Lesson Using proportions to solve Chemistry problems
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<H2>Using proportions to solve Chemistry problems </H2> This lesson is the continuation of the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A> of this module. The lesson presents some typical word problems related to <B>Chemistry</B> that can be solved using proportions. You will see how useful the proportions are in solving chemistry problems. <H3>Problem 1. Mass of oxygen in water</H3><BLOCKQUOTE>Chemical formula for water is {{{H[2]O}}} (two atoms of Hydrogen and one atom of Oxygen). Gram molecular weight of oxygen {{{O[2]}}} 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?</BLOCKQUOTE> <B>Solution</B> Let {{{mO[2]}}} be the mass of oxygen in 3600 grams of water. Note that gram molecular weight of water {{{H[2]O}}} 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[2]/3600}}} = {{{16/18}}}. This proportion has the unknown extreme term {{{mO[2]}}}. Use the rule for solving proportions from the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A>: 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[2]}}} {{{mO[2]}}} = {{{3600*16/18}}} = {{{3200}}}. <B>Answer</B>. The mass of oxygen in 3600 grams of water is 3200 grams. <H3>Problem 2. Mass of nitrogen in ammonia</H3><BLOCKQUOTE>Chemical formula for ammonia is {{{NH[3]}}} (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?</BLOCKQUOTE> <B>Solution</B> Let {{{mN}}} be the mass of nitrogen in 680 grams of ammonia. Note that gram molecular weight of ammonia {{{NH[3]}}} 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/680}}} = {{{14/17}}}. This proportion has the unknown extreme term {{{mN}}}. Use the rule for solving proportions from the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A>: 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*14/17}}} = {{{560}}}. <B>Answer</B>. The mass of nitrogen in 680 grams of ammonia is 560 grams. <H3>Problem 3. Mass of methane consumed in the burning reaction</H3><BLOCKQUOTE>Consider the burning of natural gas, which is composed mostly of methane {{{CH[4]}}}. The chemical formula for the methane burning reaction is {{{CH[4] + 2O[2] = CO[2] + 2H[2]O}}} (one molecule of methane {{{CH[4]}}} and two molecules of oxygen {{{O[2]}}} form one molecule of carbon dioxide {{{CO[2]}}} and two molecules of water {{{H[2]O}}}). If you have got 500 grams of water when burned methane, how much methane was consumed?</BLOCKQUOTE> <B>Solution</B> Let {{{mCH[4]}}} be the mass of the methane consumed in the reaction. The molecular weight of water {{{H[2]O}}} is 2*(gram atomic weight of Hydrogen) + 1*(gram atomic weight of Oxygen) = 2*1 + 16 = 18. The molecular weight of methane {{{CH[4]}}} 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[4]/500}}} = {{{18/(2*18)}}}. This proportion has the unknown extreme term {{{mCH[4]}}}. Use the rule for solving proportions from the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A>: 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[4]}}} {{{mCH[4]}}} = {{{500*18/(2*18)}}} = {{{250}}}. <B>Answer</B>. The mass of the consumed methane is 250 grams. <H3>Problem 4. Redox reaction</H3><BLOCKQUOTE>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(s) + CuSO[4]}}} ---> {{{ZnSO[4] + Cu(s)}}}. 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?</BLOCKQUOTE> <B>Solution</B> 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/10}}} = {{{63.546/65.38}}}. This proportion has the unknown extreme term {{{mCu}}}. Use the rule for solving proportions from the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A>: 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*63.546/65.38 = 9.719}}} grams. <B>Answer</B>. The mass of copper replaced in this reaction is equal to 9.719 gram. <H3>Problem 5. Photosyntesis reaction</H3><BLOCKQUOTE><B>Photosynthesis</B> is a chemical process in green plants that converts <B>carbon dioxide</B> gas of the atmoshere into <B>organic compounds</B>, especially <B>sugars</B>, using the energy from sunlight (from <A HREF= http://en.wikipedia.org/wiki/Photosynthesis>Wikipedia</A>). The Photosynthesis reaction formula is {{{6 CO[2] + 6 H[2]O}}} + light energy ---> {{{C[6]H[12]O[6] + 6 O[2]}}} (carbon dioxide gas {{{CO[2]}}} from the atmosphere and water {{{H[2]O}}} from the soil are consumed by plants, as well as the sun light energy, producing carbohydrate and releasing oxygen gas {{{O[2]}}} 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?</BLOCKQUOTE> The atomic weight of hydrogen is approximately 1. The atomic weight of carbon is approximately 12. The atomic weight of oxygen is approximately 16. <B>Solution</B> Let {{{mCO[2]}}} be the mass of carbon dioxide consumed in the reaction, and let {{{mO[2]}}} be the mass of oxygen released to the atmosphere in the reaction. The gram molecular weight of carbon dioxide is {{{12 + 2*16 = 44}}}. The gram molecular weight of carbohydrate {{{C[6]H[12]O[6]}}} is {{{6*12 + 12 + 6*16 = 180}}}. Since, in accordance to the photosynthesis reaction formula, 6 molecules of carbon dioxide {{{CO[2]}}} are consumed to produce one molecule of carbohydrate {{{C[6]H[12]O[6]}}}, you can write the proportion {{{mCO[2]/10000}}} = {{{6*44/180}}}, where 10000 represents 10000 kg or 10 tons of mass of carbohydrate. This proportion has the unknown extreme term {{{mCO[2]}}}. Use the rule for solving proportions from the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A>: 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[2]}}} = {{{6*44*10000/180}}} = {{{14666.7}}} kilograms. Since, according to the photosynthesis reaction formula, 6 molecules of oxygen {{{O[2]}}} are released to the atmosphere when one molecule of carbohydrate {{{C[6]H[12]O[6]}}} is produced, you can write the proportion {{{mO[2]/10000}}} = {{{6*32/180}}}, where 10000 represents same 10000 kg or 10 tons of mass of carbohydrate. This proportion has the unknown extreme term {{{mO[2]}}}. Use the rule for solving proportions from the lesson <A HREF= http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A>: 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[2]}}} = {{{6*32*10000/180}}} = {{{10666.7}}} kilograms. <B>Answer</B>. 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 <B>proportions</B> in this site are - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/-Proprtions.lesson>Proportions</A> - <A HREF = http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-word-problems.lesson>Using proportions to solve word problems</A> - <A HREF = http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-word-problems-in-Physics.lesson>Using proportions to solve word problems in Physics</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Typical-problems-on-proportions.lesson>Typical problems on proportions</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-estimate-the-number-of-fish-in-a-lake.lesson>Using proportions to estimate the number of fish in a lake</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/HOW-TO-algebreze-and-solve-this-problem-using-proportions.lesson>HOW TO algebraize and solve these problems using proportions</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-word-problems-in-Geometry.lesson>Using proportions to solve word problems in Geometry</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Using-proportions-to-solve-some-nice-simple-Travel-and-Distance-problems.lesson>Using proportions to solve some nice simple Travel and Distance problems</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Miscellaneous-problems-on--proportions.lesson>Advanced problems on proportions</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Problems-on-proportions-for-mental-solution.lesson>Problems on proportions for mental solution</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/Selected-problems-on-proportions-from-the-archive.lesson>Selected problems on proportions from the archive</A> - <A HREF=https://www.algebra.com/algebra/homework/proportions/lessons/Entertainment-problems-on-proportions.lesson>Entertainment problems on proportions</A> - <A HREF=http://www.algebra.com/algebra/homework/proportions/lessons/OVERVIEW-of-lessons-on-proportions.lesson>OVERVIEW of lessons on proportions</A>
