Question

When the sugar glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is burned in air, carbon dioxide and water vapor are produced. Write the balanced chemical equation for this process, and calculate the theoretical yield of carbon dioxide when \(1.00 \mathrm{g}\) of glucose is burned completely.

Short answer

The balanced chemical equation for the combustion of glucose is \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O}\). The theoretical yield of carbon dioxide when 1.00 g of glucose is burned completely is 1.46 g.

Step-by-step solution

Write the unbalanced equation

First, we'll write the unbalanced chemical equation for the process. Glucose (\(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\)) reacts with oxygen (\(\mathrm{O}_2\)) to form carbon dioxide (\(\mathrm{CO}_2\)) and water vapor (\(\mathrm{H}_2\mathrm{O}\)): $$ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} $$

Balance the chemical equation

We need to balance the chemical equation to ensure that the same number of each type of atom appears on both sides. We will balance each element one at a time: 1. There are 6 carbon atoms in glucose, so we need 6 carbon atoms in carbon dioxide. We'll put a coefficient of 6 in front of \(\mathrm{CO}_2\): $$ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + \mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} $$ 2. There are 12 hydrogen atoms in glucose, so we need 12 hydrogen atoms in water. We'll put a coefficient of 6 in front of \(\mathrm{H}_2\mathrm{O}\): $$ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + \mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} $$ 3. Finally, let's balance the oxygen atoms. There's a total of 18 oxygen atoms on the right side of the equation (12 oxygen atoms in carbon dioxide and 6 oxygen atoms in water). We'll put a coefficient of 6 in front of \(\mathrm{O}_2\) on the left side to balance the 18 oxygen atoms: $$ \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 + 6\mathrm{O}_2 \rightarrow 6\mathrm{CO}_2 + 6\mathrm{H}_2\mathrm{O} $$ Now the equation is balanced.

Calculate the theoretical yield of carbon dioxide

To find how much carbon dioxide is produced when 1.00 g of glucose is burned, we'll use stoichiometry. First, we need to determine the molar mass of glucose, which is: $$ 180.16 \frac{\text{g}}{\text{mol}} = 6(12.01)+12(1.01)+6(16.00) $$ Now we can use the balanced equation to determine the mole ratio between glucose and carbon dioxide: $$ \frac{6\text{ mol CO}_2}{1\text{ mol C}_6\text{H}_{12}\text{O}_6} $$ Next, we'll convert the 1.00 g of glucose to moles and then use the mole ratio to find the moles of carbon dioxide produced: $$ 1.00 \frac{\text{g}}{180.16 \frac{\text{g}}{\text{mol}}} = 0.00555\text{ mol }\text{C}_6\text{H}_{12}\text{O}_6 $$ Now we multiply the moles of glucose by the mole ratio to find the moles of carbon dioxide produced: $$ 0.00555\text{ mol }\text{C}_6\text{H}_{12}\text{O}_6 \times \frac{6\text{ mol CO}_2}{1\text{ mol C}_6\text{H}_{12}\text{O}_6} = 0.0333\text{ mol }\text{CO}_2 $$ Finally, we'll convert the moles of CO2 to grams using its molar mass (44.01 g/mol): $$ 0.0333\text{ mol }\text{CO}_2 \times 44.01\frac{\text{g}}{\text{mol}} = 1.46\text{ g CO}_2 $$

Conclusion

The theoretical yield of carbon dioxide when 1.00 g of glucose is burned completely is 1.46 g.

Key concepts

Balancing Chemical Equations

Understanding how to balance chemical equations is foundational to mastering chemical reactions. When we balance equations, we ensure that the same number of each type of atom is present on both sides of the equation, upholding the law of conservation of mass. For instance, the combustion of glucose mentioned in our exercise involves aligning the number of carbon, hydrogen, and oxygen atoms on the reactant side (glucose and oxygen) with those on the product side (carbon dioxide and water vapor).

This step is crucial because it reflects the reality of chemical processes—matter is neither created nor destroyed. Beginners might start with one element at a time, adjusting coefficients before compounds to achieve balance. A common tip is to start with the most complex molecule, typically the one with the most elements, and progress to the simplest last. To improve your skill in balancing chemical equations, practice with varied scenarios and molar relationships between reactants and products.

Stoichiometry

Stoichiometry is the mathematical relationship between reactants and products in a chemical reaction, based on the balanced equation. It provides a quantitative measure of how much reactant is needed to produce a desired amount of product, or conversely, how much product can be produced from a given amount of reactant. In our example with glucose, stoichiometry enables us to calculate the theoretical yield of carbon dioxide. Using the balanced equation, we establish a mole ratio which acts as a conversion factor—offering a clear path from the mass of glucose to the mass of carbon dioxide produced.

To enhance comprehension, think of stoichiometry as a recipe that dictates precise proportions. Just as a cook needs to know the amount of each ingredient for a dish, a chemist uses stoichiometry to figure out the quantities needed for a reaction. Pay attention to the mole ratio, as it's the key to connecting the reactants and products.

Molar Mass

Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). This is a critical concept in stoichiometry because it relates the mass of a substance to the amount in moles, allowing chemists to convert between mass and moles efficiently. Knowing the molar mass of glucose (180.16 g/mol) in our exercise, we could figure out how many moles of glucose are present in a given weight, and then how many moles of carbon dioxide will form. The molar mass of each substance in a reaction must be used to convert between mass and moles.

Calculation of molar mass involves the sum of the atomic masses of all atoms in the molecule (as taken from the periodic table). For instructional purposes, simplifying examples to diatomic molecules can make the concept easier to grasp before tackling more complex compounds. Always assure the accuracy of your molar masses, as they are crucial to obtaining correct stoichiometric calculations.