AP Chemistry Unit 1: Atomic Structure and Properties

Explore AP Chemistry Unit 1 questions on moles, mass spectra, elemental composition, electron configuration, photoelectron spectroscopy, periodic trends, and ionic compounds.

Covered syllabus points

1.1 Moles and Molar Mass 1.2 Mass Spectra of Elements 1.3 Elemental Composition of Pure Substances 1.4 Composition of Mixtures 1.5 Atomic Structure and Electron Configuration 1.6 Photoelectron Spectroscopy 1.7 Periodic Trends 1.8 Valence Electrons and Ionic Compounds

[Maximum number: 6]

Answer the following questions related to manganese compounds.

(a)

Manganese has several common oxidation states.

[ 2 ]

(i) For the correct answer: 1 point
Accept one of the following:

- $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{5}$

- $[\mathrm{Ar}] 4 s^{2} 3 d^{5}$

(i)

Write the complete electron configuration for an Mn atom in the ground state.

[ 1 ]

(i) For the correct answer: 1 point
Accept one of the following:

- $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{5}$

- $[\mathrm{Ar}] 4 s^{2} 3 d^{5}$

(ii)

When manganese forms cations, electrons are lost from which subshell first? Identify both the number and letter associated with the subshell.

A student performs an experiment to produce a manganese salt of unknown composition, $\mathrm{Mn}_{x} \mathrm{Cl}_{y}(a q)$ , and determine its empirical formula. The student places a sample of Mn(s) in a beaker containing excess HCl(a q), as represented by the following equation.

x \mathrm{Mn}(s)+y \mathrm{HCl}(a q) \rightarrow \mathrm{Mn}_{x} \mathrm{Cl}_{y}(a q)+\frac{y}{2} \mathrm{H}_{2}(g)

The student heats the resulting mixture until only $\mathrm{Mn}_{x} \mathrm{Cl}_{y}(s)$ remains in the beaker. The data are given in the following table.

[ 1 ]

(b)

Calculate the mass of Cl in the sample of $\mathrm{Mn}_{x} \mathrm{Cl}_{y}(s)$ remaining in the beaker.

[ 1 ]

For the correct calculated value:
$62.673 \mathrm{~g}-61.262 \mathrm{~g}=1.411 \mathrm{~g} \mathrm{Cl}$

1 point

(c)

Calculate the number of moles of Cl in the sample of $\mathrm{Mn}_{x} \mathrm{Cl}_{y}(s)$ remaining in the beaker.

[ 1 ]

For the correct calculated value, consistent with part (b):
$1.411 \mathrm{~g} \mathrm{Cl} \times \frac{1 \mathrm{~mol} \mathrm{Cl}}{35.45 \mathrm{~g} \mathrm{Cl}}=0.03980 \mathrm{~mol} \mathrm{Cl}$

1 point

(d)

The student determines that 0.0199 mol of Mn was used in the experiment. Use the data to determine the empirical formula of the $\mathrm{Mn}_{x} \mathrm{Cl}_{y}(s)$ .

[ 1 ]

For the correct answer, consistent with part (c):
$\frac{0.03980 \mathrm{~mol} \mathrm{Cl}}{0.0199 \mathrm{~mol} \mathrm{Mn}}=\frac{2 \mathrm{~mol} \mathrm{Cl}}{1 \mathrm{~mol} \mathrm{Mn}} \Rightarrow \mathrm{MnCl}_{2}$

1 point

(e)

The student repeats the experiment using the same amounts of Mn and HCl and notices that some of the $\mathrm{Mn}_{x} \mathrm{Cl}_{y}$ splatters out of the beaker as it is heated to dryness. Will the number of moles of Cl calculated for this trial be greater than, less than, or equal to the number calculated in part (c) ? Justify your answer.

[ 1 ]

For the correct answer and a valid justification:

Less than. If some of the mass of aqueous $\mathrm{Mn}_{x} \mathrm{Cl}_{y}$ is lost due to splattering, the final mass of the dry beaker and $M n_{x} C l_{y}$ will be decreased, which will decrease the calculated mass and number of moles of chlorine in the dry solid.

1 point

[Maximum number: 4]

Answer the following questions about magnesium.

(a)

An incomplete mass spectrum for magnesium is shown in the diagram.

The percent abundance of magnesium-24 is 79\%. The percent abundances of the other two

natural isotopes of magnesium, magnesium-25 and magnesium-26, are approximately equal.

[ 2 ]

(i)

Complete the mass spectrum in part A by drawing thick lines in the appropriate locations

to represent the percent abundance of magnesium-25 and magnesium-26.

[ 1 ]

(ii)

Describe the difference in atomic structure that accounts for the difference in mass

between magnesium-25 and magnesium-26.

[ 1 ]

(b)

A student prepares a $1.85 \times 10^{-3} M$ solution of $\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}(a q)$ in beaker 1 and a $2.80 \times 10^{-4} M$

solution of $\mathrm{NaOH}(a q)$ in beaker 2, as shown.

The particle diagram shown represents a magnesium ion, $\mathrm{Mg}^{2+}$ , in beaker 1. A sodium

ion, $\mathrm{Na}^{+}$ , in beaker 2 has a weaker attraction to water than the $\mathrm{Mg}^{2+}$ does. Explain this

phenomenon using Coulomb's law and each of the following.

