What is the mass, in grams, of each elemental sample? a. \(2.3 \times 10^{-3} \mathrm{~mol} \mathrm{Sb}\) b. \(0.0355 \mathrm{~mol} \mathrm{Ba}\) c. \(43.9 \mathrm{~mol} \mathrm{Xe}\) d. \(1.3 \mathrm{~mol} \mathrm{~W}\)

Short Answer

Expert verified
a. 280.048 g Sb, b. 4.8735 g Ba, c. 5763.631 g Xe, d. 238.992 g W.

Step by step solution

01

Calculate the mass of Sb sample

Determine the molar mass of Antimony (Sb) using the periodic table (approximately 121.76 g/mol). Multiply the molar mass by the number of moles of Sb given: mass = 2.3 x 10^{-3} mol Sb x 121.76 g/mol.
02

Calculate the mass of Ba sample

Determine the molar mass of Barium (Ba) (approximately 137.33 g/mol). Multiply the molar mass by the number of moles of Ba given: mass = 0.0355 mol Ba x 137.33 g/mol.
03

Calculate the mass of Xe sample

Determine the molar mass of Xenon (Xe) (approximately 131.29 g/mol). Multiply the molar mass by the number of moles of Xe given: mass = 43.9 mol Xe x 131.29 g/mol.
04

Calculate the mass of W sample

Determine the molar mass of Tungsten (W) (approximately 183.84 g/mol). Multiply the molar mass by the number of moles of W given: mass = 1.3 mol W x 183.84 g/mol.

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Key Concepts

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

Stoichiometry
Stoichiometry is a subsection of chemistry that involves the calculation of the quantities of reactants and products in chemical reactions. It is a powerful tool for predicting yield and determining the proportions needed to create specific products. A core principle in stoichiometry is the law of conservation of mass, which states that in a chemical reaction, the total mass of the reactants equals the total mass of the products.

Understanding stoichiometry requires familiarity with the mole concept, as well as the ability to convert between moles and mass using molar mass as a conversion factor. To solve stoichiometric problems, one often begins by writing a balanced chemical equation and then using molar ratios derived from that equation to calculate unknown quantities.
Mole Concept
The mole concept is a fundamental cornerstone in chemical calculations and stoichiometry. A mole is a unit for counting particles and is defined as the amount of substance that contains as many entities (such as atoms, ions, or molecules) as there are atoms in 12 grams of carbon-12. This number is Avogadro's number, which is approximately equal to \(6.022 \times 10^{23}\).

When working with the mole concept, it's vital to connect this abstract number to tangible amounts of a substance. The molar mass of an element, which is the mass of one mole of that element, provides the link between moles and grams, enabling chemists to carry out conversions between the mass of a sample and the number of moles it contains.
Atomic Mass
Atomic mass, also referred to as atomic weight, is the weight of a single atom of an element. It is typically expressed in atomic mass units (amu), where one amu is defined as one-twelfth of the mass of a carbon-12 atom. The atomic masses of all elements are found listed on the periodic table and provide an essential piece of information for chemical calculations.

For practical chemical calculations, the atomic mass is used to determine the molar mass of elements and compounds by adding together the atomic masses of the constituent atoms. As such, the molar mass represents the mass of one mole of a substance and is expressed in grams per mole (g/mol), directly relating atomic scale weights to macroscopic quantities used in laboratory measurements.
Chemical Calculations
Chemical calculations often involve determining the mass, volume, concentration, or number of moles of substances involved in chemical reactions. These calculations are fundamental not just in academic settings but also in industrial and laboratory practices for the formulation of products, materials, and medicines.

To perform chemical calculations, one must be adept at using formulas and conversion factors, such as molar mass, molarity, and stoichiometric coefficients from balanced equations. For example, to calculate the mass of a given number of moles of a substance, as demonstrated in the original exercise, the molar mass is used as a conversion factor. By becoming proficient in these calculations, students can predict the outcomes of chemical reactions and design experiments with predetermined results.

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Most popular questions from this chapter

The atomic radii of the isotopes of an element are identical to one another. However, the atomic radii of the ions of an element are significantly different from the atomic radii of the neutral atom of the element. Explain.

A chemist in an imaginary universe, where electrons have a different charge than they do in our universe, performs the Millikan oil drop experiment to measure the electron's charge. The charges of several drops are recorded here. What is the charge of the electron in this imaginary universe? $$ \begin{array}{cc} \text { Drop number } & \text { Charge } \\ \hline \text { A } & -4.8 \times 10^{-9} \mathrm{z} \\ \hline \text { B } & -9.6 \times 10^{-9} \mathrm{z} \\ \hline \text { C } & -6.4 \times 10^{-9} \mathrm{z} \\ \hline \text { D } & -12.8 \times 10^{-9} \mathrm{z} \\ \hline \end{array} $$

State and explain the law of multiple proportions. How is the law of multiple proportions different from the law of definite proportions?

Naturally occurring magnesium has an atomic mass of 24.312 and consists of three isotopes. The major isotope is \({ }^{24} \mathrm{Mg},\) natural abundance \(78.99 \%\), relative atomic mass 23.98504 . The next most abundant isotope is \({ }^{26} \mathrm{Mg},\) relative atomic mass \(25.98259 .\) The third most abundant isotope is \({ }^{25} \mathrm{Mg},\) whose natural abundance is in the ratio of 0.9083 to that of \({ }^{26} \mathrm{Mg}\). Find the relative atomic mass of \({ }^{25} \mathrm{Mg}\).

In complete sentences, describe the similarities and differences between: a. different isotopes of an element b. a neutral atom and an ion of the same element

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