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6: Counting Molecules Through Measurements

  • Page ID
    283231
  • Atoms and the Mole

    Exercise \(\PageIndex{1.a}\)

    What is the total mass (amu) of carbon in each of the following molecules?

    1. CH4
    2. CHCl3
    3. C12H10O6
    4. CH3CH2CH2CH2CH3
    Answer a

    \(12.01\;amu\)

    Answer b

    \(12.01\;amu\)

    Answer c

    \(12(12.01)\;=\;144.12\;amu\)

    Answer d

    \(5(12.01)\;=\;60.05\;amu\)

    Exercise \(\PageIndex{1.b}\)

    Calculate the molecular or formula mass of each of the following:

    1. CH4
    2. CHCl3
    3. C12H10O6
    4. CH3CH2CH2CH2CH3
    Answer a

    \(12.01 \;+\;4(1.007)\;=\;16.04\;amu\)

    Answer b

    \(12.01 \;+\;1.007\;+\;3(35.45)\;=\;119.37\;amu\)

    Answer c

    \(12(12.01)\;+\;10(1.007)\;+\;6(16.00)\;=\;250.19\;amu\)

    Answer d

    \(5(12.01)\;+\;12(1.007)\;=\;72.13\;amu\)

    Exercise \(\PageIndex{1.c}\)

    Calculate the molecular or formula mass of each of the following:

    1. P4
    2. H2O
    3. Ca(NO3)2
    4. CH3CO2H (acetic acid)
    5. C12H22O11 (sucrose, cane sugar)
    Answer a

    \(4(30.974)\;=\;123.896\;amu\)

    Answer b

    \(2(1.007)\;+\;15.999\;=\;18.015\;amu\)

    Answer c

    \(40.078\;+\;2(14.007\;+\;3(15.99))\;=\;164.086\;amu\)

    Answer d

    \(2(12.011)\;+\;2(15.999)\;+\;4(1.007)\;=\;60.052\;amu\)

    Answer e

    \(12(12.011)\;+\;22(1.007)\;+\;11(15.999)\;=\;342.297\;amu\)

    Exercise \(\PageIndex{1.d}\)

    Determine the molecular mass of the following compounds:

    1. A structure is shown. Two C atoms form double bonds with each other. The C atom on the left forms a single bond with two H atoms each. The C atom on the right forms a single bond with an H atom and with a C H subscript 2 C H subscript 3 group.
    2. A structure is shown. There is a C atom which forms single bonds with three H atoms each. This C atom is bonded to another C atom. This second C atom forms a triple bond with another C atom which forms a single bond with a fourth C atom. The fourth C atom forms single bonds with three H atoms each.
    3. A structure is shown. An S i atom forms a single bond with a C l atom, a single bond with a C l atom, a single bond with an H atom, and a single bond with another S i atom. The second S i atom froms a single bond with a C l atom, a single bond with a C l atom, and a single bond with an H atom.
    4. A structure is shown. A P atom forms a double bond with an O atom. It also forms a single bond with an O atom which forms a single bond with an H atom. It also forms a single bond with another O atom which forms a single bond with an H atom. It also forms a single bond with another O atom which forms a single bond with an H atom.
    Answer a

    \(4(12.011)\;+\;8(1.007)\;=\;56.107\;amu\)

    Answer b

    \(4(12.011)\;+\;6(1.007)\;=\;54.091\;amu\)

    Answer c

    \(2(28.085)\;+\;4(35.45)\;+\;2(1.007)\;=\;199.998\;amu\)

    Answer d

    \(30.974\;+\;4(15.99)\;3(1.007)\;=\;97.995\;amu\)

    Mass, Moles and Number of Particles

    Exercise \(\PageIndex{2.a}\)

    Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2H5OH), 0.60 mol of formic acid (HCO2H), or 1.0 mol of water (H2O)? Explain why.

    Answer

    Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.

