Question

For parts of the free-response question that require calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Examples and equations may be included in your answers where appropriate.


A sample of a pure element is analyzed using a mass spectrometer. The results are shown below.

The figure presents a bar graph. The horizontal axis is labeled Atomic Mass, in a m u, and the numbers 24, 25, and 26, are indicated. The vertical axis is labeled Percent Abundance, and the numbers 0 through 100, in increments of 20, are indicated. There are 3 bars of data. The data presented by the bars are as follows. Note that all values are approximate. Bar 1, 24 a m u, 79 percent abundance. Bar 2, 25 a m u, 10 percent abundance. Bar 3, 26 a m u, 11 percent abundance.

(a) How many different isotopes of the element were in the sample?

(b) Describe how to use the information from the mass spectrum to determine the average atomic mass of the element.


(c) Identify the element.


(d) Write the ground-state electron configuration of an atom of the element that you identified in part (c).

Answer 1

Answer:

so long

Explanation:

please mark me brainliest and follow me

Answer 2

Given:

A graph of percentage abundance vs Atomic Mass

Bar 1 - 24 amu, 79% abundance

Bar 2 - 25 amu, 10 % abundance

Bar 3 - 26 amu, 11% abundance

To Find:

(a) How many different isotopes of the element were in the sample?

(b) Describe how to use the information from the mass spectrum to determine the average atomic mass of the element.

(c) Identify the element.

(d) Write the ground-state electron configuration of an atom of the element that you identified in part (c).

Solution:

• Isotopes are two or more types of atoms that have the same atomic number but differ in mass number.

(a)

• Since 3 different bars were recorded, we can conclude that 3 different isotopes of the element were found in the sample.
• The atomic weight of the different isotopes is 24 amu, 25 amu, and 26 amu respectively.

(b)

• The average atomic mass of the element can be calculated from the formula:

      Average Mass = ∑$\frac{M . abundance}{100}$

where M is the atomic mass of the isotope.

Substituting known values:

      Average Mass = $\frac{24.79 + 25.10 + 26.11}{100}$

     Average mass = $\frac{2432}{100}$

     Average Mass of Element = 24.32 amu

(c)

• Based on the average atomic mass of the element and its isotopes, we can conclude that the element is Magnesium(Mg).
• Its atomic number is 12 and its atomic mass is ≈24.30 amu.

(d)

• For Mg, Z=12. Therefore its ground state configuration is:

                                 Mg =  1s²2s²2p⁶3s²

Therefore, the answers to the questions are:

(a) 3 different isotopes were present.

(b) Average Atomic Mass of the element = 24.32 amu

(c) The element is Magnesium(Mg).

(d) The ground state electronic configuration of Mg is 1s²2s²2p⁶3s².