Example 3.1 Calculate the molecular mass of glycerol (1,2,3-propanetriol). Strategy To determine a...

Example 3.1

Calculate the molecular mass of glycerol (1,2,3-propanetriol).

StrategyTo determine a molecular mass, we must start with the molecular formula, which is not stated, but we can deduce the formula either from the name or from the molecular model (see Figure 2.18).

SolutionOne –OH group replaces one H atom on each of the three C atoms in propane, leading to the condensed structural formula CH2OHCHOHCH2OH, which translates to the molecular formula C3H8O3. To obtain the molecular mass, we must add together three times the atomic mass of carbon, eight times the atomic mass of hydrogen, and three times the atomic mass of oxygen:

3 x atomic mass of C = 3 x 12.011 u = 36.033 u 8 x atomic mass of H = 8 x 1.00794 u = 8.06352 u 3 x atomic mass of O = 3 x 15.9994 u = 47.9982 u Molecular mass of C3H8O3 = 92.095 u

Calculate, to three significant figures, the molecular mass of (a) P4, (b) N2O4, (c) H2SO4,and (d) CH3CH2COOH.

Exercise 3.1A

Calculate, to five significant figures, the molecular mass of (a) phosphorus pentachloride,(b) dinitrogen pentoxide, (c) butanoic acid, and (d) methyl butyl ketone (CH3COCH2CH2CH2CH3).

Exercise 3.1B

Example 3.2

Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by homegardeners.

StrategyBefore we can determine a formula mass, we must first write the correct chemical formula. Then we can sum the masses of the atoms represented in the formula.

SolutionAmmonium sulfate is an ionic compound consisting of ammonium ions (NH4

+) and sulfate ions (SO4

2–); its formula is therefore (NH4)2SO4. To derive a formula mass from a complex formula like this one, we must make certain that all the atoms in the formula unit are accounted for, which means paying particular attention to all the subscripts and parentheses in the formula. Let’s first note the relevant atomic masses and the way in which they must be combined:

Solution continuedSumming the atomic masses

(NH4)2 : [14.0067 u + 14 x 1.00794 u)] x 2 = 36.0769 uSO4 : 32.066 u + (4 x 15.9994 u) = 96.064 u Formula mass of (NH4)2SO4 = 132.141 u

Example 3.2 continued

AssessmentAs long as every atom in the formula unit is accounted for, we can check our answer by using a different summation, for example, by considering each element separately:

Formula mass = (2 x atomic mass N) + (8 x atomic mass H) + (1 x atomic mass of S) + (4 x atomic mass O)

= (2 x 14.0067 u) + (8 x 1.00794 u) + (1 x 32.066 u) + (4 x 15.9994 u)

= 132.141 u

Calculate, to three significant figures, the formula mass of (a) Li2O, (b) Mg(NO3)2,(c) Ca(H2PO4)2 and (d) K2SbF5.

Exercise 3.2A

Calculate, to five significant figures, the formula mass of (a) sodium hydrogen sulfite,(b) ammonium perchlorate, (c) chromium(III) sulfate, and (d) copper(II) sulfate pentahydrate.

Exercise 3.2B

Example 3.3

Determine (a) the mass of a 0.0750-mol sample of Na, (b) the number of moles of Na in a 62.5-g sample, (c) the mass of a sample of Na containing 1.00 x 1025 Na atoms, and (d) the mass of a single Na atom.

StrategyTo perform these calculations, we will use a conversion factor derived from molar mass to relate moles and grams, and a conversion factor derived from Avogadro’s number to relate moles and number of atoms. These are the relationships embodied in Equation (3.1).

Solution(a) To convert from moles to grams, we need to use the first and third terms in

Equation (3.1), written as a conversion factor: 22.99 g Na/1 mol Na:

(b) Again, we need the first and third terms in Equation (3.1), but this time we write the conversion factor as the inverse—1 mol Na/22.99 g Na—because we are to convert from grams to moles:


Solution continued(c) Here we can use all three terms in Equation (3.1), written as two conversion

factors, to convert first from number of atoms to number of moles, and then from moles to the mass in grams:

Alternatively, we can use the second and third terms in Equation (3.1), written as the conversion factor 22.99 g Na/6.022 x 1023 Na atoms, to convert directly from number of atoms to mass in grams:

