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Chapter 2: Problem 161

Identify, from the list below, the information needed to calculate the number of atoms in \(1.00 \mathrm{cm}^{3}\) of iron. Outline the procedure used in this calculation. (a) the structure of solid iron (b) the molar mass of iron (c) Avogadro's number (d) the density of iron (e) the temperature (f) iron's atomic number (g) the number of iron isotopes

Short Answer

Expert verified
(b) molar mass, (c) Avogadro's number, (d) density of iron are needed.

Step by step solution

01

Analyze the Volume of Iron

We need to start with the given volume of iron, which is \(1.00 \ \text{cm}^3\). The volume is crucial as it will help us relate this information to the mass using density, and eventually to the number of atoms.
02

Identify Necessary Information for Mass Calculation

To find the mass of a \(1.00 \ \text{cm}^3\) of iron block, we need the density of iron. This is because density (\(\rho\)) is defined as mass (m) divided by volume (V). The equation \(\rho = \frac{m}{V}\) can be rearranged to find mass: \(m = \rho \times V\). Hence, we need option (d), the density of iron.
03

Relate Mass to Moles Using Molar Mass

Once we have the mass, we convert it to moles of iron using the molar mass. The molar mass allows us to convert grams to moles using the formula: \(\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\). Therefore, option (b), the molar mass of iron, is also necessary.
04

Calculate Number of Atoms Using Avogadro's Number

After obtaining the amount in moles, we convert moles to atoms. Avogadro's number (\(6.022 \times 10^{23} \ \text{atoms/mol}\)) is used for this conversion, by multiplying the number of moles by Avogadro's number. This requires option (c), Avogadro's number.
05

sIdentify Irrelevant Information

For this calculation, we do not need (a) the structure of solid iron, (e) the temperature, (f) iron's atomic number, or (g) the number of iron isotopes. These pieces of information are not necessary for determining the number of atoms in a given volume.

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Key Concepts

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

Density of Iron
Density is a fundamental property of materials that relates their mass to volume. For iron, understanding its density is crucial in calculating the number of atoms in a given volume.
The density of iron is typically around 7.87 grams per cubic centimeter (g/cm鲁).
This value tells us that for each cubic centimeter of iron, there is a mass of 7.87 grams.
  • Mass is directly derived from density and volume:
    ewline Mass (m) = Density (蟻) 脳 Volume (V)
Knowing the density allows you to convert between the volume and mass of iron, a key first step in finding the number of atoms in the substance.
Molar Mass
Molar mass is another essential concept commonly used in chemistry to relate mass to moles.
For iron, its molar mass is approximately 55.85 grams per mole (g/mol).
This means one mole of iron atoms weighs 55.85 grams. To calculate moles from a given mass:
  • Use the formula:
    Moles = Mass (g) / Molar Mass (g/mol)
The conversion from grams to moles using molar mass is vital for any sample to determine how many particles, like atoms, are present, helping bridge the mass to the microscopic scale.
Avogadro's Number
Avogadro's number is a vital constant in chemistry, helping convert moles of a substance into actual numbers of molecules or atoms.
This number is approximately 6.022 脳 10虏鲁 atoms per mole.
  • To find the number of atoms, multiply the moles of iron by Avogadro's number:
    Number of Atoms = Moles 脳 Avogadro's Number (6.022 脳 10虏鲁 atoms/mol)
This is the step that takes us from the macroscopic amount of a substance to the atomic scale, highlighting the enormous number of atoms present even in small amounts of material.
Convert Volume to Mass
Converting volume to mass is a straightforward process once the density is known. For instance, if you have a volume of iron, you can calculate its mass by using the density value.
  • The calculation is simple:
    Mass (g) = Volume (cm鲁) 脳 Density (g/cm鲁)
In our case, with a volume of 1.00 cm鲁 of iron:
Mass = 1.00 cm鲁 脳 7.87 g/cm鲁 = 7.87 grams.
This conversion is foundational for further calculations, including determining the moles of iron present.
Convert Moles to Atoms
After determining the moles of a substance, converting these to the number of atoms is the last step in quantifying individual entities.
This conversion is possible through Avogadro's number.
  • Number of Atoms = Moles 脳 6.022 脳 10虏鲁 atoms/mol
This step is crucial in revealing the vast number of atoms in any sample, showing just how vast microscopic quantities really are. Combined with prior conversions from mass to moles, this final step completes the analysis to determine the number of atoms in a defined amount of iron.

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Most popular questions from this chapter

Calculate the molar mass and the mass percent of each element in the blue solid compound \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{SO}_{4} \cdot \mathrm{H}_{2} \mathrm{O} .\) What is the mass of copper and the mass of water in \(10.5 \mathrm{g}\) of the compound?

The action of bacteria on meat and fish produces a compound called cadaverine. As its name and origin imply, it stinks! (It is also present in bad breath and adds to the odor of urine.) It is \(58.77 \%\) C, \(13.81 \%\) H and \(27.40 \%\) N. Its molar mass is \(102.2 \mathrm{g} / \mathrm{mol}\) Determine the molecular formula of cadaverine.

Cacodyl, a compound containing arsenic, was reported in 1842 by the German chemist Robert Wilhelm Bunsen. It has an almost intolerable garlic-like odor. Its molar mass is \(210 \mathrm{g} / \mathrm{mol},\) and it is \(22.88 \% \mathrm{C}\) \(5.76 \% \mathrm{H},\) and \(71.36 \%\) As. Determine its empirical and molecular formulas.

Which of the following statements about \(57.1 \mathrm{g}\) of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) is (are) not true? (a) \(57.1 \mathrm{g}\) is 0.500 mol of octane. (b) The compound is \(84.1 \%\) C by weight. (c) The empirical formula of the compound is \(\mathbf{C}_{4} \mathbf{H}_{9}\) (d) \(57.1 \mathrm{g}\) of octane contains \(28.0 \mathrm{g}\) of hydrogen atoms.

Which of the following are correct formulas for ionic compounds? For those that are not, give the correct formula. (a) \(\mathrm{AlCl}_{2}\) (b) \(\mathrm{KF}_{2}\) (c) \(\mathrm{Ga}_{2} \mathrm{O}_{3}\) (d) MgS
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