Converting from One Unit to Another

Base and Derived Units

For most quantities, a unit is absolutely necessary to communicate values of that physical quantity. Imagine you need to buy some rope to tie something onto the roof of a car. How would you tell the salesperson how much rope you need without using some unit of measurement?

However, not all quantities require a unit of their own. Using physical laws, units of quantities can be expressed as combinations of units of other quantities. Therefore, only a small set of units is required. These units are called base units, and other units are derived units. Derived units are a matter of convenience, as they can be expressed in terms of basic units.

Different systems of units are based on different choices of base units. The most widely used system of units is the International System of Units, or SI. There are seven SI base units, and all other SI units can be derived from these base units.

The seven base SI units are: [Physical Quantity: unit symbol (unit name)]

• Length: m (meter)
• Mass: kg (kilogram)
• Time: s (second)
• Electric Current: A (Ampere)
• Thermodynamic Temperature: K (degrees Kelvin)
• Amount of Substance: mol (mole)
• Luminous Intensity: cd (candela)

The base units of SI are actually not the smallest set possible; smaller sets have been defined. For example, there are unit sets in which the electric and magnetic field have the same unit. This is based on physical laws that show that electric and magnetic fields are actually different manifestations of the same phenomenon.

Derived units are based on units from the SI system of units. For example, volume is a derived unit because volume is based on length. To calculate the volume of something, you multiply the width x length x height, all in meters. Therefore, the derived unit for volume is m³. Here is a list of some commonly derived units:

• Area: m²
• Volume: m³
• Velocity: m/s
• Acceleration: m/s²
• Density: g/mL or g/cm³
• Force: $kg\cdot m/s^2$, or the Newton (N)
• Energy: $N\cdot m$, or the Joule (J)

Dimensional Analysis

Sometimes, it is necessary to deal with measurements that are very small (as in the size of an atom) or very large (as in numbers of atoms). In these cases, it is often necessary to convert between units of metric measurement. For example, a mass measured in grams may be more convenient to work with if it was expressed in mg (10^–3 g). Converting between metric units is called unit analysis or dimensional analysis.

Unit analysis is a form of proportional reasoning where a given measurement can be multiplied by a known proportion or ratio to give a result having a different unit or dimension. Algebraically, we know that any number multiplied by one will be unchanged. If, however, the number has units, and we multiply it by a ratio containing units, the units in the number will multiply and divide by the units of the ratio, giving the original number (remember you are multiplying by one) but with different units.

This method can be generalized as: multiply or divide a given number by a known ratio to find your answer. The given number is a numerical quantity (with its units). The ratios used are based upon the units and are set up so that the units in the denominator of the ratio match the numerator units of the given and the units in the numerator of the ratio match those in either the next ratio or the final answer. When these are multiplied, the given number will now have the correct units for your answer.

Example 1

For example, say you were trying to convert 3.41 grams of He to a number of atoms of He. You would identify 3.41 grams as the given. The first step is always to place the given out front of your equation. Then find a ratio that will help you convert the units of grams to atoms. As you probably have already guessed, you need to use a couple of ratios to help you in this problem. The ratio that 4.002 g of He = 1 mole (molar mass) will help you in this problem. Avogadro's number, 6.022 x 10²³ atoms = 1 mole, will also help you in this problem. Then you set up your ratios so that your units will cancel successfully (the same unit must be in the numerator of the equation and also in the denominator of the equation). Lastly, multiply through to get your final answer. As always, your final answer should contain the correct number of sig figs and the correct units.

$3.41\:g \cdot \frac {1\:mole}{4.002\:g} \cdot \frac {6.022\cdot 10^{23} atoms}{1\:mole} = 5.13\:\cdot \:10^{23} atoms$

Example 2

If you had a sample of a substance with a mass of 0.0034 grams, and you wanted to express that mass in mg, you could use the following dimensional analysis. The given quantity is the mass of 0.0034 grams. The quantity that you want to find is the mass in mg, and we know that 1 mg = 10^-3 g. Expressing this as a proportion or ratio, there is one mg per 10^-3 grams, or 1000 mg/1 g.

Therefore, 0.0034g x (1000 mg/1 g) = 3.4 mg