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Metric conversion where only one unit is converted

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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 fact that the numerator and the denominator in a conversion 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 the hallmark of dimensional analysis. 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/s) to km/year.

**Solution:**

1) We'll start with the numerator, since that can be done in a one step conversion.

3.00 x 10 ^{8}m1 km –––––––––– x –––––– = 3.00 x 10 ^{5}m/s1 s 1000 m

2) Now, we turn to converting seconds to years. This I will do in a step-by-step manner. I happen to have memorized that there are 3600 seconds in one hour. So, we start with that conversion.

3.00 x 10 ^{5}km3600 s ––––––––––– x ––––– = 1.08 x 10 ^{9}km/hr1 s 1 hr

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

1.08 x 10 ^{9}km24 hr ––––––––––– x ––––– = 2.592 x 10 ^{10}km/day1 hr 1 day

2.592 x 10 ^{10}km365.25 day ––––––––––––– x ––––––––– = 9.47 x 10 ^{12}km/yr (to three sig figs)1 day 1 yr

4) 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}m1 km 3600 s 24 hr 365.25 day –––––––––– x ––––––– x ––––––– x ––––––– x ––––––– = 9.47 x 10 ^{12}km/yr1 s 1000 m 1 hr 1 day 1 yr On the Internet, it may also be seen this way:

3.00 x 10

^{8}m/s x (1 km / 1000 m) x (3600 s / hr) x (24 hr / day) x (365.25 day / yr) = 9.47 x 10^{12}km/yr

5) 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 365.25) on your calculator and then round off.

6) 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 dimensional analysis method right from the start of instruction.

Notes on variations of the above problem:

1) Notice that I used 365.25 days rather than 365. Using the latter figure results in an answer of 9.4608 x 10

^{12}km/yr, which rounds off to 9.46 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. Consequently, the question you could see might ask for the conversion from cm/s to km/hr. As in, right now . . . .

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

2) Convert seconds to hours:

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

3) A slightly more compact version:

(3.00 x 10^{10}cm/s) (1 km / 10^{5}cm) = 3.00 x 10^{5}km/s(3.00 x 10

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

4) Done in DA (dimensional analysis) style:

3.00 x 10 ^{10}cm1 m 1 km 60 s 60 min –––––––––– x ––––––– x ––––––– x ––––––– x ––––––– = 1.08 x 10 ^{9}km/hr1 s 100 cm 1000 m 1 min 1 hr The two length conversions could be combined (1 km = 10

^{5}cm) and the two time conversions could be combined (1 hr = 3600 s).

**Example #3:** Convert 6.43 g/mL to its equivalent in kg/L.

**Solution:**

1) Convert grams to kilograms:

6.43 g/mL x (1 kg/1000 g) = 0.00643 kg/mL

2) Convert mL to L:

0.0643 kg/mL x (1000 mL/L) = 6.43 kg/L

3) Dimensional analysis:

6.43 g 1 kg 1000 mL –––––– x –––––– x ––––––– = 6.43 kg/L 1 mL 1000 g 1 L

Comment: teachers like to ask this question on the test.

**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}

**Solution to (a):**

1) 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, convert both measurements of the cylinder to cm::

4.50 dm 10 cm ––––––– x ––––– = 45.0 cm 1 1 dm

5.50 x 10¯ ^{5}km10 ^{5}cm––––––––––– x ––––– = 5.50 cm 1 1 km

2) Plug our numbers into the volume formula provided to get cm^{3}.

V = π r^{2}hV = (3.14159) (5.50 cm)

^{2}(45.0 cm)V = 4276.5 cm

^{3}Rounding to three sig figs and noting that 1 cm

^{3}= 1 mL, we have this for the final answer:V = 4280 mL = 4.28 x 10

^{3}mL

**Solution to (b):**

1) 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 volume formula:

V = (3.14159) (55.0 mm)^{2}(450 mm)V = 4276489 mm

^{3}To three sig figs, we have 4.28 x 10

^{6}mm^{3}

2) 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} (because one of factor of 10 increase was squared to give a factor of 100 increase). That means the answer to part (b) is the answer to part (a) times 1000, resulting in 4.28 x 10^{6} mm^{3}.

