Chapter 3 Stoichiometry. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question,...

Chapter 3

StoichiometryStoichiometry

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 3–2

QUESTIONCurrently we can measure relative atomic mass using an instrument known as a mass spectrometer. However, the masses of atoms are still not reported as whole numbers. What is the chief reason for this?

1. Atoms are so small that even modern techniques have a limitedaccuracy that keeps the value from being a whole number.

2. The atomic mass scale is based on carbon-12 and since thisatom does not have a whole number mass the comparisons willnot be able to have whole number masses also.

3. Isotopic abundance is averaged to provide the mass of atoms.4. Atomic mass units figure in here, but I am not sure how.


ANSWERChoice 3 properly explains that the masses of elements are based on the weighted average of the isotopes for each element. Therefore, even carbon will not have a whole number mass.

Section 3.2: Atomic Masses


QUESTIONSuppose an element consisted of three naturally occurring isotopes. The reported average mass for the element was 75.42 amu. The following information is known: isotope I has a mass of 80.45 and makes up 2.4% of the element’s atoms. Isotope II has a mass of 72.22 and makes up 30.0% of the element’s atoms. What would you estimate as the mass of isotope III?

1. 742. 773. 674. With only this information, I can’t really make a

reasonable estimate.


ANSWERChoice 2 provides a reasonable estimate. Since the average is approximately 75 the third isotope must be slightly higher than that to balance the 72 value, but since the 72 mass isotope is only 30% of the atoms, isotope III, with approximately 70% of the atoms would not have to be very much greater.



QUESTIONSince it is extremely difficult to “see” atoms without elaborate equipment, chemists rely on the atomic mass scale when we need to determine the number of atoms in a sample. An analogy for this type of counting would be as follows: A U.S. nickel coin has a mass of approximately 5.0 grams. How many nickels make up $100? What would be the mass of the money? (By the way, could you lift $100 worth of nickels?)

1. 2,000 nickels; 10,000 grams; yes2. 2,000 nickels; 10,000 grams; probably not3. 20,000 nickels; 100,000 grams; yes4. 20,000 nickels; 100,000 grams; not likely


ANSWERChoice 1 provides all three correct responses. $100 20 nickels/$1 = 2000 nickels; Then 5 grams/nickel = 10,000 grams. This is 10 kg or about 22 pounds.



QUESTIONConsider that a diamond in a jewelry item is made of only carbon atoms. If the diamond had a mass of 0.20 grams (about 1 carat) how many atoms of carbon would be present?

1. 122. 2.43. 1.0 1022

4. 1.0 1023


ANSWERChoice 3 correctly gives the atomic make up of the 0.20 gram diamond. Assuming that one carbon atom has a mass = 12 amu; (or 12 amu (1 g/6.02 1023 amu) = 2.0 10–23 grams)then, 0.20 grams (1 atom/2.0 10–23 grams) = 1.0 1022 atoms.

Section 3.2: Atomic MassesSection 3.3: The Mole


QUESTIONSince atoms are so very small there will be a large number even in a small sample of an element. For example, if you just received 10 billion atoms of gold would this be better than winning a lottery prize? Prove by reporting the mass of gold.

1. The mass of gold would be 1970 grams.2. The mass of gold would be 3 10–22 grams.3. The mass of gold would be 1.7 10–17 grams.4. The mass of gold would be 3 10–12 grams.


ANSWERAlthough it is not better than winning the lottery, choice 4 reflects the correct mass of ten billion gold atoms.

10 109 atoms (197 amu/1 atom) (1 g/6.02 1023 amu)

Section 3.3: The Mole


QUESTIONThe metal gallium (atomic number 31) has the properties you would expect in a metal, but it is so soft it will melt in your hand. If you had 10.0 grams of solid gallium in your palm and watched it turn into liquid how many moles of liquid gallium would you soon have?

1. 6.97 moles2. 69.7 moles3. 0.143 moles4. 8.6 1022 moles


ANSWERChoice 3 represents the mole equivalent for 10.0 grams of gallium.

