Download Problem Set 7

Problem Set 7
CP Chemistry
Answer Key
Chapters 11 and 12
The Mole and Stoichiometry
The mole
1) How many atoms of carbon are there in 3 moles of diamond?
(A diamond is solid carbon). (Cd = carbon as the allotrope of diamond)
3 moles Cd x (6.02 x 1023 atoms Cd / 1 mole Cd) = 1.806 x 1024 atoms Cd
2) You love donuts. Every school day you buy two donuts. How many donuts will
you buy in on school year (180 days)? How many dozen donuts is that? How
many gross is that? (1 gross is 12 dozens, or 144 donuts)
180 days x (2 donuts / day) = 360 donuts
360 donuts x (1 dozen / 12 donuts) = 30 dozen
30 dozen x ( 1 gross / 12 dozen) = 2.5 gross… that is gross!
3) How many atoms of hydrogen are there in 1 mole of water molecules?
1 mole H2O x (2 moles H atoms / 1 mole H2O molecules) = 2 moles H atoms
2 moles H atoms x (6.02 x1023 atoms / 1 mole H atoms) = 1.204 x 1024 H atoms
4) The decimal at the end of a sentence has 3.00 x 1018 atoms of graphite in it. How
many moles are there in that many atoms? (Cg = carbon as the allotrope graphite)
3.00 x 1018 atoms Cg x (1 mole Cg / 6.02 x 1023) = 4.98 x 10-6mole Cg
Relative Atomic Mass and the Periodic Table
5) Give the three sub-atomic particles that make up atoms with their relative masses
in amu. Be certain to explain which particles contribute the most to the mass of
an atom and where those particles are found. Define amu, too! The proton has a
mass of 1 amu, the electron has a mass near 1/2000th an amu, the neutron has
a mass of 1 amu. The protons and neutrons are typically found in the
nucleus and they make up most of the mass of the atom. The electrons are
typically located in orbitals surrounding the nucleus . An amu is the atomic
mass unit used to show the relative mass of sub-atomic particles that make
up atomic weight.
6) There are 100 students all chosen to be part of a shoe size measurement
experiment. 60 students have size 6 shoes. 30 students have size 7 shoes and 10
students have size 10 shoes. What is the weighted average shoe size for the
students?
60/100 x 6 = 3.6
30/100 x 7= 2.1
10/100 x 10 = 1
weighted average shoe size is 3.6 + 2.1 + 1 = 6.7
7) Calculate the average atomic mass of iron using the following data:
(We will assume the atomic mass of each isotope is its mass number.)
a. Iron-54 has a relative abundance of 5.90% 54 x 0.059 = 3.19
b. Iron-56 has a relative abundance of 91.72% 56 x .9172 = 51.36
c. Iron-57 has a relative abundance of 2.10% 57 x 0.021 = 1.20
d. Iron-58 has a relative abundance of 0.280%. 58 x .0028= 0.16
Weighted average atomic mass = 55.91 amu
8) What is the difference between the average atomic mass of an element and the
atomic weight? (Think Units)
The average atomic mass is given in amu, while atomic weight is typically
expressed in grams per mole. More typically the term molar mass is used to
express grams per mole amounts.
9) Convert the following masses into moles for each element given.
a. 3.45 grams of tin x (1 mol Sn / 118.71 g Sn)=0.291 mol Sn
b. 2.54 x 10-3 grams of sodium x (1 mol Na/ 22.9 g Na)= 1.10 x 10-4 mol Na
c. 3.92 grams of oxygen gas (O2) x (1 molO2/32.0 gO2) = 0.123 mol O2
10) Convert the following amount of each element into grams of the element.
a. 1.23 moles of gold x (196.96 gAu/1mol Au) =242 g Au
b. 4.35 x 10-3 moles of zirconium x (91.224g Zr/1mol Zr) =0.397 g Zr
c. 3.24 moles of nitrogen gas (N2) oops! Freebie
3.24 mol N2 x (28.0 gN2/1 mol N2) = 90.72 g N2
11) One mole of any gas at STP is 22.4 L. What are the conditions of STP? Why is
this previous statement true if CO2 has a different size than radon, but both are
gases? The conditions at STP are 0oC and 1 atm of pressure. CO2 and Rn are
certainly different sized particles, but because gases have no intermolecular
attraction and very high kinetic energy, the volume they occupy as a sample
does not relate to the volume they occupy as very small molecules or atoms.
