Mole-Gram Calculations
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Mole-Gram 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 carbon-12. 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 unit--a mole--that 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 Mole-Gram 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 Wrap-Up for this lesson.  Click here to go there now.)


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 102 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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