Annotated Solution for USNCO National Exam 2022 Part I

Annotated Solution for USNCO National Exam 2022 Part I

Author: Luke Chen

Proofreader: Dr. Qiang ¡°Steven¡± Chen

Updated on July 15, 2022

Question 1:

Difficulty: Medium

The first step to any stoichiometry problem is to write the balanced chemical equation. In this case,

Learning Objectives:

Sr(s) + 2NH3(l) = Sr(NH2)2(s) + H2(g). We are given that Sr is the limiting reagent, and we can

stoichiometry

calculate that 0.125 g / 87.62

g¡¤mol-1

= 0.00143 mol Sr reacts. By the balanced chemical equation (1

mole Sr reacts to form 1 mol H2) we get that 0.00143 moles of H2 are produced. B is the answer.

If you are not sure about what is strontium amide, you may still figure out the stochiometric ratio of

Sr and H2 by knowing that each Sr loses two electrons during the reaction. Moreover, the amide here

is the conjugate base (NH2¨C) of ammonia, which is different from the amide (-C(=O)¨CNH-) in organic

chemistry.

Question 2:

Difficulty: Hard

One strategy for this question is to go through each answer choice and verify the % arsenic by mass.

Learning Objectives:

A) %As = (1¡Á74.92) / (12.01 + 3 + 74.92 + 14.01) ¡Á 100% = 72.1%.

B) %As = (1¡Á74.92) / (12.01¡Á2 + 6 + 74.92) = 71.4%.

C) %As = (2¡Á74.92) / (12.01¡Á2 + 6 + 74.92¡Á2 + 14.01¡Á2) = 72.1%

D) %As = (2¡Á74.92) / (12.01¡Á4 + 12 + 74.92¡Á2) = 71.4%

We are between B and D for the correct answer, and it comes down to whether each compound is para

or diamagnetic. Typically, MO theory is preferred for issues of magnetism, but since each of these

compounds is relatively complex, we will use Lewis structures instead. Drawing the Lewis structure

of each molecule, we notice that 3 unbonded electrons are found on the central As in the Lewis

structure for C2H6As. This means that one unpaired electron exists in compound B and B is

paramagnetic. No radical electrons exist on compound D. D is thus our answer since it has the

right %As and is also diamagnetic. You may interpret C4H12As2 as the dimer of C2H6As, which pair

the single electron on As through the dimerization.

The compound C4H12As2 is actually called cacodyl, or tetramethyldiarsine with the structure and

geometry shown below.

Source: Wikipedia

mass percent and empirical

formula; paramagnetic vs

diamagnetic; prediction of

unpaired electron(s)

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/ Annotated Solution for USNCO National Exam 2022 ¨C Part I

Difficulty: Medium

Question 3:

Learning Objectives:

Let us go through each answer choice.

stoichiometry; limiting

reactant; graph reading

A) A seems reasonable. When only 1.00 g of Fe was added, the graph is still upward sloping meaning

adding more Fe would increase product yield. This implies that Br 2 is still present in the system and

Fe is limiting.

B) From the graph, we can easily see that 10.6 g product is obtained when 2.00 g of Fe is added. If

FeBr2

is the

(55.85+77.90¡Á2) g

1 mol FeBr2

product, let¡¯s

check the stoichiometry,

2.00 g Fe ¡Á

1 mol Fe

55.85 g

¡Á

1 mol FeBr2

1 mol Fe

¡Á

= 7.72 g. This is not 10.6 g, thus answer choice B is incorrect. The product is actually

FeBr3, which is reasonable as Br2 is a strong oxidant and oxidizes Fe(s) to a higher oxidation state.

C) Looking at the graph, when 2.50 g of Fe are added, this is already well into the flat region of the

graph (the vertex exists when 2.00 g of Fe are added). The flat region means that adding additional Fe

here will not produce any more product because no more bromine is left to react (Br 2 is the limiting

reagent). Adding any more Fe past 2.00 g will be in excess, thus an excess 0.50 g Fe exists when 2.50

g Fe is added. C is incorrect.

