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Chapter 7: Problem 95

Fiber in the Diet Dietary fiber is a mixture of many compounds, including xylose \(\left(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\right)\) and methyl galacturonate \(\left(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7}\right)\) a. Do these compounds have the same empirical formula? b. Write balanced chemical equations for the complete combustion of xylose and methyl galacturonate.

Short Answer

Expert verified
Provide the balanced chemical equations for their complete combustion reactions. Answer: No, xylose and methyl galacturonate do not have the same empirical formula. The empirical formula for xylose is C鈧匟鈧佲個O鈧�, while the empirical formula for methyl galacturonate is C鈧嘓鈧佲倐O鈧�. The balanced chemical equations for their complete combustion reactions are: For xylose: C鈧匟鈧佲個O鈧� + 5O鈧� 鈫� 5CO鈧� + 5H鈧侽 For methyl galacturonate: C鈧嘓鈧佲倐O鈧� + 5O鈧� 鈫� 7CO鈧� + 6H鈧侽

Step by step solution

01

a. Finding empirical formulas and comparison

First, we need to find the empirical formula for xylose: Xylose has a molecular formula of \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\). Since there is no common divisor for the subscripts, the empirical formula remains the same: \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\). Next, we find the empirical formula for methyl galacturonate: Methyl galacturonate has a molecular formula of \(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7}\). Again, there is no common divisor for the subscripts. Thus, the empirical formula remains the same: \(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7}\). As we can see, the empirical formulas of xylose and methyl galacturonate are different, which means they do not have the same empirical formula.
02

b. Writing and balancing combustion reactions

For the complete combustion of xylose: The reaction involves oxygen (O2) and forms carbon dioxide (CO2) and water (H2O) as products. The unbalanced equation is: \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5} + O_{2} \rightarrow CO_{2} + H_{2}O\) Now, we balance the chemical equation: \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5} + 5O_{2} \rightarrow 5CO_{2} + 5H_{2}O\) For the complete combustion of methyl galacturonate: Again, the reaction involves oxygen (O2) and forms carbon dioxide (CO2) and water (H2O) as products. The unbalanced equation is: \(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7} + O_{2} \rightarrow CO_{2} + H_{2}O\) Now, we balance the chemical equation: \(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7} + 5O_{2} \rightarrow 7CO_{2} + 6H_{2}O\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equations
Chemical equations are a symbolic representation of chemical reactions. They show the substances involved and the products formed. The reactants are on the left, the products on the right, separated by an arrow indicating the direction of the reaction.

Balancing these equations is crucial, as it ensures the law of conservation of mass is followed. This means that the number of atoms for each element must be the same on both sides of the equation. It often involves adding coefficients to ensure this balance is achieved.
  • Reactants: substances that undergo a reaction.
  • Products: substances formed as a result of the reaction.
  • Coefficients: numbers added to balance the equation, ensuring equal numbers of each type of atom on both sides.
In the case of the combustion reactions of xylose and methyl galacturonate, each element's atoms are counted and adjusted using coefficients to create balanced chemical equations.
Combustion Reaction
A combustion reaction is a type of chemical reaction where a substance combines with oxygen to release energy in the form of heat and light. This process typically produces carbon dioxide and water as products when it involves organic compounds like xylose and methyl galacturonate.

Combustion reactions are essential in various applications, such as energy production, cooking, and vehicles. In chemistry, these reactions help us understand how different compounds react with oxygen.

In our example:
  • Xylose undergoes combustion to form carbon dioxide and water according to the balanced equation: \[\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5} + 5O_{2} \rightarrow 5CO_{2} + 5H_{2}O\]
  • Methyl galacturonate also combusts to form carbon dioxide and water: \[\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7} + 5O_{2} \rightarrow 7CO_{2} + 6H_{2}O\]
Balancing these reactions involves adjusting the coefficients to ensure the same number of atoms for each element, maintaining the balance of mass.
Dietary Fiber
Dietary fiber is an integral part of a healthy diet. It consists of plant-based compounds that are not completely broken down by digestive enzymes in the human gastrointestinal tract. Despite this, it plays several vital roles in maintaining good health.