12) Make an educated guess as to the volume (in liters) of gas you exhale. Calculate
these liters into moles and into molecules.
Each breath is about 2.5 Liters. (my educated guess… having never been a biology major…)
2.5 L gas x (1 mol/22.4L) = 0.11 mol gas
0.11 mol gas x (6.02 x 1023 molecules/1 mol) = 6.72 x 1022 molecules
13) There are 100.0 mL of Neon gas inside a neon light at the local pool hall.
Assuming the light was produced at STP, how many moles of gas are inside the
bulb? What is the mass of that many moles of neon?
100.0 mL x (1L / 1000 mL) = 0.1000 L Ne
0.1000L x (1 mol / 22.4 L) = 0.00446 mol Ne
0.00446 mol Ne x ( 20.17 g Ne / 1mol Ne) = 0.0900 g Ne
14) The density of argon gas is reported to be 1.784 g/L at STP. Use the molar mass
of argon and the conditions at STP to determine the density of argon to confirm
this value.
(39.948 g Ar / 1 mol Ar) x ( 1 mol Ar / 22.4 L Ar) = 1.783 g/L Ar – close enough!
15) Calculate the average mass of one atom of uranium. (gram / atom)
(238.0289 g U / 1 mole U) x (1 mole U/6.02 x 1023 atoms U) = 3.9539 x 10-22 g/atom U
Chemical Formulas and Moles
16) Calculate the molar mass of the following compounds:
a. zinc chloride, ZnCl2 65.39 g/mol Zn + 2 x 35.45 g/mol Cl = 136.29 g/mol ZnCl2
b. water, H2O 2 x 1.00 g/mol H + 16.00 g/mol O = 18.00 g/mol H2O
c. strontium nitrate, Sr(NO3)2
87.62 g/mol Sr + 2 x 14.00 g/mol N + 6x 16.0 g/mol O = 211.62 g/mol Sr(NO3)2
17) Every can of Jolt Cola contains 200 mg of caffeine, C8N4O2H10. If you were to
drink two cans of Jolt right now to stay awake while working on this problem set,
how many moles of caffeine would you ingest?
200 mg/can x 2 cans = 400 mg x 1 g/ 1000 mg = 0.400 g caffeine
0.400 g caffeine x (1 mol caffeine / 194 g caffeine) = 2.06 x 10-3 moles caffeine
18) Give the molar mass of each of these substances:
i. Mg(OH)2 24.305 g/mol Mg + 2 x 16.0 g/mol O + 2 x 1.00 g/mol H =
58.305 g/mol Mg(OH)2
ii. Chromium (IV) nitrate 52.00 g/mol Cr +4x 14.0 g/mol N+ 12 x 16.0
g/
g
mol O = 300 /mol Cr(NO3)4
iii. Bromic acid
1.00 g/molH + 79.9 g/mol Br + 3x16.0 g/mol O = 128.9 g/mol HBrO3
19) Give the mass of the following samples
i. 0.25 moles of copper (II) sulfate
0.25 mol CuSO4 x (159.55 gCuSO4 / 1mol CuSO4 )= 40. g CuSO4
ii. 1.2 x 10-3 moles of sodium oxalate
-3
1.2x 10 mol Na2C2O4 ( 134 g Na2C2O4 / 1 mol Na2C2O4) = 0.16 g Na2C2O4
20) Give the number of moles of the following samples
i. 0.0045 grams of magnesium hydroxide
0.0045 g Mg(OH)2 x (1 mol Mg(OH)2 / 58.305 gMg(OH)2 ) = 7.72 x 10-5 mole Mg(OH)2
ii. 2.56 mg of calcium phosphate
2.56 g Ca3(PO4)2 ( 1mol Ca3(PO4)2 / 310.15 g Ca3(PO4)2) = 8.25 x 10-3 mol Ca3(PO4)2
Using Analytical Data
21) Use water as an example to explain what percent composition tells you.
Percent composition tells me the percent of each element in a compound. The
percent composition of water is 11.11 % hydrogen and 88.89 % oxygen.
