Question

Reference Section \(3.2\) to find the atomic masses of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\), the relative abundance of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) in natural carbon, and the average mass (in amu) of a carbon atom. If you had a sample of natural carbon containing exactly 10,000 atoms, determine the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms present. What would be the average mass (in amu) and the total mass (in amu) of the carbon atoms in this 10,000 -atom sample? If you had a sample of natural carbon containing \(6.0221 \times 10^{25}\) atoms, determine the number of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms present. What would be the average mass (in amu) and the total mass (in amu) of this \(6.0221 \times 10^{23}\) atom sample? Given that \(1 \mathrm{~g}=6.0221 \times 10^{23} \mathrm{amu}\), what is the total mass of \(1 \mathrm{~mol}\) of natural carbon in units of grams?

Short answer

In the 10,000-atom sample of natural carbon, there are 9893 ${ }^{12} \mathrm{C}$ atoms and 107 ${ }^{13} \mathrm{C}$ atoms, with an average mass of 12.01104 amu and a total mass of 120110.4 amu. In the \(6.0221 \times 10^{23}\)-atom sample, there are \(5.9633 \times 10^{23}\, { }^{12} \mathrm{C}\) atoms and \(6.1435 \times 10^{21}\, { }^{13} \mathrm{C}\) atoms, with an average mass of 12.01104 amu and a total mass of \(7.23378 \times 10^{24}\, \mathrm{amu}\). The total mass of 1 mole of natural carbon is 12.011 g.

Step-by-step solution

Find Atomic Masses

Here, we look up to reference section \(3.2\) to find the atomic masses of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\). The atomic masses are \({ }^{12} \mathrm{C}=12.00000 \mathrm{amu}\) and \({ }^{13} \mathrm{C}=13.00335 \mathrm{amu}\).

Find Relative Abundance

Next, we find the relative abundance of \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\). \({ }^{12} \mathrm{C}=98.93 \%\) and \({ }^{13} \mathrm{C}=1.07 \%\).

Calculate Average Mass of Carbon Atom

To find the average mass of a carbon atom in amu, we calculate: \[\text{Average mass} = (\% \mathrm{Abundance_{12C}} \times \mathrm{mass_{12C}} ) + (\% \mathrm{Abundance_{13C}} \times \mathrm{mass_{13C}}) = \] \[ (0.9893 \times 12.00000) + (0.0107 \times 13.00335) = 12.01104 \ \text{amu} \]

Calculate Number of C Isotopes in a 10,000-Atom Sample

In the 10,000-atom sample, we will have: Number of \({ }^{12} \mathrm{C}\) atoms = \(\mathrm{Total \ atoms} \times \% \mathrm{Abundance_{12C}} = 10000 \times 0.9893 = 9893\). Number of \({ }^{13} \mathrm{C}\) atoms = \(\mathrm{Total \ atoms} \times \% \mathrm{Abundance_{13C}} = 10000 \times 0.0107 = 107\).

Calculate Average and Total Mass of Carbon Atoms in a 10,000-Atom Sample

The average mass of carbon atoms in this sample is calculated in Step 3, which is \(12.01104 \ \mathrm{amu}\). The total mass of carbon atoms in this 10,000-atom sample is: \[ \text{Total mass} = \text{Total atoms} \times \text{Average mass} = 10000 \times 12.01104 = 120110.4 \ \mathrm{amu} \]

Calculate Number of C Isotopes in a \(6.0221 \times 10^{23}\)-Atom Sample

In the \(6.0221 \times 10^{23}\)-atom sample, we will have: Number of \({ }^{12} \mathrm{C}\) atoms = \(\mathrm{Total \ atoms} \times \% \mathrm{Abundance_{12C}} = (6.0221 \times 10^{23}) \times 0.9893 = 5.9633 \times 10^{23}\). Number of \({ }^{13} \mathrm{C}\) atoms = \(\mathrm{Total \ atoms} \times \% \mathrm{Abundance_{13C}} = (6.0221 \times 10^{23}) \times 0.0107 = 6.1435 \times 10^{21}\).

