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From the discussion and examples in the mole section, it is clear that there is a need to convert between units. The emphasis there was between the number of objects and moles of objects. There were, however, other examples of conversions. An important one is between moles and grams of a substance. The ability to go between units of different properties, like the moles of atoms and the grams of atoms, is crucial in several areas of chemistry, including being able to determine measurable properties, like the mass of a substance produced in a chemical reaction.The focus here is to provide a methodology that can be used to convert between units.
A unit conversion expresses an original measurement in new units such that the magnitude of the original measurement is reflected in the new units.
Unit conversions can be within one type of measurement or between different types of measurement. Conversion between liters and milliliters doesn't change what is being reported, in this case volume. A conversion between liters and grams, on the other hand, would convert between volume and mass. The quantity that allows the conversion is called the conversion factor. In the liters to gram example the conversion factor would be the density, which gives the number of grams per liter. The conversion factor between the number of objects and moles of objects is 6.02 x 1023 objects/mole.
Conversions Using Balanced Equations
As an introduction consider the last lesson on balanced chemical equations. The balanced equations provide mole ratios between reactants, between reactants and products, and between products. These ratios can be used to convert from the moles of reactants used to the moles of products formed. They can do conversions between reactants or between products as well, but the focus here will be between reactants and products.
In the water example two moles of hydrogen molecules reacted with one mole of oxygen molecules to form two moles of water molecules:
2H2 + O2 → 2H2O
The O2:H2O ratio is 1:2 in this case. That is to say, one mole of oxygen molecules will produce two moles of water molecules. It could also be written:
1 mole O2 2 mole H2O Or
2 mole H2O 1 mole O2 If you produced four moles of H2O, how many moles of O2 would be used up? From the balanced chemical equation it is seen that the number of water molecules produced is twice the number in the equation. That suggests that twice as many oxygen molecules will be needed. Two moles of oxygen molecules would be needed to produce four moles of water molecules. This can also be obtained mathematically as follows:
( 4 mole H2O ) (
1 mole O2 2 mole H2O ) = 2 mole O2 Notice how "mole H2O" is on both the top and bottom of this mathematical expression, canceling out and leaving "mole O2" as the unit in the answer. The balanced chemical equation has given us a conversion factor between moles of reactants and moles of products. This is the key to unit conversions.
Another quick example before giving the long version. If you used up two moles of O2, how many moles of H2O would be formed?
( 2 mole O2 ) (
2 mole H2O 1 mole O2 ) = 4 mole H2O Again notice the cancellation of units in this example and notice that the balanced chemical equation has provided a conversion factor for this calculation.
Same Type Conversions
Conversions within a single type of measurement (volume, mass, length, etc.) have conversion factors that are usually standard multipliers. Some common multipliers are kilo (103), milli (10-3), centi (10-2), etc. A kilometer, for instance, is 1000 meters (1 x 103 meters), a milligram is 0.001 gram (1 x 10-3 gram), etc.
The conversions can be stated several ways. There are 1000 grams in a kilogram. Or there are 1 x 10-3 kilograms in a gram. Equivalently, one kilogram is 1000 grams or 1 x 10-3 kilograms is one gram. It is important to be able to recognize that all of these expressions say the same thing. Any of them could be used as a conversion factor (see below).
In a mathematical expression the conversion factors are set up in such a way as to cancel out all of the units except the one of interest. So, one way of deciding which conversion factor statement to use is to see which units need to be canceled out.
The example of converting 1.2 kg (kilograms) to grams (g) is given below. In this case the kg units cancel out of the expression and leave grams. Notice that the units cancel out, but the numbers are multiplied or divided as required by the mathematical expression.
( 1.2 kg ) (
1000 g 1 kg ) = 1200 g or
( 1.2 kg ) (
1 g 1 x 10-3 kg ) = 1200 g Do you see that the two conversion factors are equal? You could use your calculator to divide one by 1 x 10-3 to see if it is equal to 1000. To convert from grams to kilograms would require the reciprocal conversion factors. Notice how the grams cancel out to leave kilograms in this case.
( 1200 g ) (
1 kg 1000 g ) = 1.2 kg or
( 1200 g ) (
1 x 10-3 kg 1 g ) = 1.2 kg It is best to use the smallest number of conversion factors possible, but sometimes a direct conversion factor is not known and more steps are required. These steps could be done one at a time or grouped together in one expression.
