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?

Step by step solution

01

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 \({ }^{13} \mathrm{C}=13.00335 \mathrm{amu}\).

02

Find Relative Abundance

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

03

Calculate Average Mass of Carbon Atom

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

04

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 = \(\mathrm{Total \ atoms} \times \% \mathrm{Abundance_{13C}} = 10000 \times 0.0107 = 107\).

05

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: \[ \text{Total mass} = \text{Total atoms} \times \text{Average mass} = 10000 \times 12.01104 = 120110.4 \ \mathrm{amu} \]

06

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 = \(\mathrm{Total \ atoms} \times \% \mathrm{Abundance_{13C}} = (6.0221 \times 10^{23}) \times 0.0107 = 6.1435 \times 10^{21}\).

07

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: \[ \text{Total mass} = \text{Total atoms} \times \text{Average mass} = (6.0221 \times 10^{23}) \times 12.01104 = 7.23378 \times 10^{24} \ \mathrm{amu} \]

08

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 \(12.011 \, \mathrm{g}\).

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