    Exercise \(\PageIndex{2.b}\)

    Calculate the molar mass of each of the following:

    1. S8
    2. C5H12
    3. Sc2(SO4)3
    4. CH3COCH3 (acetone)
    5. C6H12O6 (glucose)
    Answer a

    \(8(32.070)\;=\;256.560\;g/mol\)

    Answer b

    \(5(12.011)\;+\;12(1.008)\;=\;72.150\;g/mol\)

    Answer c

    \(2(44.956)\;+\;3(32.07+4(15.999)\;=\;378.110\;g/mol\)

    Answer d

    \(3(12.011)\;+\;6(1.008)\;+\;15.999\;=\;58.080\;g/mol\)

    Answer e

    \(6(12.011)\;+\;12(1.008)\;+\;6(15.999)\;=\;180.156\;g/mol\)

    Exercise \(\PageIndex{2.c}\)

    Calculate the molar mass of each of the following:

    1. the anesthetic halothane, C2HBrClF3
    2. the herbicide paraquat, C12H14N2Cl2
    3. caffeine, C8H10N4O2
    4. urea, CO(NH2)2
    5. a typical soap, C17H35CO2Na
    Answer a

    \(2(12.011)\;+\;1.008\;+\;79.900\;+\;35.45\;+\;3(18.998)\;=\;197.382\;g/mol\)

    Answer b

    \(12(12.011)\;+\;14(1.008)\;+\;2(14.007)\;2(35.45)\;=\;257.158 g/mol\)

    Answer c

    \(8(12.011)\;+\;10(1.008)\;4(14.007)\;+\;2(15.999)\;=\;194.194 g/mol\)

    Answer d

    \(12.011\;+\;15.999\;+\;2(14.007\;+\;2(1.008))\;=\;60.056 g/mol\)

    Answer e

    \(18(12.011)\;+\;35(1.008)\;+\;2(15.999)\;+\;22.990\;=\;306.466 g/mol\)

    Exercise \(\PageIndex{2.d}\)

    Determine the mass of each of the following:

    1. 0.0146 mol KOH
    2. 10.2 mol ethane, C2H6
    3. 1.6 × 10⁻³mol Na2SO4
    4. 6.854 × 10³ mol glucose, C6H12O6
    5. 2.86 mol Co(NH3)6Cl3
    Answer a

    \(38.098\;+\;15.999\;+\;1.008\;=\;55.105\;g/mol\\55.105\;g/mol\;\times\;.0146\;mol\;=\;0.805\;g\)

    Answer b

    \(2(12.011\;+\;6(1.008)\;=\;30.070\;g/mol\\30.070\;g/mol\;\times\;10.2\;mol\;=306.714\;g\)

    Answer c

    \(2(22.990)\;+\;32.07\;+\;4(15.999)\;=\;142.046\;g/mol\\142.046\;g/mol\;\times\;1.6\;\times10^{-3}\;mol\;=\;.227\;g\)

    Answer d

    \(6(12.011)\;+\;12(1.008)\;+\;6(15.999)\;=\;180.156\;g/mol\\180.156\;g/mol\;\times\;6.854\;\times\;10^{3}\;mol\;=\;1.235*10^6\;g\)

    Answer e

    \(58.933\;+\;6(14.007\;+\;3(1.007))\;+\;3(35.45)\;=\;267.451\;g/mol\\267.451\;g/mol\;\times\;2.86mol\;=\;764.909\;g\)

    Exercise \(\PageIndex{2.e}\)

    Determine the mass of each of the following:

    1. (a) 2.345 mol LiCl
    2. (b) 0.0872 mol acetylene, C2H2
    3. (c) 3.3 × 10−2 mol Na2 CO3
    4. (d) 1.23 × 103 mol fructose, C6 H12 O6
    5. (e) 0.5758 mol FeSO4(H2O)7
    Answer a

    \(7.0\;+35.45\;=\;42.45\;g/mol\\42.45\;g/mol\;\times\;2.345\;mol\;=99.545\;g\)

    Answer b

    \(2(12.011)\;+\;2(1.008)\;=\;26.038\;g/mol\\26.038\;g/mol\;\times\;0.0872\;mol\;=2.271\;g\)

    Answer c

    \(2(22.99)\;+\;12.011\;+\;3(15.999)\;=\;105.988\;g/mol\\105.988\;g/mol\;\times\;3.3\;\times\;10^{-2}\;mol\;=\;3.498\;g\)

    Answer d

    \(6(12.011)\;+\;12(1.008)\;+\;6(15.999)\;=\;180.156\;g/mol\\180.156\;g/mol\;\times\;1.23\;\times\;10^3\;mol\;=2.216\;\times\;10^5\;g\)

