Answer Key, Problem Set 4 Complete solutions

Chemistry 121 Mines, Fall 2019

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Answer Key, Problem Set 4—Complete solutions

1. NT1; 2. NT2; 3. 2.108 (Also calculate how many times longer or shorter this distance is than the distance from Earth to Proxima Centauri [given

in the problem]. In other words, how many stacks of pennies (or what fraction of a stack, if the stack won’t make it that far) could you make stretching from

Earth to Proxima Centauri with Avogadro’s number of pennies?); 4. 2.122 (also stop in the middle of the calculation to report/determine the mass, in

grams, of all the lead atoms); 5. NT3; 6. NT4; 7. NT5; 8. NT6; 9. 3.94; 10. 3.122; 11. 3.116; 12. 3.118; 13. 3.100

-------------------------------- Moles of atoms / monatomic elements, molar mass, and # of atoms (“interconversions”):

1. NT1. Avogadro’s “number” is 6.022 x 1023. (a) What are the most general units for this number (use the word “thing” somewhere, I’m serious!)? Take some time to think about this, as you always want to make sure to use units when you are using Avogadro’s number in a calculation! (b) Do you generally use Avogadro’s number when you want the mass of a sample of something? Think carefully about this! What do you (generally) use Avogadro’s number “for”—calculating what from what?

Answers:

(a) “things per mole of things”

(b) No. Avogadro’s number is used to calculated (number of) “things” from “moles” (of things) or “moles” (of things) from (number of) things. It is almost never used to calculate a mass of any kind. (Yes, it is true that the reason Avogadro’s number is what it is is because a gram is Avogadro’s number of times larger than an amu, but the number itself is almost never used to make a conversion between amus and grams. Note that an amu is not a “thing”.)

2. NT2. (variant of 2.86 in Tro) Consider a sample of 4.91 x 1021 atoms of Pt (on Earth). Show the setup, including units

(which you should be doing for every calculation in this course!), and execution of the calculation of each of the following:

(a) mass of the sample, in amu (use the (average) atomic mass of Pt)

Answer: 9.578…x 1023 amu Pt

Work: 4.91 x 1021 Pt atoms x atom Pt

amu 195.089.578…x 1023 amu Pt

(b) moles of Pt in the sample

Answer: 0.00815 mol Pt

Work: 4.91 x 1021 Pt atoms x 3...50.0081atoms Pt 10 x 6.022

atoms Pt mol 123

0.00815 mol Pt

(c) mass of the sample, in g (use the molar mass of Pt)

Answer: 1.59 g Pt

Work: 0.008153… mol Pt atoms 05...91.5Pt mol 1

Pt g 195.08x 1.59 g Pt

(d) Divide the mass of the sample, in amu [(a)], by the mass of the sample, in grams [(c)]. What are the units of the result of this calculation? What does this say about the amount of mass represented by a “gram” relative to the mass represented by “an amu”? Is a gram the same as an amu? Explain.

amu/g 10 x 2..26.0Pt g 0...91.5

Pt amu 108...x 79.5 2323

The units are “amus per gram”. This result means that a g is 6.022 x 1023 times as much mass as an amu! A gram is most definitely not the same as an amu! An amu is a very tiny amount (a nanoscopic amount) of mass—a gram is a macroscopic amount of mass.

Answer Key, Problem Set 4

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3. 2.108. A penny has a thickness of approximately 1.0 mm. If you…

Also calculate how many times longer or shorter this distance is than the distance from Earth to Proxima Centauri. In other words, how many stacks of pennies (or what fraction of a stack, if the stack won’t make it that far) could you make stretching from Earth to Proxima Centauri with Avogadro’s number of pennies?

Answer: The stack would extend for 6.02 x 1023 km, which is about 15000 times farther than the distance to Proxima Centauri! As such, you could make about 15000 separate stacks of these pennies on Earth (next to each other), all of which would make it all the way to that star!? That’s crazy huge!

Work:

km 10 x 6.02m 10

km 1x

mm 1

m10 x

penny

mm 1 x pennies 10 x 6.02 17

3

-323

This is way more than 40 x 1012 km!!! How many times? Just divide the larger by the smaller:

05051 km 10 x 40

km 10 x 6.0212

17

This is crazy! This means that you could make about 15000 of these stacks between Earth and Proxima Centauri with a mole of pennies! (Or you could back and forth between these bodies 15000 times before you run out of pennies!) That’s way more than I was thinking when I first saw this problem (and I’ve thought about how huge Avogadro’s number is for a long time!!!). Man is that huge!!

