Lesson Using proportions to solve Chemistry problems

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Using proportions to solve Chemistry problems


This lesson is the continuation of the lesson  Proportions  of this module.
The lesson presents some typical word problems related to  Chemistry  that can be solved using proportions.
You will see how useful the proportions are in solving chemistry problems.

Problem 1. Mass of oxygen in water

Chemical formula for water is  H%5B2%5DO  (two atoms of Hydrogen and one atom of Oxygen).
Gram molecular weight of oxygen  O%5B2%5D  is  32  (gram atomic weight of Oxygen is  16), gram atomic weight of Hydrogen  H  is  1.
What is the mass of oxygen in  3600  grams of water?

Solution
Let  mO%5B2%5D  be the mass of oxygen in 3600 grams of water.
Note that gram molecular weight of water  H%5B2%5DO  is  2*(gram atomic weight of Hydrogen) + 1*(gram atomic weight of Oxygen) = 2*1 + 16 = 18.
Since the mass of oxygen is proportional to the total mass of water,  you can write the proportion
mO%5B2%5D%2F3600 = 16%2F18.
This proportion has the unknown extreme term  mO%5B2%5D.
Use the rule for solving proportions from the lesson  Proportions:  the unknown extreme term is equal to the product of mean terms divided by the known extreme term.
Applying this rule you get the unknown value of  mO%5B2%5D
mO%5B2%5D = 3600%2A16%2F18 = 3200.

Answer.  The mass of oxygen in 3600 grams of water is 3200 grams.


Problem 2. Mass of nitrogen in ammonia

Chemical formula for ammonia is  NH%5B3%5D  (one atom of Nitrogen and three atoms of Hydrogen).
Gram atomic weight of Nitrogen  N  is 14,  gram atomic weight of Hydrogen  H  is 1.
If we have  680  grams of ammonia,  what is the mass of nitrogen?

Solution
Let  mN  be the mass of nitrogen in  680  grams of ammonia.
Note that gram molecular weight of ammonia  NH%5B3%5D  is  1*(gram atomic weight of Nitrogen) + 3*(gram atomic weight of Hydrogen) = 1*14 + 3*1 = 17.
Since the mass of nitrogen is proportional to the total mass of ammonia,  you can write the proportion
mN%2F680 = 14%2F17.
This proportion has the unknown extreme term  mN.
Use the rule for solving proportions from the lesson  Proportions:  the unknown extreme term is equal to the product of mean terms divided by the known extreme term.
Applying this rule you get the unknown value of  mN
mN = 680%2A14%2F17 = 560.

Answer.  The mass of nitrogen in  680  grams of ammonia is  560 grams.


Problem 3. Mass of methane consumed in the burning reaction

Consider the burning of natural gas,  which is composed mostly of methane  CH%5B4%5D.
The chemical formula for the methane burning reaction is
CH%5B4%5D+%2B+2O%5B2%5D+=+CO%5B2%5D+%2B+2H%5B2%5DO
(one molecule of methane  CH%5B4%5D  and two molecules of oxygen  O%5B2%5D  form one molecule of carbon dioxide  CO%5B2%5D  and two molecules of water H%5B2%5DO).
If you have got  500  grams of water when burned methane,  how much methane was consumed?

Solution
Let  mCH%5B4%5D  be the mass of the methane consumed in the reaction.
The molecular weight of water  H%5B2%5DO  is  2*(gram atomic weight of Hydrogen) + 1*(gram atomic weight of Oxygen) = 2*1 + 16 = 18.
The molecular weight of methane  CH%5B4%5D  is  1*(gram atomic weight of Carbon) + 4*(gram atomic weight of Hydrogen) = 1*14 + 4*1 = 18.
Since one molecule of methane produces two molecules of water,  you can write the proportion
mCH%5B4%5D%2F500 = 18%2F%282%2A18%29.
This proportion has the unknown extreme term  mCH%5B4%5D.
Use the rule for solving proportions from the lesson  Proportions:  the unknown extreme term is equal to the product of mean terms divided by the known extreme term.
Applying this rule you get the unknown of  mCH%5B4%5D
mCH%5B4%5D = 500%2A18%2F%282%2A18%29 = 250.

Answer.  The mass of the consumed methane is  250  grams.


Problem 4. Redox reaction

The following is an example of a redox reaction.
Zinc metal reacts with  copper(II) sulfate solution  (aqueous)  to give copper metal and a solution of zinc sulfate  (aqueous).
Zn%28s%29+%2B+CuSO%5B4%5D ---> ZnSO%5B4%5D+%2B+Cu%28s%29.
Actually,  one atom of  Zink  displaces and replaces one atom of  Copper  in the reaction.
If  10  grams of zink  (metal)  was the input,  how much copper  (metal)  was replaced in this reaction?