10.0 grams (1 mole/69.72 grams) = 0.143 moles

Section 3.2: The Mole


QUESTIONThe fuel in small portable lighters is butane (C4H10). Suppose after using such a lighter for a few minutes (perhaps to encourage your favorite concert performer to play one more encore) you had used 1.0 gram of fuel. How many moles of butane would this be?

1. 58 moles2. 0.077 moles3. 1.7 10–24 moles4. 0.017 moles


ANSWERChoice 4 shows the answer for a proper grams to mole conversion. You need to know the molar mass of butane: 4 12.01 g/mol = 48.04 for carbon + 10 1.008 g/mol = 10.08 for hydrogen. Total = 48.04 + 10.08 = 58.12 g/mol. Next; 1.0 gram of butane 1 mol/58.12 g = 0.017 mol

Section 3.4: Molar Mass


QUESTIONSuppose you were very thirsty, but the only source of water was administered by a chemist… . Which would help satisfy your thirst more…42.0 grams of water or 6.02 1024 molecules of water? How much greater is your selection over the one you did not select?

1. I pick 6.02 1024 molecules; it is 7.7 moles greater.2. I pick 6.02 1024 molecules; it is 24 grams greater.3. I pick 42.0 grams; it is 24 grams greater.4. I pick 42.0 grams; it is 6.02 1023 molecules greater.


ANSWERChoice 1 correctly selects the most thirst-quenching choice and expresses how much greater it is than the other choice.

42.0 g (1 mol/18.02 g) = 2.33 moles

6.02 1024 molecules (1 mol/6.02 1023 molecules) = 10.0 moles

10.0 moles – 2.33 moles = 7.7 moles greater



QUESTIONThe compound Cr2O3 (chromium (III) oxide) is one of the key components responsible for the red color of ruby gems. If you had 34.8 grams of Cr2O3, how many grams of chromium (atomic number = 24) metal would be present?

1. 11.9 grams of Cr 2. 5.85 grams of Cr3. 23.8 grams of Cr4. 69.8 grams of Cr


ANSWERChoice 3 correctly predicts the grams of Cr found in 34.8 grams of Cr2O3.

34.8 grams Cr2O3 1 mol/152.0 g = 0.229 moles of Cr2O3

0.229 moles Cr2O3 2 mol Cr/1 mol Cr2O3 = 0.458 moles of Cr0.458 moles Cr 52.00 g/1 mol Cr = 23.8 gramsOR, 2 (52.00)/152.0 100 = 68.42 % Cr0.684 34.8 = 23.8 g Cr



QUESTIONMorphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound?

1. 42.5%2. 27.9%3. 71.6%4. This cannot be solved until the mass of the sample is given.


ANSWERChoice 3 is correct. First determine the molar mass of the compound, then divide that into the total mass of carbon present, and finally, multiply that by 100.

(17 12.01) / ((17 12.01) + (19 1.008) + (1 14.01) + 3 16.00)) = 0.7160.716 100 = 71.6 %

Section 3.5: Percent Composition of Compounds


QUESTIONThe dye indigo is a compound with tremendous economic importance (blue jeans wouldn’t be blue without it.) Indigo’s percent composition is: 73.27% C; 3.84% H; 10.68%N and 12.21% O. What is the empirical formula of indigo?

1. C6H4NO2. C8H3NO3. C8H5NO4. I know this should be whole numbers for each atom, but I do not

know how to accomplish that.


ANSWERChoice 3 is the smallest whole number ratio of the atoms that make up a molecule of indigo. The percentage must be converted to a mass, then the mass is converted to moles of the atoms and finally, the smallest is divided into the others to obtain the proper ratio.

Section 3.6: Determining the Formula of a Compound


QUESTIONWithout further information, which of the following is more likely to be both a molecular formula and an empirical formula for a compound? Briefly explain why. H2O2; C6H6; C2H6O

1. H2O2 . The whole numbers for each atom are reduced as far aspossible and still retain the identity of the molecule.

2. C6H6. The whole number of atoms are equal for each element.3. C2H6O . One element shows only one atom so the formula

ratio could not be further reduced, but the molecular formula could just have one atom.

