Moles – Avogadro’s Number

  I.                    The chemical formula represents 1 mole of that substance.

II.                 The formula mass (expressed in grams) is the mass of 1 mole of that substance.

III.               1 mole of a substance contains 6.02 E 23 particles (atoms, mlcl, ions, electrons, etc.)

IV.              1 mole of any gas at STP (1 atmosphere of pressure and 0 °C) occupies 22.4 liters of volume.

 Name                           Formula                        Formula mass               # of particles

Atomic nitrogen            N                                 14.0                             6.02 E 23 atoms

Nitrogen gas                 N₂                                28.0                             6.02 E 23 mlcl 
                                                                                                                 (1.20 E 24 atoms)

Silver ions                     Ag⁺¹                             108                              6.02 E 23 ions

Sodium chloride            NaCl                            58.5                             6.02 E 23 ion pairs

Ammonium sulfate        (NH₄)₂SO₄                   132                              6.02 E 23 mlcl

 

Problems:

1 mole Mo  =  _________ g

1 mole Mo  =  _________ atoms

7 moles Mo = _________ g

7 moles Mo = _________ atoms

 1 mol Th₃(PO₄)₄  =  _________ mlcl  =  _________ g

0.5 mol CO₂  =  ________g   =  ___________ mlcl

1.5 mol MgCl₂  =  _________g  =  ___________ ion pairs

 1 mol KNO₃   =  _________ mol K, _________ mol N, ________mol O

1 mol KNO₃   =  _________ g K, __________g N, ___________ g O

10 g KNO₃  =  __________ mol KNO₃, __________ g K, ___________ atoms O

 2.2 mol of H₃PO₄  there are ___________ g H, __________mol p, __________atoms O

 Molarity

 Molarity:  the ratio between the moles of dissolved substance (solute) and the volume of the solution (in liters or cubic decimeters)

 Example:          1 M HNO₃  =  1 mole of HNO₃ in 1 L of solution

                    0.273 M Ba(NO₃)₂ contains 0.372 moles of barium nitrate in 1 L of solution

 Sample Problem:

            What is the molarity of a 250 mL solution containing 9,46 B CsBr?

Solution:

            9.46 g CsBr__|  1 mole  CsBr__  =  0.0444 mol CsBr
                                     |   213 g CsBr

 

            250 mL__|__1 L__       =  0.250 L
                             |    1000 mL

 

            molarity =  mole    =    0.044 4 mol  CsBr  =  0.178 M CsBr
                                  liter             0.250 L

 

 Problems: 

1.      145 g  (NH₄)₂C₄H₄O₆ in 500 mL of solution

2.      13.2 g MnSeO₄ in 500 mL of solution

3.      45.1 g cobalt (II) sulfate in 250 mL of solution

4.      41.3 g iron (II) nitrate in 100 mL solution

5.      49.9 g Pb(ClO₄)₂ in 200 mL of solution

6.      35.0 g MnSiF₆ in 50.0 mL of solution

 

 Sample test problems:

1.      7.25 E 4 grams of carbon (IV) tellurate contains ________grams of oxygen

2.      9.45 E – 15 moles of carbon (IV) tellurate contains _________ atoms of carbon

3.      5.50 E 16 atoms of zinc weighs _________ grams

4.      5.50 E 16 atoms of manganese is _________ moles of manganese

5.      8.35 moles of chromium (III) arsenate weighs __________ grams.

6.      8.35 moles of chromium (III) arsenate contains _________atoms of arsenic

 

  More Mole Problems