D) Once again looking at the graph, 3.50 g added Fe also is in the flat region where Br2 is the limiting

reagent. D is clearly incorrect.

A is thus our answer.

Note: The chemical equation is Fe(s) + 3/2 Br2(l) = FeBr3(s).

Difficulty: Medium

Learning Objectives:

stoichiometry; mixture

analysis

Question 4:

We know the masses of Cu2S and CuS in the ore must add to 89.0 g. We also know that there must be

in total 67.5 g / 63.55 g¡¤mol-1 = 1.06 moles of copper total between Cu2S and CuS. Calling the moles

of Cu2S x and the moles of CuS y, we can set up the following system of equations.

2x + y = 1.06 (balancing the number of moles of copper)

(63.55¡Á2 + 32.07)x + (63.55 + 32.07)y = 89.0 (balancing mass)

Solving the system gives us y = 0.290 mol of CuS, multiplying that by the molar mass of (63.55+32.0)

g¡¤mol-1 gives us 27.7 g of CuS, approximately B.

You may set up the mass of each compound and use the Cu% in each to obtain a similar system of

equations.

Difficulty: Medium

Question 5:

Learning Objectives:

This is a simple stoichiometry problem, although it may look complicated. We know that there are

solution stoichiometry

0.02123 L ¡Á 0.0235 M = 4.98¡Á10¨C4 mol S2O32¨C added. Since I3¨C and S2O32¨C react in a 1:2 ratio (by the

2nd equation), we know that there must be 4.98¡Á10¨C4 /2 = 2.49 ¡Á 10¨C4 mol I3¨C. Now looking at the first

equation, 1 mol of I3¨C forms when 1 mol OCl¨C reacts. Thus, the 2.49¡Á10¨C4 mol OCl¨C react and are

present in the bleach sample. To find the concentration of hypochlorite, we divide 2.49¡Á10¨C4 by the

volume of 0.0750 to get a concentration of 0.00332 M, or C.

Annotated Solution for USNCO National Exam 2022 ¨C Part I / 3

Question 6:

Difficulty: Medium

This question tests your understanding of colligative properties. The boiling point elevation is directly

Learning Objectives:

proportional to the number of particles in solution. We can rank the boiling points of each aqueous

solution by taking the product of the salt concentration and the number of ions formed per formula

unit of salt dissolving.

A.

0.10 ¡Á 4 ions (3 Na+ and 1 PO43¨C) = 0.40

B.

0.30 ¡Á 3 ions (2 Na+ and 1 HPO42¨C) = 0.90

C.

0.50 ¡Á 2 ions (1 Na+ and 1 H2PO4¨C) = 1.00

D.

0.70 ¡Á appx. 1 ¡°particle¡± (H3PO4 does not fully ionize) = 0.70

colligative properties of

solutions; dissociation of

strong and weak

electrolytes

Though H3PO4 is triprotic, even the first dissociation is very limited as indicated by a small Ka1 value,

similarly, the dissociation of HPO42¨C and H2PO4¨C is more negligible.

C forms the greatest number of particles and thus C has the highest boiling point.

(For those wanting the mathematical explanation, ¦¤T=imKb. i represents the number of ions formed

per formula unit of salt, m is the molality of the salt, and K b is a constant that depends on the solvent

identity. Kb is constant for all the answer choices since the solvent is water for all trials. We can

approximate molality using molarity for a dilute aqueous solution since 1 L of solution typically uses

very close to 1 kg of solvent.)

Question 7:

Flame test colors for IA and IIA metals should be memorized. Al does not have a clear flame test, B

is green typically, Na is bright yellow, and Sr is red. D is our answer.