(d) The answer must have the unit grams per sodium atom (g/Na atom). Thus, if we know the mass of a certain number of Na atoms, our answer is simply that mass divided by the corresponding number of atoms. And we know these quantities from the molar mass (22.99 g Na/mol Na) and Avogadro’s number

(1 mol Na/6.022 x 1023 atoms). Our answer is the product of these two factors:


AssessmentIn these examples, note the common practice of expressing molar mass and Avogadro’s number with at least one significant figure more than the number of significant figures in the least precisely known quantity. Doing this ensures that the precision of the calculated results is limited only by the least precisely known quantity. In part (d), this simple fact should deter you from mistakenly multiplying instead of dividing by Avogadro’s number: Individual atoms are exceedingly small and possess masses that are many orders of magnitude less than one gram. It is also worth noting that the calculated mass is the true mass of a sodium atom—23Na has no isotopes. For elements with two or more isotopes, the mass calculated for an atom is a weighted average.

Calculate (a) the mass in milligrams of 1.34 x 10–4 mol Ag and (b) the number of oxygen atoms in 20.5 mol O2.

Exercise 3.3A

Calculate (a) the number of moles of Al in a cube of aluminum metal 5.5 cm on an edge(d = 2.70 g/cm3) and (b) the volume occupied by 4.06 x 1024 Br atoms present as Br2 molecules in liquid bromine (d = 3.12 g/mL).

Exercise 3.3B

Example 3.4

Determine (a) the number of NH4+ ions in a 145-g sample of (NH4)2SO4 and (b) the

volume of 1,2,3-propanetriol (glycerol, d = 1.261 g/mL ) that contains 1.00 mol O atoms.

StrategyAs before, we will use a conversion factor derived from molar mass to relate moles and grams, and a conversion factor derived from Avogadro’s number to relate moles and number of ions or atoms.

Solution(a) In this calculation, we need the relationships given in Equation (3.3). Also, we

must use the relationship between the NH4+ ion and the (NH4)2SO4 formula

unit shown on page 81; that is, 2 mol NH4+/1 mol (NH4)2SO4. The path to the

desired answer is

Solution continued(b) The conversion factors needed for this calculation are (1) the relationship

between moles of O atoms and C3H8O3 molecules (diagram on page 81), (2) the molar mass of C3H8O3 (Equation 3.2), and (3) the inverse of the density of C3H8O3:


AssessmentIn Example 3.3d, we cited the need to divide by Avogadro’s number because the mass of an individual atom (or ion or molecule) is exceedingly small. By constrast, in part (a) of this example we must multiply (not divide) by Avogadro’s number. The number of atoms, ions, or molecules in a macroscopic sample of matter is exceedingly large.


Determine (a) the total number of Cl atoms in 125 mL of liquid CCl4 (d = 1.589 g/mL) and (b) the number of grams of carbon in 215 g of sucrose, C12H22O11.

Exercise 3.4A

You need to obtain 1.00 mol C atoms, and your source for these C atoms is the sucrose (C12H22O11) contained in an aqueous solution having a density of 1.0181 g/mL and 5.05% of its mass present as sucrose. How many milliliters of the solution are required?

Exercise 3.4B

Example 3.5 An Estimation Example

Which of the following is a reasonable value for the number of atoms in 1.00 g of helium?

(a) 4.1 x 10–23 (c) 1.5 x 1023

(b) 4.0 (d) 1.5 x 1024

Analysis and ConclusionsResponse (a) is what we get if, in error, we divide the number of moles of He, 0.25, by Avogadro’s number. Clearly, we can’t have anything less than one atom. Response (b) expressed as 4.0 u is the atomic mass of He; expressed as 4.0 g/mol He, it is the molar mass. In either case, it is far too small to be the number of atoms for any macroscopic sample. Response (c) is the correct order of magnitude, as is response (d). However, (d) is larger than Avogadro’s number, while 1.00 g is less than one mole of helium. We would expect a fraction of a mole of helium to have fewer than Avogadro’s number of atoms. Response (c) is the correct answer, obtained from the calculation: 0.25 x NA.

Which of the following is a reasonable value for the mass of 1.0 x 1023 magnesiumatoms? (Try to reason through the problem, and avoid using your calculator if possible.Your goal is to find a reasonable answer rather than to calculate a specific number.)

