3.5 Mole Definition and Mass Percent Calculations and Reviews

A. Mole Definition and Reviews

The Mole is the ratio between the mass of the matter (element, chemical compound or molecule) to its atomic mass or molar mass.

Mole =  [mass of the matter] / [atomic mass or molar mass]

Molar mass = sum of all atomic masses of the elements making the matter of the molecule of compound

Also the Mole is defined in connection to the particles of the matter (atoms, molecules and formula units (ions)):

The Mole is defined as the mass of the compound or the element that contains the same number of fundamental units as there are found in 12.000 grams of ¹²C (Carbon – 12 isotope).

This means that the atomic mass of Carbon atom of (12 grams) is equal exactly equal 1 mole of Carbon.

Using Avogadro’s number, one can define the mole to be:

Examples:

1. How many moles are in 40.0 grams of water?

40.0 g H₂O x 1 mole H₂O = 2.22 mole H₂O

18.01 g H₂O

2. How many grams are in 3.7 moles of Na₂O?

3. How many atoms are in 14 moles of cadmium?

4. How many moles are in 4.3 x 1022 molecules of H₃P0 4?

 4.3  x 10 ²² molecules H₃PO₄ x        1 mole H₃PO₄              = 7.1 x 10-² moles H₃PO₄

6.022 x 10²³ molecules H₃PO₄

5. How many molecules are in 48.0 grams of NaOH?

6. How many grams are in 4.63 x 1024 molecules of CCl4?

4.63 x 10²⁴ molecules CCl₄ x        1 mole CCl₄x 153.8 g CCI₄_ = 1180 g CCl₄

6.022 x 10²³ molecules CCl₄  1 mole CCI₄

7. How many moles are in 15 grams of lithium? (molar mass of lithium is 6.94 g/mole)

8) How many grams are in 2.4 moles of sulfur? (molar mass of sulfur is 32.07 g/ mole)

9. How many grams are in 4.5 moles of sodium fluoride, NaF?

(molar mass of NaF is 22.99 + 19.00 = 41.99 g/ mole)

    10. How many moles are in 98.3 grams of aluminum hydroxide, Al(OH)₃?

(molar mass of Al(OH)₃ is 26.98 + (3 x 16.00) + (3 x 1.01) = 78.01 g/ mole)

11.  How many moles are in 25.0 grams of water?

1.39 moles

1 mole H₂O = 18.0 g H₂O

    12. How many grams are in 4.500 moles of Li₂O?

134.6 grams

1 mole Li₂O = 29.90 g Li₂O

   13. How many molecules are in 23.0 moles of oxygen?

1.38 x 10²⁵ molecules

 1 mole oxygen molecules = 6.02 x 10²³ oxygen molecules

    14.  How many moles are in 3.4 x 10²³ molecules of H₂SO₄?

0.56 moles

1 mole anything = 6.02 x 10²³ anything

   15. How many molecules are in 25.0 grams of NH₃?

8.85 x 10²³ molecules

1 mole NH₃ = 17.0 g NH₃

1 mole anything = 6.02 x 10²³ anything

= 8.85 x 10²³ molecules

  16.  How many grams are in 8.200 x 10²² molecules of N₂I₆?

107.5 grams

1 mole N₂I₆ = 789.4 g N₂I₆

1 mole anything = 6.02 x 10²³ anything

   17.  How many moles are in 15 grams of lithium?

          ? moles Li = 15 g Li X [ 1 mol Li / 6.941 g Li] = 2.2 mol Li

   18. How many grams are in 2.4 moles of sulfur?

? grams S = 2.4 mol S X [32.066 g S / 1 mol S] = 77 g S

   19.  How many moles are in 3.5 grams of K₂Cr₂O₇?

First determine the molar mass of K₂Cr₂O₇

= 2 K + 2 Cr + 7 O = [2x (39.0983)] + [2x(51.996)] + [7x(15.9994)] = 294.185 g /mol K₂Cr₂O₇

