Question

Explain which of the following statement(s) is (are) correct concerning glucose (blood sugar), \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (a) The percentages, by mass, of \(\mathrm{C}\) and \(\mathrm{O}\) are the same as in CO. (b) The ratio of \(\mathrm{C}: \mathrm{H}: \mathrm{O}\) atoms is the same as in dihydroxyacetone, \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2} \mathrm{CO}\) (c) The proportions, by mass, of \(C\) and \(O\) are equal. (d) The highest percentage, by mass, is that of H.

Short answer

Only statement (b) is correct.

Step-by-step solution

Understanding the Compound Structure

Glucose has a molecular formula of \(C_6H_{12}O_6\) and dihydroxyacetone has a molecular formula of \((CH_2OH)_{2}CO\). The total atomic masses of carbon (C), hydrogen (H), and oxygen (O) are 12 g/mol, 1 g/mol, and 16 g/mol respectively.

Evaluating Statement (a)

In CO, there is one atom each of carbon and oxygen. As the atomic masses of carbon and oxygen are 12 g/mol and 16 g/mol respectively, the percentage mass of carbon is approximately 43% and that of oxygen is approximately 57%. However, in glucose, one must first calculate the total mass of glucose molecule by adding the atomic masses of all constituent atoms together, which equals 180 g/mol. Then, the percentage mass of carbon and oxygen in glucose are calculated. For carbon, it is \( (6 \cdot 12 / 180) \cdot 100 = 40% \) and for oxygen, it is \( (6 \cdot 16 / 180) \cdot 100 = 53.3% \). These percentages are not the same as in CO. Therefore, statement (a) is incorrect.

Evaluating Statement (b)

For glucose, the ratio of C:H:O is 6:12:6 or 1:2:1 when simplified. For dihydroxyacetone or \((CH_2OH)_{2}CO\), there are 3 Carbon atoms, 6 Hydrogen atoms and 3 Oxygen atoms, so the ratio of C:H:O is 3:6:3 or 1:2:1 when simplified. Therefore, statement (b) is correct.

Evaluating Statement (c)

In glucose, the proportion mass of carbon and oxygen has been calculated previously in Step 2 as 40% and 53.3% respectively. The proportions are not equal. Therefore, statement (c) is incorrect.

Evaluating Statement (d)

In glucose, the proportion mass of hydrogen is calculated as \( \left(\frac{12 \times 1 }{180}\right) \times 100 = 6.7% \) which is not greater than the proportions of either carbon or oxygen. Therefore, statement (d) is incorrect.

Key concepts

Molecular Formula

A molecular formula provides crucial information about the exact number and types of atoms present in a molecule. For glucose, the molecular formula is \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \). This means each glucose molecule contains 6 carbon (C), 12 hydrogen (H), and 6 oxygen (O) atoms.

The molecular formula helps us understand the composition and structure, providing insight into the molecule's possible chemical reactions and properties.

It differs from the empirical formula, which only shows the simplest ratio of elements, by showing the actual number of atoms.

• Glucose Molecular Formula: \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \)
• Dihydroxyacetone Molecular Formula: \( (\mathrm{CH}_{2}\mathrm{OH})_2 \mathrm{CO} \)

Understanding molecular formulas is like having a detailed recipe—it tells us exactly what ingredients are needed and in what amounts to create the final product.

Atomic Mass

Atomic mass represents the average mass of an element's atoms, usually in grams per mole (g/mol). For chemistry calculations, such as those involving glucose, atomic masses are essential.

Knowing the atomic mass of each element allows us to compute the mass of the molecule as a whole.

• Carbon (C): 12 g/mol
• Hydrogen (H): 1 g/mol
• Oxygen (O): 16 g/mol

By considering glucose's molecular formula \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), we calculate the total molecular mass:

\[ 6(12) + 12(1) + 6(16) = 180 \text{ g/mol} \]
Understanding atomic mass is fundamental to creating accurate calculations and understanding the substance's properties.

Chemical Ratios

Chemical ratios reflect the proportion of different types of atoms in a compound. Such ratios are vital for understanding the compound's makeup and potential behavior.

In glucose, with a formula of \( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \), the ratio of carbon to hydrogen to oxygen atoms is 6:12:6.
Simplifying this ratio, we get 1:2:1.

This tells us that in every glucose molecule, for every carbon atom there are two hydrogen atoms and one oxygen atom.

• Glucose C:H:O Ratio: 1:2:1
• Dihydroxyacetone C:H:O Ratio: 1:2:1

Chemical ratios are important for predicting reactions and the potential products, as they maintain consistency across different conditions of a reaction.

Percentage Composition

Percentage composition represents the mass percentage of each element within a compound. This concept is very helpful when we want to compare the content of elements in different compounds.

For glucose:\[ \%C = \left(\frac{6 \times 12}{180}\right) \times 100 = 40\% \] \[ \%H = \left(\frac{12 \times 1}{180}\right) \times 100 = 6.7\% \] \[ \%O = \left(\frac{6 \times 16}{180}\right) \times 100 = 53.3\% \]
The calculations above demonstrate how each element's contribution to the entire mass of glucose is considered.

• Higher proportion of oxygen at 53.3%.
• Carbon contributes substantially with 40%.
• Hydrogen makes up a smaller 6.7%.

Understanding percentage composition allows us to investigate chemical reactions further, knowing which elements contribute more or less to a compound's mass.