Difficulty: Easy

Learning Objectives:

flame test

Question 8:

Difficulty: Medium

A) Chlorine is more reactive than Bromine and thus the Br2 cannot reduce and displace Cl¨C. Looking

Learning Objectives:

at a reduction potential table, indeed Cl 2 has a more positive reduction potential than Br2. No reaction

and thus no color change occurs.

description chemistry /

reactions, color of

ions/gases

B) The chemical reaction that occurs is CO32¨C(aq) + 2H+(aq) = H2O(l) + CO2(g). None of the reactants

or products are colored, thus no color change occurs.

C) A complexation reaction occurs 3C2O42¨C(aq) + Fe3+(aq) = [Fe(C2O4)3]3¨C(aq). Complex ion reactions

often involve color changes. Even if you didn¡¯t know that trioxalatoferrate(III) is green (compared to

the yellow Fe3+ ion in the reactants and colorless oxalate ion), you could guess C is the correct answer.

D) No reaction occurs for this answer choice and no color change occurs. Note that Co 2+ is a pink ion

in solution.

Question 9:

Difficulty: Medium

Three elements are known to form protective oxide coatings: Al, Si, and Cr. A is aluminum and the

Learning Objectives:

correct answer. We can eliminate the other answer choices since B) and D) are alkaline earth metals

and oxidize to form oxides in air, which further reacts with water and CO2 from the air, and C) is iron

which is known to rust by oxidizing in air (in the presence of water).

properties of metals

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/ Annotated Solution for USNCO National Exam 2022 ¨C Part I

Difficulty: Medium

Question 10:

Learning Objectives:

Determining the number of components in a mixture is best done by chromatography. Boiling point

colligative properties of

solutions; dissociation of

strong and weak

electrolytes

determination does not tell us the number of compounds in a mixture if the boiling points of the

components are relative close. Combustion analysis is typically used to identify the empirical formula

of a sample, this will not tell us the number of compounds in a mixture. In this case, nail polish remover

consists of many volatile organic compounds, such as acetone. Since they vaporize at low temperatures,

gas chromatography is preferred to paper chromatography (works best for non¨Cvolatile and colored

species such as dyes). A is the answer.

Difficulty: Hard

Question 11:

Learning Objectives:

Knowing solubility rules, we can eliminate A and D since alkaline earth metal sulfides tend to be

solubility rules; HSAB

theory

soluble. If you are familiar with the cation qualitative separation scheme, you know that CdS

precipitates out in acidic conditions while MnS only precipitates out once base is added. C

You may use hard-soft acid-base (HSAB) theory to get a more detailed understanding of this problem.

Here the annotation to a similar problem in Local 2018 (Q8) is quoted with minor revision for your

reference:

According to hard-soft acid-base theory (HSAB), hard acids bind strongly with hard bases, soft acids

bind strongly with soft bases. The order of the hardness of the four cations from hard to soft is Ba2+ >

Ca2+ > Mn2+ > Cd2+, as higher charged and smaller sized cations are harder Lewis acids; more

electronegative, lower charged, and smaller sized anions are harder Lewis bases. S2¨C, less

electronegative, larger in size, is a typical soft base, which binds strongly with soft acids and binds

weakly with hard acids. So CdS, a soft acid-soft base combination, has the strongest binding affinity

and the lowest solubility in water. Therefore, CdS is the only sulfide precipitate here which is not

soluble even in strong acids.

In general, metal sulfide precipitates tend to be more soluble in acid as the formation of the weak

electrolyte H2S, indicated by MS(s) + 2H+(aq) ¡ú M2+(aq) + H2S(aq). So, if the precipitate has a

smaller Ksp (less soluble), it is harder to dissolve in both water and in acid.

Difficulty: Hard

Learning Objectives:

laboratory and error

analysis; ideal gas law;

partial pressure of water

vapor

Question 12:

Let us write a skeleton equation for the reaction: M(s) + H2SO4(aq) = MSO4(aq) + H2(g). The

collection of gas over water is an essential laboratory technique. A reaction occurs in a water through

such that the gasses produced are collected in a flask or graduated cylinder. Their volume is determined

and, by using the ideal gas law, the number of moles of gas produced can be calculated. P(H2) for the

ideal gas law is calculated by subtracting the vapor pressure of water from the atmospheric pressure.

