Mass 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:

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

2) We are given grams of hydrogen gas:

10.0 g H _{2}–––––––– x ––––––––––– x ––––––––––––––– x ––––––––––––– = answer

3) We convert grams to moles:

10.0 g H _{2}1 mol H _{2}–––––––– x –––––––– x ––––––––––––––– x ––––––––––––– = answer 2.016 g H _{2}

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

10.0 g H _{2}1 mol H _{2}1 mol H _{2}O–––––––– x –––––––– x ––––––––– x ––––––––––––– = answer 2.016 g H _{2}2 mol H _{2}Remember, the molar ratio is constructed so as to cancel the moles of the given substance (the H

_{2}) and introduce the moles of the target substance (the H_{2}O). 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 H _{2}1 mol H _{2}1 mol H _{2}O6.022 x 10 ^{23}molecules H_{2}O–––––––– x –––––––– x ––––––––– x –––––––––––––––––––––– = 1.49 x 10 ^{24}molecules H_{2}O2.016 g H _{2}2 mol H _{2}1 mol H _{2}O

**Example #2:** 1.08 x 10^{24} molecules of hydrogen gas react. How many grams of ammonia are produced?

**Solution:**

1) Write the balanced chemical equation:

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

2) Convert molecules to moles:

1.08 x 10 ^{24}molecules H_{2}1 mol H _{2}–––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––– = answer 6.022 x 10 ^{23}molecules H_{2}

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

1.08 x 10 ^{24}molecules H_{2}1 mol H _{2}2 mol NH _{3}–––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––– = answer 6.022 x 10 ^{23}molecules H_{2}3 mol H _{2}

4) Determine grams of NH_{3} produced:

1.08 x 10 ^{24}molecules H_{2}1 mol H _{2}2 mol NH _{3}17.0307 g NH _{3}–––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––– = 20.36 g NH _{3}6.022 x 10 ^{23}molecules H_{2}3 mol H _{2}1 mol NH _{3}

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

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

When 4.550 x 10^{24} 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 10 ^{24}molecules H_{2}O1 mol H _{2}O1 mol O _{2}31.9994 g O _{2}––––––––––––––––––––––– x ––––––––––––––––––––––– x ––––––––– x ––––––––––– = 120.9 g O _{2}(to four sig figs)6.022 x 10 ^{23}molecules H_{2}O2 mol H _{2}O1 mol O _{2}(a) (b) (c) (d)

2) Notice that everything cancels on the diagonal:

(a) 'molecules H_{2}O' (numerator) with 'molecules H_{2}O' (denominator)

(b) 'mol H_{2}O' (num) with 'mol H_{2}O' (denom)

(c) 'mol O_{2}' (num) with 'mol O_{2}' (denom)

(d) 'g O_{2}' (num) is the only remaining unit and it's the one we want.

**Example #4:** How many grams of H_{2} are required to produce 1.230 x 10^{24} molecules of water?

**Solution:**

1) Balanced chemical equation:

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

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

1.230 x 10 ^{24}molecules H_{2}O1 mol H _{2}O2 mol O _{2}2.016 g H _{2}––––––––––––––––––––––– x ––––––––––––––––––––––– x ––––––––– x ––––––––––– = 17.18 g H _{2}(to four sig figs)6.022 x 10 ^{23}molecules H_{2}O2 mol H _{2}O1 mol H _{2}

**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:

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

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

28.82 g NH _{3}1 mol NH _{3}3 mol H _{2}6.022 x 10 ^{23}molecules H_{2}––––––––––– x ––––––––––– x –––––––– x ––––––––––––––––––––– = 1.528 x 10 ^{24}molecules H_{2}(to four sig figs)17.0307 g NH _{3}2 mol NH _{3}1 mol H _{2}

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

(1.528 x 10^{24}molecules H_{2}) (2 atoms/molecule) = 3.056 x 10^{24}atoms HThis step is incorporated into the DA set up to solve Example #7.

**Example #6:** 1.85 x 10^{25} __atoms__ of hydrogen react. How many grams of oxygen are consumed?

**Solution:**

1) Here's the balanced chemical equation:

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

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 10 ^{25}atoms H1 molecule H_{2}1 mol H _{2}1 mol O _{2}31.9994 g O _{2}––––––––––––––––– x ––––––––– x ––––––––––––––––––––––– x ––––––––– x ––––––––––– = 246 g O _{2}2 atoms H6.022 x 10 ^{23}molecules H_{2}2 mol H _{2}1 mol O _{2}

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:

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

2) Do some dimensional analysis:

25.0 g H _{2}O1 mol H _{2}O2 mol H _{2}6.022 x 10 ^{23}molecules H_{2}2 atoms H––––––––––– x ––––––––––– x –––––––– x ––––––––––––––––––––– x ––––––––––– = 1.67 x 10 ^{24}atoms H18.015 g H _{2}O2 mol H _{2}O1 mol H _{2}1 molecule H_{2}

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 10^{24} 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 10 ^{24}molecules––––––––––––––––––––––– = 8.96712 mol H _{2}6.022 x 10 ^{23}molecules/mol

2) The balanced chemical equation:

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

3) There is a 3 to 2 molar ratio between H_{2} and NH_{3}. The ratio and proportion to be used is this:

3 mol H _{2}8.96712 mol H _{2}–––––––– = –––––––––––– 2 mol NH _{3}x x = 5.97808 mol NH

_{3}

4) Convert moles to molecules:

(5.97808 mol NH_{3}) (6.022 x 10^{23}molecules/mol) = 3.60 x 10^{24}molecules NH_{3}

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 10 ^{24}molecules H_{2}1 mol H _{2}2 mol NH _{3}6.022 x 10 ^{23}molecules NH_{3}––––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––––––––––––– = 3.60 x 10 ^{24}molecules NH_{3}6.022 x 10 ^{23}molecules H_{2}3 mol H _{2}1 mol NH _{3}

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 10^{23} molecules of water was produced. How many __molecules__ (see example #10) of oxygen were required?

**Solution:**

1) Balanced chemical equation:

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

2) Dimensional analysis (with bonus cancellation):

2.84 x 10 ^{23}molecules H_{2}O1 mol H _{2}O1 mol O _{2}~~6.022 x 10~~molecules O^{23}_{2}––––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––––––––––––– = 1.42 x 10 ^{23}molecules O_{2}~~6.022 x 10~~molecules H^{23}_{2}O2 mol H _{2}O1 mol O _{2}

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 10^{23} 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 10 ^{23}molecules H_{2}O1 mol H _{2}O1 mol O _{2}6.022 x 10 ^{23}molecules O_{2}2 atoms O ––––––––––––––––––––– x ––––––––––––––––––––– x ––––––––– x –––––––––––––––––––––– x ––––––––––– = 2.84 x 10 ^{23}atoms O6.022 x 10 ^{23}molecules H_{2}O2 mol H _{2}O1 mol O _{2}1 molecule O _{2}

Notice that the 2 in '2 mol H_{2}O' 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.