Regardless of the sample size, the mass of the water will always be composed of
11.11% hydrogen and 88.89% oxygen.
(2gH/mol H2O/18g/mol H2O x 100 = 11.11 % H, 16gO/molH2O/18g/molH2O x 100 = 88.89 % O)
22) What is the percent composition of TNT: C7H10(NO2)3?
232 g/mol TNT
% C = 84 / 232 x 100 = 36.2% C
% H = 10 /232 x 100= 4.3 % H
% N = 42 / 232 x 100 =18.1 % N
% O =96 /232 x 100 = 41.4 % O
23) What does the empirical formula tell you about a substance? What is the
empirical formula of glucose, C6H12O6? All values are divisible by 6, sot he
empirical formula of glucose is CH2O.
24) You have worked diligently in lab to determine that the percent composition of a
binary ionic compound is 53.7 % oxygen and 46.3 % lithium. What is the empirical
formula of this substance? Does the periodic table support your data?
53.7 % O  53.7 g O x (1mol O/16.0 gO) =3.35 mol O
46.3% Li  46.3 g Li (1 mol Li / 6.94 g Li) = 6.70 mol Li
6.70 mol Li / 3.35 mol O = 2 mol Li to 1 mol O  Li2O
The periodic table does support my findings…Li+1 and O-2  Li2O
25) Acetylene, a gas used in welding, has an empirical formula of CH. If the
molecular mass of acetylene gas is 26.04 g/mol, what is its molecular formula?
12.0 g/mol C + 1.01 g/mol H = 13.02 g/mol CH 26.04 / 13.02 = 2 CH x 2  C2H2
26) You and your lab partner worked just as diligently on a second lab to determine that
the molecular mass of an organic compound was 120.0 g/mol and the substance was
40.0% carbon, 6.66% hydrogen and 53.33 % oxygen. What is the empirical formula
for the substance? What is the molecular formula for the substance?
40.0 % C  40.0 g C x (1 mol C / 12.0 g C) = 3.33 mol C
6.66 % H  6.66 g H x (1 mol H / 1.00 g H ) = 6.66 mol H
53.33 % O  53.33 g O x (1 mol O /16.0 g O) = 3.33 mol O
empirical formula = CH2O because 6.66 / 3.33 = 2
empirical mass = 30.0 g/mol
27) A 175.0 gram sample of Monosodium Glutamate (MSG) has the following
components: 56.15 g C, 9.43 g H, 74.81 g O, 13.11 g N and 21.49 g Na. What is
the empirical formula of this commercial flavor enhancer?
56.15 g C / 175.0 g MSG = 32.1 % C  32.1 g C x (1 mol C / 12.0 g C) = 2.67 mol C
9.43 g H / 175.0 g MSG = 5.39 % H  5.39 g H x (1 mol H / 1.0 g H) = 5.39 mol H
74.81 g O / 175 g MSG = 42.7% O  42.7 g O x (1 mol O / 16.0 g O) = 2.67 mol O
13.11 g N / 175 g MSG = 7.49 % N  7.49 g N x(1 mol N / 14.0 g N) = 0.53 mol N
21.49 g Na / 175 g MSG = 12.3% Na 12.3g Na x(1 mol Na / 23.0 g Na) = 0.53 mol Na
ratios all with sodium
2.67 mol C / 0.53 mol Na  4 mol C to 1 mole Na
5.39 mol H / 0.53 mol Na  10 mol H to 1 mole Na
empirical formula
2.67 mol O / 0.53 mol Na  4 mol O to 1 mole Na
= NaC4H10O4N
0.53 mol N / 0.53 mol Na  1 mol N to 1 mole Na
Solutions
28) Define the term molarity (M). Give its units and how it is useful to chemists.
Molarity is the concentration of solutions in moles of solute in liters of solution.
Chemists use Molar concentrations more often than any other concentration
unit because it relates a molar amount of the compound in a solution vs the total
volume of the solution, not the mass or other measurement, but the VOLUME is
the key.
29) Define the term molality (m). Give its units and how it is useful to chemists.
The unit molality is used to measure the concentration of a solution in moles of
solute in kilograms of solvent. This unit is useful for solubility of compounds in
a yet to be produced solution or in terms of environmental contamination or
biological process.