4. 2.122. The U.S. Environmental Protection Agency (EPA) sets limits on healthful levels of air pollutants. The maximum

level …. (Also stop in the middle of the calculation to report/determine the mass, in grams, of all the lead atoms).

Pb g 10 x 528.g 1

g 10x

m 1

g 1.5x

cm 1

m 10x

cm 1

m 10x

cm 1

m 10x

mL 1

cm 1x

L 1

mL 1000 x L 5.50 9-

-6

3

-2-2-23

g 10 x 8.3 9-

Pb) (of atoms 9732.atoms mol 1

atoms 10 x 6.022x

g 207.2

Pb mol 1 x Pb g 10 x 528.

239- 1310 x 2.4 ...

→ Note that although the number of atoms seems enormous (24 trillion!), it is an extremely tiny amount of mass (about 10,000 times smaller than 0.1 mg, which is the detection limit of the extremely sensitive balances in our lab). Wow!

Moles-related "interconversions" (moles↔g, moles↔fu, atom/ion↔fu, etc.) Associated with compounds or molecular elements (includes subscript ratios and relation to chemical formulas):

5. NT3 3.68(a&c)* Calculate the mass (in g) of each sample. Add: (c’) calculate the (average) mass of a water molecule in amus; and (d) Calculate the mass, in grams, of 0.00000010 mol of water;

(a) 4.5 x 1025 O3 molecules Answer: 3.6 x 103 g

4.5 x 1025 O3 molecules x ...8653.O mol 1

O g 48.00x

molecules O 10 x 6.022

molecules O mol 1

3

3

323

3 3.6 x 103 g

(c) 1 water molecule (average) Answer: 2.992 x 10-23 g

MM H2O: 2(1.008) + 1(16.00) = 18.016 g/mol

1 H2O molecule x ...612.99OH mol 1

OH g 6118.0x

molecules OH 10 x 6.022

molecules OH mol 1

2

2

223

2 2.992x 10-23 g

(c’) 1 water molecule (in amu) Answer: 18.016 18.02 amu (on average)


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Note that the masses in (c) and (c’) are for the same sample and are thus the same amount of mass. (c) is a way smaller “number” because a g is such an enormously bigger amount of mass (6.022 x 1023 times more) than an amu. Same idea as in Q2 above (part d). Do you see this better now?

(d) 0.00000010 mol water Answer: 0.0000018 g (or 1.8 x 10-6 g or 1.8 g)

02...80.000001OH mol 1

g 18.02 x OH mol 0.00000010

2

2 0.0000018 g

6. NT4. In each case below, which sample has the greater mass? NOTE: If the two samples have essentially the same

mass (within experimental uncertainty), say "SAME". Assume that all the "1 g" or "1 mol" quantities in this problem represent "exact" amounts (not 1 SF). The focus of this problem is NOT on uncertainty--it is about understanding and distinguishing terms related to amounts of chemical substances and entities (mol, g, atoms, molecules, amus, etc.)

Answers in bold, with very brief reasoning (detailed reasoning follows after):

(a) 1 mol of iron or 1 mol of aluminum

Each Fe atom weighs more than each Al atom [~56 amu > ~27 amu]; OR 55.85 g (1 mol Fe) > ~26.98 g (1 mol Al)

(b) 6.022 x 1023 lead atoms or 1 mol of lead ~SAME

6.022 x 1023 “things” is 1 mol of “things”

(c) 1 mol of copper atoms or 1 copper atom

6.022 x 1023 Cu atoms >> 1 Cu atom !!! (and so it will have way more mass!)

(d) 1 mol of Cl2 or 1 mol of Cl

There are twice as many (total) Cl atoms in the sample of Cl2 (b/c each FU has 2 vs. 1); OR 70.9 g > 35.45 g

(e) 1 g of oxygen atoms or 1 g of oxygen molecules SAME

1 g = 1 g ! (“Which weighs more, a pound of feathers or a pound of bricks?” The same! But there will be lot more feathers than bricks, because each feather weighs less.)

(f) 24.3 g of Mg or 1 mol of Mg ~SAME

The molar mass of Mg is 24.3 g (to 3 SF)

(g) 1 mol of NH3 or 1 g of NH3

17 g > 1 g

(h) 1 mol of I2 or 1 molecule of I2

6.022 x 1023 molecules >> 1 molecule !!! (a mol is not a thing; it’s a “moleion” things!)

