# Unit Conversion

Skills to Develop

- The skill to convert a quantity into various units is important.
**Unit conversion**and**dimensional analysis**are also scientific methods. There are many examples in Chemistry, and you will encounter them later.

Example 1

The world's ocean is estimated to contain \(\mathrm{1.4 \times 10^9\; km^3}\) of water.

- What is the volume in liters?
- What is the weight if the specific density is 1.1?
- How many moles of water are present if all the weight is due to water?
- How many moles of \(\ce{H}\) atoms (not \(\ce{H2}\)) are there in the ocean?
- How many \(\ce{H}\) atoms are present in the ocean?

*Solution*

\(\mathrm{1.4e9 \left (\dfrac{1000\: m}{1\: km} \right )^3 \left (\dfrac{10\: dm}{1\:m} \right )^3\\

= 1.4e21\: dm^3 \left (\dfrac{1\: L}{1\: dm^3} \right )\\

= 1.5e21\: L \left (\dfrac{1.1\: kg}{1\: L} \right )\\

= 1.5e21\: kg \left (\dfrac{1000\: g}{1\: kg} \right )\left (\dfrac{1\: mol}{18\: g} \right )\\

= 8.3e22\: mol\: H_2O \left (\dfrac{2\: mol\: H\: atoms}{1\: mol\: H_2O} \right )\\

= 1.7e23\: mol\: H \left (\dfrac{6.02e23\: atoms}{1\: mol} \right )\\

= 5.0e46\: H\: atoms}\)

In this example, a quantity has been converted from a unit for volume into other units of volume, weight, amount in moles, and number of atoms. Every factor used for the unit conversion is a unity. The numerator and denominator represent the same quantity in different ways.

Even in this simple example, several concepts such as the quantity in moles, Avogadro's number, and specific density (or specific gravity) have been applied in the conversion. If you have not learned these concepts, you may have difficulty in understanding some of the conversion processes. Identify what you do not know and find out in your text or from a resource.

Example 2

A typical city speed for automobiles is 50 km/hr. Some years ago, most people believed that 10 seconds to dash a 100 meter race was the lowest limit. Which speed is faster, 50 km/hr or 10 m/s?

**SOLUTION**

For comparison, the two speeds must be expressed in the same unit. Let's convert 50 km/hr to m/s.

\[ \mathrm{50 \;\dfrac{\cancel{km}}{\cancel{hr}} \left(\dfrac{1000\; m}{1\; \cancel{km}}\right) \left(\dfrac{1\; \cancel{hr}}{60\;\cancel{min}}\right) \left(\dfrac{1\;\cancel{min}}{ 60\; s}\right) =13.89\; m/s} \]

Thus, 50 km/hr is faster.

Note: a different unit can be selected for the comparison (e.g., miles/hour) but the result will be the same (test this out if interested).

Exercise 2

The speed of a typhoon is reported to be 100 m/s. What is the speed in km/hr and in miles per hour?

Example 3: Convert Quantities into SI units

Mr. Smart is ready for a T-bone steak. He went to market A and found the price to be 4.99 dollars per kilograms. He drove out of town to a roadside market, which sells at 2.29 per pound. Which price is better for Mr. Smart?

**SOLUTION**

To help Mr. Smart, we have to know that 1.0 kg is equivalent to 2.206531 lb or 1 lb = 453.2 g. By the way, are these the same?

\[ \mathrm{4.99\; \dfrac{$}{\cancel{kg}} \left( \dfrac{1\; \cancel{kg}}{2.206532\; lb} \right) = 2.26468 \;\dfrac{$}{lb}}\]

Of course, with the money system in Canada, there is no point quoting the price as detailed as it is given above. This brings about the significant digit issue, and the quantization. The price is therefore 2.26 $/lb, better for Mr. Smart than the price of 2.29 $/lb.