Mole-Mass Problems

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The solution procedure used below involves making two ratios and setting them equal to each other. This is called a proportion. One ratio will come from the coefficients of the balanced equation and the other will be constructed from the problem. The ratio set up from data in the problem will almost always be the one with an unknown in it.

You will then cross-multiply and divide to get the answer.

However, there is one addition to the above technique. One of the values will need to be expressed in moles. This could be either a reactant or a product. In either case, moles will have to be converted to grams or the reverse.

Suppose you are given a mass in the problem. You will need to convert this to moles FIRST. You do this by dividing the mass given by the molar mass of the substances. This technique is covered in the mole section of the ChemTeam. Click this link to go to the proper mole file for review.

Suppose you are asked for a mass as an answer. You will convert the moles you calculated in the proportion to grams. You do this by multiplying the moles by the molar mass of the substance. This technique is covered in the mole section of the ChemTeam. Click this link to go to the proper mole file for review.


Here is the equation we'll use for the first three examples:

2KClO3 ---> 2KCl + 3O2

Example #1: 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced?

Solution:

1) Let's use this ratio to set up the proportion:

KClO3
––––––
O2

2) That means the ratio from the equation is:

2
––
3

3) The ratio from the data in the problem will be:

1.50
––––
x

4) The proportion (setting the two ratios equal) is:

1.50   2
–––– = ––
x   3

Cross-multiplying and dividing gives x = 2.25 mol of O2 produced.

5) The last step is to convert from moles to grams:

2.25 mol x 32.0 g/mol = 72.0 grams.

The 32.0 g/mol is the molar mass of O2


Example #2: If 80.0 grams of O2 was produced, how many moles of KClO3 decomposed?

1) Let's use this ratio to set up the proportion:

O2
––––––
KClO3

2) That means the ratio from the equation is:

3
––
2

3) The ratio from the data in the problem will be:

2.50
––––
x

The 2.50 mole came from 80.0 g ÷ 32.0 g/mol. The 32.0 g/mol is the molar mass of O2. Be careful to keep in mind that oxygen is O2, not just O.

4) The proportion (setting the two ratios equal) is:

2.50   3
–––– = ––
x   2

Solving by cross-multiplying and dividing gives x = 1.67 mol of KClO3 decomposed.


Example #3: We want to produce 2.75 mol of KCl. How many grams of KClO3 would be required?

Solution:

1) Let's use this ratio to set up the proportion:

KCl
––––––
KClO3

2) The ratio from the equation is:

2
––
2

3) The ratio from the data in the problem will be:

2.75
––––
x

4) The proportion (setting the two ratios equal) is:

2.75   2
–––– = ––
x   2

Solving the above gives 2.75 mol of KClO3.

5) However, the question wants grams for the answer:

2.75 mol times 122.55 g/mol = 337 g

The 122.55 g/mol is the molar mass of KClO3.


Here's the equation to use for the next three examples:

2H2 + O2 ---> 2H2O

Example #4: How many grams of H2O are produced when 2.50 moles of oxygen are used?

Solution:

1) Here are the two substances in the molar ratio I used:

O2
––––––
H2O

2) The ratio from the equation is:

1
––
2

3) The molar ratio from the problem data is:

2.50
––––
x

4) The proportion to use is:

2.50   1
–––– = ––
x   2

5.00 mol of water is produced, but since the problem asks for grams, we multiply by 18.0 g/mol (the molar mass of water) to get the final answer of 90.0 g.


Example #5: If 3.00 moles of H2O are produced, how many grams of oxygen must be consumed?

Solution:

1) Here are the two substances in the molar ratio I used:

O2
––––––
H2O

2) The ratio from the equation is:

1
––
2

3) The molar ratio from the problem data is:

x
––––
3.00

4 The proportion to use is:

x   1
–––– = ––
3.00   2

We know that 1.50 mol of O2 was consumed, so multiplying that by 32.0 g/mol gives 48.0 g.


Example #6: How many grams of hydrogen gas must be used, given the data in example #5?

1) Here are the two substances in the molar ratio I used:

H2
––––
O2

2) The ratio from the equation is:

2
––
1

3) The molar ratio from the problem data is:

x
––––
1.50

4) The proportion to use is:

x   2
–––– = ––
1.50   1

The H2 / H2O ratio of 2/2 could have been used also. In that case, the ratio from the problem would have been 3.00 over x, since you were now using the water data and not the oxygen data.

3.00 mol times 2.02 g/mol (the molar mass of hydrogen) gives 6.06 g.


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