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

**Solution:**

1) 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}

2) 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}

3) Here's the most-common way DA solutions are formatted:

4.09 x 10¯ ^{6}kg10 ^{6}mg1 L 1 cm ^{3}––––––––––– x –––––– x ––––––– x ––––– = 0.00409 mg/cm ^{3}1 L 1 kg 1000 mL 1 mL

4) Some teachers prefer the DA method for homework and test answers. Some prefer the steps to be separated (with intermediate answers shown). Others do not care. Be sure to check what your teacher desires.

**Example #6a:** Convert 303.0 mi/hr to feet/min.

**Solution:**

V = (303.0 mi/hr) (1 hr / 60 min) (5280 ft / mile) = 26600 ft/min

303 mi 1 hr 5280 ft –––––– x ––––– x –––––– = 26660 ft/min (to 4 sig figs) 1 hr 60 min 1 mi

Comments:

The hr/min factor converts 303.0 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 26660 ft/s. It has been rounded off to four significant figures.

Note that this example uses English units. The principles of converting are the same as with metric units.

The hr/min conversion as well as the foot/mile conversion are defined amounts. As such, they play no role in determining significant figures.

**Example #6b:** Convert 303.0 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 ––––––– = 444.4 ft/s <--- 4 sig figs 1 hr 60 min 60 sec 1 mi

**Example #7:** Convert 2113 km/h into cm/s.

**Solution:**

1) Do the hour to second conversion:

2113 km 1 hr ––––––– x ––––– = 0.5869444 km/s 1 hr 3600 s

2) Convert km to m, then convert m to cm (as opposed to converting km directly to cm in one step):

0.5869444 km 1000 m ––––––––––– x –––––– = 586.9444 m/s s 1 km

3) Now, the m to cm conversion:

586.9444 m 100 cm –––––––––– x –––––– = 58694.44 cm/s = 5.869 x 10 ^{4}cm/s (to 4 sig figs)s 1 m

4) Here's the full conversion, with km being directly converted to cm:

2113 km 1 hr 10 ^{5}cm––––––– x ––––– x ––––– = 5.869 x 10 ^{4}cm/s1 hr 3600 s 1 km

**Example #8:** How many grams of lead are there in a lead brick 5.00 cm by 13.0 cm by 24.0 cm? The density of lead is 11300 kg/m^{3}.

**Discussion:**

We could change the cm to m and calculate the volume in m^{3}, then use the density to get the mass of lead. Another path would be to change the density to use cm^{3}and then calculate the volume in cm^{3}and thence to the mass.I think I will do both!!

**Solution where cm is changed to m first:**

1) Change cm to m:

(5.00 cm) (1 m / 100 cm) = 0.0500 m

(13.0 cm) (1 m / 100 cm) = 0.130 m

(24.0 cm) (1 m / 100 cm) = 0.240 m

2) Calculate the density in m^{3}:

(0.0500 m) (0.130 m) (0.240 m) = 0.00156 m^{3}

3) Determine mass in kg, then g:

(11300 kg/m^{3}) (0.00156 m^{3}) = 17.628 kg(17.628 kg) (1000 g / kg) = 17628 g

to three sig figs, 17600 g

**Solution where m ^{3} is changed to cm^{3} first:**

1) Convert density to g/cm^{3}:

11300 kg 1 m ^{3}1000 g ––––––– x ––––––– x –––––– = 11.3 g/cm ^{3}m ^{3}(100 cm) ^{3}1 kg Notice that I included both conversions into this step.

2) Determine volume of the lead brick:

(5.00 cm) (13.0 cm) (24.0 cm) = 1560 cm^{3}

3) Determine mass in grams:

(11.3 g/cm^{3}) (1560 cm^{3}) = 17628 gTo three sig figs, 17600 g

**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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