If isotopic distribution of \(\mathrm{C}^{12}\) and \(\mathrm{C}^{14}\) is \(98.0 \%\) and \(2.0 \%\), respectively, then the number of \(\mathrm{C}^{14}\) atoms in \(12 \mathrm{~g}\) of carbon is (a) \(1.032 \times 10^{22}\) (b) \(1.20 \times 10^{22}\) (c) \(5.88 \times 10^{23}\) (d) \(6.02 \times 10^{23}\)

Short Answer

Expert verified
The number of \(\mathrm{C}^{14}\) atoms in 12 g of carbon is \(1.20 × 10^{22}\) atoms.

Step by step solution

01

Determine the mole quantity of carbon

Calculate the number of moles of carbon in 12 grams by using the molar mass of carbon (approximately 12 g/mol for \(\mathrm{C}^{12}\)). The number of moles \(n\) is equal to the mass \(m\) divided by the molar mass \(M\): \[\frac{m}{M} = \frac{12 \: \text{g}}{12 \: \text{g/mol}} = 1 \text{mol}.\]
02

Calculate the total number of carbon atoms

Use Avogadro's number to find the total number of atoms in 1 mole of carbon, which is \(\frac{6.022 × 10^{23} \text{ atoms}}{1 \text{ mol}} \). Multiplying by the number of moles gives you the total number of carbon atoms: \[1 \text{ mol} × 6.022 × 10^{23} \text{ atoms/mol} = 6.022 × 10^{23} \text{ atoms.}\]
03

Find the proportion of \(\text{C}^{14}\) atoms

Since \(\text{C}^{14}\) makes up 2.0% of the carbon atoms, multiply the total number of carbon atoms by 0.02 (2.0%) to find the number of \(\text{C}^{14}\) atoms: \[6.022 × 10^{23} \text{ atoms} × 0.02 = 1.2044 × 10^{22} \text{ atoms.}\]
04

Round the answer to the correct number of significant figures

The percentage of \(\text{C}^{14}\) was given to 2 significant figures, so the final answer should be rounded accordingly to 2 significant figures. Thus you have: \[1.20 × 10^{22} \text{ atoms of } \text{C}^{14}.\]

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate molar mass is a fundamental chemistry skill. It refers to the mass (in grams) of one mole of a substance. The molar mass of an element is determined by its atomic weight listed on the periodic table and is expressed in grams per mole (g/mol).

For isotopes, the atomic weight is an average, taking into account the relative abundance of each isotope. In practical terms, you can calculate the number of moles of an element by dividing the provided mass of the element by its molar mass. For instance, the molar mass of carbon-12 is approximately 12 g/mol, so 12 grams of carbon-12 equals exactly one mole.

By finding the moles, one can then proceed to use this information to relate mass to the number of atoms or molecules, bridging the gap between the macroscopic world we can measure and the microscopic world of atoms and molecules.
Avogadro's Number
Avogadro's number is a key constant in chemistry, representing the number of constituent particles (typically atoms or molecules) found in one mole of a substance. The value is 6.022 x 10^23 particles per mole.

It's used to convert between the number of atoms or molecules and moles, allowing chemists to work with amounts of a substance that can be physically measured. When we calculate the number of atoms in a sample, we multiply the number of moles by Avogadro's number. As shown in the exercise solution, calculating the total number of carbon atoms involves applying Avogadro's number to find how many atoms are in a mole of carbon.
Significant Figures
The concept of significant figures in measurements reflects the precision of those measurements. They are the digits in a number that carry meaningful information about its precision. When performing calculations, the number of significant figures in your final answer should correspond to the number in the least precise measurement you have.

In the example provided, given that the isotopic distribution is known to 2 significant figures, the final answer should be expressed with 2 significant figures as well. This is why the number of C^14 atoms is rounded to 1.20 x 10^22, maintaining the clarity and correctness of our calculation's precision according to the rules of significant figures.

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Most popular questions from this chapter

The volume of one mole of water at \(277 \mathrm{~K}\) is \(18 \mathrm{ml}\). One \(\mathrm{ml}\) of water contains 20 drops. The number of molecules in one drop of water will be \(\left(N_{\mathrm{A}}=6 \times 10^{23}\right)\) (a) \(1.07 \times 10^{21}\) (b) \(1.67 \times 10^{21}\) (c) \(2.67 \times 10^{21}\) (d) \(1.67 \times 10^{20}\)

The empirical formula of an organic gaseous compound containing carbon and hydrogen is \(\mathrm{CH}_{2}\). The volume occupied by certain mass of this gas is exactly half of the volume occupied by the same mass of nitrogen gas under identical conditions. The molecular formula of the organic gas is (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{CH}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

A certain mixture of \(\mathrm{MnO}\) and \(\mathrm{MnO}_{2}\) contains \(66.67\) mol per cent of \(\mathrm{MnO}\). What is the approximate mass per cent of \(\mathrm{Mn}\) in it? \((\mathrm{Mn}=55)\) (a) \(66.67\) (b) \(24.02\) (c) \(72.05\) (d) \(69.62\)

How much \(\mathrm{BaCl}_{2}\) would be needed to make \(250 \mathrm{ml}\) of a solution having the same concentration of \(\mathrm{Cl}^{-}\) as one containing \(\begin{array}{llll}3.78 & \mathrm{~g} & \mathrm{NaCl} \text { per } 100 \mathrm{ml} \text { ? }\end{array}\) \((\mathrm{Ba}=137)\) (a) \(16.8 \mathrm{~g}\) (b) \(67.2 \mathrm{~g}\) (c) \(33.6 \mathrm{~g}\) (d) \(22.4 \mathrm{~g}\)

The average density of the universe as a whole is estimated as \(3 \times 10^{-29} \mathrm{~g} / \mathrm{ml}\). If we assume that the entire mass is only \(\mathrm{H}\) atoms, what is the average volume of space that contains one \(\mathrm{H}\) atom? (a) \(111.111\) (b) \(1.8 \times 10^{-5} 1\) (c) \(55.561\) (d) \(3.6 \times 10^{-5} 1\)

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