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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, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), 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

Expert verified
In conclusion, the formula weights are: (a) Nitrous oxide (N鈧侽): \(44.02 \: g/mol\) (b) Benzoic acid (HC鈧嘓鈧匫鈧): \(122.13 \: g/mol\) (c) Mg(OH)鈧: \(58.33 \: g/mol\) (d) Urea ((NH鈧)鈧侰O): \(60.07 \: g/mol\) (e) Isopentyl acetate (CH鈧僀O鈧侰鈧匟鈧佲倎): \(133.24 \: g/mol\)

Step by step solution

01

(a) Nitrous oxide - N鈧侽

Elements present: Nitrogen (N) and Oxygen (O) Atomic weights: N = 14.01, O = 16.00 Formula weight of N鈧侽 = (2 * atomic weight of N) + (1 * atomic weight of O) = (2 * 14.01) + (1 * 16.00) = \(28.02 + 16.00\) = \(44.02 \: g/mol\)
02

(b) Benzoic acid - HC鈧嘓鈧匫鈧

Elements present: Hydrogen (H), Carbon (C), and Oxygen (O) Atomic weights: H = 1.01, C = 12.01, O = 16.00 Formula weight of HC鈧嘓鈧匫鈧 = (1 * atomic weight of H) + (7 * atomic weight of C) + (5 * atomic weight of H) + (2 * atomic weight of O) = (1 * 1.01) + (7 * 12.01) + (5 * 1.01) + (2 * 16.00) = \(1.01 + 84.07 + 5.05 + 32\) = \(122.13 \: g/mol\)
03

(c) Mg(OH)鈧

Elements present: Magnesium (Mg), Oxygen (O), and Hydrogen (H) Atomic weights: Mg = 24.31, O = 16.00, H = 1.01 Formula weight of Mg(OH)鈧 = (1 * atomic weight of Mg) +(2 * (atomic weight of O + atomic weight of H)) = (1 * 24.31) +(2 * (16.00 + 1.01)) = \(24.31 + 2 * 17.01\) = \(24.31 + 34.02\) = \(58.33 \: g/mol\)
04

(d) Urea - (NH鈧)鈧侰O

Elements present: Nitrogen (N), Hydrogen (H), Carbon (C), and Oxygen (O) Atomic weights: N = 14.01, H = 1.01, C = 12.01, O = 16.00 Formula weight of (NH鈧)鈧侰O = (2 * atomic weight of N) + (4 * atomic weight of H) + (1 * atomic weight of C) + (1 * atomic weight of O) = (2 * 14.01) + (4 * 1.01) + (1 * 12.01) + (1 * 16.00) = \(28.02 + 4.04 + 12.01 + 16.00\) = \(60.07 \: g/mol\)
05

(e) Isopentyl acetate - CH鈧僀O鈧侰鈧匟鈧佲倎

Elements present: Hydrogen (H), Carbon (C), and Oxygen (O) Atomic weights: H = 1.01, C = 12.01, O = 16.00 Formula weight of CH鈧僀O鈧侰鈧匟鈧佲倎 = (6 * atomic weight of H) + (7 * atomic weight of C) + (2 * atomic weight of O) + (11 * atomic weight of H) = (6 * 1.01) + (7 * 12.01) + (2 * 16.00) + (11 * 1.01) = \(6.06 + 84.07 + 32 + 11.11\) = \(133.24 \: g/mol\) In conclusion, the formula weights are: (a) Nitrous oxide (N鈧侽): \(44.02 \: g/mol\) (b) Benzoic acid (HC鈧嘓鈧匫鈧): \(122.13 \: g/mol\) (c) Mg(OH)鈧: \(58.33 \: g/mol\) (d) Urea ((NH鈧)鈧侰O): \(60.07 \: g/mol\) (e) Isopentyl acetate (CH鈧僀O鈧侰鈧匟鈧佲倎): \(133.24 \: g/mol\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Mass Calculation
The molecular mass of a compound, often referred to as the molecular weight, is a critical concept in chemistry. It is the sum of the atomic weights of all atoms in a molecule. This value is usually reported in grams per mole (g/mol), representing how much one mole of a substance weighs.

To calculate molecular mass, one must know the chemical formula of the substance and the atomic weights of the elements that comprise it. The steps involve multiplying the atomic weight of each element by the number of times that element appears in the molecule, then summing these values. For instance, when calculating the molecular mass of water (H鈧侽), the process involves the sum of twice the atomic weight of hydrogen and once the atomic weight of oxygen.

For accuracy, using the most up-to-date atomic weights provided by reliable resources such as the International Union of Pure and Applied Chemistry (IUPAC) is key.
Stoichiometry in Chemistry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It's grounded in the law of conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction.

Understanding the stoichiometry of a reaction allows chemists to predict the amounts of products that will form from given reactants. This is critical for scaling up reactions for industrial processes, ensuring reactions go to completion, and preventing wastage of chemicals.

An essential aspect of stoichiometry is the mole concept, which is a way to group atoms and molecules to talk about their amounts just as the dozen is used to group things by twelve. Therefore, studying stoichiometry involves working with the mole ratio derived from a balanced chemical equation and involves the use of formula weights to ensure proper proportions of substances.
Atomic Weights
Atomic weight, or atomic mass, is a fundamental property of an element's atoms. It's a weighted average of the masses of the isotopes of an element, as found in nature, relative to the mass of a carbon-12 atom, which is set as 12 atomic mass units (amu).

The periodic table lists the atomic weight for each element, allowing for the calculation of the molecular mass of compounds as seen in the provided exercise. For example, the atomic weight of carbon is approximately 12.01 amu, which means a carbon atom on average is slightly heavier than 12 times the mass of a carbon-12 atom.

This value is crucial because it forms the basis of calculating the formula weights of compounds by summing the atomic weights of the constituent atoms, as illustrated in the textbook solution. Keeping atomic weights in context is important for understanding the macroscopic quantities of chemicals used in experiments and industrial applications.

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Most popular questions from this chapter

Sodium hydroxide reacts with carbon dioxide as follows: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ Which reagent is the limiting reactant when \(1.85 \mathrm{~mol}\) \(\mathrm{NaOH}\) and \(1.00 \mathrm{~mol} \mathrm{CO}_{2}\) are allowed to react? How many moles of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\) can be produced? How many moles of the excess reactant remain after the completion of the reaction?

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resultant combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

Calculate the percentage by mass of oxygen in the following compounds: (a) morphine, \(\mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3} ;\) (b) codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\) (c) cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\) (d) tetracycline, \(\mathrm{C}_{22} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{8} ;\) (e) digitoxin, \(\mathrm{C}_{4} \mathrm{H}_{64} \mathrm{O}_{13} ;\) (f) vancomycin, \(\mathrm{C}_{66} \mathrm{H}_{75} \mathrm{Cl}_{2} \mathrm{~N}_{9} \mathrm{O}_{24}\)

(a) What is the mass, in grams, of \(0.0714\) mol of iron(III) sulfate? (b) How many moles of ammonium ions are in \(8.776 \mathrm{~g}\) of ammonium carbonate? (c) What is the mass, in grams, of \(6.52 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium \(^{\otimes}\) ) if \(0.05570\) mol weighs \(15.86 \mathrm{~g}\) ?

Calculate the following quantities: (a) mass, in grams, of \(0.105\) moles sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) (b) moles of \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) in \(143.50 \mathrm{~g}\) of this substance (c) number of molecules in \(1.0 \times 10^{-6} \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) number of \(\mathrm{N}\) atoms in \(0.410 \mathrm{~mol} \mathrm{NH}_{3}\)

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