? moles K₂Cr₂O₇ = 3.5 g K₂Cr₂O₇ X [ 1 mol K2Cr2O7 / 294.185 g K₂Cr₂O₇] = 0.012 mol K₂Cr₂O₇

  20. How many grams are found in 2.5 moles of Al(OH)₃?

First determine the molar mass of Al(OH)₃

= 1 Al + 3 O + 3 H

= [1x(26.98154)] + [3x(15.9994)] + [3x(1.00797)] = 78.0036 g Al(OH)₃ / mol Al(OH)₃

? grams Al(OH)₃ = 2.5 mol Al(OH)₃ X [78.0036 g Al(OH)₃ / mol Al(OH)₃] = 195.009 g Al(OH)₃ = 2.0 X 10² g Al(OH)₃ 

Note that the answer is rounded to sig. figs.

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B) Mass Percent Calculations Review

Mass Percent covers the mass of an elements to the some of the elements within a molecule or a compound.

It covers the mass of the solute to the total mass of the solution which includes the mass of the solute and the solvent.

The mass % is defined as illustrated in the figure below:

Examples:

1. 5.0 grams of sugar are dissolved in 150. g of water What is the mass percent of sugar in the solution?

{ (5.0 g) / [(5.0 g + 150. g)] } * 100 % =

{ (5.0 g) / [ (155. g) ] }  * 100 % = 3.2 %

2. 200.0 gram solution of alcohol contains 180.0 mL of water. What is the mass percent of alcohol? (Remember water’s density is 1.000 g /mL)

The mass of 180. mL of water = [180.0 mL * 1.000 g / mL] = 180.0 g

 Mass of the solute = 200.0 g – 180.0 g = 20.0 g

            [ (20.0 g) / (200.0 g) ] * 100 % ] = 10.0%

3. How many grams of NaBr are needed to make 50. g of a 5.0% solution?

           Set the formula up:

            5.0 %   =  X / 50. g

            5.0 g NaBr / 100 g solution = X / 50. g solution

           Solving for X

           Cross multiply yields:

          {  [ 5.0 g NaBr / 100 g solution ] * [ 50. g solution] } = X

          Cancelling g solution:

          {  [ 5.0 g NaBr / 100 g solution ] * [ 50. g solution] } = X

          {  [ 5.0 g NaBr / 100 ] * [ 50. ] }  = 2.5 g NaBr

4. How many grams of LiOH are needed to make 25 g of a 4.0 % solution?

       Set the formula up:

        4.0 %   =  X / 25 g solution

       [ 4.0 g LiOH / 100 g solution] =  X / 25 g solution

       Solving for X

           Cross multiply yields:

           [ 4.0 g LiOH / 100 g solution] * [ 25 g solution ] = X

           [ 4.0 g LiOH / 100 g solution] * [ 25 g solution ] = X

           [ 4.0 g LiOH / 100 g solution] * [ 25 g solution ] = 1.0 g LiOH

5. What mass of NaF must be mixed with 25 mL of water to create a 3.5% by mass solution?

(Remember water’s density is 1.000 g /mL)

             Set the formula up:

           X = mass of NaF in g

           Mass of water = 25 mL = 25 g

           Mass of the solution = (X + 25) g

           3.5 %   =  X / X + 25 g solution

          [ 3.5 g Na F / 100 g solution ] = X / (X + 25 g) solution

          Cancelling g solution from both sides:

         [ 3.5 g Na F / 100 g solution ] = X / (X + 25 g) solution

         [ 3.5 g Na F / 100 ] = X / (X + 25)

         3.5 g NaF        =      X

          100                       (X + 25)

        (X + 25) * (3.5 g Na F) = 100 X

        3.5 X + 87.5 g = 100 X

       Moving 3.5 X to the right side of the equation:

