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Calculate the molar masses (in grams per mole) of (a) cane sugar, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (b) laughing gas, \(\mathrm{N}_{2} \mathrm{O}\). (c) vitamin A, \(\mathrm{C}_{20} \mathrm{H}_{30} \mathrm{O}\).

Short Answer

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Answer: (a) The molar mass of cane sugar is 342 g/mol. (b) The molar mass of laughing gas is 44 g/mol. (c) The molar mass of vitamin A is 286 g/mol.

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01

Find the atomic weights of the elements in each compound

Using a periodic table, find the atomic weights of the elements that make up each compound - carbon (C), hydrogen (H), oxygen (O), and nitrogen (N). Their atomic weights are as follows (rounded to nearest whole number for simplicity): Carbon (C) = 12 g/mol Hydrogen (H) = 1 g/mol Oxygen (O) = 16 g/mol Nitrogen (N) = 14 g/mol
02

Calculate the molar mass of cane sugar (\(\rm C_{12}H_{22}O_{11}\))

Using the formula mentioned above and the atomic weights of the elements, we can calculate the molar mass of cane sugar as follows: Molar mass of cane sugar = (12 atoms of C x 12 g/mol) + (22 atoms of H x 1 g/mol) + (11 atoms of O x 16 g/mol) = 144 g/mol + 22 g/mol + 176 g/mol = 342 g/mol
03

Calculate the molar mass of laughing gas (\(\rm N_{2}O\))

Now, we'll use the formula to calculate the molar mass of laughing gas: Molar mass of laughing gas = (2 atoms of N x 14 g/mol) + (1 atom of O x 16 g/mol) = 28 g/mol + 16 g/mol = 44 g/mol
04

Calculate the molar mass of vitamin A (\(\rm C_{20}H_{30}O\))

Finally, we will calculate the molar mass of vitamin A using the same formula: Molar mass of vitamin A = (20 atoms of C x 12 g/mol) + (30 atoms of H x 1 g/mol) + (1 atom of O x 16 g/mol) = 240 g/mol + 30 g/mol + 16 g/mol = 286 g/mol Now we have the molar masses of each compound: (a) Molar mass of cane sugar = 342 g/mol (b) Molar mass of laughing gas = 44 g/mol (c) Molar mass of vitamin A = 286 g/mol

Key Concepts

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

Molar Mass Formula
The molar mass of a substance is the mass of one mole of that substance, which is the mass of 6.022 x 10^23 particles (Avogadro's number) of the substance. In simple terms, it tells us how much one mole of any substance weighs. The unit for molar mass is grams per mole (g/mol).

To calculate the molar mass of a compound, we use the molar mass formula where the molar mass of the compound is equal to the sum of the molar masses of each constituent element, multiplied by the number of atoms of that element in the compound. In mathematical terms:
\[\text{Molar mass} = \text{sum of} (n \times \text{atomic weight of element})\]
where 'n' represents the number of atoms of each element in the molecule. This formula requires a basic understanding of the periodic table and the atomic weight of elements, which are usually listed in atomic mass units (amu). To find the molar mass in grams per mole, one must use the atomic weight equivalent in grams per mole.
Atomic Weight
Atomic weight, also known as relative atomic mass, is the weighted average mass of the atoms in a naturally occurring element, relative to the mass of a carbon-12 atom, which is taken as exactly 12 atomic mass units (amu).

On the periodic table, the atomic weight is usually displayed below the elemental symbol and represents the average mass of all the isotopes of that element, taking into account their abundance. For the purpose of calculating molar mass, we often use a simplified version of the atomic weight rounded to the nearest whole number, since most common elements' isotopes have masses close to a whole number in amu.

For example:
  • Carbon would have an atomic weight of approximately 12 amu, which for molar mass calculations converts to 12 g/mol.
  • Oxygen's atomic weight is approximately 16 amu, translating to 16 g/mol for molar mass.
Using these weights in grams per mole allows us to accurately calculate the molar mass of compounds constructed from these elements.
Stoichiometry
Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is the calculation of the quantities of substances involved in chemical reactions based on the conservation of mass.

In stoichiometry, the molar mass serves as a conversion factor between the mass of a substance and the number of moles. Since reactions occur according to mole ratios as depicted in balanced chemical equations, the knowledge of molar masses allows chemists to determine the amounts of each substance needed to start a reaction and the amounts of products that will be formed.
Moreover, with stoichiometry, one can also calculate the yields (theoretical and actual), limiting reactants (the reactant that runs out first), and percent composition of compounds. Using stoichiometry in conjunction with the molar mass formula and atomic weight allows students and chemists alike to perform precise chemical calculations and is fundamental to the practice of chemistry.

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Most popular questions from this chapter

Determine the simplest formulas of the following compounds: (a) the food enhancer monosodium glutamate (MSG), which has the composition \(35.51 \%\) C, \(4.77 \% \mathrm{H}, 37.85 \% \mathrm{O}, 8.29 \% \mathrm{~N}\), and \(13.60 \% \mathrm{Na} .\) (b) zircon, a diamond-like mineral, which has the composition \(34.91 \%\) \(\mathrm{O}, 15.32 \% \mathrm{Si}\), and \(49.76 \% \mathrm{Zr}\) (c) nicotine, which has the composition \(74.0 \%\) C, \(8.65 \% \mathrm{H}\), and \(17.4 \% \mathrm{~N} .\)

Determine the simplest formulas of the following compounds: (a) saccharin, the artificial sweetener, which has the composition \(45.90 \% \mathrm{C}, 2.75 \% \mathrm{H}, 26.20 \% \mathrm{O}, 17.50 \% \mathrm{~S}\), and \(7.65 \% \mathrm{~N}\) (b) allicin, the compound that gives garlic its characteristic odor, which has the composition \(6.21 \% \mathrm{H}, 44.4 \% \mathrm{C}, 9.86 \% \mathrm{O}\), and \(39.51 \% \mathrm{~S}\). (c) sodium thiosulfate, the fixer used in developing photographic film, which has the composition \(30.36 \% \mathrm{O}, 29.08 \% \mathrm{Na}\), and \(40.56 \% \mathrm{~S}\).

The meat from one hazelnut has a mass of \(0.985 \mathrm{~g}\). (a) What is the mass of a millionth of a mole \(\left(10^{-6}\right)\) of hazelnut meats? (A millionth of a mole is also called a micromole.) (b) How many moles are in a pound of hazelnut meats?

Write balanced equations for the reaction of aluminum metal with the following nonmetals: (a) sulfur (b) bromine (c) nitrogen (d) oxygen (forming \(\mathrm{O}^{2-}\) ions) (e) oxygen (forming \(\mathrm{O}_{2}{ }^{2-}\), or peroxide ions)

Convert the following to moles. (a) \(4.00 \times 10^{3} \mathrm{~g}\) of hydrazine, a rocket propellant (b) \(12.5 \mathrm{~g}\) of tin(II) fluoride, the active ingredient in fluoride toothpaste (c) \(13 \mathrm{~g}\) of caffeine, \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}\)

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