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.

In this tutorial, there will be an addition to the technique used in mole-mole problems. One of the values involved will be expressed in grams. This could happen two different ways:

(Given Grams, Calculate Moles) suppose you are given a mass in the problem statement. 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.(Given Moles, Calculate Grams) suppose you are asked for a mass as an answer to the problem. You will will first use the ratio and proportion to calculate an answer in moles. Then, that answer in moles will be converted to grams and this will be the final answer. 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 mol KClO _{3}––––––––––– 3 mol O _{2}

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

1.50 mol KClO _{3}––––––––––––– x

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

1.50 mol KClO _{3}2 mol KClO _{3}–––––––––––– = ––––––––––– x 3 mol O _{2}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) (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 mol O _{2}––––––––––– 2 mol KClO _{3}

3) Before creating the molar ratio from information in the problem, we must convert the grams given in the problem to moles:

80.0 g ÷ 32.0 g/mol = 2.50 mol of O_{2}Remember that oxygen is O

_{2}.

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

2.50 mol O _{2}––––––––– x

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

2.50 mol O _{2}3 mol O _{2}–––––––––– = –––––––––– x 2 mol KClO _{3}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 mol KCl –––––––––– 2 mol KClO _{3}

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

2.75 mol KCl ––––––––––– x

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

2.75 mol KCl 2 mol KCl –––––––––––– = –––––––––– x 2 mol KClO _{3}Solving the above gives 2.75 mol of KClO

_{3}.

5) The question wants grams for the answer:

(2.75 mol) (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 mol O _{2}––––––––– 2 mol H _{2}O

3) The molar ratio from the problem data is:

2.50 mol O _{2}–––––––––– x

4) The proportion to use is:

2.50 mol O _{2}1 mol O _{2}–––––––––– = –––––––– x 2 mol H _{2}Ox = 5.00 mol H

_{2}O

5) Convert moles to grams:

(5.00 mol) (18.0 g/mol) = 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 mol O _{2}––––––––– 2 mol H _{2}O

3) The molar ratio from the problem data is:

x ––––––––––– 3.00 mol H _{2}O

4 The proportion to use is:

x 1 mol O _{2}––––––––––– = –––––––– 3.00 mol H _{2}O2 mol H _{2}Ox = 1.50 mol O

_{2}

5) Convert moles to grams:

(1.50 mol) (32.0 g/mol) = 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 mol H _{2}––––––– 1 mol O _{2}

3) The molar ratio from the problem data is:

x ––––––––– 1.50 mol O _{2}

4) The proportion to use is:

x 2 mol H _{2}––––––––– = ––––––– 1.50 mol O _{2}1 mol O _{2}x = (3.00 mol) (2.016 g/mol) = 6.05 g (to three sig figs)

5) 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.

**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 mol NH _{3}x ––––––––– = ––––––––––––––––– 3 mol H _{2}0.000086806 mol H _{2}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 dimensional analysis:

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

**Example #11:** What mass of carbon is present in 4.806 x 10^{26} molecules of C_{2}H_{5}OH?

**Solution:**

1) Determine moles of ethyl alcohol present:

4.806 x 10^{26}molecules / 6.022 x 10^{23}molecules/mol = 798.07373 mol

2) There are 2 moles of C in every mole of C_{2}H_{5}OH:

798.07373 mol x 2 = 1596.1475 mol of C

3) Determine mass of C:

(1596.1475 mol) (12.011 g/mol) = 19171.3276225 g19.17 kg

**Example #12:** In exactly 1 mole of the hydrate CuSO_{4} **⋅** 5H_{2}O, how many grams are present of (a) the hydrate, (b) the anhydrate, and (c) water.

**Solution:**

1) In one mole of the hydrate, there is present this:

one mole of CuSO_{4}

five moles H_{2}O

2) Determine the molar mass for the hydrate. Remember to include the five waters.

249.681 gThis is the answer to (a)

3) Determine the molar mass of the anhydrate:

159.607 gThis is the answer to (b)

Remember, the anhydrate does not have any water in it, hence the formula is CuSO

_{4}.

4) Determine the total mass of water in one mole of the hydrate:

(18.015 g/mol) (5 mol) = 90.074 gThis is the answer to (c).

You could have also done this:

249.681 g − 159.607 g = 90.074 g