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Chapter 4: Problem 45

In photosynthesis, green plants convert \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) to glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). How many grams of \(\mathrm{CO}_{2}\) are required to produce \(5.1 \mathrm{g}\) of glucose? \(6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\ell) \frac{\text { Photosynthesis }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{aq})+6 \mathrm{O}_{2}(\mathrm{g})\)

Short Answer

Expert verified
7.47 grams of \(\mathrm{CO}_{2}\) are required.

Step by step solution

01

Determine the molar mass of glucose

To find the molar mass of glucose, \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), add the atomic masses of its atoms: \(6\times12.01\,\mathrm{g/mol}\) for carbon, \(12\times1.01\,\mathrm{g/mol}\) for hydrogen, and \(6\times16.00\,\mathrm{g/mol}\) for oxygen. The molar mass is calculated as follows:\[\text{Molar mass of } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} = (6\times12.01) + (12\times1.01) + (6\times16.00) = 180.18\,\mathrm{g/mol}\]
02

Determine moles of glucose produced

Using the mass of glucose produced, \(5.1\,\mathrm{g}\), and its molar mass, \(180.18\,\mathrm{g/mol}\), calculate the number of moles of glucose:\[\text{Moles of glucose} = \frac{5.1}{180.18} \approx 0.0283\,\text{mol}\]
03

Use stoichiometry to find moles of CO2 needed

From the balanced equation \(6\,\mathrm{CO}_{2} + 6\,\mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6} + 6\,\mathrm{O}_{2}\), we see 6 moles of \(\mathrm{CO}_2\) are required to produce 1 mole of \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\). Thus, for 0.0283 moles of glucose:\[\text{Moles of } \mathrm{CO}_2 = 0.0283 \times 6 = 0.1698\,\text{mol}\]
04

Calculate the mass of CO2 required

Determine the molar mass of \(\mathrm{CO}_{2}\), which involves one carbon \(12.01\,\mathrm{g/mol}\) and two oxygen atoms \(16.00\,\mathrm{g/mol}\) each:\[\text{Molar mass of } \mathrm{CO}_{2} = 12.01 + 2\times16.00 = 44.01\,\mathrm{g/mol}\]Using the number of moles of \(\mathrm{CO}_2\), \(0.1698\,\text{mol}\), calculate the mass:\[\text{Mass of } \mathrm{CO}_2 = 0.1698 \times 44.01 \approx 7.47\,\mathrm{g}\]

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Key Concepts

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

Molar Mass Calculation
To perform calculations involving chemical compounds, understanding molar mass is crucial. The molar mass of a compound represents the mass of one mole of that compound. For glucose, or \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), this is determined by adding up the atomic masses of each element within the molecule.
The steps are as follows:
  • Carbon ( Cdata-) in glucose has 6 atoms, each with a mass of 12.01 g/mol.
  • Hydrogen ( Hdata-) is present in 12 atoms, each weighing 1.01 g/mol.
  • Oxygen ( Odata-) which consists of 6 atoms, each having a mass of 16.00 g/mol.
When these are all added, the molar mass of glucose comes to: \( 180.18 \, \text{g/mol} \).This value helps us to convert between grams and moles in chemical reactions.
Stoichiometry
Stoichiometry allows us to quantify the relationship between reactants and products in a chemical reaction. By looking at the balanced chemical equation for photosynthesis, \(6\, \mathrm{CO}_2 + 6\, \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\, \mathrm{O}_2\), we can see the ratio of substances involved.
Here's how it works in our exercise:
  • The formation of 1 mole of glucose requires 6 moles of carbon dioxide.
  • With this stoichiometric ratio (6:1), we can determine how many moles of carbon dioxide are needed for any given amount of glucose.

