Special note on a particular conversion not mentioned below.

Metric conversions where two units (numerator and denominator) are converted

Return to Metric Table of Contents

Skills you need to do this include:

1) memorize the metric prefixes names and symbols

2) determine which of two prefixes represents a larger amount

3) determine the exponential "distance" between two prefixes

4) significant figure rules

5) scientific notation

Here are two typical metric conversion problems:

1) Convert 2.50 μg to picograms.

2) Convert 0.080 cm to km.

The explanation below will focus on the first problem. Try to set up the second on as you read through the first example. The answers are provided below.

There is second type of metic conversion, one that involves converting both the numerator and the denominator. You can go to another tutorial which discusses the second type.

Also:

Special note on a particular conversion that is sometimes not mentioned in class.

It has to do with the fact that 1 mL equals 1 cm^{3}and 1 L equals 1 dm^{3}.

Just below I discuss how to construct a conversion factor. There is an important point about the numerator and the denominator of the conversion factor. Here it is:

The numerator and the denominator of a conversion factor BOTH describe the same amount.In essence, a conversion factor is equal to one. This is because the numerator and the denominator both describe exactly the same amount. Here's an example:

5280 feet / 1 mile

Both 5280 feet and 1 mile describe exactly the same distance.

Here's a metric example:

1 kg / 1000 g

1000 g and 1 kg both have exactly the same amount of mass.

So, when you are done making a given conversion, you haven't changed the amount, you've just changed the way it is written. For example, 1 kg is differently written from 1000 g, but they both describe the same amount of stuff.

The key skill in solving these problems is to construct a conversion factor. This conversion factor will make the old unit go away (micrograms and km in the top two examples) and create the new unit (pm and cm) in its place. Along with this change, there will be a change in the value of the number.

Let's focus on the first example: Convert 2.50 μg to picograms

**STEP ONE:** Write the value (and its unit) from the problem, then in order write: 1) a multiplication sign, 2) a fraction bar, 3) an equals sign, and 4) the unit in the answer. Put a gap between 3 and 4. All that looks like this:

The fraction bar will have the conversion factor. There will be a number and a unit in the numerator and the denominator.

**STEP TWO:** Write the unit from the problem in the denominator of the conversion factor, like this:

**STEP THREE:** Write the unit expected in the answer in the numerator of the conversion factor.

**STEP FOUR:** Examine the two prefixes in the conversion factor. In front of the LARGER one, put a one.

There is a reason for this. I'll get to it in a second.

**STEP FIVE:** Determine the absolute distance between the two prefixes in the conversion unit. Write it as a positive exponent in front of the other prefix.

Now, multiply and put into proper scientific notation format. Don't forget to write the new unit. Sometimes, the exponential number is in the denominator. You must move it to the numerator and when you do so, remember to change the sign. Also, DO NOT move the unit with it. That unit has been cancelled and is no longer there.

Here are all five steps for the second example, put into one image:

Note that the old unit cancels, since it appears in the numerator and denominator of two parts of a multiplication problem.

Why a one in front of the larger unit? I believe it is easier to visualize how many small parts make up one bigger part, like 1000 m making up one km. Going the other way, visualizing what part a larger unit is of one smaller unit, is possible, but requires more sophistication. For example, how many meters are in one nanometer? The answer is 0.000000001 or 10¯^{9}. You may be able to handle a conversion involving what part a larger unit is of a smaller unit and that is just fine.

The conversion factor I have been discussing above is sometimes called a "unitary rate." Unitary in this case simply means that the conversion factor equals one. Look at the conversion factor in the example, the 10^{6} pg / 1 μm. The numerator and the denominator both describe the same amount, as in 10^{6} pg **equals** 1 μm. The word unitary is used to identify the fact that both values describe the same amount.

What I have done is describe a system where the unit with the one (1 μm in my example) can be in either the numerator or denominator of the conversion unit. You may have a teacher that forces the one to be only in the denominator. What that means is that you will have to decide if you are going from a large unit to a small unit (making the numerator exponent positive) or going from a small one to a large one (making the numerator unit negative).

**Three Comments**

1) If you do the conversion correctly, the numerical part and the unit will go in opposite directions. If the unit goes from smaller (mm) to larger (km), then the numerical part goes from larger to smaller. There will never be a correct case where number and unit both go larger or both go smaller.2) A common mistake is to put the one in front of the SMALLER unit. This results in a wrong answer. Put the one in front of the LARGER unit.

3) Sometimes you see this:

10In that case, there is an assumed one in front of the unit with no number. In other words, 10^{6}pg / μm^{6}pg / 1 μm means the same thing as 10^{6}pg / μm. In closing: many times, the 1 will be assumed when it is in the denominator. Very seldom is the 1 assumed when in the numerator. Almost always, the 1 is used explicitly when it is in the numerator.

The answers to the two examples above are 2.50 x 10^{6} pg and 8.0 x 10¯^{7} km.

**Five More Conversion Examples**

1) 0.75 kg to milligrams

2) 1500 millimeters to km

3) 2390 g to kg

4) 0.52 km to meters

5) 65 kg to g

Did you read that special note about a conversion sometimes not mentioned in class? Here's the link.

6) Convert 1.80 mL to cm^{3}.

1.80 mL times (1 cm^{3}/ 1 mL) = 1.80 cm^{3}

7) Convert 1.80 L to dm^{3}

1.80 L times (1 dm^{3}/ 1 L) = 1.80 dm^{3}

Be careful with the next two problems.

8) Convert 1.80 L to cm^{3}

1.80 L times (1000 cm^{3}/ 1 L) = 1.80 x 10^{3}cm^{3}

9) Convert 1.80 mL to dm^{3}.

1.80 mL times (1 dmOne dm^{3}/ 1000 mL) = 0.00180 dm^{3}

Special note on a particular conversion not mentioned above.

Metric conversions where two units (numerator and denominator) are converted