(i) 1 oxygen molecule or 1 oxygen atom

two O atoms have more mass than just one

(j) 24.00 amus of C-12 OR 24.00 g of C-12

a gram >> an amu !!! (the “thing” here (C-12) is irrelevant b/c “gram” and “amu” are both mass units)


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More Detailed Reasoning: IMPORTANT NOTES:

1) a mole is a general unit, like a “dozen”; it should ALWAYS be followed by some “thing” in order to give it some meaning. A mole of eggs means 6.022 x 1023 eggs. **You must always get a picture of what the “thing” is that you have x moles “of”; that will clarify what you actually have and so then you can confidently answer questions like these.** For example:

2) a “molecule” of something, by definition, contains MORE THAN ONE ATOM. Therefore, a “mole of oxygen molecules” MUST contain more total atoms than a “mole of oxygen atoms”. Since each atom of

oxygen has the same (average) mass, a mole of oxygen molecules will have a greater mass than a mole

of oxygen atoms. 3) a mole of atoms contains 6.022 x 1023 times more atoms than a single atom; therefore a mole of atoms

has a mass that is 6.022 x 1023 times the mass of a single atom, and is therefore MUCH, MUCH, MUCH more massive than a single atom!

(a) 1 mol of iron or 1 mol of aluminum. Since a single atom of iron has a greater mass than a single

atom of aluminum (55.847 amu vs. 26.9815 amu, on average), a mole of iron will have a greater mass than a mole of aluminum. Why? There are the same number of items in a mole of any type of “item”. Clearly if you have the same number of bowling balls as ping pong balls, the sample of bowling balls will have a greater mass since each bowling ball weighs more than each ping pong ball.

(b) 6.022 x 1023 lead atoms or 1 mol of lead. These two will have essentially the SAME MASS since they represent essentially the same quantity of lead.

(c) 1 copper atom or 1 mol of copper atoms. See Note (3) above! A mole is a macroscopic amount

of a substance; an atom is a nanoscopic amount of a substance. THESE ARE HUGELY DIFFERENT!!!!

(d) 1 mol of Cl or 1 mol of Cl2. See note 2 above! There are twice as many atoms of Cl in a mol of

Cl2 as in a mol of Cl because there are two atoms of Cl in each molecule of Cl2. So clearly the sample of Cl2 will have a greater mass.

(e) 1 g of oxygen atoms or 1 g of oxygen molecules. VERY TRICKY! You must look carefully here to

see that they say 1 gram of each. This is not the same as (d) where they say 1 mol of each! If both samples have a mass of 1 g, then by definition they have the SAME MASS!! (What this means is that there must be half as many oxygen molecules in the molecular sample as the number of atoms in the first sample.)

(f) 24.3 g of Mg or 1 mol of Mg. SAME MASS. This is essentially the same as in (b). One mole of

Mg should have a mass of 24.3050 g, which is essentially the same as 24.3 g. (g) 1 mol of NH3 or 1 g of NH3. A mole of NH3 (molecules) has a mass of about 14.0 + 3(1.0) = 17.0

grams (because it contains one mole of N atoms, and 3 moles of H atoms), so clearly it is more massive than 1 g of NH3.

(h) 1 molecule of I2 or 1 mol of I2. See answer to (c)! A mole of I2 means “a mole of I2 molecules”

which means 6.022 x 1023 I2 molecules; a macroscopic amount vs. a nanoscopic amount. (i) 1 oxygen molecule or 1 oxygen atom. An oxygen molecule has two atoms of oxygen in it, so it

clearly will have twice the mass of 1 oxygen atom.

(j) 24.00 amus of C-12 OR 24.00 g of C-12. The mass in a gram is enormously greater than the mass in an amu (1 g = 6.022 x 1023 amu) so 24 g is enormously greater in mass than 24 amus. A gram is a macroscopic amount of mass and an amu is a nanoscopic amount of mass. A gram is about the mass of half a penny, and an amu is roughly the mass of a single proton or neutron!


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7. NT5 3.82(a&c) Determine the number of … Add: For b&d, calculate the number of O atoms in each sample)

NOTES: Going from “moles of AxBy” to “moles of B” is like going from “dozens of bicycles” to “dozens of wheels”. [which you do NOT need any MASSES for, right?]. You must use the subscript of “B” to create the “interconversion factor”.

