convenient shortcut.

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