6: Counting Molecules Through Measurements

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Atoms and the Mole

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

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

CH₄
CHCl₃
C₁₂H₁₀O₆
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:

CH₄
CHCl₃
C₁₂H₁₀O₆
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:

P₄
H₂O
Ca(NO₃)₂
CH₃CO₂H (acetic acid)
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:

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:

S₈
C₅H₁₂
Sc₂(SO₄)₃
CH₃COCH₃ (acetone)
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:

the anesthetic halothane, C₂HBrClF₃
the herbicide paraquat, C₁₂H₁₄N₂Cl₂
caffeine, C₈H₁₀N₄O₂
urea, CO(NH₂)₂
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:

0.0146 mol KOH
10.2 mol ethane, C₂H₆
1.6 × 10⁻³mol Na₂SO₄
6.854 × 10³ mol glucose, C₆H₁₂O₆
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:

(a) 2.345 mol LiCl
(b) 0.0872 mol acetylene, C₂H₂
(c) 3.3 × 10⁻² mol Na₂ CO₃
(d) 1.23 × 10³ mol fructose, C₆ H₁₂ O₆
(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:

0.600 mol of oxygen atoms
0.600 mol of oxygen molecules, O₂
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?

20.0 g of H₂O (18.02 g/mol)
77.0 g of CH₄ (16.06 g/mol)
68.0 g of CaH₂ (42.09 g/mol)
100.0 g of N₂O (44.02 g/mol)
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:

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

Answer

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

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

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

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:

15.8% carbon and 84.2% sulfur
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: \(C\;240\;\times\;.7595\;=\;182.28/12.011\;=15\\ H\;240\;\times\;.0633\;=\;15.19/1.008\;=15\\N\;240\;\times\;.1772\;=\;42.53/14.007\;=4\\C_{15}H_{15}N_{3}\)