Going from “moles of B” to “number of B atoms” is like going from “dozens of wheels” to “wheels”. Instead of multiplying by “12” (things per dozen things), you multiply by NA (things per mol things)

(a) 4.88 mol H2O2 Answer: 4.88 mol H2O2 x 22OH mol 1

O mol 29.76 mol O (atoms)

(c) 0.0237 mol H2CO3 Answer: 0.0237 mol H2CO3 x 32COH mol 1

O mol 30.0711 mol O (atoms)

Be careful not to mix up “moles of O” with “number of O atoms”! The following two parts (as I amended them) need the “extra step” to get to number of atoms

(b’) 2.15 mol N2O

Answer: 2.15 mol N2O x atoms) (of mol 1

atoms 10 x 6.022x

ON mol 1

O mol 1 23

2

1.294… 1.29 x 1024 O atoms

(d’) 24.1 mol CO2

Answer: 24.1 mol CO2 x atoms) (of mol 1

atoms 10 x 6.022x

CO mol 1

O mol 2 23

2

2.902… 2.90 x 1025 O atoms

8. NT6. (From Zumdahl, 7th ed.) A given sample of a xenon fluoride compound contains molecules of the type XeFn

where n is some whole number. If a sample containing 9.03 x 1020 molecules of XeFn weighs 0.368 g, determine

the value for n in the formula.

Answer: n = 6

Strategies: There are several strategies one could pursue here! Most chemists (including myself) would do this at the "macroscopic" level and deal with grams and moles. However, you should realize that a perhaps more "visualizable" means of doing this would be to do it at the molecular/nanoscopic level. I'll show two macro approaches and one nano approach:

#1: Use molecules to get moles of compound; use moles and grams to get molar mass; recognize that although the subscript of F is unknown, the subscript for Xe is “one”, and thus there is one mole of Xe in one mole of XeFn; thus, assume one mole of compound and subtract away the mass of one mole of Xe to get grams of F in one mole of compound; divide by 19.0 g/mol F to get the number of moles of F per mole of compound (n). An abbreviated flow diagram showing this is:

molecules XeFn → moles XeFn → g/mol XeFn → g F (in a mole of XeFn) → mol F (in a mole of XeFn)

Execution of #1:

20

23

9.03 x 10 molecules

6.022 x 10 molecules/mol n0.00150 mol XeF ; n

n

0.368 g XeF

0.00150 mol XeF 245.3 g/mol

245.3 g XeFn – 131.3 g Xe = 114 g F (in one mole of XeFn); 114 g F

6.00 mol F 19.00 g/mol F

= 6n

--------------------------------------


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#2: Use molecules to get moles of compound; this equals the moles of Xe in the 0.368 g sample; use moles of Xe with 131.3 g/mol Xe to get the mass of Xe in the sample; subtract away the mass of Xe from 0.368 g to get grams of F in the sample; divide by 19.0 g/mol F to get the number of moles of F in the sample; divide moles of F by moles of Xe to get n. In abbreviated form:

molecules XeFn → moles XeFn → mol Xe → g Xe → g F → mol F → mol F / mol Xe ( n)

Execution of #2:

20

23

9.03 x 10 molecules 1 mol Xe 131.3 g x x

1 mol XeF 1 mol Xe6.022 x 10 molecules/mol n

n0.00150 mol XeF 0.197 g Xe ;

0.368 g XeFn – 0.197 g Xe = 0.171 g F; 0.171 g F

19.00 g/mol F 0.00900 mol F ;

0.00900 mol F6

0.00150 mol Xe n = 6

-------------------------------------- #3: Divide mass by #molecules to get the mass per molecule; convert to amu; subtract away the

(average) mass of one Xe atom (131.3 amu) to get the mass of F atoms in one molecule; divide by the (average) atomic mass to get n. In abbreviated form:

g / molecule XeFn → amu / molecule XeFn → amu F (in a molecule of XeFn) → atoms of F (in a molecule of XeFn)

Execution of #3

23

20

0.368 g 6.022 x 10 amu x = 245.4

1 9.03 x 10 molecules g -224.075 x 10 g (per molecule) amu (per molecule)

245.3 amu XeFn – 131.3 amu Xe = 114 amu F (in 1 molecule); 114 amu F

6 atoms F 19.00 amu/atom F

= 6n

Mass %: [Only in Mastering]

Empirical Formula and Molecular Formulas from experimental data:

9. 3.94. A 45.2-mg sample of phosphorus …

Answer: P4Se3

Strategy:

1) (Optional: Recognize that since the mass ratios and mass % of elements are fixed for a compound, you can use gram amounts rather than mg amounts [saves a conversion step])

2) Recognize that in order to determine EF, you need masses (or % masses) of all elements (see

prior problem’s “Strategy”), and that you can obtain mSe by: mSe msample mP

(NOTE: the “selenide” here is the compound formed when P and Se react)

3) Calculate moles of each element(‘s atoms), reduce the ratio, and express as a formula.


