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How many grams of gold (Au) are there in 15.3 moles of Au?

Short Answer

Expert verified
The mass of 15.3 moles of gold is 3013.1 grams.

Step by step solution

01

Understand the Problem

We need to find out the weight in grams of 15.3 moles of gold. From the periodic table, we know that the molar mass of gold (Au) is 197 grams/mole.
02

Use the Conversion Factor

We will use the molar mass of gold as a conversion factor. The conversion factor is 197 grams/mole. This means that one mole of gold has a mass of 197 grams.
03

Perform the Calculation

To get the total mass, multiply the number of moles by the molar mass. That is, \( 15.3 \, moles \times 197 \, \frac{grams}{mole}\). The 'moles' will cancel out and the answer will be in grams.

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Key Concepts

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

Molar Mass
The concept of molar mass is pivotal when dealing with chemical quantities. It is the mass of one mole of a given substance and is expressed in grams per mole (g/mol). Think of it as the 'atomic weight' for a compound, which takes into account the individual atomic masses of each element that composes it, as found in the periodic table.

Molar mass effectively bridges the gap between the micro (atoms, molecules) and the macro (grams) world, making it possible to convert from moles, a count of particles, to grams, which is a measure of mass we can practically sense. For instance, the molar mass of gold (Au) is 197 g/mol, meaning each mole of gold weighs 197 grams.
Stoichiometry
Stoichiometry is an essential topic in chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. This field of study is not only about balancing equations but also about understanding the relationships and conversions between different substances involved in a reaction.

Through stoichiometry, the concept of the mole serves as a central unit, allowing chemists to work with amounts in the laboratory without having to count every atom or molecule. Stoichiometry leverages the laws of conservation of mass and fixed proportions to ensure chemical equations represent the science accurately.

To navigate stoichiometric calculations, you'll often convert moles of one substance to moles of another using the balanced chemical equation, and then use molar masses to convert moles into grams if needed.
Chemical Calculations
Chemical calculations encompass a wide array of mathematical techniques used in chemistry to relate the quantities of reactants and products in a chemical reaction. They are built upon stoichiometry and molar masses, allowing chemists to predict yields, determine reactant quantities, and more.

For any such calculation, understanding the role of the mole is crucial. It serves as a bridge between the atomic scale and the macroscopic world. In performing chemical calculations, the accuracy of molar masses and the careful balancing of equations are essential to achieve precise and reliable results.

In the conversion from moles to grams, the steps are straightforward: identify the molar mass, treat it as a conversion factor, and multiply it by the number of moles to get the mass. This process not only applies to pure substances but is also the backbone of calculations in reactions where multiple substances interact.

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Most popular questions from this chapter

Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a component of gasoline. Complete combustion of octane yields \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{CO}_{2} .\) Incomplete combustion produces \(\mathrm{H}_{2} \mathrm{O}\) and CO, which not only reduces the efficiency of the engine using the fuel but is also toxic. In a certain test run, 1.000 gal of octane is burned in an engine. The total mass of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) produced is \(11.53 \mathrm{~kg} .\) Calculate the efficiency of the process; that is, calculate the fraction of octane converted to \(\mathrm{CO}_{2}\). The density of octane is \(2.650 \mathrm{~kg} / \mathrm{gal}\)

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

A compound X contains 63.3 percent manganese (Mn) and 36.7 percent O by mass. When \(X\) is heated, oxygen gas is evolved and a new compound Y containing 72.0 percent \(\mathrm{Mn}\) and 28.0 percent \(\mathrm{O}\) is formed. (a) Determine the empirical formulas of X and Y. (b) Write a balanced equation for the conversion of \(\mathrm{X}\) to \(\mathrm{Y}\)

Without doing any detailed calculations, estimate which element has the highest percent composition by mass in each of the following compounds: (a) \(\mathrm{Hg}\left(\mathrm{NO}_{3}\right)_{2}\) (b) \(\mathrm{NF}_{3}\) (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{C}_{2952} \mathrm{H}_{4664} \mathrm{~N}_{812} \mathrm{O}_{832} \mathrm{~S}_{8} \mathrm{Fe}_{4}\)

An iron bar weighed \(664 \mathrm{~g}\). After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust \(\left(\mathrm{Fe}_{2} \mathrm{O}_{3}\right) .\) Calculate the final mass of the iron bar and rust.

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