Problem 24 - Chemistry The Central Science

Question

Determine the formula weights of each of the following compounds: (a) nittrous oxide, \(\mathrm{N}_{2} \mathrm{O}\) , known as laughing gas and used as an anesthetic in dentistry; (b) benzoic acid; \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) a substance used as a food preservative; \((c) \mathrm{Mg}(\mathrm{OH})_{2},\) the active ingredient in milk of magnesia; (d) urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO},\) 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.

Step-by-Step Solution

Answer

The formula weights of the given compounds are: (a) Nitrous oxide, \(N_2O\): 44.02 g/mol (b) Benzoic acid, \(C_6H_5COOH\): 160.19 g/mol (c) Magnesium hydroxide, \(Mg(OH)_2\): 58.33 g/mol (d) Urea, \((NH_2)_2CO\): 60.07 g/mol (e) Isopentyl acetate, \(CH_3CO_2C_5H_{11}\): 130.19 g/mol

Step 1: (a) Nitrous oxide, N₂O

: To compute the molecular weight of nitrous oxide, \(N_2O\), we first find the atomic weights (molar masses) of each element, using the periodic table: - Nitrogen (N) has an atomic weight of 14.01 g/mol - Oxygen (O) has an atomic weight of 16.00 g/mol The molecular weight of nitrous oxide is thus: \[M_{N_2O} =2 \cdot 14.01 + 16.00 = 44.02~\text{g/mol}\]

Step 2: (b) Benzoic acid, C₆H₅COOH

: Next, we find the molecular weight of benzoic acid, \(C_6H_5COOH\), by considering the atomic weights of its elements: - Carbon (C) has an atomic weight of 12.01 g/mol - Hydrogen (H) has an atomic weight of 1.01 g/mol - Oxygen (O) has an atomic weight of 16.00 g/mol The molecular weight of benzoic acid is thus: \[M_{C_6H_5COOH} = 6\cdot 12.01 + 5\cdot 1.01 + 12.01 + 1.01 + 2 \cdot 16.00 = 122.12 + 5.05 + 33.02 = 160.19~\text{g/mol}\]

Step 3: (c) Magnesium hydroxide, Mg(OH)₂

: For magnesium hydroxide, \(Mg(OH)_2\), we use the atomic weights of its elements: - Magnesium (Mg) has an atomic weight of 24.31 g/mol - Oxygen (O) has an atomic weight of 16.00 g/mol - Hydrogen (H) has an atomic weight of 1.01 g/mol The molecular weight of magnesium hydroxide is thus: \[M_{Mg(OH)_2} = 24.31 + 2 \cdot (16.00 + 1.01) = 24.31 + 34.02 = 58.33~\text{g/mol}\]

Step 4: (d) Urea, (NH₂)₂CO

: Now, for urea, \((NH_2)_2CO\), we use the atomic weights of its elements: - Nitrogen (N) has an atomic weight of 14.01 g/mol - Hydrogen (H) has an atomic weight of 1.01 g/mol - Carbon (C) has an atomic weight of 12.01 g/mol - Oxygen (O) has an atomic weight of 16.00 g/mol The molecular weight of urea is thus: \[M_{(NH_2)_2CO} = 2 \cdot (14.01 + 2\cdot 1.01) + 12.01 + 16.00 = 60.07~\text{g/mol}\]

Step 5: (e) Isopentyl acetate, CH₃CO₂C₅H₁₁

: Finally, we find the molecular weight of isopentyl acetate, \(CH_3CO_2C_5H_{11}\), using the atomic weights of its elements: - Carbon (C) has an atomic weight of 12.01 g/mol - Hydrogen (H) has an atomic weight of 1.01 g/mol - Oxygen (O) has an atomic weight of 16.00 g/mol The molecular weight of isopentyl acetate is thus: \[M_{CH_3CO_2C_5H_{11}} = 3\cdot 12.01 + 3\cdot 1.01 + 2\cdot 16.00 + 2\cdot 12.01 + 5\cdot 1.01 + 5 \cdot 12.01 + 11\cdot 1.01 = 130.19~\text{g/mol}\]

Key Concepts

Formula Weights of Compounds

Understanding the formula weights of compounds is integral to mastering chemistry. The formula weight, also referred to as molecular weight, is the sum of the atomic weights of all atoms in a chemical formula. Each element's atomic weight is multiplied by the number of times the element appears in the formula.

For instance, in a compound like water, H₂O, we would find the formula weight by adding the atomic weight of two hydrogen atoms with that of one oxygen atom. It's a straightforward calculation method that can be applied to any chemical compound, from simple diatomic molecules to complex organic structures.

In exercises, the computation of formula weights helps students familiarize themselves with the chemical make-up of substances and lays the foundation for further studies in stoichiometry, reactivity, and properties of substances.

Atomic Weight

The atomic weight, also known as the relative atomic mass, is essentially the mass of an atom on a scale where the mass of a carbon-12 atom is 12 units. Thanks to the periodic table, which lists the average atomic weight for each element based on isotopic composition and abundance, we can easily find this value for any given element.

These weights are averages because most elements exist as a mixture of different isotopes, each with its own atomic mass. For the sake of simplicity and practicality in calculations, we use these average atomic weights. This is pivotal in calculating the molecular weights of compounds, as seen in the exercises.

Periodic Table

The periodic table is not just a chart of elements; it's a comprehensive tool that organizes the chemical behavior of elements and provides vital data for chemical calculations. It presents elements in order of increasing atomic number and groups them into categories that exhibit similar chemical properties.

More importantly for our uses, the periodic table provides the atomic weights essential for molecular weight calculations. With this table, we relate the micro world of atoms to the macro world of grams and moles, making sense of quantities in chemical reactions.

Molar Mass

The molar mass is the bridge that connects the atomic scale with the real-life scale of chemistry. It is defined as the mass of one mole (approximately 6.022 x 10²³ entities) of a substance and is expressed in grams per mole (g/mol).

The molar mass of each element can be found on the periodic table and is roughly equal to the atomic weight. It's a fundamental concept that ties in with the idea of formula weights, as we consider the molar masses of individual elements to find the formula weight of the entire compound. Understanding how to apply the molar mass in calculations enables us to quantify how much of a substance we have in a reaction or solution.