Word Problems in Chemistry and Physics

This article is about word problems that occur in quantitative sciences like chemistry and physics. Word problems in chemistry and physics are a crucial point for students. Mastering them will allow students to progress in chemistry and physics from high school onwards, even when confronting tricky word problems.

There are several “ key phrases” that are useful in order to start unraveling a word problem. The word “of” indicates an amount, an object, and a unit of measure nearby.  For example, you might see a word problem that says “7 buckets of lemons” or “0.5 moles of iron (III) oxide.” In these cases, “0.5” or “7” are the quantities, “buckets” or “moles” are the units of measurement, and “lemons” or iron (III) oxide particles are the objects.

These are often followed by sentences containing the word “per” or  “for each”: “there are 9 lemons per bucket.” In word problems, the word “per” indicates a way to change between 2 different types of units. To change units of buckets into units of individual lemons, write the number of lemons per bucket as a fraction, replacing the word “per” with the division sign: 9 lemons/bucket. This is called a conversion factor. Multiply the number of buckets by your conversion factor; units will cancel the same way that numbers do. Seven buckets times 9 lemons/bucket gives you 63 lemons in total.

In chemistry, these “per” sentences are often not given. They must be figured out from the context of the problem. In this case, there are 6.02 x 10^23 particles per mole.  Since iron (III) oxide has the formula Fe2O3, there are 2 atoms of iron per particle of iron (III) oxide or 2 moles of iron per mole of iron (III) oxide. Using the strategy outlined above, your conversion factor could be:

6.02 x 10^23 particles/mole iron (III) oxide,

2 atoms Fe / particle of iron (III) oxide, or

2 moles of iron/mole of iron (III) oxide.

Choose the lower portion of the fraction according to what you have, and the higher portion of the fraction according to what you want. Since we have moles of iron (III) oxide, we can only start with the 1st or the 3rd conversion factor, not the 2nd. In the top portion of those conversion factors, we can access information about particles of iron (III) oxide or moles of iron by multiplying.

Notice that particles of iron (III) oxide appear in the lower portion of the 2nd conversion factor. If we are willing to use two conversion factors in a row, we can convert our starting amount into particles of iron (III) oxide, then convert that into atoms of iron. You can visualize this type of problem as a game of dominoes, played by matching tiles end-to-end, where each domino represents a conversion factor. If you can start from the information in the problem and construct a bridge of dominoes to the information you want, then you can solve the problem. To create a “domino,” find a conversion factor in the text of the problem, using reference information like molar mass, or using the ratios of the reactions and chemicals involved. To add it to your “bridge,” multiply by it.

Notice that each of our “dominoes” has a direction. If you multiply 100 miles by 25 miles per gallon, you end up with 2500 miles times miles per gallon. The problem is that you failed to cancel units properly.  To reverse a domino, turn the fraction upside down: 25 miles/gallon turns into 1 gallon/25 miles or 1/25 gallon per mile. The true solution here is that you used 100 miles times 1/25 gallon per mile, or 4 gallons.

A Complete Example

The following example is typical of chemistry: the initial amounts are given in grams, but the calculations must be carried out in moles. This is an important strategy, because chemistry on paper is done with mole ratios but chemistry and practices done with scales and masses.

Suppose a 0.257 kg block of iron (Fe) is completely combusted. How many grams of oxygen gas (O2) were used?

First, we outline some conversion factors. To help do this, we set up the combustion reaction: 4Fe + 3O2 gives 2Fe2O3.

Second, we identify our target units: grams of oxygen.   Finally, we arrange our conversion factors and multiply. A sample pair of canceling units is in bold.

0.257 kg iron *

1000 g/kg *

(1/55.847 mol iron/g iron)*

(2 moles iron (III) oxide per mol iron)*

(3/2 moles oxygen gas per mol iron (III) oxide)*

(2 atoms oxygen per mole oxygen gas)*

(15.999grams/atom of oxygen)

=  110.4g oxygen.

It’s a good idea to check your answer intuitively at the end.  For example, here we used about 40% as much oxygen as iron.  This makes sense because oxygen (though more moles are used) is much lighter.