### StoichiometryMass Examples which also use Avogadro's Number

Avogadro's Number will be involved in every example below. If you are given grams in the problem, the Avogadro's Number will be used at the end. If you are given a number of molecules, then Avogadro's Number will be used at the start.

Example #1: 10.0 g of hydrogen gas react. How many molecules of water are formed?

Solution:

1) Write the balanced chemical equation:

2H2 + O2 ---> 2H2O

2) We are given grams of hydrogen gas:

 10.0 g H2 –––––––– x ––––––––––– x ––––––––––––––– x ––––––––––––– = answer

3) We convert grams to moles:

 10.0 g H2 1 mol H2 –––––––– x –––––––– x ––––––––––––––– x ––––––––––––– = answer 2.016 g H2

4) Construct the molar ratio (hydrogen and water are involved):

 10.0 g H2 1 mol H2 1 mol H2O –––––––– x –––––––– x ––––––––– x ––––––––––––– = answer 2.016 g H2 2 mol H2

Remember, the molar ratio is constructed so as to cancel the moles of the given substance (the H2) and introduce the moles of the target substance (the H2O). The target substance is the one that is associated with the answer.

5) What we have calculated to this point is th moles of water. Just one more step to get to molecules of water:

 10.0 g H2 1 mol H2 1 mol H2O 6.022 x 1023 molecules H2O –––––––– x –––––––– x ––––––––– x –––––––––––––––––––––– = 1.49 x 1024 molecules H2O 2.016 g H2 2 mol H2 1 mol H2O

Example #2: 1.08 x 1024 molecules of hydrogen gas react. How many grams of ammonia are produced?

Solution:

1) Write the balanced chemical equation:

N2 + 3H2 ---> 2NH3

2) Convert molecules to moles:

 1.08 x 1024 molecules H2 1 mol H2 –––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––– = answer 6.022 x 1023 molecules H2

3) Create the molar ratio. It's between hydrogen and ammonia:

 1.08 x 1024 molecules H2 1 mol H2 2 mol NH3 –––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––– = answer 6.022 x 1023 molecules H2 3 mol H2

4) Determine grams of NH3 produced:

 1.08 x 1024 molecules H2 1 mol H2 2 mol NH3 17.0307 g NH3 –––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––– = 20.36 g NH3 6.022 x 1023 molecules H2 3 mol H2 1 mol NH3

Example #3: Water is formed according to the following chemical equation:

2H2 + O2 ---> 2H2O

When 4.550 x 1024 molecules of water are formed, how many grams of oxygen were consumed?

Solution:

1) Here's the entire dimensional analysis set up:

 4.550 x 1024 molecules H2O 1 mol H2O 1 mol O2 31.9994 g O2 ––––––––––––––––––––––– x ––––––––––––––––––––––– x ––––––––– x ––––––––––– = 120.9 g O2 (to four sig figs) 6.022 x 1023 molecules H2O 2 mol H2O 1 mol O2 (a) (b) (c) (d)

2) Notice that everything cancels on the diagonal:

(a) 'molecules H2O' (numerator) with 'molecules H2O' (denominator)
(b) 'mol H2O' (num) with 'mol H2O' (denom)
(c) 'mol O2' (num) with 'mol O2' (denom)
(d) 'g O2' (num) is the only remaining unit and it's the one we want.

Example #4: How many grams of H2 are required to produce 1.230 x 1024 molecules of water?

Solution:

1) Balanced chemical equation:

2H2 + O2 ---> 2H2O

2) Here's the entire dimensional analysis set up:

 1.230 x 1024 molecules H2O 1 mol H2O 2 mol O2 2.016 g H2 ––––––––––––––––––––––– x ––––––––––––––––––––––– x ––––––––– x ––––––––––– = 17.18 g H2 (to four sig figs) 6.022 x 1023 molecules H2O 2 mol H2O 1 mol H2

Example #5: 28.82 g of ammonia decompose. (a) How many molecules of hydrogen are produced? (b) How many atoms of hydrogen is this?

Solution:

1) The chemical equation:

2NH3 ---> N2 + 3H2

2) The dimensional analysis set up follows. Notice where Avogadro's Number goes.

 28.82 g NH3 1 mol NH3 3 mol H2 6.022 x 1023 molecules H2 ––––––––––– x ––––––––––– x –––––––– x ––––––––––––––––––––– = 1.528 x 1024 molecules H2 (to four sig figs) 17.0307 g NH3 2 mol NH3 1 mol H2

3) For (b), we note that there are 2 atoms of H for every one molecule of H2 Therefore:

(1.528 x 1024 molecules H2) (2 atoms/molecule) = 3.056 x 1024 atoms H

This step is incorporated into the DA set up to solve Example #7.

