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

Short Answer

Expert verified
The number of grams in 15.3 moles of Au (gold) is around 3011 grams. This result is obtained by multiplying the number of moles given (15.3) by the molar mass of gold which is 197.0 g/mol.

Step by step solution

01

Determine the Molar Mass of Au

The first step is to determine the molar mass of Au. For gold (Au), the molar mass is 197.0 grams per mole. This data can be found on any standard periodic table.
02

Use the Molar Mass Conversion Factor

Next, the molar mass of Au (197.0 grams/mole) should be employed as a conversion factor. We have 15.3 moles of Au and we know each mole of Au weights 197.0 grams. Therefore, the calculation will be as follows: \(15.3 \, \text{moles of Au} \times 197.0 \, \text{grams/mole}\).
03

Calculate the Result

By doing the calculation from step 2, the product gives the amount of gold by weight, which is in grams. It is also worth mentioning, due to the multiplication operation, the mole unit gets canceled out leaving the final answer in grams.

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Key Concepts

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

Molar Mass of Gold
Understanding the concept of molar mass is crucial when converting between moles and grams of a substance. The molar mass is defined as the weight in grams of one mole of any chemical element or compound. For gold, which is denoted as Au on the periodic table, the molar mass is specifically 197.0 grams per mole. This figure is essential for converting moles of gold to grams.

When dealing with mole-to-grams conversions, it's important to remember that the molar mass serves as a conversion factor, allowing us to bridge moles, which is a unit of quantity, with grams, a unit of mass. To help visualize this, think of molar mass as the price per item. If you know the price for one item and you have multiple items, you can calculate the total price. Similarly, knowing the molar mass of gold allows you to calculate the total mass if you have more than one mole.
Stoichiometry
Stoichiometry is akin to a recipe for chemistry. It is the branch that deals with the relationship between the quantities of reactants and products in a chemical reaction. In the context of mole to grams conversion, stoichiometry provides a methodical way to use these relationships for calculating masses, volumes, and the number of particles.

In practice, stoichiometry can be seen as a two-step process. First, use the balanced chemical equation to find the mole ratio between reactants and products. Second, use the molar mass of the substances to convert these ratios into measurable quantities like grams. Even when not dealing directly with a reaction, the principles of stoichiometry, such as the conservation of mass and mole-to-gram conversions, are applicable to understand the scale and proportions of the substances involved.
Avogadro's Number
At the heart of understanding moles is Avogadro's number, which is famously 6.022 x 10^23. This number represents the quantity of particles, like atoms or molecules, in one mole of any substance. It serves as a bridge between the macroscopic world we live in and the atomic or molecular scale.

Avogadro's number allows chemists to count particles by weighing them. For instance, if you weighed out 197.0 grams of gold, you would have exactly Avogadro's number of gold atoms. This concept underpins the calculations we make when converting moles to grams or vice versa. Knowing that a mole is a specific quantity allows for accurate and meaningful conversions, much like knowing the exact number of eggs in a dozen ensures you can follow a recipe correctly.

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Most popular questions from this chapter

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 \(\left.\pi r^{2} .\right)\)

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?

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of aluminum metal. It is prepared by the reaction $$ \mathrm{CaF}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{CaSO}_{4}+2 \mathrm{HF} $$ In one process \(6.00 \mathrm{~kg}\) of \(\mathrm{CaF}_{2}\) are treated with an excess of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and yield \(2.86 \mathrm{~kg}\) of \(\mathrm{HF}\). Calculate the percent yield of HF.

What is the mass (in amu) of a carbon-12 atom? Why is the atomic mass of carbon listed as 12.01 amu in the table on the inside front cover of this book?

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ How much sulfur (in tons), present in the original materials, would result in that quantity of \(\mathrm{SO}_{2} ?\)

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