Problem 61
Question
Superman comes from the planet Krypton. If you have \(0.00789 \mathrm{g}\) of the gaseous element krypton, how many moles does this represent? How many atoms?
Step-by-Step Solution
Verified Answer
0.00789 g of krypton is approximately 9.41 x 10^-5 moles, which is about 5.67 x 10^19 atoms.
1Step 1: Identify the Molar Mass of Krypton
The molar mass of krypton is approximately 83.8 g/mol, which is derived from the periodic table.
2Step 2: Calculate the Moles of Krypton
To find the number of moles, use the formula: \[\text{moles} = \frac{\text{mass of krypton} \ (\mathrm{g})}{\text{molar mass of krypton} \ (\mathrm{g/mol})}\]Substitute the known values: \[\text{moles} = \frac{0.00789 \ \mathrm{g}}{83.8 \ \mathrm{g/mol}} \approx 9.41 \times 10^{-5} \ \text{moles}\]
3Step 3: Calculate the Number of Atoms of Krypton
To find the number of atoms, multiply the moles of krypton by Avogadro's number \(6.022 \times 10^{23} \text{atoms/mol}\):\[\text{atoms} = 9.41 \times 10^{-5} \ \text{moles} \times 6.022 \times 10^{23} \text{atoms/mol} \approx 5.67 \times 10^{19} \text{atoms}\]
Key Concepts
Molar MassAvogadro's NumberElement Measurement
Molar Mass
The concept of molar mass is central to understanding chemical reactions and conversions between grams and moles. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol).
It is equivalent to the relative atomic mass of an element, which can be found on the periodic table. For a single element like krypton, its molar mass is essentially the atomic weight listed on the table.
In our example, krypton's molar mass is approximately 83.8 g/mol. Understanding this allows us to compute how much of the substance, in moles, a given sample mass represents. To find moles, we divide the mass of the krypton sample by its molar mass, which tells us how much of the substance, in moles, we have. This holistic comprehension enables detailed elemental analysis and calculations in various scientific fields.
It is equivalent to the relative atomic mass of an element, which can be found on the periodic table. For a single element like krypton, its molar mass is essentially the atomic weight listed on the table.
In our example, krypton's molar mass is approximately 83.8 g/mol. Understanding this allows us to compute how much of the substance, in moles, a given sample mass represents. To find moles, we divide the mass of the krypton sample by its molar mass, which tells us how much of the substance, in moles, we have. This holistic comprehension enables detailed elemental analysis and calculations in various scientific fields.
Avogadro's Number
Avogadro's number is a key term in chemistry that links macroscopic and atomic-scale measurements. It is defined as the number of atoms, ions, or molecules in one mole of a substance.
The constant value is 6.022 x 10^{23}, and it allows us to convert between the number of moles and the number of individual particles, like atoms, in a sample.
Simply put, if you have one mole of krypton atoms, you have exactly 6.022 x 10^{23} krypton atoms. In our exercise, with krypton, we multiply the amount of krypton measured in moles by Avogadro's number to ascertain how many atoms are contained in the sample.
The constant value is 6.022 x 10^{23}, and it allows us to convert between the number of moles and the number of individual particles, like atoms, in a sample.
Simply put, if you have one mole of krypton atoms, you have exactly 6.022 x 10^{23} krypton atoms. In our exercise, with krypton, we multiply the amount of krypton measured in moles by Avogadro's number to ascertain how many atoms are contained in the sample.
- This helps bridge the gap between the relatively small scale of individual atoms and the larger scale we deal with in practical scenarios.
- It enables precise measurements and can be applied across various scientific investigations.
Element Measurement
Measuring elements precisely is crucial to chemical experimentation and industrial processes. We measure elements in medicine, environmental science, and many other fields.
Typically, elemental measurement involves weighing a sample and using molar mass and Avogadro’s number to convert this weight into moles or atoms. This conversion is a foundational skill in chemistry.
To perform these measurements, follow these simplified steps:
1. **Weigh the Sample:** Take the mass of your element. In our case, it was 0.00789 grams of krypton.
2. **Use Molar Mass:** Use the periodic table to find molar mass, like krypton's 83.8 g/mol, to convert the weight to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)
3. **Multiply by Avogadro's Number:** Convert moles to number of atoms with: \(\text{number of atoms} = \text{moles} \times 6.022 \times 10^{23}\)
These steps form the backbone of chemical quantification and are essential to accurate experimentation.
Typically, elemental measurement involves weighing a sample and using molar mass and Avogadro’s number to convert this weight into moles or atoms. This conversion is a foundational skill in chemistry.
To perform these measurements, follow these simplified steps:
1. **Weigh the Sample:** Take the mass of your element. In our case, it was 0.00789 grams of krypton.
2. **Use Molar Mass:** Use the periodic table to find molar mass, like krypton's 83.8 g/mol, to convert the weight to moles using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)
3. **Multiply by Avogadro's Number:** Convert moles to number of atoms with: \(\text{number of atoms} = \text{moles} \times 6.022 \times 10^{23}\)
These steps form the backbone of chemical quantification and are essential to accurate experimentation.
Other exercises in this chapter
Problem 58
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You are given \(15 \mathrm{g}\) each of yterium, boron, and copper. Which sample represents the largest number of atoms?
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The recommended daily allowance (RDA) of iron in your diet is 15 mg. How many moles is this? How many atoms?
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Put the following elements in order from smallest to largest mass: (a) \(3.79 \times 10^{24}\) atoms Fe (e) 9.221 mol \(\mathrm{Na}\) (b) \(19.921 \mathrm{mol}
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