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How many grams of gold \((\mathrm{Au})\) are there in 15.3 moles of \(\mathrm{Au}\) ?

Short Answer

Expert verified
There are 3014.1 grams of gold in 15.3 moles of Au.

Step by step solution

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01

Identify Given Information

We are given that there are 15.3 moles of gold (\( \mathrm{Au} \)). The goal is to find out how many grams this is.
02

Determine the Molar Mass of Gold

The molar mass of gold (\( \mathrm{Au} \)) is \( 197.0 \) grams per mole. This value comes from the periodic table.
03

Use the Molar Mass to Convert Moles to Grams

To convert moles to grams, use the formula: \( \text{Grams} = \text{Moles} \times \text{Molar Mass} \). In this case, substitute the values we know: \( 15.3 \text{ moles} \times 197.0 \text{ g/mol} \).
04

Perform the Calculation

Multiply the number of moles by the molar mass to get the total grams: \( 15.3 \times 197.0 = 3014.1 \) grams.

Key Concepts

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

Moles
In chemistry, a mole is a fundamental concept that helps us understand the quantity of substance involved in reactions and other processes. Simply put, a mole is a unit of measurement used to express amounts of a chemical substance.

One mole is defined as containing exactly 6.022 x 1023 (Avogadro's number) of particles, atoms, ions, or molecules. This large number may seem daunting, but it helps in dealing with the macroscopic quantities of material that we encounter in the chemical world. For instance, in our exercise, we are dealing with 15.3 moles of gold (Au). This means we have 15.3 times 6.022 x 1023 gold atoms involved.

A few key points about moles:
  • Provides a bridge between atomic scale and real-world scale quantities.
  • Crucial for understanding stoichiometry in reactions.
  • Allows chemists to simplify complex equations using a standard unit of measure.
Understanding moles is essential for converting between different chemical units, such as going from moles to grams, as we do in this exercise.
Molar Mass
Molar mass is the mass of one mole of a given element or compound, and it is a helpful tool to connect the microscopic world of atoms to the macroscopic world of grams and kilograms. Molar mass is usually expressed in grams per mole (g/mol), which indicates how many grams a single mole of a substance weighs.

For gold, its molar mass is notably 197.0 g/mol. This value is crucial because it allows chemists to accurately convert moles of a substance into grams by simply multiplying the number of moles by the molar mass.

Why is molar mass important?
  • Essential for converting moles to grams and vice versa.
  • Helps determine the proportions of elements in chemical reactions.
  • Derived from the periodic table, ensuring accurate and consistent use worldwide.
In our exercise, the conversion of moles of gold to grams is achieved by using its molar mass (197.0 g/mol), highlighting how indispensable molar mass is in such calculations.
Gold
Gold (Au) is one of the most valued and well-known elements due to its lustrous appearance and rarity. With an atomic number of 79, it is a heavy metal with a significant historical and economic impact.

In scientific terminology, gold's attractiveness goes beyond its physical properties; its malleability and conductivity make it an essential component in electronics and various industries. When studying the chemistry of gold, its molar mass (197.0 g/mol) particularly stands out as a vital factor, as we have seen in our example, where we converted moles to grams.

Noteworthy characteristics of gold:
  • Inert and does not tarnish easily, maintaining its sheen over time.
  • Atomic structure makes it dense yet malleable.
  • Widely used in jewelry, currency, and high-tech industries due to its unique properties.
Understanding gold's characteristics helps us appreciate both its aesthetic and practical value, while also grasping its role in chemical calculations like the one shown in the exercise.

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Most popular questions from this chapter

Lysine, an essential amino acid in the human body, contains \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{N}\). In one experiment, the complete combustion of \(2.175 \mathrm{~g}\) of lysine gave \(3.94 \mathrm{~g}\) \(\mathrm{CO}_{2}\) and \(1.89 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In a separate experiment, \(1.873 \mathrm{~g}\) of lysine gave \(0.436 \mathrm{~g} \mathrm{NH}_{3}\). (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is \(150 \mathrm{~g}\). What is the molecular formula of the compound?

A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\). Assuming complete decomposition, calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from \(46.0 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\). (The products are \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\).)

Each copper(II) sulfate unit is associated with five water molecules in crystalline copper(II) sulfate pentahydrate \(\left(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\right) .\) When this compound is heated in air above \(100^{\circ} \mathrm{C},\) it loses the water molecules and also its blue color: $$ \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CuSO}_{4}+5 \mathrm{H}_{2} \mathrm{O} $$ If \(9.60 \mathrm{~g}\) of \(\mathrm{CuSO}_{4}\) is left after heating \(15.01 \mathrm{~g}\) of the blue compound, calculate the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) originally present in the compound.

The molar mass of caffeine is \(194.19 \mathrm{~g}\). Is the molecular formula of caffeine \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}\) or \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?\)

The following is a crude but effective method for estimating the order of magnitude of Avogadro's number using stearic acid \(\left(\mathrm{C}_{18} \mathrm{H}_{36} \mathrm{O}_{2}\right)\). When stearic acid is added to water, its molecules collect at the surface and form a monolayer; that is, the layer is only one molecule thick. The cross-sectional area of each stearic acid molecule has been measured to be \(0.21 \mathrm{nm}^{2}\). In one experiment, it is found that \(1.4 \times 10^{-4} \mathrm{~g}\) of stearic acid is needed to form a monolayer over water in a dish of diameter \(20 \mathrm{~cm}\). Based on these measurements, what is Avogadro's number? (The area of a circle of radius \(r\) is \(\pi r^{2}\).)

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