PS4-7

Execution:

mSe= 131.6 g of PxSey (the “selenide”) – 45.2 g P = 86.4 g Se

P mol 951.4P mol / g 30.97

P g 245.

491.0951.4 SeP Divide all subscripts by the smaller to get:

Se mol 491.0Se mol / g 78.96

Se g 486.

491.0

491.0

491.0

951.4 SeP P1.333Se1 P4.001Se3 P4Se3

10. 3.122. A hydrate of copper(II) chloride has the following formula: CuCl2 • xH2O. The water…

Answer: x = 2; formula of hydrate is CuCl2• 2H2O

NOTE: Please try to recognize the similarity of this problem to what you did in Experiments 4 and 6! (Although the way I

made you work up Experiment 6, you found the ratio of the actual number of FU’s of each in the sample rather than the moles (since we had not learned about moles yet and I wanted to stress that you could do the problem without “moles”)

Strategy:

1) Recognize that x here represents the ratio of “waters” : “FU’s of CuCl2”, and thus it is also the ratio of moles of H2O (molecules) : moles of CuCl2 (FU’s).

2) This means you need to figure out the two mole values in #1 to calculate x.

3) Recognize that 2.69 g represents the mass of the “anhydrous compound”, which is CuCl2! Thus, you can calculate the molar mass of CuCl2 [= 1(63.55) + 2(35.45) = 134.45 g/mol)] and use it with the mass of CuCl2 (2.69 g) to calculate “moles of CuCl2”. [You are half done with the problem! Now you need to get “moles of H2O”]

4) Recognize that mwater = mhydrate – manhydrous. Once you have this mass, use it, along with the molar mass of H2O [= 2(1.008) + 16.00 = 18.02 g/mol] to calculate “moles of H2O”.

5) Divide the moles of H2O by moles of CuCl2 (and round to nearest whole number) to get x.

Execution:

2

2

2 CuClCuCl mol / g 134.45

CuCl g 2.69 mol 0.0200 (in sample of hydrate)

Mass of water in original hydrate = 3.41 g – 2.69 g = 0.72 g

OHOH mol / g 18.02

OH g 0.722

2

2 mol 9690.03 (in sample of hydrate)

x = 297791.CuCl mol 00.020

OH mol 9690.03

2

2 There are 2 molecules of H2O per FU of hydrate compound

Cumulative problems applying above concepts and skills:

11. 3.116. A metal (M) forms an oxide with the formula M2O. If ….

Answer: M is K (potassium)

Strategy(ies) and Solutions:

In this problem, you ultimately have to realize that the only way to determine M’s identity is to figure out the molar mass of M. Then you can use the Periodic Table to match the molar mass to a single element. There are many correct approaches to determining the molar mass from the info given. Most will involve assuming a 100-g (exact) sample so that you “have” 16.99 g of O and (100-16.99) = 83.01 g of M. Let’s assume that. Also, you must realize that you can calculate

Multiply all subscripts by 3

(because 0.333 1/3)


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molar mass of a substance “experimentally” by knowing the grams and moles of a single sample of the substance:

MM = sample in FUs) (of moles of #

sample in grams of #

I’ll now give two approaches for moving forward:

(A) “My” approach:

1) Calculate the moles of O in the sample (by dividing g O by MM of O).

2) Use the subscript ratio in the formula to calculate moles of M in the sample.

3) Divide mass of M by moles of M to get MM.

Execution of (A):

(atoms) M mol 7532.12O mol 1

M mol 2x

O mol / g 16.00

O g 16.99 ; MM(of M) = g/mol 39.09

M mol 7532.12

M g 83.01

The element on the Periodic Table with MM closest to 39.09 g/mol is K (MM = 39.10 g/mol)

(B) A former student’s approach that I found very clever:

1) Let MM = the MM of M

2) Because of the subscript ratio, the mass ratio of M:O must be: 1(16.00)

x 2

O

M MM

g

g

3) Set the mass ratio in (2) equal to 83.01 g / 16.99 g (mass values from problem), and solve for MM!

Execution of (B):

K is M g/mol 39.0916.99 x 2

16.00 x 83.01

1(16.00)

x 2

O 16.99

M 83.01 MM

MM

g

g (potassium)

12. 3.118. Fructose is a common sugar found in fruit. Elemental analysis of fructose gives the following….

