How do you find the first conversion in the factor label method?

1 Answer

You can start anywhere, but it is easiest to start with a measurement that has only one unit in it.

For example, you are given a problem:

21.5 cal = _ J

We know that 1 cal = 4.184 J

Using this information, we make two conversion factors. We take the ratio of the two units.

#"1 cal"/"4.184 J"# =1 or #"4.184 J"/"1 cal"# =1

Now you multiply by the conversion factor gives the final answer in joule.

If you multiply by the second conversion factor, the "cal" will cancel out and give the answer in "J".

21.5 cal × #"4.184 J"/"1 cal"# = 90.0 J

Now let's apply this method do a more complicated problem.

The label on a bag of chips reads 245 Cal. If the body stores 1 lb of fat for each
14.6 × 10³ kJ of excess energy consumed, how many bags of chips contain enough energy to result in 1 lb of body fat?

Where to start? You must convert the word problem into a mathematics problem. You must somehow get from

1 lb of body fat → → … → → #x# bags of chips (crisps?).

You have the conversion factors

  • 1 bag chips = 245 cal
  • 1 lb fat = 14.6 × 10³ kJ

You can convert lb fat → kJ and cal → bags of chips. There are some gaps, but you know some more conversion factors

  • 1 kJ = 1000 J
  • 1 cal = 4.184 J
  • 1 Cal = 1000 cal

We now have enough conversion factors to string together.

We start with what they asked us to convert, "1 lb of body fat". It contains only one unit, "lb". We write

1 lb fat × #("14.6 × 10"^3" kJ")/"1 lb fat" × "1000 J"/"1 kJ" × "1 cal"/"4.184 J" × "1Cal"/"1000 cal" × "1 bag chips"/"245 Cal"# = 14.2 bags of chips

Notice how each conversion factor cancels the units in the preceding one to give the desired units in the answer.