Example #6: 1.85 x 1025 atoms of hydrogen react. How many grams of oxygen are consumed?

Solution:

1) Here's the balanced chemical equation:

2H2 + O2 ---> 2H2O

2) Please note that we measure how much hydrogen reacts based on how many hydrogen molecules there are, not how many atoms there are. This is the very first step of the calculation and you see it in the boldfaced part of the DA set up:

 1.85 x 1025 atoms H 1 molecule H2 1 mol H2 1 mol O2 31.9994 g O2 ––––––––––––––––– x ––––––––– x ––––––––––––––––––––––– x ––––––––– x ––––––––––– = 246 g O2 2 atoms H 6.022 x 1023 molecules H2 2 mol H2 1 mol O2

3) The boldfaced part of the DA set up is where the change from number of atoms to number of molecules takes place.

4) Compare that boldfaced part to the boldfaced part in Example #7

Example #7: 25.0 g of water decomposes. How many atoms of hydrogen are produced?

Solution:

1) The balanced equation:

2H2 + O2 ---> 2H2O

2) Do some dimensional analysis:

 25.0 g H2O 1 mol H2O 2 mol H2 6.022 x 1023 molecules H2 2 atoms H ––––––––––– x ––––––––––– x –––––––– x ––––––––––––––––––––– x ––––––––––– = 1.67 x 1024 atoms H 18.015 g H2O 2 mol H2O 1 mol H2 1 molecule H2

3) Note that the molecules to atoms conversion is incorporated into the DA. In Example #5, it was broken out into a separate step.

4) Note the placement of the molecule to atom conversion: in #6, it is at the start and in #7 it is at the end.

Example #8: 5.40 x 1024 molecules of hydrogen gas react. How many molecules of ammonia are produced?

Solution:

I will solve this problem by steps, then show the dimensional analysis at the end.

1) Convert molecules to moles:

 5.40 x 1024 molecules ––––––––––––––––––––––– = 8.96712 mol H2 6.022 x 1023 molecules/mol

2) The balanced chemical equation:

N2 + 3H2 ---> 2NH3

3) There is a 3 to 2 molar ratio between H2 and NH3. The ratio and proportion to be used is this:

 3 mol H2 8.96712 mol H2 –––––––– = –––––––––––– 2 mol NH3 x

x = 5.97808 mol NH3

4) Convert moles to molecules:

(5.97808 mol NH3) (6.022 x 1023 molecules/mol) = 3.60 x 1024 molecules NH3

5) Notice that no molar masses were used. Avogadro's Number is "per mole" not "per gram."

6) Dimensional analysis set up:

 5.40 x 1024 molecules H2 1 mol H2 2 mol NH3 6.022 x 1023 molecules NH3 ––––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––––––––––––– = 3.60 x 1024 molecules NH3 6.022 x 1023 molecules H2 3 mol H2 1 mol NH3

7) Notice anything about the DA set up? Hint: there is something other than units that could be cancelled? (The correct answer is yes.)

Example #9: 2.84 x 1023 molecules of water was produced. How many molecules (see example #10) of oxygen were required?

Solution:

1) Balanced chemical equation:

2H2 + O2 ---> 2H2O

2) Dimensional analysis (with bonus cancellation):

 2.84 x 1023 molecules H2O 1 mol H2O 1 mol O2 6.022 x 1023 molecules O2 ––––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––––––––––––– = 1.42 x 1023 molecules O2 6.022 x 1023 molecules H2O 2 mol H2O 1 mol O2

3) I cancelled the numerical value for Avogadro's Number, one value being in the numerator and one in the denominator. Notice that I left the units untouched with the strikeout line. All the units cancel in the normal diagonal way.

4) If you encounter a problem like this and never notice that the Avogadro Number values can be cancelled, no harm will be done. You'll simply by Avogadro's Number at the start and multiply by it at the end. No harm done, just a few extra keystrokes on the calculator.

Example #10: 2.84 x 1023 molecules of water was produced. How many atoms (see example #9) of oxygen were required?

Solution:

1) The complete DA set up:

 2.84 x 1023 molecules H2O 1 mol H2O 1 mol O2 6.022 x 1023 molecules O2 2 atoms O ––––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––––––––––––– x ––––––––––– = 2.84 x 1023 atoms O 6.022 x 1023 molecules H2O 2 mol H2O 1 mol O2 1 molecule O2

Notice that the 2 in '2 mol H2O' cancels with the 2 in '2 atoms O.' All the units cancel in the normal fashion, leaving the desired 'atoms O' for our final unit.