Answer: C6H12O6

Strategy:

Although you could solve this problem by assuming 100 g of sample and finding the empirical formula (as in #14 above), and then dividing MM by EM (as in #16 above) [see NOTE at end of the

“execution” paragraph below], the more straightforward approach here is to just

1) assume a sample of 180.16 grams, since that must be one mole of the compound.

2) Use the %’s to calculate masses of each element in the sample (as “usual”, although here you actually have to do the multiplication since the answer is not equal to the %!).

3) Calculate moles of each element in the sample (as “usual”).

4) Recognize the following: Since there must be a whole number of moles of each atom-type in a mole of a substance (equal to the subscript in the formula), when you calculate the number of moles of each element from the masses now (i.e., #3 above), you will get whole numbers that are the subscripts in the molecular formula (i.e., you don’t ever calculate an empirical formula here—you go “directly” to the molecular formula)!


PS4-9

Execution:

Mass of C (in a one-mole [i.e., 180.16 g] sample) = 180.16 g x 0.4000 [i.e., 40.00%] = 72.064 g C

Mass of H (in a one-mole [i.e., 180.16 g] sample) = 180.16 g x 0.0672 [i.e., 6.72%] = 12.106 g H

Mass of O (in a one-mole [i.e., 180.16 g] sample) = 180.16 g x 0.5328 [i.e., 53.28%] = 95.989 g O

C mol 306.00C mol / g 12.01

C g 4672.0

H mol 099012.H mol / g 1.008

H g 06112. C6H12O6 (is the molecular formula)

O mol 395.99O mol / g 16.00

O g 9895.9

NOTE: If you had gone the “traditional” route of assuming 100 g of sample, you would have ended up with ~3.33 moles of C, 6.666 mol H, and 3.33 mol O, which would have given an empirical formula of CH2O. The EM of this is 30.03 g/mol. 180.16 / 30.03 = 6.00, so the molecular formula would be C6H12O6 (same as result shown above). Either way is fine!

13. 3.100. Tartaric acid is the white, powdery substance that coats tart candies such as Sour Patch Kids. Combustion

analysis of a …

Answer: C2H3O3

Strategy: As in Q9, we need ratios of “moles of atoms” of each element (here, it’s C, H, and O atoms. In a combustion analysis, all the carbon ends up in the CO2, so the number of moles of C atoms simply equals the number of moles of CO2 (because there’s one mole of C atoms per mole of CO2). All of the H ends up in the H2O, so the number of moles of H atoms in the sample is simply equal to the number of moles of H atoms in the H2O. This will be TWO times the number of moles of H2O (because there are two moles of H atoms per mole of H2O. What about the oxygen? As in Experiment 4 (CuxCly∙z H2O), you must realize that once you know the mass of the C and H in the sample (which you can calculate from the moles of atoms of each, by multiplying by the molar masses of C and H, respectively), you can figure out the mass of O in it by difference (i.e., all the “remaining” mass that isn’t C or H must be O!). Once you have that mass, calculate moles of O by dividing by its molar mass. Then you can reduce the mole ratio of atoms and express it as a formula. Execution:

From CO2: atoms C mol 290.319 2

2

2

2

CO mol 1

C mol 1x CO mol 290.319

CO mol / g 44.01

CO g 14.08

From H2O: atoms H mol 4690.47 OH mol 1

H mol 2x OH mol 790.23

OH mol / g 18.02

OH g 4.32

2

2

2

2

The masses of the C atoms and the H atoms are thus:

H&C g H mol 4690.47 C mol 290.319 H g 2930.48H mol 1

H g 1.008x C g 223.84

C mol 1

C g 12.01x 4954.32

So, mass of O atoms in the sample = 12.01 g (total) – 4.32549 g (C + H) = 7.684... g O

which means there are atoms O mol 200.48O mol / g 16.00

O g 487.6

So the empirical formula must be:

101.5891.41

290.319

200.48

290.319

590.47

290.319

290.319200.48590.47290.319 OHCOHCOHC C2H3O3


(because 0.498 ½ )


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====================== END OF WRITTEN SET =======================

Some miscellaneous answers from old keys (many of these problems are assigned now in Mastering, though the

values may differ there).