Calculate Average and Total Mass of Carbon Atoms in a \(6.0221 \times 10^{23}\)-Atom Sample

The average mass of carbon atoms in this sample is calculated in Step 3, which is \(12.01104 \ \mathrm{amu}\). The total mass of carbon atoms in this \(6.0221 \times 10^{23}\)-atom sample is: \[ \text{Total mass} = \text{Total atoms} \times \text{Average mass} = (6.0221 \times 10^{23}) \times 12.01104 = 7.23378 \times 10^{24} \ \mathrm{amu} \]

Calculate the Total Mass of 1 Mole of Natural Carbon in Grams

Finally, we'll calculate the total mass of 1 mole of natural carbon in grams: \[ \text{Total mass of 1 mole} = \frac{\text{Total mass of }6.0221 \times 10^{23} \text{-atom sample}}{\text{Mass conversion factor}} = \frac{7.23378 \times 10^{24} \, \mathrm{amu}}{6.0221 \times 10^{23} \, \mathrm{amu} \, \mathrm{per} \, \mathrm{g}} = 12.011 \, \mathrm{g} \] So the total mass of 1 mole of natural carbon is \(12.011 \, \mathrm{g}\).

Key concepts

Carbon Isotopes

Carbon isotopes are different forms of carbon atoms that vary in their number of neutrons. The two most common isotopes are \({ }^{12} \text{C}\) and \({ }^{13} \text{C}\). This means, while they both have the same number of protons (6), \({ }^{12} \text{C}\) has 6 neutrons and \({ }^{13} \text{C}\) has 7 neutrons. This difference in neutron number contributes to their different atomic masses: 12.00000 amu for \({ }^{12} \text{C}\) and 13.00335 amu for \({ }^{13} \text{C}\). \({ }^{12} \text{C}\) is by far the most abundant, making up about 98.93% of natural carbon, whereas \({ }^{13} \text{C}\) is much less common, presenting only about 1.07%.
Understanding these isotopes helps us comprehend why atoms of the same element can have different masses, which is crucial for calculating atomic weights used in various scientific equations.

Relative Abundance

Relative abundance refers to the proportion of each isotope present in a natural sample of an element. For carbon, as previously mentioned, we have \({ }^{12} \mathrm{C}\) at 98.93% and \({ }^{13} \mathrm{C}\) at 1.07%. Knowing the relative abundance of isotopes helps us predict the overall behavior of a sample, as well as accurately calculate the average atomic mass.

For example, when calculating how many \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) atoms you might have in a given sample, their relative abundances directly influence the results. This is particularly useful when dealing with large numbers of atoms - like in amounts we encounter with the mole concept.

Average Atomic Mass

The average atomic mass of an element takes into account all its isotopes and their relative abundances. For carbon, this value is influenced by the mass and abundance of both \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\). The formula to calculate the average atomic mass is as follows:

• Multiply the atomic mass of each isotope by its relative abundance (expressed as a decimal).
• Add these values together to find the average.

Using this method for carbon, we get: \(0.9893 \times 12.00000 + 0.0107 \times 13.00335 = 12.01104 \ ext{ amu}\).
The value 12.01104 amu represents the mass you would expect to measure if you randomly picked one carbon atom out of any natural sample of carbon. This averaged figure is critical as it is often used in chemistry for mass-based calculations and reactions.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the amount of substance to its mass, useful for counting particles like atoms and molecules. One mole is defined as exactly \(6.0221 \times 10^{23}\) particles of a substance, known as Avogadro's number. This makes it possible to scale up from the atomic or molecular scale to gram quantities that can be measured in a lab.

For instance, if you had \(1 \text{mol}\) of natural carbon, it would contain \(6.0221 \times 10^{23}\) atoms, and its mass would be calculated using its average atomic mass. Based on the average atomic mass of carbon calculated earlier (12.011 amu), the mass of 1 mole of carbon atoms would be 12.011 grams. This conversion is crucial because it links the atomic world to the macroscopic world, enabling chemists to measure out precise quantities of elements for reactions.