Mole-Mass Examples

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 that is set up using 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.

The above is the technique used in mole-mole problems.

Here will now be an addition to the technique used in mole-mole problems. 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:

2KClO_{3}---> 2KCl + 3O_{2}

**Example #1:** 1.50 mol of KClO_{3} decomposes. How many grams of O_{2} will be produced?

**Solution:**

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

KClO _{3}–––––– O _{2}

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 O

_{2}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 O

_{2}

**Example #2:** If 80.0 grams of O_{2} was produced, how many moles of KClO_{3} decomposed?

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

O _{2}–––––– KClO _{3}

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 O

_{2}. Be careful to keep in mind that oxygen is O_{2}, 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 KClO

_{3}decomposed.

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

**Solution:**

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

KCl –––––– KClO _{3}

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 KClO

_{3}.

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

2.75 mol times 122.55 g/mol = 337 gThe 122.55 g/mol is the molar mass of KClO

_{3}.

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

2H_{2}+ O_{2}---> 2H_{2}O

**Example #4:** How many grams of H_{2}O are produced when 2.50 moles of oxygen are used?

**Solution:**

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

O _{2}–––––– H _{2}O

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 H_{2}O are produced, how many grams of oxygen must be consumed?

**Solution:**

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

O _{2}–––––– H _{2}O

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 O

_{2}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:

H _{2}–––– O _{2}

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 H

_{2}/ H_{2}O 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.

**Example #7:** How many molecules (not moles) of NH_{3} are produced from 1.75 x 10¯^{4} g of H_{2}? Express your answer numerically as the number of molecules.

**Solution:**

1) Write the balanced chemical equation for the reaction:

3H_{2}+ N_{2}---> 2NH_{3}

2) Determine moles of H_{2} present:

1.75 x 10¯ ^{4}g––––––––––– = 0.000086806 mol 2.016 g/mol

3) Use the 2:3 NH_{3} to H_{2} molar ratio to determine moles of NH_{3}:

2 x ––– = ––––––––––––– 3 0.000086806 mol x = 0.0000578707 mol of NH

_{3}

4) Use Avogadro's Number to determine number of molecules:

(0.0000578707 mol) (6.022 x 10^{23}molecules/mol) = 3.48 x 10^{19}molecules

5) Set up using the style of dimensional analysis, we have this:

1 mol H _{2}2 mol NH _{3}6.022 x 10 ^{23}molecules NH_{3}1.75 x 10¯ ^{4}g H_{2}x ––––––– x ––––––– x –––––––––––––––––––––– = 3.48 x 10 ^{19}molecules of NH_{3}2.016 g H _{2}3 mol H _{2}mol NH _{3}

**Example #8:** The balanced chemical equation for the reaction between sodium carbonate and hydrochloric acid is:

CaCO_{3}(s) + 2HCl(aq) ---> CaCl_{2}(s) + CO_{2}(g) + H_{2}O(ℓ)

If the equation is interpreted on the mole level, what is the total mass of products in the balanced equation?

**Solution:**

1) "interpreted on the mole level" means this:

one mole of CaCO_{3}reacts with two moles of HCl to produce one mole of CaCl_{2}, one mole of CO_{2}, and one mole of H_{2}O

2) We want the combined weight of the reactants: one mole of CaCl_{2}, one mole of CO_{2}, and one mole of H_{2}O:

CaCl_{2}---> 110.984 g/mol

one mole of CO_{2}---> 44.009 g/mol

one mole of H_{2}O ---> 18.015 g/molThe sum is 173.008 g

3) By the way, the sum of the reactants (one mole of CaCO_{3} and two moles of HCl) will also add up to 173.008. You may check that for yourself, if you so desire.

**Example #9:** 1.5503 grams of magnesium reacts with hydrochloric acid according to this balanced reaction:

Mg(s) + 2HCl(aq) ---> MgCl_{2}(aq) + H_{2}(g)

How many moles of hydrogen gas will be produced?

**Solution:**

Comment: you will first convert grams of Mg to moles of Mg, then, using the Mg to H_{2} ratio of moles (found in the balanced chemical equation) you convert to moles of H_{2}.

1) Grams to moles:

1.5503 g / 24.305 g/mol = 0.063785 mol of Mg

2) Use molar ratio:

The Mg:H_{2}molar ratio is 1:11 is to 1 as 0.063785 mol is to x

x = 0.063785 mol of H

_{2}

3) Many teachers (and textbooks) set up problems like the above problem using "dimensional analysis" (also called the factor-label method or the unit-factor method). Here's the set up, formatted two different ways:

1.5503 g Mg x (1 mol Mg / 24.305 g Mg) x (1 mol H_{2}/ 1 mol Mg) = 0.063785 mol H_{2}

1 mol Mg 1 mol H _{2}1.5503 g x ––––––– x ––––––– = 0.063785 mol H _{2}24.305 g Mg 1 mol Mg

**Example #10:** Calculate the moles of NO produced by the reaction of 27.85 grams of N_{2} according to the following chemical equation:

N_{2}+ O_{2}---> 2NO

Comment: nothing is said about the oxygen. The usual assumption is there is sufficient oxygen for all the nitrogen to be used up. if you assume insufficient oxygen is present, then the problem cannot be solved, so you don't have to do it.

Have fun justifying the second assumption to your teacher!

**Solution:**

1 mol N _{2}2 mol NO 27.85 g N _{2}x–––––––––– x ––––––– = 1.988 mol NO 28.014 g N _{2}1 mol N _{2}

This link to an outside source that contains two mole-mass problems along with the solutions, as well as some introductory explanation.