30) What is the molar concentration (M) of 0.54 moles of NaHCO3 dissolved in 0.45
L of solution?
M = 0.54 mol/0.45 L = 1.2 M NaHCO3
31) What volume of solution is prepared for a 0.87 M concentration and 0.35 moles of
sodium chloride?
0.87M = 0.35 mol / x L
x L = 0.35 mol/0.87 M
x L = 0.402 L
32) How many moles of solute are in 0.037 liters of a 4.3 M solution?
4.3 M = x mol / 0.037 L
4.3 M x 0.037 L= x mol
x mol = 0.16 mol
33) How many grams of solid silver nitrate (gfm = 169.8 g/mol) are needed to prepare
0.100 L of a 0.25M concentration?
0.25M = x mol / 0.100 L
x mol = 0.025 mol AgNO3= x (169.8 g/mol) = 4.25 g AgNO3
34) What is the molal concentration (m) of a solution that contains 3.0 moles of
lead(II) nitrate in 2.01 x 103 g of water?
m = 3.0 mol/2.01 kg = 1.49 m
35) How many moles of sodium hydroxide are dissolved in 4.65 kg of water if the
solution has a 0.200 molal concentration?
0.200 m = x mol / 4.65 kg = 0.93 mol NaOH
36) How many kilograms of benzene are needed as the solvent to dissolve 0.45 moles
of sucrose (C6H12O6) to make a 3.5 molal solution?
3.5 m = 0.45 mol / x kg
 0.13 kg benzene
Dilutions M1V1 = M2V2 and logic
37) How many milliliters of 14 M HCl are needed to produce 100mL of a diluted
solution of 5.0M HCl? Approximately how many mL of water should be added
14 M (V1) = 5.0M (100 mL)
V1 = 35.7 mL g concentrated HCl
and (100 mL – 35.7 mL) = about 64.3 mL H2O
38) How many milliliters of water were added to a 3.7M solution, if you have 34.1
mL of a dilute solution at 1.3M solution?
3.7 M (V1) = (34.1 ml)(1.3 M) V1 = 11.9 mL
so 34.1 – 11.9 = 22.1 mL H2O were added
39) How many milliliters of 18.0 M H2SO4 are needed to produce 100.00 mL of a
diluted solution of 1.50 M HCl? Approximately how many mL of water should
be added?
(18.0 M) (V1) = (1.50 M) (100.0 mL)
V1 = 8.33 mL
and 100-8.33 = 91.66 mL H2O
40) How many milliliters of water were added to a 5.00 M solution, if you have 35.0
mL of a dilute solution at 0.250 M solution?
(5.00M) (V1) = (35.0 mL)(0.250M)
V1 =1.75 mL so 35 -1.75 = 33.25 mL H2O
41) Dissociate the following ionic compounds if they are soluble in water.
a. Na2CO3  2Na+1 + CO3-2
d. Ca(OH)2  insoluble
b. KCl K+1+Cl-1
e. Mg3(PO4)2  insoluble
+3
-2
c. Al2(SO4)3  2Al + 3 SO4
42) Given the following data determine the density, molarity and molality of the
saturated solution of potassium nitrate.
a. Mass of empty evaporating dish: 35.45 g
b. Volume of saturated solution: 5.00 mL
c. Mass of dish with solution: 41.87 g
d. Mass of dish with solid salt: 39.77 g
Density = g/mL
(41.87g-35.45g)/5.00mL = 1.28 g/mL
Molarity = mol KNO3/L solution 39.77g-35.45g = 4.32 g KNO3
4.32 g KNO3 x (1 molKNO3 / 101.1 g KNO3) =0.0427 mol KNO3
0.0437 mol KNO3 / 0.005 L = 8.55 M KNO3
molality = mol KNO3/kg H2O
39.77g-35.45g = 4.32 g KNO3
4.32 g KNO3 x (1 molKNO3 / 101.1 g KNO3) =0.0427 mol KNO3
41.87g -39.77g = 2.1 g H2O x 1 kg /1000g = 0.0021 kg H2O
0.0427 mol KNO3 / 0.0021 kg H2O = 20.3m
Stoichiometry
43) Balance these reactions
a. C7H6O3 (aq) + C4H6O3 (aq)  C9H8O4 (s) + HC2H3O2 (l) balanced!
