6: Counting Molecules Through Measurements


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

$$12.01\;amu$$

$$12.01\;amu$$

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

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

Exercise $$\PageIndex{1.b}$$

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

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

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

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

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

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

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)

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

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

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

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

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

Exercise $$\PageIndex{1.d}$$

Determine the molecular mass of the following compounds:

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

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

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

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

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.

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)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$$15.999\;g/mol\;\times\;0.600\;mol\;=\;9.599\;g$$

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

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

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.

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

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

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

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?

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

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?

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)

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

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.

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

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

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

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

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?

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

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.

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.

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

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