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 Oxygen and one atom of Hydrogen).
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 us denote as mO%5B2%5D 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 us denote as mN 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 us denote as mCH%5B4%5D 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 us denote as mCu 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 us denote as mCO%5B2%5D the mass of carbon dioxide consumed in the reaction, and
denote as mO%5B2%5D 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.
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