2.82. How many moles of aluminum do 5.8 x 1024 aluminum atoms represent?

Answer: 9.6 mol Al (reasonable, because 6 x 1024 atoms is more (almost 10 times more!) than one mole)

Work / Reasoning:

A mole of any “item” is, by definition, Avogadro’s number (NA) of items (which, to 4 SF, is 6.022 x 1023 items). Thus:

5.8 x 1024 Al atoms x atoms Al10 x 6.022

atoms Almol 123

9.63… 9.6 mol Al

2.84. What is the mass, in grams, of each elemental sample?

Strategy: Use the periodic table to determine how much mass one mole of that element’s atoms will contain (“molar mass”). Then you can use that value to figure out how much mass is in any number of moles of it (by multiplying). This is like calculating “cost” from “price per lb” and “lb” values.

(a) 2.3 x 10-3 mol Sb Answer: 0...80.2Sb mol 1

Sb g 121.76 x Sb mol 10 x 32. 3- 0.28 g Sb

(c) 43.9 mol Xe Answer: 3.6...657Xe mol 1

Xe g 131.29 x Xe mol 943. 5760 or 5.76 x 103 g Xe

2.88. Calculate the number of atoms in each sample. [Again, assume “natural” samples from Earth]

(b) 39.733 g S Answer: 39.733 g S x atoms907.46atoms mol 1

atoms 10 x 6.022x

g 32.07

S mol 1 23

10 x 7.461 23 ... *

(d) 97.552 g Sn Answer: 97.552 g Sn x atoms674.948atoms mol 1

atoms 10 x 6.0221x

g 118.71

Sn mol 1 23

10 x 4.9488 23 ... **

*This result only has 4 SF because of the molar mass of S found in your text. It’s typically better to use more SF’s in a molar mass than what is “needed” in a problem to maintain precision, but in this case, your text only provided 4 so I used that value.

**This result can only have 5 SF if you use NA from the back cover of your text , rounded to 5 SF (NA is not an exact quantity!). If you used 6.022 x 1023, you technically should have gotten 4.949 x 1023 (rounded to 4 SF) (which would still have been “correct”).

2.136. Without doing any calculations, determine which of the samples contains the greatest amount of the element in

moles. Which contains the greatest mass of the element?

(a) 55.0 g Cr (b) 45.0 g Ti (c) 60.0 g Zn

Answers: 55.0 g has the greatest amount of moles (of atoms); 60.0 g of Zn has the greatest amount of mass

Reasoning: From the Periodic Table, you can see that (left two columns):

One mole of ___ has a mass of: Sample in Problem is

Cr 52.00 g 55.0 g > one mole

Ti 47.87 g 45.0 g < one mole

Zn 65.38 g 60.0 g < one mole

(You could also just divide the # of atoms by NA)


PS4-11

Since Cr is the only sample that has a mass greater than its molar mass, it is the only one that has greater than one moles’ worth of atoms in it. Thus it has the “most moles”. Since the values given are in grams (a unit of mass), clearly the 60.0 g sample has the “most mass”.

3.64. How many moles (of molecules or formula units) are in each sample?

(a) 55.98 g CF2Cl2 Answer: 0.4630 mol (of CF2Cl2 molecules)

MM of CF2Cl2: 1(12.01) + 2(19.00) + 2(35.45) = 120.91 g/mol

2222

22 ClCF8990.462g 1120.9

ClCF mol 1 x ClCF g 855.9 mol 0.4630

(c) 0.1187 g C8H18 Answer: 0.001039 (or 1.039 x 10-3) mol (of C8H18 molecules)

MM of C8H18: 8(12.01) + 18(1.008) = 114.22 g/mol

188188

188 HC2...90.00103g 2114.2

HC mol 1 x HC g 70.118 mol 0.001039

3.70. A salt crystal has a mass of 0.12 mg. How many NaCl formula units does it contain?

Answer: 1.2 x 1018 FU NaCl

...3621.FUs mol 1

FU 10 x 6.022x

g 58.44

NaCl mol 1x

mg 1

g 10 x NaCl mg 0.12

233

1.2 x 1018 FU (of NaCl)

3.74. Iron from the earth is in the form of iron ore. Common ores include Fe2O3 (hematite), Fe3O4 (magnetite), and FeCO3

(siderite). Calculate the mass percent composition of iron for each of these iron ores. Which ore has the highest iron content?

Answers: 69.94%, 72.36% and 48.20% respectively. Fe3O4 has the highest iron content.