b. Mg(OH)2 (aq) + H2CO3  2H2O (l) +MgCO3 (aq)
c. C3H6 (g) + 4.5 O2 (g)  3 CO2 (g) + 3 H2O (aq)
d. 2NaN3 (s)  2Na (s) +3 N2 (g)
e. NCl3 (aq) + 3H2O (l)  NH3 + 3HOCl (aq)
f. P4O10 (s) + 6H2O (l)  4H3PO4 (l)
44) Write a balanced chemical equation based for the reaction of aqueous aluminum
hydroxide reacting with aqueous sulfuric acid to produce water and solid
aluminum sulfate. 2 Al(OH)3 (aq) + 3 H2SO4 (aq)  Al2(SO4)3 (aq) + 6 H2O (l)
45) Write a sentence describing in detail this reaction: (see the previous question as an
example)
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
2 moles of solid potassium chlorate decompose into 2 moles of solid potassium
chloride and 3 moles of oxygen gas.
46) Draw a picture showing how 12 molecules of hydrogen gas will react with 3
molecules of nitrogen gas to produce the maximum molecules of ammonia (NH3)
gas. Use blue for hydrogen and red for nitrogen. Give the balanced synthesis
reaction and remember your diatomic elements.
+
3H2 (g) + N2 (g)  2NH3 (g)
There are 3 molecules of hydrogen that do not react – they are in excess.
47) The last few questions are from the last unit, why are they here? Relate what
balanced chemical equations tell you about a reaction.
Stoichiometric ratios from balanced chemical reactions allow scientists to relate
quantities of one substance to another. Without the ratios from equations, the
Stoichiometry is useless.
48) Define Excess Reactant and Limiting Reactant. Why are these two terms
important in industrial production of compounds?
The reactant is excess is the substance that remains after the limiting reactant runs
out. The limiting reactant is the reactant that runs out before any other in a
chemical reaction – it limits the amount of product made. The limiting reactant in
most industrial reactions are typically the most expensive compounds.
49) Explain why “Theoretical Yield” is more than the “Actual Yield” for most
reactions. The theoretical yield is typically more than actual yield because of
human errors, laboratory conditions are not always ideal, the concentrations
of products slows reactions down by getting in the way… others
50) Determine the Percentage Yield if a reaction has the theoretical yield of 3.45
grams of product and the actual yield of 3.25 grams of product.
3.25 g / 3.45 g = 94.2 % yield
Use this reaction for the next four questions:
Cu (s) + 4 HNO3 (aq)  Cu(NO3)2 (aq) + 2 NO2 (g) + 2 H2O (l)
51) 3.0 moles of copper react completely in a reaction.
a. How many moles of nitric acid are needed?
3.0 mol Cu x (4 mol HNO3 / 1 mol Cu) = 12.0 mol HNO3
b. How many moles of copper (II) nitrate are produced?
3.0 mol Cu x (1 mol Cu(NO3)2 / 1 mol Cu) = 3.0 mol Cu(NO3)2
c. How many moles of water are produced?
3.0 mol Cu x (2 mol H2O / 1 mol Cu) = 6.0 mol H2O
d. How many moles of nitrogen dioxide are produce?
3.0 mol Cu x (2 mol NO2 / 1 mol Cu) = 6.0 mol NO2
e. How many liters of NO2 are produced assuming the lab was performed at
STP? 6.0 mol NO2 x 22.4 L/ 1mol = 134.4 L
52) If 1.2 moles of copper react with 4.0 moles of HNO3, what is the limiting
reactant?
HAVE
NEED
1.2 mol Cu x (4mol HNO3 / 1 mol Cu) = 4.8 mol HNO3
4.0 mol HNO3 x (1 mol Cu / 4 mol HNO3) = 1 mol Cu
You don’t have enough HNO3 – it runs out (need 4.8 mol, have 4.0 mol) so HNO3 is
the limiting reactant.
53) If 1.2 moles of copper are placed into 4.9 moles of HNO3, how much of the
reactant in excess remains at the end of the reaction?
