Mass-Mass Calculation


  

Under going construction.

A Mass-Mass Calculation is a Law of Conservation of Mass dependent relationship used to determine how much of a product/reactant will be made/consumed during a chemical reaction. When chemicals are reacted with each other, the products of the reaction are formed without a measureable loss or a gain of total mass.  

The above video demonstrates how many grams of potassium chloride, KCl(s), can be produced when 17.2g of oxygen, O(g), is produced? 

The first step is to convert oxygen’s gram quantity to moles of oxygen. The second step uses the stoichiometry of the balanced equation

2KClO(s)  3O(g) + 2KCl(s) 

to find the moles of potassium chloride, KCl(s), produced. In the third step the molar amount of potassium chloride, KCl(s), is then converted to grams of potassium chloride, KCl(s).

Step 1: One mole of oxygen, O(g), is 32.00 grams/mole and is found using the gram atomic mass(16.00g) of oxygen from the periodic table. The formula mass calculation is as follows.

Element   Subscript   X   Element Mass = Formula Mass

       O            2            X       16.00g          =    32.00g/m

Dividing the given 17.2 grams oxygen, O(g), by the 32.00 grams/mole for the oxygen, O(g), produces moles of oxygen, O(g), gas formed during reaction.

17.2g O(g) ÷ (32.00g O(g)/1 m O(g)) = 0.538m O(g)

17.2 grams oxygen, O(g), is equal to 0.538 moles of oxygen, O(g).

Step 2: The 0.538 moles of oxygen, O(g), must be converted to moles potassium chloride, KCl(s), using the balanced equation

2KClO(s)  3O(g) + 2KCl(s)

0.538m O(g) X (2mKCl(s)/3mO(g) = 0.359m KCl(s)

Step 3: The 0.359m potassium chloride, KCl(s), must be converted to grams. The formula mass for potassium chloride, KCl(s), is found by the following formula mass calculation using gram atomic masses for each potassium(39.10g) and chlorine(35.45g) from the periodic table.

Element   Subscript    X   Element Mass = Formula Mass

    K                  1          X       39.10g          =    39.10g

    Cl                 1          X       35.45g          =    35.45g

The molar mass of potassium chloride, KCl(s), is 39.10g + 35.45g = 74.55g/m.

0.359m KCl(s) X (74.55g/1m KCl(s)) = 26.76g KCl(s)

Therefore, 26.76g potassium chloride, KCl(s), is produced at the same time 17.2g of oxygen gas, O(g), is produced.


Mass-Mass Calculation Example One: Potassium Nitride Decomposition

How many grams of potassium nitride, KN(s), can be consumed when 20.1g of nitrogen gas, N(g), is produced? 

The first step is to convert nitrogen’s gram quantity to moles of nitrogen gas, N(g). The second step uses the stoichiometry of the balanced equation

2KN(s)  1N(g) + 6K(s) 

to find the moles of potassium nitride, KN(s), consumed. In the third step the molar amount of potassium nitride, KN(s),  is then converted to grams of potassium nitride, KN(s).

Step 1: One mole of nitrogen gas, N(g), is 28.02 grams/mole and is found using the gram atomic mass(14.01g) of nitrogen from the periodic table. The formula mass calculation is as follows.

Element   Subscript   X   Element Mass = Formula Mass

    N                2           X      14.01g           =   28.02g/m

Dividing the given 20.1 grams nitrogen gas, N(g), by the 28.02 grams/mole for the nitrogen gas, N(g), produces moles of nitrogen, N(g), gas formed during reaction.

20.1 N(g) ÷ (28.02g N(g)/1 m N(g)) = 0.717m N(g)

20.1 grams nitrogen gas, N(g), is equal to 0.717 moles of nitrogen gas, N(g).

Step 2: The 0.717m moles of nitrogen gas, N(g), must be converted to moles potassium nitride, KN(s), using the balanced equation

2KN(s)  1N(g) + 6K(s)

0.717m N(g) X (2m KN (s)/1mN(g) = 1.43m KN(s)

Step 3: The 1.43 m potassium nitride, KN(s), must be converted to grams. The formula mass for potassium nitride, KN(s),  is found by the following formula mass calculation using gram atomic masses for each potassium(39.10g) and nitrogen(14.01g) from the periodic table.

Element   Subscript    X   Element Mass = Formula Mass

    K                 3           X       39.10g          =    117.30g

    N                 1           X       14.01g          =    14.01g

The molar mass of potassium nitride, KN(s), is 117.30g + 14.01g = 131.31g/m.

1.43m KN (s) X (131.31g KN (s)/1m KN (s)) = 188g KN (s)

Therefore, 188g  potassium nitride, KN(s), is produced at the same time 20.1g of nitrogen gas, N(g),  is produced.


Mass-Mass Calculation Example Two: Potassium Iodide Reacts with Antimony(IV) Chloride

How many grams of antimony(III) chloride, SbCl(s), can be precipitated from solution when 15.7g of potassium iodide, KI(aq), in solution is consumed? 