Strategy: Pretend you had one mole of each compound. Find the mass of the sample (molar mass) and the mass of (just) the iron in the sample. % is “partial over total” x 100, and here “partial” means “mass of Fe” and “total” means “mass of whole sample [Fe + O]”

hence: %Fe 100x sample mass

Fe mass

Execution of Strategy:

Fe2O3: MM = 2(55.85) + 3(16.00) 159.7 g %Fe Fe 3...469.9100x g 7159.

g 2(55.85)69.94%

Fe3O4: MM = 3(55.85) + 4(16.00) 231.55 g %Fe Fe 0...672.3100x g 55261.

g 3(55.85) 72.36%

FeCO3: MM = 1(55.85) + 1(12.01) + 3(16.00) 115.86 g %Fe Fe 4...048.2100x g 115.86

g 1(55.85) 48.20%

3.78. The American Dental Association recommends that an adult female should consume 3.0 mg of fluoride (F-) per day

to prevent tooth decay. If the fluoride is consumed in the form of sodium fluoride (45.24% F), what amount of sodium fluoride contains the recommended amount of fluoride?

Answer: 0.0066 g (or 6.6 mg) NaF

(Reasonable: 121 g of CF2Cl2 is one mole, so 56 g should be about half a mole)


PS4-12

Reasoning: 45.24%F means NaF g 0100.

F g 45.24. Use this like a “conversion factor” to calculate “g NaF”

from “g F”. Most chemists would convert mg to g first and report the answer in grams, although you

could also use the factor NaF mg 0100.

F mg 45.24 to go directly to mg (since the Q does not specify “grams”!)

Setup: ) (or 3.60.006 F g 45.24

NaF g 0100.x

mg 1

g 0 1 xF mg 3.0

-3

mg 6.6g 0.0066 ..

3.90. Calculate the empirical formula for each natural flavor based on its elemental mass percent composition.

(a) methyl butyrate (component of apple taste and smell): C 58.80%, H 9.87%, O 31.33%

Answer: C5H10O2

Strategy:

Execution: Assuming a 100-gram sample of the compound, you’d have 58.80 g of C (atoms), 9.87 g of H (atoms), and 31.33 g O (atoms).

C mol 954.89C mol / g 12.01

C g 58.80

H mol 9.7916H mol / g 1.008

H g 9.87 181.951699.7954.89 OHC Divide all subscripts by the smallest to get:

O mol 181.95O mol / g 16.00

O g 31.33 1105.002.50

181.95

181.95

181.95

1699.7

181.95

954.89 OHCOHC

3.96. The molar mass and empirical formula of several compounds are listed below. Find the molecular formula of each

compound.

(a) C4H9, 114.22 g/mol Answers: (a) C8H18 (b) C6Cl6 (c) C18H12N6

(b) CCl, 284.77 g/mol

(c) C3H2N, 312.29 g/mol

Strategy:

Recognize that the molecular formula is always a multiple of the empirical formula (because the empirical formula is a reduced version of the molecular formula). It follows then that the molar mass must be a multiple of the empirical (molar) mass (your text calls this the “empirical formula molar mass”)

Thus, to determine the molecular formula from the empirical formula and molar mass:

1) Calculate the molar mass of the empirical formula (think of this as the molar mass of a hypothetical substance

whose molecular formula IS the empirical formula given).

2) Divide the molar mass of the compound by the empirical (molar) mass to see how many times larger the formula units are. Round this to a whole number (call it "n").

3) Multiply all the subscripts in the empirical formula by "n". (This makes the formula units n times bigger than

those given by the empirical formula.)

Execution:

masses of each element in any specific

mass of compound [100 g is most convenient

unless molar mass is provided]

mass %'s of each element moles of each element

reduced

mole ratio (and formula)


(because 0.5 is ½ ) C5H10O2


PS4-13

(a) C4H9, 114.22 g/mol

Empirical (molar) mass: 4(12.01) + 9(1.008) 57.112 g/mol

2 9...91.99(EM) g/mol 2157.1

(MM) g/mol 114.22 Formula units are 2 x “larger” (C4H9)2 C8H18

(b) CCl, 284.77 g/mol

Empirical (molar) mass: 1(12.01) + 1(35.45) 47.46 g/mol

6 6.0002...(EM) g/mol 47.46

(MM) g/mol 284.77 Formula units are 6 x “larger” (CCl)6 C6Cl6

(c) C3H2N, 312.29 g/mol

Empirical (molar) mass: 3(12.01) + 2(1.008) + 1(14.01) 52.056 g

6 9...95.99(EM) g/mol 6552.0

(MM) g/mol 312.29 Formula units are 6 x “larger” (C3H2N)6 C18H12N6

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