In this problem, the stoichiometry helps translate the moles of glucose required (0.0283 mol) into moles of \( \mathrm{CO}_2 \), which is \( 0.1698 \text{ mol} \). This means moles of reactants are not just abstract numbers; they directly relate to the quantities of substances we deal with practically.
Chemical Reactions
Chemical reactions鈥攍ike photosynthesis鈥攆acilitate the conversion of reactants into products. The reaction here involves \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \mathrm{O} \), which transform into glucose (\( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \)) and oxygen.
Notably, this reaction is:
  • Balanced: Each side of the equation has equal numbers of each type of atom.
  • Guided by stoichiometry: It tells us how many molecules of \( \mathrm{CO}_{2} \) need to react with \( \mathrm{H}_{2} \mathrm{O} \).

In balancing and understanding chemical reactions, it becomes manageable to determine the necessary quantities of substances involved, which is integral to calculations like our problem demands. Thus, observing stoichiometric coefficients helps streamline the process of scaling up or down in chemical synthesis.
Glucose Production
The ultimate goal of the photosynthesis reaction is the production of glucose. Glucose serves as a primary energy source for plants and, indirectly, for many other organisms. For this exercise, the task was to determine the mass of \( \mathrm{CO}_2 \) needed to produce a specific mass of glucose鈥擻( 5.1 \, \mathrm{g} \).
By understanding the balanced equation and converting measured values through stoichiometry, we get to the root of how glucose quantities are tied to amounts of reactants.
  • The moles calculated for glucose help derive equivalent moles for \( \mathrm{CO}_2 \) needed.
  • Then, use the molar mass of \( \mathrm{CO}_2 \) (44.01 g/mol) to find the mass in grams.
This step-by-step relationship shows how intertwined these concepts are, where starting with just a few grams in the lab scales up to understanding a grand natural process.

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Most popular questions from this chapter

Diethyl ether is made from ethanol according to the following reaction: \(2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\ell) \longrightarrow\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{O}(\ell)+\mathrm{H}_{2} \mathrm{O}(\ell)\) In an experiment, 517 g of ethanol gave 391 g of diethyl ether. What was the percent yield in this experiment?

The two major sources of energy in our diets are fats and carbohydrates. Palmitic acid, one of the major components of both animal fats and vegetable oils, belongs to a group of compounds called fatty acids. The metabolism of fatty acids is responsible for the energy from fats. The major carbohydrates in our diets are sucrose (table sugar; Section \(19.4 \mathrm{A}\) ) and starch (Section 19.5A). Both starch and sucrose are first converted in the body to glucose, and then glucose is metabolized to produce energy. The heat of combustion of palmitic acid is 2385 kcal/mol, and that of glucose is \(670 .\) kcal/mol. Below are unbalanced equations for the metabolism of each body fuel: (a) Balance the equation for the metabolism of each fuel. (b) Calculate the heat of combustion of each in kcal/g. (c) In terms of kcal/mol, which of the two is the better source of energy for the body? (d) In terms of kcal/g, which of the two is the better source of energy for the body?

Calculate the number of grams in: (a) \(1.77 \mathrm{mol}\) of nitrogen dioxide, \(\mathrm{NO}_{2}\) (b) \(0.84 \mathrm{mol}\) of 2 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) (rubbing alcohol) (c) 3.69 mol of uranium hexafluoride, UF \(_{6}\) (d) 0.348 mol of galactose, \(C_{6} H_{12} O_{6}\) (e) \(4.9 \times 10^{-2}\) mol of vitamin \(C, C_{6} H_{8} O_{6}\)

Which of these reactions are exothermic, and which are endothermic? (a) \(2 \mathrm{NH}_{3}(\mathrm{g})+22.0 \mathrm{kcal} \longrightarrow \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g})+124\) kcal (c) \(\quad \mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+94.0 \mathrm{kcal}\) (d) \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+9.80 \mathrm{kcal} \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) (e) \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+\) \(4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+531 \mathrm{kcal}\)

The heat of combustion of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is \(670 \mathrm{kcal} /\) mol. The heat of combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\), is 327 kcal/mol. The heat liberated by oxidation of each compound is the same whether it is burned in air or metabolized in the body. On a kcal/g basis, metabolism of which compound liberates more heat?
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