      87.5  g = 100 X – 3.5 X

      87.5 g = 96.5 X

      X = (87.5 g / 96.5) = 0.906 g = 0.91 g

6. An 800.0 g solution of Kool-Aid contains 780.0 g of water. What is the mass percent of solute in this solution?

Mass of solute (Kool – Aid) = 800.0 – 780.0 = 20.0 g

Mass of the solution = 800.0 g

Mass % = [(20.0 g) / (800.0 g)] * 100% = 2.50 %

7. What is the mass percent of a solution created by adding 10.0 g of olive oil to 90.0 g of vegetable oil?

Mass of the solute = 10.0 g olive oil

Mass of the solvent = 90.0 g of vegetable oil.

Mass of the solution = 10.0 + 90.0 = 100.0 g solution

Mass % = 10.0 g olive oil / 100.0 g solution = 10.0 %

8. If 4000.0 g solution of salt water contains 40.0 g of salt, what is its mass percent?

Mass of the solute = 40.0 g salt water

Mass of the solution = 4000.0 g solution

Mass % = 40.0 g salt water / 4000.0 g solution = 1.00 %

Percentage Composition of Water in Salts (Hydrates Percentage Composition)

Example:

9. What is the composition of water in copper sulfate pentahydrate?

copper sulfate pentahydrate has the following formula:

CuSO₄ 5H₂O

From the formula one can deduce that in 1 mol of CuSO₄ 5H₂O has 5 mol of water.

First we have to calculate the molar mass of the whole molecule/compound:

H₂O = 5 x [ 2H + 1 O] = 2 [2 x 1.00797 g/mol  +  1 x 15.9994 g/mol] = 18.01534 g/mol

CuSO₄ = [1 Cu + 1 S + 4 O] = [ 1 x 63.546 g/mol  + 1 x 32.06 g/mol  + 4 x 15.9994 g/mol]

CuSO₄ = 159.609 g/mol

The molar mass of the CuSO₄ 5H₂O = [5 x18.01534 g/mol + 159.609 g/mol] = 249.686 g/mol

 Mass % of H₂O     = { [5 X 18.01534 g/mol] x [249.686 g/mol] } * 100% = 36.07599% = 36.076 %

Mass composition of an Element within its Compound

Example:

10.  Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate NaHCO₃.

First, look up the atomic masses for the elements from the Periodic Table. The atomic masses are found to be:

Na is 22.99 g/mol

H is 1.01 g/mol

C is 12.01 g/mol

O is 16.00 g/mol

Next, determine how many grams of each element are present in one mole of NaHCO₃:

22.99 g in 1 mol of Na

1.01 g in 1 mol of H

12.01 g in 1 mol C

48.00 g (3 mol x 16.00 g/mol) in 3 mol O

The mass of one mole of NaHCO₃ is:

22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g in 1 mol NaHCO₃

Mass percentages of each element are

mass % Na = 22.99 g / 84.01 g x 100 = 27.36 %

mass % H = 1.01 g / 84.01 g x 100 = 1.20 %

mass % C = 12.01 g / 84.01 g x 100 = 14.30 %

mass % O = 48.00 g / 84.01 g x 100 = 57.14 %

Verification: Adding all the mass % it should make 100%

27.36 + 14.30 + 1.20 + 57.14 = 100.00 %

Mass Percent of an Element within its Compound

Example:

               11.  What the mass % of S and O in SO₃?

Atomic mass of S = 32.06 g/mol

Atomic mass of O = 16.00 g/mol

The molar mass of SO₃ = 1 S + 3 O = [32.06 g/mol + 3 x 16.00 g/mol] = 80.06 g/mol

Mass % S = [32.06 g/mol / 80.06 g/mol] x 100% = 40.05 %

Mass % ) = { [3 x 16.00 g/mol] / [80.06] } x 100% = 59.95%

Verification:

Adding both mass % should yield 100% = 40.05 % + 59.95 % = 100.00 %