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How Many Atoms Are You? Estimate the number of atoms in the body of a \(50-\mathrm{kg}\) physics student. Note that the human body is mostly water, which has molar mass 18.0 \(\mathrm{g} / \mathrm{mol}\) and that each water molecule contains three atoms.

Short Answer

Expert verified
A 50 kg person has approximately \(3.012 \times 10^{27}\) atoms in their body.

Step by step solution

01

Identify the composition of the body

The human body is primarily composed of water, which makes up about 60% of the body's weight. Therefore, we can estimate that a 50 kg person is made up of approximately 30 kg of water.
02

Convert the mass of water to grams

To perform calculations in chemistry, it's common to convert mass into grams. Therefore, convert 30 kg of water to grams: \(30 \text{ kg} = 30,000 \text{ g}\).
03

Calculate moles of water

Using the molar mass of water (18.0 g/mol), calculate the number of moles of water in the body. \(\text{Number of moles} = \frac{30,000 \text{ g}}{18.0 \text{ g/mol}} = 1666.67 \text{ moles}\).
04

Determine number of molecules of water

Use Avogadro's number \(6.022 \times 10^{23}\) molecules/mol to find the number of molecules in the body. \(1666.67 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mol} \approx 1.004 \times 10^{27}\text{ molecules}\).
05

Calculate the total number of atoms

Since each water molecule contains three atoms (2 hydrogen and 1 oxygen), multiply the number of molecules by 3 to find the total number of atoms: \(1.004 \times 10^{27} \text{ molecules} \times 3 \approx 3.012 \times 10^{27}\text{ atoms}\).

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Key Concepts

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

Molar Mass of Water
The concept of molar mass is essential in understanding chemical reactions and the composition of substances. In the context of water, the molar mass is the total mass of one mole of water molecules, which consists of two hydrogen atoms and one oxygen atom. The molar mass of hydrogen is approximately 1.0 g/mol, and for oxygen, it's 16.0 g/mol. Therefore, the molar mass of water (Hâ‚‚O) is calculated as follows:
  • Molar mass of Hâ‚‚: 2 x 1.0 g/mol = 2.0 g/mol
  • Molar mass of O: 1 x 16.0 g/mol = 16.0 g/mol
  • Total molar mass of Hâ‚‚O: 2.0 g/mol + 16.0 g/mol = 18.0 g/mol
This calculation is important because it helps determine the number of moles in a given mass of water, which is crucial for further calculations such as determining the number of molecules and ultimately, the number of atoms in a sample like the human body.
Avogadro's Number
Avogadro's number is a fundamental constant used in chemistry to describe the number of units (usually atoms or molecules) in a mole of any substance. Its value is approximately 6.022 x 10²³. This means that one mole of any element or compound contains exactly 6.022 x 10²³ atoms or molecules.

This constant is vital because it acts as a bridge between the atomic scale and the macroscopic scale that we can measure. For instance, when calculating the number of molecules in a given mass, as in the body's water content, Avogadro's number allows us to convert from moles to molecules. In our example of a 50 kg person, once we know the number of moles, we can use Avogadro's number to calculate the precise number of water molecules, leading us closer to finding the total number of atoms.
Human Body Composition
Human body composition plays a key role in understanding how many atoms comprise a person. Primarily composed of water, which makes up roughly 60% of body weight, the body's chemical composition also includes other elements like carbon, nitrogen, and calcium, among others.

For a simplified calculation, focusing on water is a helpful approach because the majority of our body's atoms are found there. By estimating that 60% of a 50 kg human body is water, we calculate that there are about 30 kg of water. This estimate helps simplify the process of determining the number of moles, molecules, and atoms, giving a substantial figure. However, it's crucial to remember that while water dominates, every other molecule in the body contributes to the total atom count, albeit in smaller quantities.
Chemical Calculations
Chemical calculations involve a series of mathematical steps to understand the composition of substances at a molecular level. These calculations often require converting mass to moles, determining the number of molecules, and finally, calculating the total number of atoms.
  • First, convert the person's mass of body water from kilograms to grams, as it's easier to work in smaller units in chemistry.
  • Next, use the molar mass of water to determine the number of moles of water. For example, 30,000 grams of water divided by 18.0 g/mol gives the moles of water available.
  • Avogadro's number then helps convert moles to molecules, allowing us to see how many water molecules are present in the body.
  • Finally, knowing each molecule has three atoms, we multiply the total number of molecules by 3 to find the number of atoms. This step-by-step approach breaks down complex processes into manageable calculations.
Understanding these steps gives insight not only into the specific problem of calculating atoms in the body but also into broader principles applicable in many areas of chemistry.

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Most popular questions from this chapter

(a) Compute the increase in gravitational potential energy for a nitrogen molecule (molar mass 28.0 \(\mathrm{g} / \mathrm{mol} )\) for an increase in elevation of 400 \(\mathrm{m}\) near the earth's surface. (b) At what temperature is this equal to the average kinetic energy of a nitrogen molecule? (c) Is it possible that a nitrogen molecule near sea level where \(T=15.0^{\circ} \mathrm{C}\) could rise to an altitude of 400 \(\mathrm{m} ?\) Is it likely that it could do so without hitting any other molecules along the way? Explain.

An automobile tire has a volume of 0.0150 \(\mathrm{m}^{3}\) on a cold day when the temperature of the air in the tire is \(5.0^{\circ} \mathrm{C}\) and atmospheric pressure is 1.02 atm. Under these conditions the gauge pressure is measured to be 1.70 atm (about 25 \(\mathrm{lb} / \mathrm{in.} .\) ). After the car is driven on the highway for 30 min, the temperature of the air in the tires has risen to \(45.0^{\circ} \mathrm{C}\) and the volume has risen to 0.0159 \(\mathrm{m}^{3}\) . What then is the gauge pressure?

Gaseous Diffusion of Uranium. (a) A process called gaseous diffusion is often used to separate isotopes of uranium that is, atoms of the elements that have different masses, such as 235 \(\mathrm{U}\) and 238 \(\mathrm{U} .\) The only gaseous compound of uranium at ordinary temperatures is uranium hexafluoride, UF \(_{6}\) . Speculate on how 235 \(\mathrm{UF}_{6}\) and \(^{238} \mathrm{UF}_{6}\) molecules might be separated by diffusion. (b) The molar masses for \(^{235} \mathrm{UF}_{6}\) and 238 \(\mathrm{UF}_{6}\) molecules are 0.349 \(\mathrm{kg} / \mathrm{mol}\) and \(0.352 \mathrm{kg} / \mathrm{mol},\) respectively. If uranium hexafluoride acts as an ideal gas, what is the ratio of the root-meansquare speed of \(^{235} \mathrm{UF}_{6}\) molecules to that of \(^{238} \mathrm{UF}_{6}\) molecules if the temperature is uniform?

For diatomic carbon dioxide gas \(\left(\mathrm{CO}_{2},\) molar mass \right. 44.0 \(\mathrm{g} / \mathrm{mol} )\) at \(T=300 \mathrm{K}\) , calculate (a) the most probable speed \(v_{\mathrm{mp}} ;\) (b) the average speed \(v_{\mathrm{av}} ;(\mathrm{c})\) the root-mean-square speed \(v_{\mathrm{rms}}\)

The size of an oxygen molecule is about 2.0 \(\times 10^{-10} \mathrm{m}\) Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures \((T=300 \mathrm{K})\) .

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