Question

Determine the formula weights of each of the following compounds: (a) nitrous oxide, \(\mathrm{N}_{2} \mathrm{O},\) known as laughing gas and used as an anesthetic in dentistry; (b) benzoic acid, \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\), a substance used as a food preservative; (c) \(\mathrm{Mg}(\mathrm{OH})_{2}\), the active ingredient in milk of magnesia; (d) urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}, \mathrm{a}\) compound used as a nitrogen fertilizer; (e) isopentyl acetate, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{5} \mathrm{H}_{11}\), responsible for the odor of bananas.

Short answer

The formula weights of the given compounds are as follows: (a) Nitrous oxide, $\mathrm{N}_{2} \mathrm{O}$: 44.013 g/mol (b) Benzoic acid, $\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}$: 122.123 g/mol (c) Magnesium hydroxide, $\mathrm{Mg}(\mathrm{OH})_{2}$: 58.319 g/mol (d) Urea, $\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}$: 60.056 g/mol (e) Isopentyl acetate, $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{C}_{5} \mathrm{H}_{11}$: 114.188 g/mol

Step-by-step solution

(a) Formula weight of Nitrous oxide (N2O)

: 1. Identify elements: Nitrogen (N) and Oxygen (O) 2. Atomic weights: Nitrogen (N) = 14.007, Oxygen (O) = 15.999 3. Multiply by the number of atoms: (2 × 14.007) + (1 × 15.999) 4. Formula weight of N2O = 28.014 + 15.999 = 44.013 g/mol

(b) Formula weight of Benzoic acid (HC7H5O2)

: 1. Identify elements: Hydrogen (H), Carbon (C), and Oxygen (O) 2. Atomic weights: Hydrogen (H) = 1.008, Carbon (C) = 12.011, Oxygen (O) = 15.999 3. Multiply by the number of atoms: (1 × 1.008) + (7 × 12.011) + (5 × 1.008) + (2 × 15.999) 4. Formula weight of HC7H5O2 = 1.008 + 84.077 + 5.040 + 31.998 = 122.123 g/mol

(c) Formula weight of Mg(OH)2

: 1. Identify elements: Magnesium (Mg), Oxygen (O), and Hydrogen (H) 2. Atomic weights: Magnesium (Mg) = 24.305, Oxygen (O) = 15.999, Hydrogen (H) = 1.008 3. Multiply by the number of atoms: (1 × 24.305) + (2 × 15.999) + (2 × 1.008) 4. Formula weight of Mg(OH)2 = 24.305 + 31.998 + 2.016 = 58.319 g/mol

(d) Formula weight of Urea ((NH2)2CO)

: 1. Identify elements: Nitrogen (N), Hydrogen (H), Carbon (C), and Oxygen (O) 2. Atomic weights: Nitrogen (N) = 14.007, Hydrogen (H) = 1.008, Carbon (C) = 12.011, Oxygen (O) = 15.999 3. Multiply by the number of atoms: (2 × 14.007) + (4 × 1.008) + (1 × 12.011) + (1 × 15.999) 4. Formula weight of (NH2)2CO = 28.014 + 4.032 + 12.011 + 15.999 = 60.056 g/mol

(e) Formula weight of Isopentyl acetate (CH3CO2C5H11)

: 1. Identify elements: Hydrogen (H), Carbon (C), and Oxygen (O) 2. Atomic weights: Hydrogen (H) = 1.008, Carbon (C) = 12.011, Oxygen (O) = 15.999 3. Multiply by the number of atoms: (3 × 1.008) + (2 × 12.011) + (5 × 12.011) + (11 × 1.008) + (1 × 15.999) 4. Formula weight of CH3CO2C5H11 = 3.024 + 24.022 + 60.055 + 11.088 + 15.999 = 114.188 g/mol

Key concepts

Atomic Weight

Every element has a unique atomic weight, which is essentially the average mass of atoms of an element, calculated using the relative abundance of isotopes. This concept helps us to compute more complex measurements like formula or molecular weights. In simple terms:

• The atomic weight is measured in atomic mass units (amu), where 1 amu is defined as a twelfth of the mass of an atom of carbon-12.
• Each element's atomic weight is slightly different due to isotopic variations, meaning some naturally occurring atoms have extra neutrons.

When looking at nitrous oxide, for example, we use the atomic weights of nitrogen (14.007 amu) and oxygen (15.999 amu) to calculate its formula weight.

Chemical Compounds

Chemical compounds are substances formed from two or more elements that are chemically bonded in a fixed ratio. The properties of these compounds are distinct from their elements. Understanding chemical compounds involves examining:

• The types of bonds (ionic, covalent) holding the atoms together.
• The ratio of each element present in the compound.
• Specific examples, like nitrous oxide ( _2O), benzoic acid ( HC_7H_5O_2), and others, which demonstrate fixed proportions and unique properties.

Chemical compounds make up all the materials around us, including synthetic and natural substances. This concept is a stepping stone to more complex topics in chemistry.

Molecular Formula

A molecular formula provides the exact number of different atoms in a single molecule of a compound. This is different from an empirical formula, which shows the simplest whole-number ratio of atoms in the compound. Here are key aspects of molecular formulas:

• They are used for defining the composition of a molecule, showing each type of atom and its quantity.
• Compounds like benzoic acid have molecular formulas that not only represent the elements present but their exact numbers, like C_7H_5O_2.
• Molecular formulas are critical for determining the molecular weights necessary in chemical calculations.

Understanding these formulas is crucial for problem-solving in chemistry, revealing the detailed makeup of a compound.

Problem Solving in Chemistry

Approaching problems in chemistry, such as calculating formula weights, involves a systematic method. This method breaks down complex chemical questions into manageable steps:

• Start by identifying the elements present in the chemical formula and note their atomic weights.
• Multiply each atom's atomic weight by the number of times it appears in the formula.
• Sum these values to find the formula weight and express it in grams per mole.

For example, computing the formula weight of urea involves taking into account the correct amount of nitrogen, hydrogen, carbon, and oxygen in its molecular formula, (NH_2)_2CO. Problem-solving skills in chemistry are honed by practicing these calculations, understanding the principles backing them, and being attentive to detail. These steps are crucial for accuracy in both academic and practical chemistry applications.