[ 2 ]

(i)

The relative charge of the ions

[ 1 ]

B (i) For a correct explanation:

Point 03

The charge on the sodium ion is less than the charge on the magnesium ion. A smaller

charge results in a weaker Coulombic attraction between $\mathrm{Na}^{+}$ and water.

(ii)

The relative radii of the ions

[ 1 ]

(ii) For a correct explanation:

Point 04

The $\mathrm{Na}^{+}$ ion is larger than the $\mathrm{Mg}^{2+}$ ion, so the distance between the $\mathrm{Na}^{+}$ and the

oxygen on the water molecule will be greater. As distance increases, Coulombic attraction

decreases.

[Maximum number: 2]

Answer the following questions about ascorbic acid (vitamin C).

A student combusts a sample of ascorbic acid, $\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{O}_{x}$ , to determine its chemical

composition. The only products of the reaction are 0.2400 mol of $\mathrm{CO}_{2}$ and 2.883 g of $\mathrm{H}_{2} \mathrm{O}$ .

The mole ratio of carbon (C) to oxygen (O) is 1:1 in ascorbic acid. Based on this

information and your answer to part A (i), determine the empirical formula of ascorbic

acid.

(ii) For the correct calculated number of moles of H (may be implicit):

$0.1600 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} \times \frac{2 \mathrm{~mol} \mathrm{H}^{\mathrm{H}}}{1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}}=0.3200 \mathrm{~mol} \mathrm{H}$

For the correct empirical formula.

Examples of acceptable responses may include the following:

- x: y: z= (moles of C):(moles of H):(moles of O)

x: y: z=0.2400: 0.3200: 0.2400=3: 4: 3

Therefore, the empirical formula of ascorbic acid is $\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}$ .

- $0.2400 \mathrm{~mol} \mathrm{CO}_{2} \times \frac{1 \mathrm{~mol} \mathrm{C}}{1 \mathrm{~mol} \mathrm{CO}_{2}}=0.2400 \mathrm{~mol} \mathrm{C}$ $\frac{0.3200 \mathrm{~mol} \mathrm{H}}{0.2400 \mathrm{~mol} \mathrm{C}}=\frac{4 \mathrm{H}}{3 \mathrm{C}}$

Given that the ratio of C:O is 1:1, the empirical formula of ascorbic acid is

$\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}.$

[Maximum number: 4]

Answer the following questions relating to Fe and its ions, $\mathrm{Fe}^{2+}$ and $\mathrm{Fe}^{3+}$ .

(a)

Write the ground-state electron configuration of the $\mathrm{Fe}^{2+}$ ion.

[ 1 ]

$1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$

OR

$[\mathrm{Ar}] 3 d^{6}$

1 point is earned for a correct electron configuration.

Ion

Ionic Radius (pm)

$\mathrm{Fe}^{2+}$ $\mathrm{Fe}^{3+}$

(b)

The radii of the ions are given in the table above. Using principles of atomic structure, explain why the radius of the $\mathrm{Fe}^{2+}$ ion is larger than the radius of the $\mathrm{Fe}^{3+}$ ion.

[ 1 ]

Both ions have the same nuclear charge; however, the greater
number of electrons in the outermost shell of $\mathrm{Fe}^{2+}$ results in greater
electron-electron repulsion within that shell, leading to a larger
radius.

1 point is earned for a valid
explanation.

(c)

Calculate the percent by mass of Fe in the original sample of powdered Fe(s) with the inert impurity.

[ 1 ]

Calculate the percent by mass of Fe in the original sample of powdered Fe(s) with the inert impurity.

$\begin{gathered} 0.09431 \mathrm{~mol} \mathrm{Fe} \times \frac{55.85 \mathrm{~g} \mathrm{Fe}}{1 \mathrm{~mol}}=5.267 \mathrm{~g} \mathrm{Fe} \frac{5.267 \mathrm{~g} \mathrm{Fe}}{6.724 \mathrm{~g} \text { sample }} \times 100=78.33 \% \end{gathered}$

1 point is earned for correct calculation of the
mass percent based on the answer to part (g).

(d)

If the oxidation of the Fe(s) in the original sample was incomplete so that some of the 7.531 g of product was FeO(s) instead of $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ , would the calculated mass percent of Fe(s) in the original sample be higher, lower, or the same as the actual mass percent of $\operatorname{Fe}(s)$ ? Justify your answer.

Sulfur atom = Carbon atom = Oxygen atom =

[ 1 ]

The calculated mass percent of Fe would be lower than the actual mass percent of Fe.

A sample that contains any FeO (rather than $\mathrm{Fe}_{2} \mathrm{O}_{3}$ ) will have a higher actual mass percent of Fe than a completely oxidized sample would have. Therefore, when the moles of Fe are calculated (assuming all the mass of the sample is $\mathrm{Fe}_{2} \mathrm{O}_{3}$ ) the calculated number of moles of Fe, and hence the calculated mass percent of Fe, will be lower.

1 point is earned for the correct answer and a valid explanation.

Sulfur atom = Carbon atom = Oxygen atom =

Compound

Molecular Structure

Boiling Point at 1 atm
(K)

$\mathrm{CS}_{2}$

Figure for Answer to Question (d) — AP Chemistry

COS

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Covered syllabus points

1 Atomic Structure and Properties question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)

Question (e)

1 Atomic Structure and Properties question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)

1 Atomic Structure and Properties question 3

1 Atomic Structure and Properties question 4

Question (a)

Question (b)

Question (c)

Question (d)