 
MoleGram Calculations
One of the things you will do throughout this course is to work back and forth between
mass in grams and the amount of material in moles. In order to convert back and forth
between mass and moles, you simply need to use the same kind of conversion calculations
that you have done before. I'll work through two examples (these are also shown in your
workbook as Example 7), then you can practice the calculations yourself in Exercise 8 in
your workbook.
The task is to express 45 grams of carbon in moles. There are different
ways of accomplishing that task. We'll use the conversion factor approach. 

45 g C 

= ? moles C 
We start with 45 grams of carbon. We want to find out how many moles we
have, so we need a relationship between the number of grams of carbon and the number of
moles. 

45 g C 
x? moles C
? g C 
= ? moles C 
That relationship is simply that 12.0 grams of carbon is one mole. So we
set up the calculation with 45 grams of carbon times the conversion factor of one mole
over 12.0 grams of carbon. 

45 g C 
x 1 mole C
12.0 g C 
= ? moles C 
Carry out the calculations and we get 3.8 moles. In this case we round off
the answer to two significant digits, 3.8, because the initial value of 45 g has just two
significant digits. 

45 g C 
x 1 mole C
12.0 g C 
= 3.8 moles C 
If we had known that there were 45.0 grams of carbon, then we would have
given the answer as 3.75 moles. You still have to remember to keep working with
significant digits. 

45. 0 g C 
x 1 mole C
12.0 g C 
= 3.75 moles C 
Along that same line, if we had started with 45.00 grams of carbon, and we
wanted our answer to be more precise than 3.75, we wouldn't be able to use 12.0 grams of
carbon in the conversion factor. The precision of the atomic weights is in the
weight value rather than the mole value. When we say that one mole of carbon weighs 12.0
grams, we mean that exactly one mole of carbon weighs 12.0 grams when measured to three
significant digits. So the answer is rounded off to the number of digits shown in the
atomic weight or in the starting value, not in the 1 mole. 

45. 00 g C 
x 1 mole C
12.0 g C 
= 3.75 moles C 
To get four digits of precision in the answer, we would have to look up a
more precise atomic weight in order to get the 4th digit there, which is 12.01. So any
time that you want to work with numbers that are more precise than 3 digits, you will have
to look up an atomic weight that is more precise than just 3 digits. Remember that these
atomic weights are measured values. They are not exact, except for that one particular
isotope of carbon called carbon12. 

45. 00 g C 
x 1 mole C
12.01 g C 
= 3.747 moles C 
Next, consider the question "How much would 3.53 moles of iron
weigh?" Here we are trying to change from moles to grams, so 3.53 moles of iron is
out in front. 

3.53 moles Fe 

= ? g Fe 
The relationship between moles of iron and grams of iron is 1 mole for
every 55.8 grams of iron. So that's what goes in the conversion factor with the grams on
top and the moles on the bottom, so that the moles cancel out. 

3.53 moles Fe 
x 55.8 g Fe
1 mole Fe 
= ? g Fe 
When you multiply that through, you get 197 grams of iron. 

3.53 moles Fe 
x 55.8 g Fe
1 mole Fe 
= 197 g Fe 
These conversion calculations relating moles and grams are the same kind of
calculations that you have worked with before. It is just that now you are dealing with
another unita molethat you may have never heard of before.
As you know from lesson 1a, many units can be abbreviated; so
"gram" becomes "g", "liter" becomes "l"
or "L", and so on. Many students want to abbreviate
"mole" as "m", forgetting that we use "m" to stand
for "meter". The abbreviation for "mole" is
"mol"  not much of an abbreviation at all!
Practice with MoleGram Calculations
Before you continue with the lesson, work through the 4 problems in exercise 8. Check
your answers below before continuing. If you have any trouble, check with the instructor.
(If you would like to try some additional problems of this type, you will find
several in a "Practice Problems" page in the WrapUp for this lesson. Click here to go there now.)
Answers
OK, here are the answers that you should have for the problems in Exercise 8.
"A" is 3.6 grams of carbon. "B" is 2.2 x 10^{2} gram of
potassium. My calculations came out to 215, but since 5.5 is only good to 2 digits, I
rounded the answer off to 220. Then to show that the zero was not significant I converted
the answer to scientific notation. "C" is 1.58 moles of fluorine. "D"
is 2.4 moles of nitrogen. So, you should have those values. If you did not come up with
those answers, you should take some time now to work with the instructor and find out what
kinds of problems are keeping you from getting those answers.
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Email instructor:
Sue Eggling
Clackamas Community College
©1998, 2002 Clackamas Community College, Hal Bender