An example of a multi-step conversion of length from millimeters (mm) to centimeters (cm) going through meters (m) is given below. Notice that both mm and m cancel, leaving cm in the answer.
( 370 mm ) (
1 m 1000 mm ) (
100 cm 1 m ) = 37.0 cm A common conversion in chemistry is from milliliters to liters. How many liters is 200 milliliters (200 mL = ? L)?
( 200 mL ) (
1 L 1000 mL ) = 0.2 L All of these same-type conversions can be done in a similar way.
Converting Between Types
If the conversion factors between types of measurements are known, the same method of unit conversion discussed in the previous section can be used to convert from one type of measurement to another.
The density of water is one gram per milliliter (1.0 g/mL) under normal conditions. Knowing this allows the conversion from milliliters of water to grams of water or from grams of water to milliliters of water.
How many grams of water are in a 300 mL sample of water? In this case the milliliters will cancel and leave grams.
( 300 mL H2O ) (
1 g H2O 1 mL H2O ) = 300 g H2O How many milliliters of water correspond to 18 g of water? Again notice that the grams cancel leaving mL.
( 18 g H2O ) (
1 mL H2O 1 g H2O ) = 18 mL H2O As stated previously, converting between grams and moles is extremely important in chemistry. Fortunately the conversion factors for doing this are found on the Periodic Table of the Elements. The non-integer number associated with each atom of that element on the chart is the number of grams of that element in one mole of that element. These conversion factors have units of g/mole (or mole per gram if the reciprocal is taken). The conversion factor for lithium, Li, is 6.94 g Li atoms per one mole Li atoms (often shortened to 6.94 g Li/mole Li or 6.94 g/mole). Carbon, C, has a conversion factor of 12 g/mole (12 g C atoms/mole C atoms). Iron, Fe, has a conversion factor of 55.85 g/mole. Etc.
How many grams of calcium, Ca, are there in 1.2 moles of calcium?
( 1.2 mole Ca ) (
40.08 g Ca 1 mole Ca ) = 48.1 g Ca How many moles of carbon, C, are in 24 g C?
( 24 g C ) (
1 mole C 12 g C ) = 2 mole C How many grams of oxygen gas, O2, are there in 12 moles of oxygen gas? In this example there are two atoms in each molecule of oxygen gas or two moles of oxygen atoms in each mole of oxygen molecules. This is a conversion factor between atoms of oxygen and molecules of oxygen. Oxygen gas particles are O2 molecules.
( 12 mole O2 ) (
2 mole O 1 mole O2 ) (
16 g O 1 mole O ) = 380 g oxygen These methods work for all kinds of conversions and they can be combined in a single problem. In a certain reaction it is found that three moles of hydrogen gas, H2, completely reacts with one mole of nitrogen gas, N2 (or one mole of nitrogen gas reacts with 3 moles of hydrogen gas). The mole-gram conversion factor for hydrogen gas is 2 g/mole. If 6.5 g H2 reacts with nitrogen gas, how many moles of nitrogen gas will be consumed? This problem requires two conversions: one from grams to moles for hydrogen and one from moles of hydrogen to moles of nitrogen. Notice that grams of hydrogen gas and moles of hydrogen gas both cancel out, leaving moles of nitrogen gas.
( 6.5 g H2 ) (
1 mole H2 2 g H2 ) (
1 mole N2 3 mole H2 ) = 1.083 mole N2 Word Problems
Here are a some word problems for practice.
Here are the answers.
- The mole-grams conversion factor for water, H2O, is 18 g/mole. How many moles of water molecules are there if you have 32 g H2O molecules?
- How many grams of sodium, Na, atoms are there if you have 3.5 mole Na atoms?
- In a chemical reaction it was found that two moles of sodium, Na, completely reacted with one mole of chlorine gas, Cl2. If three moles of sodium reacted, how many moles of chlorine gas would react in this reaction?
- In a chemical reaction it was found that four moles of ammonia, NH3, completely reacted with six moles of nitrogen(II)oxide, NO. How many moles of nitrogen(II)oxide are needed to react with five moles of ammonia?
- In a chemical reaction it was found that two moles of sodium, Na, completely reacted with one mole of chlorine gas, Cl2. If 12.5 grams of sodium reacted, how many moles of chlorine gas would react in this reaction?