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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.

Short Answer

Expert verified
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 by step solution

01

(a) Nitrous oxide, N鈧侽

: 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: \[M_{N_2O} =2 \cdot 14.01 + 16.00 = 44.02~\text{g/mol}\]
02

(b) Benzoic acid, C鈧咹鈧匔OOH

: 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: \[M_{C_6H_5COOH} = 6\cdot 12.01 + 5\cdot 1.01 + 12.01 + 1.01 + 2 \cdot 16.00 = 122.12 + 5.05 + 33.02 = 160.19~\text{g/mol}\]
03

(c) Magnesium hydroxide, Mg(OH)鈧

: 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: \[M_{Mg(OH)_2} = 24.31 + 2 \cdot (16.00 + 1.01) = 24.31 + 34.02 = 58.33~\text{g/mol}\]
04

(d) Urea, (NH鈧)鈧侰O

: 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: \[M_{(NH_2)_2CO} = 2 \cdot (14.01 + 2\cdot 1.01) + 12.01 + 16.00 = 60.07~\text{g/mol}\]
05

(e) Isopentyl acetate, CH鈧僀O鈧侰鈧匟鈧佲倎

: 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: \[M_{CH_3CO_2C_5H_{11}} = 3\cdot 12.01 + 3\cdot 1.01 + 2\cdot 16.00 + 2\cdot 12.01 + 5\cdot 1.01 + 5 \cdot 12.01 + 11\cdot 1.01 = 130.19~\text{g/mol}\]

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Key Concepts

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

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.

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Most popular questions from this chapter

One of the most bizarre reactions in chemistry is called the Ugi reaction: $$ \begin{array}{l}{\mathrm{R}_{1} \mathrm{C}(=\mathrm{O}) \mathrm{R}_{2}+\mathrm{R}_{3}-\mathrm{NH}_{2}+\mathrm{R}_{4} \mathrm{COOH}+\mathrm{R}_{5} \mathrm{NC} \rightarrow} \\ {\mathrm{R}_{4} \mathrm{C}(=\mathrm{O}) \mathrm{N}\left(\mathrm{R}_{3}\right) \mathrm{C}\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) \mathrm{C}=\mathrm{ONHR}_{5}+\mathrm{H}_{2} \mathrm{O}}\end{array} $$ (a) Write out the balanced chemical equation for the Ugi reaction, for the case where \(R=C H_{3} C H_{2} C H_{2} C H_{2} C H_{2} C H_{2}-\) (this is called the hexyl group) for all compounds. (b) What mass of the "hexyl Ugi product" would you form if 435.0 \(\mathrm{mg}\) of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) was the limiting reactant?

Determine the empirical formulas of the compounds with the following compositions by mass: $$ \begin{array}{l}{\text { (a) } 10.4 \% \mathrm{C}, 27.8 \% \mathrm{S}, \text { and } 61.7 \% \mathrm{Cl}} \\ {\text { (b) } 21.7 \% \mathrm{C}, 9.6 \% \mathrm{O}, \text { and } 68.7 \% \mathrm{F}} \\ {\text { (c) } 32.79 \% \mathrm{Na}, 13.02 \% \mathrm{Al}, \text { and the remainder } \mathrm{F}}\end{array} $$

Very small semiconductor crystals, composed of approximately 1000 to \(10,000\) atoms, are called quantum dots. Quantum dots made of the semiconductor Case are now being used in electronic reader and tablet displays because they emit light efficiently and in multiple colors, depending on dot size. The density of CdSe is 5.82 \(\mathrm{g} / \mathrm{cm}^{3} .\) (a) What is the mass of one 2.5 -nm CdSe quantum dot? (b) CdSe quantum dots that are 2.5 \(\mathrm{nm}\) in diameter emit blue light upon stimulation. Assuming that the dot is a perfect sphere and that the empty space in the dot can be neglected, calculate how many Cd atoms are in one quantum dot of this size. (c) What is the mass of one 6.5-nm CdSe quantum dot? (d) CdSe quantum dots that are 6.5 \(\mathrm{nm}\) in diameter emit red light upon stimulation. Assuming that the dot is a perfect sphere, calculate how many Cd atoms are in one quantum dot of this size. (e) If you wanted to make one 6.5 -nm dot from multiple \(2.5-\)nm dots, how many 2.5 -nm dots would you need, and how many CdSe formula units would be left over, if any?

A mixture containing \(\mathrm{KClO}_{3}, \mathrm{K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3},\) and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{array}{l}{2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)} \\ {2 \mathrm{KHCO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g)} \\\ {\mathrm{K}_{2} \mathrm{CO}_{3}(s) \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)}\end{array} $$ The KCl does not react under the conditions of the reaction. If 100.0 g of the mixture produces 1.80 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}, 13.20 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and 4.00 \(\mathrm{g}\) of \(\mathrm{O}_{2},\) what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

A compound whose empirical formula is \(X F_{3}\) consists of 65\(\%\) F by mass. What is the atomic mass of \(X ?\)

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