Checking each answer choice:

A) This will cause no effect. Firstly, if you accurately take into account the new concentration of

sulfuric acid your calculations still should be accurate. Secondly, the reaction likely will occur with

sulfuric acid in excess, so the number of moles of sulfuric acid truly does not matter for the calculation.

Annotated Solution for USNCO National Exam 2022 ¨C Part I / 5

B) Since M is oxidized in the reaction, replacing some M with its already oxidized form will effectively

reduce the number of moles of M in the reaction. Less moles of gas will be produced with less M

reactant, making the calculated number of moles of M also decrease. Since the mass of ¡°M¡± (including

the oxide) is weighed to be the same, the molar mass will be too high. B is the answer.

C) This makes the pressure in the ideal gas law artificially high; since pressure is directly proportional

to moles, the number of moles of H2 calculated will also be artificially high. By the balanced equation,

this also makes the calculated number of moles of M too high while keeping the measured mass of M

the same. This makes the calculated molar mass of M too low.

D) The vapor pressure must be subtracted from the atmospheric pressure to find P(H2) used in the

ideal gas law; not subtracting the vapor pressure will make P(H2) artificially high. Just like in answer

choice C, this results in a too high calculated number of moles of M, making the calculated molar mass

too low.

Question 13:

Difficulty: Medium

A) Using the Ideal Gas Law, we can find the initial moles of Ar as 1.0 L ¡Á 1.00 atm ¡Â (0.08206 L atm

Learning Objectives:

/ mol K ¡Á 298 K) = 0.0409 mol and the initial moles of Kr as 2.0 L ¡Á 0.5 atm ¡Â (0.08206 L atm / mol

K ¡Á 298 K) = 0.0409 mol. We know the total volume of the system with the open stopcock is 1.0 +

ideal gas law; partial

pressure; average speed of

gas molecules

2.0 = 3.0 L, and that the total number of moles is 0.0409 ¡Á 2 = 0.0818 mol. Using the ideal gas law,

we can find the final pressure as P = nRT/V = 0.0818 ¡Á 0.08206 ¡Á 298 / 3.00 = 0.666 atm. This is not

1.00 atm, thus A is incorrect.

B) We calculated that the total number of moles of each gas is equal, thus B seems correct. However,

we must be careful with this answer. We know the moles of Ar and Kr in the whole system are equal,

but we cannot know how these moles are distributed (in the 2.0 L container or the connected 1.0 L

container). B is a trick answer. The number of moles of Ar in the 2.0-L container is supposed to be 2/3

of the total moles of Ar, while the number of moles of Kr in the 1.0-L container is 1/3 of the total

moles of Kr.

C) The average speed is dependent on temperature and molar mass. urms = 3RT/M. Both gasses are at

the same T but have different molar masses, thus their average speed (represented by the root mean

squared velocity, urms) cannot be equal.

D) From our last calculation, we know this is correct. Since P = nRT/V, equal number of moles of each

gas indicate that each gas has the same partial pressure. D is the answer.

Question 14:

Difficulty: Medium

Carbon and Bromine have an electronegativity difference, yet it isn¡¯t very much. By Pauling¡¯s

Learning Objectives:

electronegativity scale, Carbon and Bromine differ in electronegativity only by 0.3. Both CH 2Br2 and

CHBr3 are polar molecules, but since the electronegativity difference is slim the charge separation is

also slim. Thus, both molecules do not have strong dipole¨Cdipole interactions. Clearly, hydrogen

bonding does not occur since there is no O¨CH, F¨CH, or N¨CH very polar bond. Additionally, we know

both molecules do not have long range covalent bonds (they are molecular compounds held together

analysis of intermolecular

forces

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