    Answer e

    \(55.84\;+\;32.07\;+\;4(15.999)\;+\;7(2(1.008)\;+\;15.999)\;=\;278.011\;g/mol\\278.011\;g/mol\;\times\;0.5758\;mol\;=\;160.079\;g\)

    Exercise \(\PageIndex{2.f}\)

    Determine the mass in grams of each of the following:

    1. 0.600 mol of oxygen atoms
    2. 0.600 mol of oxygen molecules, O2
    3. 0.600 mol of ozone molecules, O3
    Answer a

    \(15.999\;g/mol\;\times\;0.600\;mol\;=\;9.599\;g\)

    Answer b

    \(2(15.999)\;g/mol\;\times\;0.600\;mol\;=\;19.199\;g\)

    Answer c

    \(3(15.999)\;g/mol\;\times\;0.600\;mol\;=\;28.798\;g\)

    Exercise \(\PageIndex{2.g}\)

    Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO4, 266 g of Al2Cl6, or 225 g of Al2S3.

    Answer

    \(26.982\;+\;30.974\;+\;4(15.999)\;=\;121.952\;g/mol\\ \frac{122\;g}{121.952\;g/mol}\;=\;1\;mol\;AlPO_4\\1\;mol\;Al\;in\;AlPO_4\\1.000\;\times\;1\;=\;1\;mol\;Al\)

     

    \(2(26.982)\;+\;6(35.45)\;=\;266.664\;g/mol\\ \frac{266\;g}{266.664\;g/mol}\;=\;0.998\;mol\;Al_2Cl_6\\2\;mol\;Al\;in\;Al_2Cl_6\\0.998\;\times\;2\;=\;1.996\;mol\;Al\)

     

    \(2(26.982)\;+\;3(32.07)\;=\;150.174\;g/mol\\ \frac{225\;g}{150.174\;g/mol}\;=\;1.498\;mol\;Al_2S_3\\2\;mol\;Al\;in\;Al_2S_3\\1.498\;\times\;2\;=\;2.996\;mol\;Al\)

    Exercise \(\PageIndex{2.h}\)

    The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone?

    Answer

    \(3104\;carats\;\times\;200\;mg\;=\;\frac{620800\;mg}{1000\;g}\;=\;620.8\;g\\\frac{620.8\;g}{12.011}\;=\;51.686\;mol\;C\\51.686\;times\;6.022\;\times\;10^{23}\;=\;3.113\;\times\;10^{25}\)

    Exercise \(\PageIndex{2.i}\)

    A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?

    Answer

    0.865 servings, or about 1 serving

    Exercise \(\PageIndex{2.j}\)

    Which of the following represents the least number of molecules?

    1. 20.0 g of H2O (18.02 g/mol)
    2. 77.0 g of CH4 (16.06 g/mol)
    3. 68.0 g of CaH2 (42.09 g/mol)
    4. 100.0 g of N2O (44.02 g/mol)
    5. 84.0 g of HF (20.01 g/mol)
    Answer

    a. 20.0 g H2O represents the least number of molecules since it has the least number of moles.

     

    Percent Composition

    Exercise \(\PageIndex{3.a}\)

    Calculate the following to four significant figures:

    1. the percent composition of ammonia, NH3
    2. the percent composition of photographic “hypo,” Na2S2O3
    3. the percent of calcium ion in Ca3(PO4)2
    Answer

    a.

    \(M_N: \;1\;\times\;14.007 \;=\;14.007  \; \\M_H:\;3\;\times\;1.008\;=\;3.024 \\M_T\;= \; 1(14.007)+ 3(1.008)=17.031\\\; \\ \%\; N\;=\;\frac{M_N}{M_T}\;=\;82.24\;\%\\\%\;H\;=\;\frac{M_H}{M_T}\;=\;17.76\;\%\)\

     

    b.