4. None of these, it is not possible for a formula to be bothempirical and molecular.


ANSWERChoice 3 shows the correct choice and proper justification. The whole numbers are reduced to their lowest value so C2H6O could be the empirical formula. The other choices could still be reduced to smaller whole numbers and maintain the same ratio.

Section 3.6: Determining the Formula of a Compound


QUESTIONAgriculturally, the following equation is important because it is used to make millions of tons of urea. When the equation is balanced, how many hydrogen atoms will be present on both sides of the equation? Also, how many moles of NH3 would be needed to react completely with 0.5 mole of CO2?

NH3 + CO2 CO(NH2)2 + H2O

1. Three; two2. Three; one3. Six; one4. Six; two


ANSWERChoice 3 answers both stoichiometry questions correctly. By placing a coefficient of 2 in front of NH3 six hydrogen atoms would be represented on the left side of the equation. This equals the six on the right side. Next, the balanced equation shows a 2:1 relationship between NH3 and CO2. So, if only 0.5 mole of CO2 were present then one mole of NH3 would be needed to maintain the 2:1 reacting ratio.

0.5 mole CO2 2 mol NH3/1mol CO2 = 1 mol NH3

Section 3.8: Balancing Chemical Equations Section 3.9: Stoichiometric Calculations: Amounts of Reactants and Products


QUESTIONCalcium carbonate can be a common antacid ingredient. The unbalanced equation shown here represents calcium carbonate reacting with hydrochloric acid (found in your stomach). CaCO3 + HCl CaCl2 + CO2 + H2O How many grams of CaCO3 are required to neutralize 25.8 grams of HCl?

1. 35.4 g2. 70.7 g3. 141 g4. 0.353 g


ANSWERChoice 1 is correct. 25.8 g 1 mol HCl/36.46 g = 0.708 mol HCl involved. Then, 0.708 mol HCl 1 mole CaCO3/2 mol HCl (from the balanced equation) = 0.354 mol CaCO3 needed. Finally, 0.354 mol 100.09 g/mol CaCO3 = 35.4 g.

Section 3.10: Calculations Involving a Limiting Reactant


QUESTIONIf two compounds are about to react, which statement about the reaction states an accurate observation?

1. If temperature conditions can be kept optimal, both compoundswill fully react with no excess.

2. The reactant present with the fewest grams will be the limitingreactant.

3. The reactant present with the fewest moles will be the limitingreactant.

4. None of the statements are accurate.


ANSWERChoice 4 should be selected. None of the statements are true. A limiting reactant can be determined only when both the moles of the two compounds AND the mole ratio by which the compounds react are known.



QUESTIONThe following balanced equation shows the reaction between zinc and sulfur to form zinc sulfide:

8 Zn + S8 8 ZnS

If you began the reaction with 15.0 grams of Zn and 10.0 grams of S8 what is the maximum, in moles, ZnS you would obtain?

1. 0.312 moles2. 0.229 moles3. 8.00 moles4. 0.258 moles


ANSWERChoice 2 is the maximum yield. The yield is limited by the moles of Zn. Even though the moles of Zn initially is larger than the initial moles of sulfur, it still limits the yield because each mole of sulfur requires 8 moles of Zn. (The moles of Zn are not 8 times greater than the moles of sulfur.)



QUESTIONTungsten metal is used to make many light bulb filaments. 18.0 grams of tungsten was made in the following reaction from 25.0 grams of WO3. What percent yield does this represent?

WO3 + 3 H2 W + 3 H2O

1. 25.8%2. 110%3. 90.8%4. I think I may have confused one of the steps in this calculation.


ANSWERChoice 3 has properly converted the initial grams of WO3 to moles and used the mole ratio of the equation to determine the maximum moles of W that would be produced if all the WO3 reacted (theoretical yield – 19.8 g). The division of the actual yield by the theoretical yield 100 produces the answer.


Chapter 3 Stoichiometry. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question,...

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