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The fat stored in a camel's hump is a source of both energy and water. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced by the metabolism of \(1.0 \mathrm{~kg}\) of fat, assuming the fat consists entirely of tristearin \(\left(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}\right)\), a typical animal fat, and assuming that during metabolism, tristearin reacts with \(\mathrm{O}_{2}\) to form only \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\).

Short Answer

Expert verified
The mass of water produced by the metabolism of 1.0 kg of tristearin fat is approximately 1.11 kg.

Step by step solution

01

Determine the molar mass of tristearin

Knowing that tristearin has a molecular formula of C57H110O6, we can calculate its molar mass by using the atomic masses of carbon, hydrogen, and oxygen: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of O = 16.00 g/mol The molar mass of tristearin is thus: \(57(12.01) + 110(1.01) + 6(16.00) = 891.27 g/mol\)
02

Convert 1 kg of fat to moles

Now that we know the molar mass, we can convert the given mass of 1.0 kg tristearin to moles: (1000 g fat) x (1 mol tristearin/891.27 g tristearin) = \(1.122 \text{ moles of tristearin}\)
03

Write and balance the chemical equation for the metabolism of tristearin

The balanced chemical equation for the metabolism of tristearin is: \(C_{57}H_{110}O_{6} + 82O_{2} \rightarrow 57CO_{2} + 55H_{2}O\)
04

Calculate moles of water produced from 1 kg of tristearin

Using stoichiometry, we can determine the moles of water produced from the 1.122 moles of tristearin: \(1.122 \text{ moles of tristearin} \times \dfrac{55 \text{ moles of water}}{1 \text{ mole of tristearin}} = 61.71 \text{ moles of water}\)
05

Convert moles of water to mass

Finally, we convert the moles of water produced to mass using the molar mass of water: Molar mass of water = (2 x 1.01 g/mol) + (1 x 16 g/mol) = 18.02 g/mol \(61.71 \text{ moles of water} \times \dfrac{18.02 \text{g of water}}{1 \text{ mole of water}} = 1112.24 \text{g of water}\) Therefore, the mass of water produced by metabolism of 1.0 kg of tristearin fat is approximately 1112.24 g or 1.11 kg.

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Key Concepts

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

Molar Mass
The molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance. It takes into account the atomic masses of each element present in a molecular formula. This concept is crucial when solving chemical reaction problems because it allows us to convert between mass and moles, facilitating calculations and understanding of chemical reactions.

In the example of tristearin, with a molecular formula of \(\mathrm{C}_{57}\mathrm{H}_{110}\mathrm{O}_{6}\), we calculate the molar mass by adding the products of the number of atoms of each element and their atomic masses:
  • Carbon (C): 57 atoms multiplied by the atomic mass of 12.01 g/mol
  • Hydrogen (H): 110 atoms multiplied by 1.01 g/mol
  • Oxygen (O): 6 atoms multiplied by 16.00 g/mol
Combining these values, we find that the molar mass of tristearin is \(891.27 \text{ g/mol}\). This calculated value is then used to convert the mass of tristearin into moles, which is a necessary step for further stoichiometric calculations.
Stoichiometry
Stoichiometry is the section of chemistry that involves using balanced chemical equations to calculate the amount of reactants or products in a chemical reaction. It is based on the principle that matter is conserved in chemical processes, meaning the number of atoms of each element is the same on both sides of the equation. This conservation allows us to relate quantities of different substances in a reaction.

The stoichiometric process begins with a balanced chemical equation. In the metabolism of tristearin, the balanced equation is: \[ \mathrm{C}_{57}\mathrm{H}_{110}\mathrm{O}_{6} + 82\mathrm{O}_{2} \rightarrow 57\mathrm{CO}_{2} + 55\mathrm{H}_{2} \mathrm{O} \]From this balanced equation, we can deduce important stoichiometric relationships. For example, one mole of tristearin yields 55 moles of \(\mathrm{H_{2}O}\). Using these stoichiometric coefficients, it's possible to convert moles of tristearin to moles of \(\mathrm{H_{2}O}\). In this case, 1.122 moles of tristearin will produce \(1.122 \times 55 = 61.71\) moles of water. These conversions are key in determining the mass of water produced in reaction scenarios.
Balanced Chemical Equation
A balanced chemical equation is essential for understanding chemical reactions because it accurately represents the conservation of mass and energy. Balancing a chemical equation ensures that the number of atoms for each element is equal on both the reactant and product sides. This balance is achieved by adjusting coefficients before each compound.

For the metabolism of tristearin, the balanced chemical equation is: \[ \mathrm{C}_{57}\mathrm{H}_{110}\mathrm{O}_{6} + 82\mathrm{O}_{2} \rightarrow 57\mathrm{CO}_{2} + 55\mathrm{H}_{2} \mathrm{O} \]This equation clearly maintains a balance by having equal numbers of carbon, hydrogen, and oxygen atoms appearing on both sides. The coefficients (e.g., 82 for \(\mathrm{O}_{2}\), 57 for \(\mathrm{CO}_{2}\), and 55 for \(\mathrm{H}_{2} \mathrm{O}\)) are critical because they denote the proportions in which reactants combine and products form.

In practical terms, if you understand and can balance chemical equations, you can determine the quantities of reactants required or products formed in a reaction. This skill is critical in disciplines such as chemistry, biochemistry, and chemical engineering.

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Most popular questions from this chapter

A sample of the male sex hormone testosterone, \(\mathrm{C}_{19} \mathrm{H}_{28} \mathrm{O}_{2}\), contains \(3.88 \times 10^{21}\) hydrogen atoms. (a) How many atoms of carbon does it contain? (b) How many molecules of testosterone does it contain? (c) How many moles of testosterone does it contain? (d) What is the mass of this sample in grams?

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O} .\) If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\). A \(0.1005-g\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

A compound whose empirical formula is \(\mathrm{XF}_{3}\) consists of \(65 \%\) \(\mathrm{F}\) by mass. What is the atomic mass of \(\mathrm{X} ?\)

Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield? (c) Can a reaction ever have \(110 \%\) actual yield?

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which is the limiting reactant when \(1.85 \mathrm{~mol} \mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

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