HAVE
NEED
1.2 mol Cu x (4mol HNO3 / 1 mol Cu) = 4.8 mol HNO3
4.9 mol HNO3 x (1 mol Cu / 4 mol HNO3) = 1.22 mol Cu
Cu limits the reaction and there is 0.10 mol of HNO3 left over. (4.9 – 4.8)
54) 4.5 moles of Cu are placed into 1.00L of 2.5M HNO3
a. Which reactant is in excess? Cu
1.00 L of 2.5 M HNO3 = 2.5 mol HNO3
HAVE
NEED
4.5 mol Cu x (4mol HNO3 / 1 mol Cu) = 4.8 mol HNO3
2.5 mol HNO3 x (1 mol Cu / 4 mol HNO3) = 0.625 mol Cu
b. How many moles of that reactant are in excess? 3.875 mol Cu
c. How many moles of water will be produced?
2.5 mol HNO x (2 mol H2O / 4 mol HNO3) = 1.25 mol H2O produced
d. How many grams of copper (II) nitrate will be produced?
2.5 mol HNO x (1 mol Cu(NO3)2 / 4 mol HNO3) = 0.625 mol Cu(NO3)2 produced
0.625 mol Cu(NO3)2 x (187.54 g Cu(NO3)2 / 1 mol Cu(NO3)2) =117. 2 g Cu(NO3)2
55) If you use 8.55 mL of 0.30 M AgCl to react with 13.45 mL of NaCl, what is the
molarity of the chloride solution? TYPO!! It should have been 0.30M AgNO3
(0.30 M AgNO3)(8.55 mL AgNO3) = (MCl)(13.45 mL)
MCl = 0.191 M
56) If your chloride concentration is 0.19 M, how many grams of chloride ions are
dissolved in 1.00 mL of solution?
0.19M =x mol / 0.001 L
1.9 x 10-4 mol Cl- x 35.45 g Cl/mol Cl =0.0067 g Cl57) If you have 6.44 mL of 0.30M HCl, how many mL of 0.60 M NaOH will you
need to run the reaction?
(0.30M)(6.44 ml) = (0.60M)(VNaOH) VNaOH = 3.22 mL
58) What concentration of NaCl exists if 32.5 mL are fully reacted with 9.45 mL of
1.2 M AgNO3?
(M1)(32.5 mL) = (1.2 M)(9.45 mL)
M1 = 3.54 M
59) You run a single replacement reaction with 0.25 L of 0.25 M of aqueous copper
(II) sulfate and 3.65 grams of iron. Solid copper is produced along with a solution
of iron (III) sulfate. How many grams of solid copper should be formed and how
many grams of the reactant in excess remain after the reaction is done?
a. Write the balanced chemical equation.
3 CuSO4 (aq) + 2 Fe (s)  3 Cu (s) + Fe2(SO4)3 (aq)
b. How many moles of each reactant are there?
0.25 L of 0.25 M Cu SO4 = 0.0625 mol CuSO4
3.65 g Fe x (1 mol Fe / 55.847 g Fe)=0.0654 mole Fe
c. Which reactant is the limiting reactant?
HAVE
NEED
0.0625 mol CuSO4 x (2 mol Fe / 3 mol CuSO4) = 0.0436 mol Fe
0.0654 mole Fe x (3 mole CuSO4/ 2 mol Fe) = 0.0981 mol CuSO4
There is not enough CuSO4 to use all the Fe up, so CuSO4 is the limiting reactant.
d. How much of the reactant in excess remains at the end of the reaction?
0.0654 mol Fe - 0.0436 mol Fe = 0.0218 mol Fe left over
e. How much copper can be produced from the reaction?
0.0625 mol CuSO4 x (3 mol Cu / 3 mol CuSO4) = 0.0625 mol Cu
60) How many grams of water can 12.5 grams of anhydrous chromium (III) sulfate
salt absorb if the formula of the hydrated salt is Cr2(SO4)3.18H2O?
12.5 g Cr2(SO4)3 x ( 1 mol Cr2(SO4)3 / 392 g Cr2(SO4)3) =0.032 mol Cr2(SO4)3
0.032 mol Cr2(SO4)3 x (18 mol H2O / 1 mol Cr2(SO4)3) = 0.574 mol H2O
0.574 mol H2O x (18.0 g H2O / 1 mol H2O) = 10.3 g H2O