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OpenStax - Chemistry 1/e (Homework)

James Finch

Chemistry - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 11 / 42

Due : Thursday, November 21, 2019 08:30 EST

Last Saved : n/a Saving...  ()

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1. –/2 points OSGenChem1 1.6.P.092. My Notes
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–/2
 
Calculate these volumes.
(a)
What is the volume (in mL) of 11.2 g rubidium, density = 1.53 g/cm3?
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7.32

 mL
(b)
What is the volume (in mL) of 39.811 g molybdenum, density = 10.18 g/cm3?
WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

3.911

 mL
†


Solution or Explanation
Process: Rearrange the known relationship for density to solve for the volume.
density = mass per unit of volume
volume = 
mass
density
(a)
volume = 
11.2 g
1.53 g/cm3
 = 7.3203 cm3 ≅ 7.32 mL
(b)
volume = 
39.811 g
10.18 g/cm3
 = 3.91071 cm3 ≅ 3.911 mL
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–/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100
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–/4
 
Identify each of the bold items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For those in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature. (Select all that apply.)
(a)
A certain molecule contains two H atoms and one S atom.

(b)
Silver wire has a density of about 10 g/cm3.

(c)
The bottle contains 16 grams of U powder.

(d)
A phosphorus molecule is composed of 4 phosphorus atoms.

†


Solution or Explanation
(a)
A certain molecule contains two H atoms and one S atom.
It is too small to be sensed directly: microscopic.
It uses specialized language: symbolic.
(b)
Silver wire has a density of about 10 g/cm3.
It is large enough to be sensed directly: macroscopic.
(c)
The bottle contains 16 grams of U powder.
It is large enough to be sensed directly: macroscopic.
It uses specialized language: symbolic.
(d)
A phosphorus molecule is composed of 4 phosphorus atoms.
It is too small to be sensed directly: microscopic.
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–/2
 
Consider the solutions presented.
(a)
Which of the following sketches best represents the ions in a solution of Bi(ClO3)3(aq)?
    
(b)
Write a balanced chemical equation showing the products of the dissolution of Bi(ClO3)3. (Include states-of-matter under the given conditions in your answer. Use the lowest possible whole number coefficients.)
(No Response) seenKey

Bi(ClO_3)_3(s) --> Bi^3+(aq) + 3 ClO_3^-(aq)
†


Solution or Explanation
(a)
Which of the following sketches best represents the ions in a solution of Bi(ClO3)3(aq)?
The given formula represents a strong electrolyte with 3 polyatomic anions for each 1 cation per formula unit. Strong electrolytes dissolve completely in water, so beaker z is the only depiction that fits the given formula.
(b)
Write a balanced chemical equation showing the products of the dissolution of Bi(ClO3)3.
Bi(ClO3)3(s) → Bi 3+(aq) + 3 ClO3−(aq)
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–/2
 
Under certain conditions, molecules of formic acid, HCOOH, form "dimers," pairs of formic acid molecules held together by strong intermolecular attractions.
Draw a dimer of formic acid, showing how two HCOOH molecules are held together. (Use the "Any" bond type to indicate intermolecular attractions.)
(No Response)
Marvin JS
State the type of IMF that is responsible.
    
†


Solution or Explanation
A dimer of formic acid is formed when two formic acid molecules combine.
The type of intermolecular force responsible for the formation of the dimer is hydrogen bonding. The molecules combine with hydrogen bonds between the H and O atoms. Two hydrogen bonds per dimer are thus established.
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5. –/24 points OSGenChem1 13.4.P.062. My Notes
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Points
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
–/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100 0/100
Total
–/24
 
Please use the values in the resources listed below instead of the textbook values.
Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.
(a)
Reaction 2 H2(g) + O2(g) equilibrium reaction arrow 2 H2O(g)
relative
change 		(No Response) -2x (No Response) -x +2x
change in
concentration
(M) 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-1.50

 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-0.750

 		1.50
(b)
Reaction CS2(g) + 4 H2(g) equilibrium reaction arrow CH4(g) + 2 H2S(g)
relative
change 		+x (No Response) +4x (No Response) -x (No Response) -2x
change in
concentration
(M) 		0.017 WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

