CK-12-Chemistry Intermediate

(Marvins-Underground-K-12) #1

3.2. Unit Conversions http://www.ck12.org


You can participate in an interactive version of this cube at http://www.dlt.ncssm.edu/core/Chapter1-Introduction/Chapter
1-Animations/M3_DM3_CM3.html.


Sample Problem 3.4: Derived Unit Conversion


Convert 3.6× 108 mm^3 to mL.


Step 1: List the known quantities and plan the problem.


Known



  • 1 m = 1000 mm

  • 1 mL = 1 cm^3

  • 1 m = 100 cm


Unknown



  • 3.6 mm^3 =? mL


This problem requires multiple steps and the technique for converting with derived units. Simply proceed one step
at a time: mm^3 to m^3 to cm^3 = mL.


Step 2: Calculate.


3 .6 mm^3 ×

( 1 m
1000 mm

) 3


×


( 100 cm
1 m

) 3


×^11 cmmL 3 = 0 .0036 mL

Numerically, the steps are to divide 3.6 by 10^9 , followed by multiplying by 10^6. Alternatively, you can shorten the
calculation by one step if you first determine the conversion factor between mm and cm. Using the fact that there
are 10 mm in 1 cm, you can avoid the intermediate step of converting to meters.


3 .6 mm^3 ×

( 1 cm
10 mm

) 3


× 11 cmmL 3 = 0 .0036 mL

Both conversion methods give the same result of 0.0036 mL.


Step 3: Think about your result.


Cubic millimeters are much smaller than cubic centimeters, so the final answer is much less than the original number
of mm^3.


Practice Problems


  1. Perform the following conversions.
    (a) 0.00722 km^2 to m^2
    (b) 129 cm^3 to L
    (c) 4.9× 105 μm^3 to mm^3


You can find more help with dimensional analysis by watching this video at http://www.chemcollective.org/stoich/dimens
ionalanalysis.php.


You can download an instructional packet that explains dimensional analysis step by step and provides practice
problems at https://docs.google.com/open?id=0B_ZuEGrhVEfMS09WeUtOMFNTaFk.

Free download pdf