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Formaldehyde, known chemically as \( \mathrm{CH}_{2} \mathrm{O} \), is one of the simplest forms of aliphatic aldehydes and is a significant industrial chemical. Formaldehyde is often used in the production of resins, and as a preservative in cosmetic and medical products. It’s a small but reactive molecule, making it very useful in polymer manufacturing, such as in the creation of Bakelite and other plastic resins. Additionally, its antiseptic properties allow it to be useful in sanitizing and embalming solutions.**Mass Percent Composition** To calculate the mass percent composition of formaldehyde, we must look at each of its constituent elements:- **Carbon (C):** The compound contains one Carbon atom, with a molar mass of 12.01 g/mol.- **Hydrogen (H):** There are two Hydrogen atoms, each contributing 1.01 g/mol, totaling 2.02 g/mol.- **Oxygen (O):** The single Oxygen atom in CH₂O adds 16.00 g/mol.The total molar mass of formaldehyde is therefore 30.03 g/mol.By evaluating the percentage of each element's contribution to the total mass, we find:- **C:** \( \frac{12.01}{30.03} \times 100 = 40.00\% \)- **H:** \( \frac{2.02}{30.03} \times 100 = 6.70\% \)- **O:** \( \frac{16.00}{30.03} \times 100 = 53.30\% \)These percentages highlight the proportions of each element in the compound, necessary for understanding its reactivity and application.

Glucose, with the formula \( C_{6} \mathrm{H}_{12} \mathrm{O}_{6} \), is a simple sugar and a primary energy source for cells in most living organisms. This monosaccharide is a key player in cellular respiration, providing energy by oxidation. Glucose is also a building block for carbohydrates and is found naturally in the bloodstream of animals and in plants through photosynthesis.**Mass Percent Composition** To delve into the composition of glucose, we count the molar mass of each element:- **Carbon (C):** Six Carbon atoms make up 72.06 g/mol (6\( \times \)12.01 g/mol).- **Hydrogen (H):** Twelve Hydrogen atoms contribute 12.12 g/mol (12\( \times \)1.01 g/mol).- **Oxygen (O):** Six Oxygen atoms add up to 96.00 g/mol (6\( \times \)16.00 g/mol).Resulting in a total molar mass of 180.18 g/mol for glucose.By turning these into percentages, we get:- **C:** \( \frac{72.06}{180.18} \times 100 = 40.00\% \)- **H:** \( \frac{12.12}{180.18} \times 100 = 6.72\% \)- **O:** \( \frac{96.00}{180.18} \times 100 = 53.28\% \)These mass percentages are essential in biochemistry, influencing how glucose is stored and metabolized in organisms.

Acetic acid is represented by the formula \( \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2} \) and is prominent as a weak organic acid found in vinegar. This compound imparts vinegar's sour taste and is widely used in food preservation, as a solvent in the production of synthetic fibers and plastics, and as a reagent in chemical processes.Understanding the mass percent composition of acetic acid will give insights into its properties and behavior in solution:- **Carbon (C):** There are 2 Carbon atoms, with a combined mass contribution of 24.02 g/mol (2\( \times \)12.01 g/mol).- **Hydrogen (H):** The molecule contains 4 Hydrogen atoms, adding up to 4.04 g/mol (4\( \times \)1.01 g/mol).- **Oxygen (O):** The 2 Oxygen atoms contribute 32.00 g/mol (2\( \times \)16.00 g/mol).This results in a molar mass of 60.05 g/mol for acetic acid.When we convert these to mass percentages, we find:- **C:** \( \frac{24.02}{60.05} \times 100 = 40.03\% \)- **H:** \( \frac{4.04}{60.05} \times 100 = 6.73\% \)- **O:** \( \frac{32.00}{60.05} \times 100 = 53.24\% \)These proportions are crucial for industrial applications, where precise concentrations determine the acidity and reactivity of acetic acid in solutions.