0.068

 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-0.017

 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-0.034
(c)
Reaction H2(g) + Cl2(g) equilibrium reaction arrow 2 HCl(g)
relative change +x (No Response) +x (No Response) -2x
change in
pressure
(atm) 		1.80 WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

1.80

 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-3.60
(d)
Reaction 2 NH3(g) + 2 O2(g) equilibrium reaction arrow N2O(g) + 3 H2O(g)
relative change (No Response) -2x / 3 (No Response) -2x / 3 (No Response) +x / 3 +x
change in
pressure
(torr) 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-44.0

 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-44.0

 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

22.0

 		66.0
(e)
Reaction NH4HS(s) equilibrium reaction arrow NH3(g) + H2S(g)
relative change +x (No Response) +x
change in
concentration
(M) 		9.7 ✕ 10−6 WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

9.7e-06
(f)
Reaction Fe(s) + 5 CO(g) equilibrium reaction arrow Fe(CO)4(g)
relative change (No Response) -5x +x
change in
pressure
(atm) 		WebAssign will check your answer for the correct number of significant figures.(No Response) seenKey

-0.095

 		0.019
†


Solution or Explanation
The relative change and change in concentration (or pressure) will have opposite signs on either side of the reaction. If the concentration/pressure of a reactant is increasing (positive relative change), any other reactants will also have a positive relative change, and the products will have a negative relative change. The opposite is true when the concentration/pressure of a product is increasing.
For each reaction, use the given relative change and the reaction coefficients to determine the other relative changes. Set the given change in concentration/pressure equal to the relative change to calculate the value of x. Substitute the value of x into the other relative changes to determine all changes in concentration/pressure.
(a)
Reaction 2 H2(g) + O2(g) equilibrium reaction arrow 2 H2O(g)
relative
change
−2x
−x
 		+2x
change in
concentration
(M) 		−1.50 −0.750 1.50
(b)
Reaction CS2(g) + 4 H2(g) equilibrium reaction arrow CH4(g) + 2 H2S(g)
relative
change 		+x
+4x
−x
−2x
change in
concentration
(M) 		0.017 0.068 −0.017 −0.034
(c)
Reaction H2(g) + Cl2(g) equilibrium reaction arrow 2 HCl(g)
relative change +x
+x
−2x
change in
pressure
(atm) 		1.80 1.80 −3.60
(d)
Reaction 2 NH3(g) + 2 O2(g) equilibrium reaction arrow N2O(g) + 3 H2O(g)
relative change
−2x
3
−2x
3
+x
3
 		+x
change in
pressure
(torr) 		−44.0 −44.0 22.0 66.0
(e)
Reaction NH4HS(s) equilibrium reaction arrow NH3(g) + H2S(g)
relative change +x
+x
change in
concentration
(M) 		9.7 ✕ 10−6 9.7 ✕ 10−6
(f)
Reaction Fe(s) + 5 CO(g) equilibrium reaction arrow Fe(CO)4(g)
relative change
−5x
 		+x
change in
pressure
(atm) 		−0.095 0.019
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(a) How many moles of CO2 contain 2.45 ✕ 1024 molecules?
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4.07

 mol CO2

(b) What number of moles is equivalent to 6.49 ✕ 1021 atoms of Hg?
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0.0108

 mol Hg


Solution or Explanation

  • Step 1 of 4

    (a) How many moles of CO2 contain 2.45 ✕ 1024 molecules?

    Write a conversion factor to change the number of molecules of CO2 to moles of CO2. The number of molecules is given. The number of moles must be determined. Therefore, the unit molecules must cancel when the given value is multiplied by the conversion factor.
    1 mol CO2 = 6.02214199 ✕ 1023 molecules CO2
    For this to occur, the number of molecules must be in the denominator (bottom) and the desired unit, moles, must be in the numerator (top).
    1 mol CO2
    6.02214199 ✕ 1023 molecules CO2

  • Step 2 of 4

    Multiply the given number of molecules by the conversion factor. Multiply the given number of molecules, 2.45 ✕ 1024, by the conversion factor equivalent to the number of molecules in one mole. To find the answer, divide the number of molecules by 6.02214199 ✕ 1023. The unit molecules will cancel.
    2.45 ✕ 1024 molecules CO2 × 
    1 mol CO2
    6.02214199 ✕ 1023 molecules CO2
     = 4.07 mol CO2

  • Step 3 of 4

    (b) What number of moles is equivalent to 6.49 ✕ 1021 atoms of Hg?