The first step is to convert potassium iodide’s gram quantity to moles of potassium iodide, KI(aq), in solution. The second step uses the stoichiometry of the balanced equation

2KI(aq)+1SbCl(aq) 1SbI(s)+2KCl(aq)+ 1I(s)

 to find the moles of antimony(III) chloride, SbCl(s),  precipitated from solution. In the third step the molar amount of antimony(III) chloride, SbCl(s), produced is then converted to grams of antimony(III) chloride, SbCl(s), precipitated from solution.

 Step 1:  One mole of potassium iodide, KI, is 166.00 grams/mole and is found using the sum of the gram atomic mass of potassium(39.10g) and iodine(126.90g) from the periodic table. The formula mass calculation is as follows.

Element   Subscript   X   Element Mass = Formula Mass

    K                1           X      39.10g            =   39.10g/m

    I                 1           X      126.90g          =  126.90g/m

The molar mass of potassium iodide, KI, is 39.10g + 126.90g = 166.00g/m.

 Dividing the given 15.7 grams potassium iodide, KI(aq), in solution by the 166.00 grams/mole for the potassium iodide, KI, produces moles of potassium iodide, KI(aq), in solution consumed during the reaction.

15.7g KI ÷ (166.00g KI/1m KI)= 0.0946m KI

15.7 grams potassium iodide, KI, is equal to 0.0946 moles of potassium iodide, KI(aq), in solution.

Step 2: The 0.0946m moles of potassium iodide, KI(aq), in solution must be converted to moles antimony(III) chloride, SbCl (s), precipitated from solution using the balanced equation

 2KI(aq)+1SbCl (aq)1SbI(s)+2KCl(aq)+1I(s)

 0.0946m KI(aq) X (1mSbCl(s)/2mKI(aq)= 0.0473m SbCl(s)

Step 3: The 0.0473m antimony(III) chloride solid, SbCl(s), must be converted to grams. The formula mass for antimony(III) chloride, SbCl(s),  is found by the following formula mass calculation using gram atomic masses for each antimony(121.76g) and chlorine(35.45g) from the periodic table.

Element   Subscript    X   Element Mass = Formula Mass

    Sb               1           X       121.76g          =    121.76g

    Cl               3            X        35.45g           =    106.35g

The molar mass of antimony(III) chloride, SbCl(s), is 121.76g + 106.35g= 228.11g/m.

0.0473m SbCl (s) X (228.11g/1m SbCl(s)) = 10.8g  SbCl(s)

Therefore, 10.8g  antimony(III) chloride, SbCl(s), is produced at the same time 15.7g of potassium iodide in solution, KI(aq), is consumed    





Mass-Mass Calculation Example Three: Aluminum Oxide Reacts with Chlorine and Carbon

How many grams of solid carbon, C(s), can be consumed when 21.9g of chlorine gas, Cl(g), is consumed?  

The first step is to convert chlorine’s gram quantity to moles of chlorine gas, Cl(g). The second step uses the stoichiometry of the balanced equation

1AlO(s)+3Cl(g)+3C(s)2AlCl(aq)+3CO(g)

to find the moles solid carbon, C(s),  consumed in the reaction. In the third step the molar amount of solid carbon, C(s), consumed is then converted to grams of solid carbon, C(s), consumed.

 Step 1: One mole of of chlorine gas, Cl(g)  is 70.90 grams/mole and is found using the sum of the gram atomic mass of 2 chlorine atoms(35.45g) from the periodic table. The formula mass calculation is as follows.

Element   Subscript   X   Element Mass = Formula Mass

      Cl               2            X        35.45g           =    70.90g

 The molar mass of chlorine gas, Cl(g), 70.90g /m is used to find the moles of chlorine gas, Cl(g), consumed.  Dividing the given 21.9 grams chlorine gas, Cl(g), by the 70.90 grams/mole for chlorine gas, Cl(g), produces moles of chlorine gas, Cl(g), consumed during reaction.

21.9g Cl(g),  ÷ (70.90g Cl(g)/1m Cl(g))= 0.309m Cl(g),

21.9 grams chlorine gas, Cl(g), is equal to 0.309 moles of grams chlorine gas, Cl(g).

Step 2: The 0.309 moles of chlorine gas, Cl(g), must be converted to moles solid carbon, C(s), consumed using the balanced equation

 1AlO(s)+ 3Cl(g)+ 3C(s)2AlCl(aq)+3CO(g)

  0.309m  Cl(g) X (3m C(s)/3m Cl(g)= 0.309m C(s)

Step 3:  The 0.309m solid carbon, C(s), must be converted to grams. The formula mass for solid carbon, C(s), is found by the following formula mass calculation using the gram atomic mass of carbon(12.01g) from the periodic table.

Element   Subscript    X   Element Mass = Formula Mass

    C               1            X        12.01g           =    12.01g

The molar mass of solid carbon, C(s), is 12.01g/m.

 0.309m C(s)X (12.01g /1m C(s)) = 3.71g  (s)

 Therefore, 3.71 grams of carbon, C(s), is consumed at the same time 21.9g of chlorine gas, Cl(g), is consumed.

               

© Pat Thayer 2014-2016