Equations and Mass Relationships in Foods/Metabolism of Dietary Sugar - ChemPRIME

Equations and Mass Relationships in Foods/Metabolism of Dietary Sugar

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We eat a lot of sugars, and in the next few sections we'll explore some of the body chemistry that explains why they may (or may not) lead to weight gain, why they're a good source of energy, and why astronauts excrete more water than they drink. As we've seen, there are many sugars, but one of the most common is sucrose, C12H22O11. A balanced overall chemical equation for the metabolism of sucrose will help us understand the questions above, and many others. A balanced chemical equation such as


2 C12H22O11(s) + 35 O2(g) → 12 CO2(g) + 11 H2O(l)      (1)


not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (1) says that 2 C12H22O11 molecules can react with 35 O2 molecules to give 12 CO2 molecules and 11 H2O molecules. It also says that 2 mol 2C12H22O11 would react with 35 mol O2 yielding 12 mol CO2 and 11 mol H2O.

In other words, the equation indicates that exactly 35 mol O2 must react for every 2 mol C12H22O11consumed. For the purpose of calculating how much O2 is required to react with a certain amount of C12H22O11 therefore, the significant information contained in Eq. (1) is the ratio


\frac{\text{35 mol O}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}


We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (1),


\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\right)=\frac{\text{35 mol O}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}

The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.



EXAMPLE 1 Derive all possible stoichiometric ratios from Eq. (1)


Solution Any ratio of amounts of substance given by coefficients in the equation may be used, so in addition to (2) above, we have:

\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{O}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{35 mol O}_{\text{2}}}

\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{CO}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{12 mol CO}_{\text{2}}}=\frac{\text{1 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{6 mol CO}_{\text{2}}}

\text{S}\left( \frac{\text{CO}_{\text{2}}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\right)=\frac{\text{12 mol CO}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}

\text{S}\left( \frac{\text{CO}_{2}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{12 mol CO}_{2}}{\text{11 mol H}_{\text{2}}\text{O}}      

\text{S}\left( \frac{\text{CO}_{2}}{\text{O}_{\text{2}}} \right)=\frac{\text{12 mol CO}_{2}}{\text{35 mol O}_{\text{2}}}      \text{S}\left( \frac{\text{O}_{\text{2}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{35 mol O}_{\text{2}}}{\text{11 mol H}_{\text{2}}\text{O}}


\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{11 mol H}_{\text{2}}\text{O}}      

There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.]



When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (1) as an example, this means that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the stoichiometric ratio S(O2/NH3):


\frac{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}}{n_{\text{O}_{\text{2}}\text{ consumed}}}=\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{O}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{35 mol O}_{\text{2}}}


Similarly, the ratio of the amount of H2O produced to the amount of NH3 consumed must be

S(H2O/C12H22O11):


\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}} = \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right) =\frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}


In general we can say that


\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{                         (3}\text{a)}


or, in symbols,


\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{                                                          (3}\text{b)}


Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.



EXAMPLE 2 Find the amount of water produced when 1 cup (roughly 8 oz or 240 g) C12H22O11 is consumed according to Eq. (1).


Solution The amount of water produced must be in the stoichiometric ratio S(H2O/C12H22O11) to the amount of sugar consumed, and the amount is n = m/M = 240 g /342.3 g mol-1 = .70 mol.


\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}} = \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right) =\frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}


Multiplying both sides by nC12H22O11 consumed, we have


n_{\text{H}_{\text{2}}\text{O produced}}=n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}\times \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right)=\text{0.70 mol}\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\times \frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}=\text{3}\text{.85 mol H}_{\text{2}}\text{O}

This calculation shows why an astronaut dirnks about 2 L of H2 per day, but excretes about 2.4 L of H2 per day! Think about it, and check your answer[1]


This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form


\text{amount of X consumed or produced}\overset{\begin{smallmatrix} 
 \text{stoichiometric} \\ 
 \text{   ratio X/Y} 
\end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}


or symbolically.


                                      n_{\text{X   consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y   consumed or produced}}

When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol C12H22O11 cancels 1 mol C12H22O11 but does not cancel 1 mol H2O.

The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.