Dietary fiber falls into two categories:
  • Soluble fiber: Dissolves in water, forming a gel-like substance. It can help lower blood cholesterol and improve blood glucose control.
  • Insoluble fiber: Does not dissolve in water. It aids in regular bowel movements and prevents constipation.
The xylose and methyl galacturonate mentioned in the exercise are types of sugar-derived compounds found in dietary fiber. They contribute various benefits such as promoting gut health and maintaining stable blood sugar levels.
Xylose
Xylose is a type of sugar, specifically a monosaccharide, commonly found in the hemicellulose component of plant cell walls. It plays a crucial role in the structure of dietary fiber.

It's part of a class of sugars known as aldoses, containing five-carbon atoms, hence called a pentose sugar. Xylose has the chemical formula of \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\) and is often used in food products as a sweetener due to its low-caloric content.

Xylose's role in dietary fiber is crucial:
  • It contributes to the overall fiber content of food, helping to regulate digestion.
  • It aids in promoting the growth of beneficial gut bacteria, improving digestive health.
Understanding xylose's interaction with other compounds and its effects on digestion helps in appreciating its nutritional benefits.
Methyl Galacturonate
Methyl galacturonate is a methyl ester of galacturonic acid, found as a component of the pectin in plant cell walls, another important constituent of dietary fiber. Structurally, it has the formula \(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7}\).

This compound is important for various reasons:
  • It contributes to the gelling properties in jams and jellies, as derived from pectin.
  • It plays a role in the hydration and structural integrity of plant cells.
In addition to its industrial uses, methyl galacturonate also has health implications when consumed as part of dietary fiber. It aids in digestion and can help to enhance the absorption of certain nutrients in the small intestine. Understanding these properties helps in utilizing dietary fiber effectively for both health benefits and culinary applications.

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Most popular questions from this chapter

A number of chemical reactions have been proposed for the formation of organic compounds from inorganic precursors, including the following: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{FeS}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{FeS}_{2}(s)+\mathrm{HCO}_{2} \mathrm{H}(\ell)$$ a. Identify the ions in \(\mathrm{FeS}\) and \(\mathrm{FeS}_{2}\) b. What are the names of \(\mathrm{FeS}\) and \(\mathrm{FeS}_{2} ?\) c. How much \(\mathrm{HCO}_{2} \mathrm{H}\) is obtained by reacting \(1.00 \mathrm{g}\) FeS, \(0.50 \mathrm{g} \mathrm{H}_{2} \mathrm{S},\) and \(0.50 \mathrm{g} \mathrm{CO}_{2}\) if the reaction results in a \(50.0 \%\) yield?

In a balanced chemical equation for the complete combustion of a hydrocarbon, what is the ratio of atoms of C in the hydrocarbon to molecules of \(\mathrm{CO}_{2}\) produced?

Two of the more common oxides of iron have the formulas \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3} .\) How much more oxygen combines with a given mass of iron to form \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) than combines with the same mass of iron to form FeO?

Mining for Gold Unlike most metals, gold occurs in nature as the pure element. Miners in California in 1849 searched for gold nuggets and gold dust in streambeds, where the denser gold could be easily separated from sand and gravel. However, larger deposits of gold are found in veins of rock and can be separated chemically in the following two-step process: (1) \(4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$4 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+4 \mathrm{NaOH}(a q)$$ (2) \(2 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+\mathrm{Zn}(s) \rightarrow\) $$2 \mathrm{Au}(s)+\mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right](a q)$$ If 23 kilograms of ore is \(0.19 \%\) gold by mass, how much Zn is needed to react with the gold in the ore? Assume that reactions 1 and 2 are \(100 \%\) efficient.

Chemistry of Soot A piece of glass held over a candle flame becomes coated with soot, which is the result of the incomplete combustion of candle wax. Elemental analysis of a compound extracted from a sample of soot gave the following results: \(7.74 \%\) H and \(92.26 \%\) C by mass. Calculate the empirical formula of the compound.
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