Chemistry » Essential Ideas in Chemistry » Mathematical Treatment of Measurement Results

# Conversion Factors

## Conversion Factors and Dimensional Analysis

A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,

$$\cfrac{2.54\text{ cm}}{1\text{ in.}} (2.54 \text{ cm} = 1\text{ in})$$$$\text{ or } 2.54 \cfrac{\text{cm}}{\text{in.}}$$

Several other commonly used conversion factors are given in the table below.

### Common Conversion Factors

LengthVolumeMass
1 m = 1.0936 yd1 L = 1.0567 qt1 kg = 2.2046 lb
1 in. = 2.54 cm (exact)1 qt = 0.94635 L1 lb = 453.59 g
1 km = 0.62137 mi1 ft3 = 28.317 L1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m1 tbsp = 14.787 mL1 (troy) oz = 31.103 g

When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:

$$\require{cancel}34\text{ in.} × \cfrac{2.54\text{ cm}}{1\mathrm{\cancel{in}.}}$$$$= 86\text{ cm}$$

Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield $$\frac{\text{in.} × \text{cm}}{\text{in.}}.$$ Just as for numbers, a ratio of identical units is also numerically equal to one, $$\frac{\text{in.}}{\text{in.}} = 1,$$ and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”)

Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.

### Using a Unit Conversion Factor

The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (see table above).

#### Solution

If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.

$$x\text{ oz}$$$$= 125\text{ g} × \text{unit conversion factor}$$

We write the unit conversion factor in its two forms:

$$\cfrac{1\text{ oz}}{28.349\text{ g}}$$ and $$\cfrac{28.349\text{ g}}{1\text{ oz}}$$

The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.

$$\require{cancel}\begin{array}{lll} x\text{ oz} & = & 125\mathrm{\cancel{g}} × \frac{1\text{ oz}}{28.349\mathrm{\cancel{g}}} \\ & = & (\frac{125}{28.349})\text{ oz} \\ & = & 4.41\text{ oz }(\text{three significant figures}) \\ \end{array}$$

Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.

### Computing Quantities from Measurement Results and Known Mathematical Relations

What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.

#### Solution

Since $$\text{density} = \frac{\text{mass}}{\text{volume}},$$ we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A × unit conversion factor. The necessary conversion factors are given in the table above: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:

$$\require{cancel}9.26 \mathrm{ \cancel{lb}} × \cfrac{453.59\text{ g}}{1 \mathrm{\cancel{lb}}}$$$$= 4.20 × 10^3 \text{ g}.$$

We need to use two steps to convert volume from quarts to milliliters.

1. Convert quarts to liters.

$$\require{cancel}4.00 \mathrm{ \cancel{qt}} × \cfrac{1 \text{ L}}{1.0567 \mathrm{\cancel{qt}}} = 3.78 \text{L}$$

2. Convert liters to milliliters.

$$\require{cancel}3.78\mathrm{ \cancel{L}} × \cfrac{1000 \text{mL}}{1\mathrm{ L}}$$$$= 3.78 × 10^3\text{ mL}$$

Then,

$$\text{density} = \cfrac{4.20 × 10^3\text{ g}}{3.78 × 10^3\text{ mL}} = 1.11\text{ g/mL}$$

Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:

$$\require{cancel}\cfrac{9.26\mathrm{ \cancel{lb}}}{4.00\mathrm{ \cancel{qt}}} × \cfrac{453.59\text{ g}}{1\mathrm{ \cancel{lb}}} ×$$$$\cfrac{1.0567\mathrm{ \cancel{qt}}}{1\mathrm{ \cancel{L}}} × \cfrac{1\mathrm{ \cancel{L}}}{1000\text{ mL}}$$$$= 1.11 \text{ g/mL}$$

### Computing Quantities from Measurement Results and Known Mathematical Relations II

(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
(b) If gasoline costs \$3.80 per gallon, what was the fuel cost for this trip?

#### Solution

(a) We first convert distance from kilometers to miles:

$$1250 \text{ km} × \cfrac{0.62137 \text{ mi}}{1\text{ km}} = 777\text{ mi}$$

and then convert volume from liters to gallons:

$$\require{cancel}213\mathrm{ \cancel{L}} × \cfrac{1.0567\mathrm{ \cancel{qt}}}{1\mathrm{ \cancel{L}}} ×$$$$\cfrac{1\text{ gal}}{4\mathrm{ \cancel{qt}}} = 56.3\text{ gal}$$

Then,

$$\text{(average) mileage} = \cfrac{777\text{ mi}}{56.3\text{ gal}}$$$$= 13.8\text{ miles/gallon} = 13.8 \text{ mpg}$$

Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:

$$\require{cancel}\cfrac{1250\mathrm{ \cancel{km}}}{213\mathrm{ \cancel{L}}} × \cfrac{0.62137\text{ mi}}{1\mathrm{ \cancel{km}}} ×$$$$\cfrac{1\mathrm{\cancel{L}}}{1.0567\mathrm{ \cancel{qt}}} × \cfrac{4\mathrm{ \cancel{qt}}}{1 \text{ gal}} = 13.8 \text{ mpg}$$

(b) Using the previously calculated volume in gallons, we find:

$$56.3\text{ gal} × \cfrac{3.80}{1\text{ gal}} = 214$$