Solution
Let  mCu  be the mass of the copper replaced in the reaction.
The gram atomic weight of  Zinc  is  65.38.
The gram atomic weight of  Copper  is  63.546.
Since,  according to the reaction formula,  one atom of  Zinc  produces one atom of  Copper  in the reaction,  you can write the proportion
mCu%2F10 = 63.546%2F65.38.
This proportion has the unknown extreme term  mCu.
Use the rule for solving proportions from the lesson  Proportions:  the unknown extreme term is equal to the product of mean terms divided by the known extreme term.
Applying this rule you get the unknown mass of copper replaced in this reaction
mCu+=+10%2A63.546%2F65.38+=+9.719 grams.

Answer.  The mass of copper replaced in this reaction is equal to  9.719  gram.


Problem 5. Photosyntesis reaction

Photosynthesis  is a chemical process in green plants that converts  carbon dioxide  gas of the atmoshere
into  organic compounds,  especially  sugars,  using the energy from sunlight  (from  Wikipedia).
The  Photosynthesis reaction  formula is
6+CO%5B2%5D+%2B+6+H%5B2%5DO + light energy ---> C%5B6%5DH%5B12%5DO%5B6%5D+%2B+6+O%5B2%5D
(carbon dioxide gas  CO%5B2%5D  from the atmosphere and water  H%5B2%5DO  from the soil are consumed by plants,  as well as the sun light energy,
producing carbohydrate and releasing oxygen gas  O%5B2%5D  to the atmosphere).
If  10  tons of carbohydrate mass was produced,  how much carbon dioxide was consumed and how much oxygen was released in this reaction?

The atomic weight of hydrogen is approximately  1.
The atomic weight of carbon is approximately  12.
The atomic weight of oxygen is approximately  16.

Solution

Let  mCO%5B2%5D  be the mass of carbon dioxide consumed in the reaction,  and let  mO%5B2%5D  be the mass of oxygen released to the atmosphere in the reaction.

The gram molecular weight of carbon dioxide is  12+%2B+2%2A16+=+44.
The gram molecular weight of carbohydrate  C%5B6%5DH%5B12%5DO%5B6%5D  is  6%2A12+%2B+12+%2B+6%2A16+=+180.
Since,  in accordance to the photosynthesis reaction formula,  6  molecules of carbon dioxide  CO%5B2%5D  are consumed
to produce one molecule of carbohydrate  C%5B6%5DH%5B12%5DO%5B6%5D,  you can write the proportion
mCO%5B2%5D%2F10000 = 6%2A44%2F180,
where  10000  represents  10000 kg  or  10 tons  of mass of carbohydrate.
This proportion has the unknown extreme term  mCO%5B2%5D.
Use the rule for solving proportions from the lesson  Proportions:  the unknown extreme term is equal to the product of mean terms divided by the known extreme term.
Applying this rule you get the unknown mass of carbon dioxide consumed from the atmosphere
mCO%5B2%5D = 6%2A44%2A10000%2F180 = 14666.7 kilograms.

Since,  according to the photosynthesis reaction formula,  6  molecules of oxygen  O%5B2%5D  are released to the atmosphere
when one molecule of carbohydrate  C%5B6%5DH%5B12%5DO%5B6%5D  is produced,  you can write the proportion
mO%5B2%5D%2F10000 = 6%2A32%2F180,
where  10000  represents same  10000 kg  or  10 tons  of mass of carbohydrate.
This proportion has the unknown extreme term  mO%5B2%5D.
Use the rule for solving proportions from the lesson  Proportions:  the unknown extreme term is equal to the product of mean terms divided by the known extreme term.
Applying this rule you get the unknown mass of oxygen released to the atmosphere
mO%5B2%5D = 6%2A32%2A10000%2F180 = 10666.7 kilograms.

Answer.  The mass of carbon dioxide consumed from the atmosphere in this reaction is equal to  14666.7 kilograms.
The mass of oxygen released to the atmosphere in this reaction is equal to  10666.7 kilograms.


My other lessons on  proportions  in this site are
    - Proportions
    - Using proportions to solve word problems
    - Using proportions to solve word problems in Physics
    - Typical problems on proportions
    - Using proportions to estimate the number of fish in a lake
    - HOW TO algebraize and solve these problems using proportions
    - Using proportions to solve word problems in Geometry
    - Using proportions to solve some nice simple Travel and Distance problems
    - Advanced problems on proportions
    - Problems on proportions for mental solution
    - Selected problems on proportions from the archive
    - Entertainment problems on proportions
    - OVERVIEW of lessons on proportions

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