Question

You have a sample of copper (Cu) and a sample of aluminum (Al) . You have an equal number of atoms in each sample. Which of the following statements concerning the masses of the samples is true? a. The mass of the copper sample is more than twice as great as the mass of the aluminum sample. b. The mass of the copper sample is more than the mass of the aluminum sample, but it is not more than twice as great. c. The mass of the aluminum sample is more than twice as great as the mass of the copper sample. d. The mass of the aluminum sample is more than the mass of the copper sample, but it is not more than twice as great. e. The masses of the samples are equal.

Short answer

The correct statement is: Statement b: \(mass_{Cu} > mass_{Al}\) and \(mass_{Cu} ≤ 2 \times mass_{Al}\).

Step-by-step solution

Find the atomic weight for copper and aluminum

To do this, we can refer to the periodic table. From the periodic table, we can find the atomic weights for these elements. The atomic weight of Copper (Cu) is 63.5 g/mol and of Aluminum (Al) is 27 g/mol.

Calculate the ratio of the masses of copper and aluminum for a given number of atoms

Given that both samples have the same number of atoms, let's say 'n', we can now calculate the ratio of their masses as follows: Ratio = \(\frac{mass_{Cu}}{mass_{Al}} = \frac{atoms_{Cu} × atomic \ weight_{Cu}}{atoms_{Al} × atomic \ weight_{Al}} = \frac{n × 63.5}{n × 27}\) As both samples have an equal number of atoms, 'n' is the same for both samples; hence, we can cancel it out: Ratio = \(\frac{63.5}{27}\)

Compare the ratio with given statements

Now, we can compare the calculated ratio to each of the statements given in the exercise to determine which one is true: a. \(\frac{mass_{Cu}}{mass_{Al}} > 2\) => \(\frac{63.5}{27} > 2\) b. \(\frac{mass_{Cu}}{mass_{Al}} > 1\) and \(\frac{mass_{Cu}}{mass_{Al}} \leq 2\) => \(\frac{63.5}{27} > 1\) and \(\frac{63.5}{27} \leq 2\) c. \(\frac{mass_{Al}}{mass_{Cu}} > 2\) => \(\frac{27}{63.5} > 2\) d. \(\frac{mass_{Al}}{mass_{Cu}} > 1\) and \(\frac{mass_{Al}}{mass_{Cu}} \leq 2\) => \(\frac{27}{63.5} > 1\) and \(\frac{27}{63.5} \leq 2\) e. \(\frac{mass_{Cu}}{mass_{Al}} = 1\) => \(\frac{63.5}{27} = 1\) By calculating the ratio, we determine that the correct statement is: Statement b: \(mass_{Cu} > mass_{Al}\) and \(mass_{Cu} ≤ 2 \times mass_{Al}\)

Key concepts

The Periodic Table and Element Information

The periodic table is not just a chart but a gateway to understanding the fundamentals of chemistry, including atomic mass comparisons. Each element on the periodic table, such as copper (Cu) and aluminum (Al) from our problem, is positioned according to its atomic number, the number of protons in its nucleus, and is associated with vital information including atomic weight.

The atomic weight (or atomic mass) listed on the table for each element is critical for calculations involving mass. In a simplified sense, it represents the average mass of atoms of the element, accounting for the various isotopes and their abundance. For the given copper and aluminum examples, their respective atomic weights are 63.5 and 27, meaning that if you were to take one mole (Avogadro's number) of atoms for each element, the mass of copper would be 63.5 grams and that of aluminum would be 27 grams.

Thus, when comparing the mass of equal numbers of atoms of different elements, the periodic table is the first stop for gathering the information you need to start your calculations.

Understanding Atomic Weight

Atomic weight, sometimes referred to as relative atomic mass, is a dimensionless physical quantity, the ratio of the average mass of atoms of an element (from a single given sample or source) to one twelfth of the mass of an atom of carbon-12. This value gives us an understanding of how one element compares to another in terms of mass.

However, it’s crucial to understand that atomic weight is an average of the isotopes of an element weighted by their relative abundances. This concept is vital because we use the atomic weight to determine the mass of an element in a sample, especially when dealing with mole calculations, as highlighted in our scenario that involves mass comparison between copper and aluminum.

Performing Mass Ratio Calculation

Mass ratio calculation is at the heart of many chemical problems and helps us to understand how mass is conserved in reactions and how elements compare in mass in a quantitative sense. The mass ratio of copper to aluminum is calculated by taking the atomic weight of copper and dividing it by the atomic weight of aluminum, with the formula:

Ratio = \(\frac{mass_{Cu}}{mass_{Al}} = \frac{atoms_{Cu} \times atomic \ weight_{Cu}}{atoms_{Al} \times atomic \ weight_{Al}} = \frac{n \times 63.5}{n \times 27}\).

Since we are considering equal numbers of atoms, the variable 'n' cancels out, simplifying our calculation significantly. What is left is a straightforward comparison between the atomic weight of copper and that of aluminum.

Having calculated that the mass of the copper sample is more than that of the aluminum sample but not more than twice as great is an application of this mass ratio concept. This kind of calculation is key for not only textbook exercises but also practical laboratory work where precise measurements are essential.