    Write a conversion factor to change the number of atoms of Hg to moles of Hg. The number of atoms is given. The number of moles must be determined. Therefore, the unit atoms must cancel when the given value is multiplied by the conversion factor.
    1 mol Hg = 6.02214199 ✕ 1023 atoms Hg
    For this to occur, the number of atoms must be in the denominator (bottom) and the desired unit, moles, must be in the numerator (top).
    1 mol Hg
    6.02214199 ✕ 1023 atoms Hg

  • Step 4 of 4

    Multiply the number of atoms by the conversion factor. Multiply the given number of atoms, 6.49 ✕ 1021, by the conversion factor equivalent to the number of atoms in one mole. The unit atoms will cancel.
    6.49 ✕ 1021 atoms Hg × 
    1 mol Hg
    6.02214199 ✕ 1023 atoms Hg
     = 0.0108 mol Hg

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–/2
 
Write the molecular and empirical formulas for the compound with the following structural formula.

molecular formula:

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C5H10

empirical formula:

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CH2



Solution or Explanation

  • Step 1 of 4

    Write the given structural formula of the compound.

  • Step 2 of 4

    Count the number of atoms of each element shown in the structural formula.
    There are 5 C atoms and 10 H atoms.

  • Step 3 of 4

    Write the molecular formula by writing the symbols for each element with a subscript indicating the number of atoms of that element determined in Step 2. Assume subscripts of "1" are understood; do not write "1" subscripts in the molecular formula.
    The five atoms of C are indicated by a subscript "5". The ten atoms of H are indicated by a subscript "10".

    The molecular formula of the compound is the following.
    C5H10

  • Step 4 of 4

    Determine the empirical formula of the compound from its molecular formula. Assume subscripts of "1" are understood; do not write "1" subscripts in the empirical formula.
    The ratio between C and H atoms is 5:10. The smallest equivalent whole-number ratio is 1:2. Therefore, the empirical formula of the compound is the following.
    CH2

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–/1
 
A compound has an empirical formula of C3H3O2 and a molar mass of 142 g/mol. What is the molecular formula?

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C6H6O4



Solution or Explanation

  • Step 1 of 5

    Recall the definition of molar mass.
    molar mass = mass of one mole of a substance (in grams)
    The molar mass of a compound is found by summing the molar masses of its constituent atoms.
    molar mass of a compound in g/mol = formula mass of that compound in amu

  • Step 2 of 5

    Recall the definition of an empirical formula.
    An empirical formula expresses the simplest whole-number ratio of distinct atoms in a compound.

  • Step 3 of 5

    Calculate the molar mass of the given empirical formula C3H3O2. (Enter unrounded values.)
    formula mass of C3H3O2 = (# C atoms)(atomic mass of C)
     + (# H atoms)(atomic mass of H)
     + (# O atoms)(atomic mass of O)
     = (3)(12.01 amu) + (3)(1.008 amu) + (2)(16.00amu)
     = 71.054 amu
    Therefore, the empirical molar mass of C3H3O2 is 71.054 g/mol.

  • Step 4 of 5

    Calculate the ratio between the given molar mass and the empirical molar mass calculated in Step 3.
    The ratio between the given molar mass of 142 g/mol and the calculated empirical molar mass of 71.054 g/mol is the following.
    142 g/mol
    71.054 g/mol
    		 = 1.998 (Enter an unrounded value. Use at least one more digit than given.)
    ≈2 (Rounded to the nearest whole number.)

  • Step 5 of 5

    Multiply the subscripts in the empirical formula by the calculated ratio from Step 4 to determine the molecular formula.
    C: 2 × 3 = 6
    H: 2 × 3 = 6
    O: 2 × 2 = 4
    Therefore, the actual molecular formula is:
    C6H6O4

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A 0.5945 g sample of an unknown acid was titrated with NaOH and found to contain 3.563 ✕ 10−3 mol acid. Another sample of the substance was subjected to elemental analysis, where it was determined that the composition of the unknown material is 57.18% C, 4.78% H, and 38.04% O.
(a) What is the molar mass of the unknown acid?
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166.9

 g/mol

(b) What is the empirical formula of the unknown acid?

chemPad

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C2H2O

(c) What is the molecular formula of the unknown acid?

chemPad

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C8H8O4


Solution or Explanation

  • Step 1 of 10

    (a) What is the molar mass of the unknown acid?

