This preview shows page 1. Sign up to view the full content.
Unformatted text preview: even perform a calculation without
requiring that you have covered that subject in class! For instance, suppose you have a
gold bar that measures about 9 inches by 5 inches by 2 inches. Given the density of gold
is 19.3 g/mL, how many pounds will this bar weigh? Dakota State University page 221 of 232 Factor Label Method General Chemistry I and II Lab Manual GIVEN: gold bar 9"x5"x2", 19.3 g/mL
FIND: pounds
CONVERSION FACTORS: We will need 1 mL = 1 cm3 , 2.54 cm/in,. 2.2046
lb/kg, 1,000 g/kg. In a problem, you would have a table of conversion factors that you
can choose from.
First, let's find the volume independently of the factor label method. We know we need
volume, because mL, milliliter, is a volume. We have a height, width and length, and the
volume of the gold bar is height times width times length, or
9inch * 5inch * 2inch = 90inch 3
Notice that even here I used labels. Now that we have our volume, let's solve this
problem using the factor label method.
3 mL 19.3g
kg
2.2046 pound 2.54cm 90inch * *
*
= 62.8 pound * 3* inch cm
mL 1000g
kg
3 A gold bar weighs over 60 pounds! Consider THAT the next time you see a movie
where they are handling gold bars single handedly!!
Notice above that we needed to convert from in3 to cm3 , but all we had was the
conversion factor 2.54 cm/in. If we had multiplied 90 in3 * (2.54 cm/in), we would have
seen that only ONE of the "inch" units would have canceled, leaving us with 228.6 in2 cm,
a meaningless number. To cancel all three inches in the in3 term, we had to cube our
conversion factor 2.54 cm/in, as shown above. Another approach would have been to
cube this number before hand and use the conversion factor 16.387 cm3 /in3 . Carry out
this calculation and be sure that your answer matches that shown in order to be sure that
you are "cubing" correctly!
Example 3: The nearest star is about 4.5 light years away (one light year is the distance
light will travel in a vacuum in one year). The Apollo spacecraft travelled about as fast
as a bullet, which we can approximate as 1,200 ft/second. Suppose that we stock a craft
which travels at 1,200 ft/sec with a crew of men and women who are to create a new
generation to take control of the ship every 20 years. How many new generations will be
required before reaching AlphaCentauri?
GIVEN: 4.5 light years, 1,200 ft/sec, 1 generation/20 years
FIND: generations
CONVERSION FACTORS: we will need 3.0x1010 cm/s (speed of light), 365.25
days/year, 24 hr/day, 60 min/hr, 60 sec/min, 2.54 cm/in, 12 in/ft
OK, first, let's see what distance we are talking, thereby converting this into a standard
distance/speed problem. Dakota State University page 222 of 232 Factor Label Method 4.5year * General Chemistry I and II Lab Manual 365.25day 24hour 60 min 60 sec 3.0x1010 cm
*
*
*
*
= 4.3x1018 cm
year
day
hour
min
sec Now, let's finish up the problem.
4.3x1018 cm * inch
ft
sec
min
hr
day
year
generation
*
*
*
*
*
*
*
=
2.54cm 12inch 1,200 ft 60 sec 60 min 24hour 365.25day
20year 186,000generations
This problem was inspired by an old episode of Star Trek....
View
Full
Document
This note was uploaded on 09/18/2012 for the course CHEMISTRY 1010 taught by Professor Kumar during the Fall '11 term at WPI.
 Fall '11
 Kumar
 Chemistry, pH, The Crucible, Dakota State University

Click to edit the document details