    \(M_{Na}: \;2\;\times\;22.99\;=\;45.98\\M_{S}:\;2\;\times\;32.07\;=\;64.14\\M_O:\;3\;\times\;15.999\;=\;47.997\\M_T\;=\;2(22.99)\;+\;2(32.07)\;=\;158.117\\\%\; Na\;=\;\frac{M_{Na}}{M_T}\;=\;29.07\;\%\\\%\; S\;=\;\frac{M_{S}}{M_T}\;=\;40.56\;\%\\\%\; O\;=\;\frac{M_{O}}{M_T}\;=\;30.36\;\%\)

     

    c.

    \(M_{Ca^{+2}}: \;3\;\times\;40.08\;=\;120.24\\P:\;2\;\times\;30.974\;=\;61.948\\O:\;8\;\times\;15.999\;=\;127.992\\M_T\;=\;3(40.08)\;+\;2(30.974)\;+8(15.999)\;=\;310.18\\\%\; Ca^{+2}\;=\;\frac{M_{Ca^{+2}}}{M_T}\;=\;38.76\;\%\)

    Exercise \(\PageIndex{3.b}\)

    Determine the percent ammonia, NH3, in Co(NH3)6Cl3, to three significant figures.

    Answer

    \(NH_3: \;6\;\times\;(14.007\;+\;3(1.008))\;=\;102.186\\Co:\;1\;\times\;58.933\;=\;58.933\\Cl:\;3\;\times\;35.45\;=\;106.35\\M_T\;=267.469\\M_{NH_{3}}\;=\;6(14.007\;+\;3(1.008))\\ \%\; NH_{3}\;=\;\frac{M_{NH_{3}}}{M_T}\;=\;38.20\;\%\)

     

    Empirical Formulas

    Exercise \(\PageIndex{4.a}\)

    Determine the empirical formulas for compounds with the following percent compositions:

    1. 15.8% carbon and 84.2% sulfur
    2. 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen
    Answer

    \(C\;100\;\times\;.158\;=\;15.8/12.011\;=\;1.315/1.315\;=\;1\\ S\;100\;\times\;.842\;=\;84.2/32.07\;=\;2.62/1.315\;=\;2\\CS_2\)

     

     

    \(C\;100\;\times\;.400\;=\;40.0/12.011\;=\;3.33/3.33\;=\;1\\ H\;100\;\times\;.067\;=\;6.7/1.008\;=\;6.64/3.33\;=\;2\\O\;100\;\times\;.533\;=\;55.3/15.999\;=\;3.46/3.33\;=\;1\\CH_2O\)

     

    Molecular Formulas

    Exercise \(\PageIndex{5.a}\)

    A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?

    Answer

    \(78.1\;\times\;.923\;=\;72.08/12.011\;=\;6\\78.1\;-\;72.08\;=\;6.02/1.008\;=\;6\\C_6H_6\)

     

    Exercise \(\PageIndex{5.b}\)

    Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.

    Answer

    Mg3Si2H3O8 (empirical formula),

    Mg6Si4H6O16 (molecular formula)

    Empirical Formula

    \(Mg\;100\;\times\;.2803\;=\;28.03/24.305\;=\;1.15/.77\;=\;1.5\;\times\;2\;=\;3\\ Si\;100\;\times\;.2160\;=\;21.60/28.085\;=\;.77/.77\;=\;1\;\times\;2\;=\;2\\H\;100\;\times\;.0116\;=\;1.16/1.008\;=\;1.15/.77\;=\;1.5\;\times\;2\;=\;3\\O\;100\;\times\;.4921\;=\;49.21/15.999\;=\;3.08/.77\;=\;4\;\times\;2\;=\;8\\Mg_3Si_2H_3O_{16}\)

    Molecular Formula

    \(Mg\;520.8\;\times\;.2803\;=\;145.98/24.305\;=6\\ Si\;520.8\;\times\;.2160\;=\;112.49/28.085\;=4\\H\;520.8\;\times\;.0116\;=\;6.04/1.008\;=6\\O\;520.8\;\times\;.4921\;=\;256.28/15.999\;=16\\Mg_6Si_4H_6O_{16}\)

    Exercise \(\PageIndex{5.c}\)

    A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.

    Answer

    \(C\;240\;\times\;.7595\;=\;182.28/12.011\;=15\\ H\;240\;\times\;.0633\;=\;15.19/1.008\;=15\\N\;240\;\times\;.1772\;=\;42.53/14.007\;=4\\C_{15}H_{15}N_{3}\)

     

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