    Calculate the molar mass of the acid. Recall that the molar mass of a compound is
    grams X
    moles X
    ;
     therefore, the molar mass of the unknown acid is:
    molar mass = 
    (0.5945 g)
    (3.563 ✕ 10−3 mol)
     = 166.9 g/mol.
    The answer is shown with four significant figures to match the precision of the measured values.

  • Step 2 of 10

    (b) What is the empirical formula of the unknown acid?

    Assume that 100 g of the substance is available. Determine the mass, in grams, of each element in that quantity of substance. (Enter an unrounded value.)
    C: 57.18% of 100 g = 57.18 g C
    H: 4.78% of 100 g = 4.78 g H
    O: 38.04% of 100 g = 38.04 g O

  • Step 3 of 10

    Use the periodic table to determine the average molar mass of each element present in the substance. Recall that average atomic mass (in amu) is reported on the periodic table and is equivalent to molar mass (in g/mol).
    average molar mass of C: 12.01 g/mol
    average molar mass of H: 1.008 g/mol
    average molar mass of O: 16.00 g/mol

  • Step 4 of 10

    Calculate the number of moles of each element present in the substance. (Enter an unrounded value. Use at least one more digit than given.)
    C: 57.18 g C × 
    1 mol C
    12.01 g C
     = 4.761 mol C
    H: 4.78 g H × 
    1 mol H
    1.008 g H
     = 4.742 mol H
    O: 38.04 g O × 
    1 mol O
    16.00 g O
     = 2.3775 mol O

  • Step 5 of 10

    Divide the number of moles present in each element by the smallest number of moles represented by any of the elements. If the results are very close to positive integers, round them to the nearest integer. The smallest number of moles from Step 4 comes from oxygen (O), 2.3775 mol.
    C: 
    4.7610 mol
     
     
    2.3775 mol
     
     
    		 = 2.003 (rounded to 4 significant figures)
     = 2 (rounded to the nearest integer)
    H: 
    4.742 mol
     
     
    2.3775 mol
     
     
    		 = 1.99 (rounded to 3 significant figures)
     = 2 (rounded to the nearest integer)
    O: 
    2.3775 mol
     
     
    2.3775 mol
     
     
    		 = 1.000 (rounded to 4 significant figures)
     = 1 (rounded to the nearest integer)

  • Step 6 of 10

    Using the relative number of atoms in the formula determined in Step 5, write the empirical formula of the substance. The C, H, and O atoms in the substance exist in the ratio 2:2:1, respectively. The empirical formula is therefore the following.
    C2H2O

  • Step 7 of 10

    (c) What is the molecular formula of the unknown acid?

    Using the periodic table and empirical formula of the compound, calculate the molar mass of the empirical formula.
    Empirical formula mass of C2H2O = (2)(atomic mass of C) + (2)(atomic mass of H)
    + (1)(atomic mass of O)
     = (2)(12.01 amu) + (2)(1.008 amu) + (1)(16.00 amu)
     = 42.036 amu
    Formula mass in amu equals molar mass in g/mol, so the molar mass of the empirical formula is 42.036 g/mol.

  • Step 8 of 10

    Divide the empirical molecular mass of the substance by the empirical molecular mass of the fragment, and then round the quotient to the nearest integer.
    166.9 g/mol
    42.036 g/mol
    		 = 3.970 (rounded to 4 significant figures)
     = 4 (rounded to the nearest integer)

  • Step 9 of 10

    Multiply each subscript in the given original empirical formula, C2H2O, by the number calculated in Step 8 to obtain the molecular formula. C8H8O4

  • Step 10 of 10

    Summarize the answer. (a) The molar mass of the unknown acid is 166.9 g/mol.

    (b) The empirical formula of the unknown acid is the following.
    C2H2O

    (c) The molecular formula of the unknown acid is the following.
    C8H8O4

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