EXAMPLE 3 It is estimated that each human exhales about 1 kg (2.2 lb)[2][3] of carbon dioxide per day. If that came entirely from glucose, what mass of glucose must be metabolized according to the equation:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O


Solution The problem asks that we calculate the mass of C6H12O6 consumed. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of CO2 to the mass of CO2. Then we can calculate the amount of C6H12O6 consumed from the amount of CO2 produced with a stoichiometric ratio, just as in Example 2. Finally, we can convert the amount of glucose to its mass using the molar mass as a conversion factor. It requires the stoichiometric ratio.

\text{S}\left( \frac{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}}

The amount of CO2 produced is n = m/M = 1000 g/44 g mol-1 = 22.7 mol

The amount of C6H12O6 consumed is then

n_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}= n_{\text{CO}_{\text{2}}\text{ produced}}\text{ }\!\!\times\!\!\text{  conversion factor}=\text{n mol CO}_{\text{2}}\times \frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}}

= 22.7 mol CO2 \times \frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}} = 3.79 mol

The mass of C6H12O6 is


\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{3.79 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}\times \frac{\text{180 g C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{682 g C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}

By similar calculations, we could show that to produce 6 mol of carbon dioxide, it takes the same amount of oxygen (22.7 mol) and the same amount (22.7 mol)of water will be produced. That's 22.7 mol x 32 g/mol = 726 g of oxygen consumed and 22.7 mol x 18 g/mol = 409 g of water produced. The reactants weigh 682 + 726 = 1408 g and the products weigh the same (within error), 1000 + 409 = 1409 g.

We notice two things: First, both products are eliminated (CO2 in breath, water in urine, so how do we gain weight? Only sugar that isn't metabolized goes into weight gain, and we'll see soon how to calculate that.

Second, this accounts for .409 kg of the total 2.4 L of water excreted per day. It's a rough estimate of the part that results from metabolism (not from ingested water).

With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of CO2 to amount (in moles) of C6H12O6 and the molar mass will convert amount of C6H12O6 to mass (in grams) of SO2. A schematic road map for the one-step calculation can be written as


n_{\text{CO}_{\text{2}}}\text{ }\xrightarrow{S\text{(C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}\text{/O}_{\text{2}}\text{)}}\text{ }n_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\xrightarrow{M_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}}\text{ }m_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}


Thus


\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{x}\text{xx mol CO}_{\text{2}}\times \text{ }\frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{11 mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{179 g}



The chemical reaction in this example is of environmental interest. If each person on Earth exhales 1000 g of CO2/day, and there are 6.7 billion people, this accounts for 6.7 x 1012 (6.7 quadrillion) g or 6.2 Tg (teragrams) of the greenhouse gas CO2. How does this compare with the amount produced by burning fuels in its effect on the atmosphere? In 2000, fossil fuel burning released 7 x 109 tons of carbon dioxide [4]. That's 7 x 109 tons x 907 kg/ton x 1000 g/kg or 6.3 x 1015. Our breath contributes about 0.1%. The mass of atmospheric carbon dioxide comes from calculations like the following:



EXAMPLE 4 What mass of oxygen and carbon dioxide would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?


Solution First, write a balanced equation


2C8H18 + 25O2 → 16CO2 + 18H2O


The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically


m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}}


m_{\text{O}_{\text{2}}}=\text{3}\text{.3 }\times \text{ 10}^{\text{15}}\text{ g }\times \text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\times \text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\times \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}}=\text{1}\text{.2 }\times \text{ 10}^{\text{16}}\text{ g }


Thus 12 Pg (petagrams) of O2 would be needed. By a similar calculation, we see that the amount of carbon dioxide would be 16/2 = 8 times the amount of octane consumed, or 2.3 x 1014 mol, which, multiplied by the molar mass 44 g/mol, is 1 x 1016 g or 10 Pg.



The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen.

References

  1. Some of the extra 0.4 L is present as moisture in food, but a large amount is produced by metabolism, where it is a product of a chemical reaction like the one in Example 2
  2. Harte, J. "Consider a Spherical Cow, A Course in Environmental Problme Solving", University Science Books, Sausalito, CA 1998, p. 263
  3. Campbell, J.A. J. Chem. Educ. 49, 181 (1972)
  4. http://www.worldwatch.org/node/1811
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