(a) 2.4 x 10–22 g (c) 2.4 g

(b) 0.17 g (d) 4.0 g

Exercise 3.5A

Without doing detailed calculations, determine which of the following has the greatest number of carbon atoms per gram of compound:

(a) CO2 (c) CH3CH2COOH

(b) CH3CH2CH3 (d) CH3CH2OCH2CH3

Exercise 3.5B

Example 3.6

Calculate, to four significant figures, the mass percent of each element in ammonium nitrate.

StrategyFirst, we will determine the molar mass of ammonium nitrate, based on the formula unit NH4NO3. Then, for one mole of compound, we can determine mass ratios and percentages.

Solution


AssessmentTo check, we add the percentages to ensure that they add up to 100.00%. (Sometimes the total may differ from 100.00% by ±0.01% due to rounding.)

Calculate the mass percent of each element in (a) ammonium sulfate and (b) urea, CO(NH2)2 . Which compound has the greatest mass percent nitrogen: ammonium nitrate(see Example 3.6), ammonium sulfate, or urea?

Exercise 3.6A

Calculate the mass percent of(a) N in triethanolamine, N(CH2CH2OH)3 (used in dry-cleaning agents and household

detergents);

(b) O in glyceryl tristearate (a saturated fat):

Exercise 3.6B

(c) H2O in copper(II) sulfate pentahydrate (a fungicide and algacide).

Example 3.7

How many grams of nitrogen are present in 46.34 g ammonium nitrate?Strategy

We first convert the mass of ammonium nitrate to moles, then use the formula NH4NO3 to obtain the ratio of moles of N to moles of NH4NO3, and finally we use the molar mass of nitrogen to calculate the required mass.

SolutionThe central factor (shown in red) in the conversion is based on the chemical formula NH4NO3. The other factors in the following setup are based on molar masses:

People with hypertension (high blood pressure) are advised to limit the amount of sodium(actually sodium ion, Na+) in their diet. Sodium hydrogen carbonate, NaHCO3, packagedas baking soda, is one of many familiar products containing sodium ions. Calculate the number of milligrams of Na+ in 5.00 g of NaHCO3.

Exercise 3.7A

A fertilizer mixture contains 12.5% ammonium nitrate, 20.3% ammonium sulfate, and11.4% urea [CO(NH2)2] by mass. How many grams of nitrogen are present in a 1.00-kg bag of this fertilizer?

Exercise 3.7B

Example 3.8 An Estimation Example

Without doing detailed calculations, determine which of these compounds contains thegreatest mass of sulfur per gram of compound: barium sulfate, lithium sulfate, sodium sulfate, or lead sulfate.

Analysis and ConclusionsTo make this comparison, we need formulas of the compounds, which we can get from their names:

BaSO4 Li2SO4 Na2SO4 PbSO4

The compound with the greatest mass of sulfur per gram of compound also has the greatest mass of sulfur per 100 g of compound—in other words, the greatest % S by mass. From the formulas, we see that in one mole of each compound there is one mole of sulfur, which means 32.066 g S. Thus, the compound with the greatest % S is the one with the smallest formula mass. Because each formula unit has oneSO4

2– ion, all we have to do is compare some atomic masses: that of barium to

twice that of lithium, and so on. With just a glance at an atomic mass table, we see the answer must be lithium sulfate, Li2SO4.

Without doing detailed calculations, determine which of these compounds has the greatest percent phosphorus by mass: lithium dihydrogen phosphate, calcium dihydrogen phosphate, or ammonium hydrogen phosphate.

Exercise 3.8A

Without doing detailed calculations, determine which of these compounds has the greatest mass percent of carbon: methanol, acetic acid, butane, or octane.

Exercise 3.8B

Example 3.9

Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula.

StrategyWe can follow the steps outlined above to determine first the number of moles of each element and then the empirical formula.

SolutionStep 1: A 100.00-g sample of phenol contains 76.57 g C, 6.43 g H, and 17.00 g O.Step 2: We convert the masses of C, H, and O to amounts in moles:

Step 3: Now we use the numbers of moles in Step 2 as subscripts in a tentative formula:

C6.375H6.38O1.063

Step 4: Next we divide each subscript by the smallest (1.063) to try to get integralsubscripts:

Step 5: The subscripts in Step 4 are all integers. We need do nothing further. The empirical formula of phenol is C6H6O.


Cyclohexanol, used in the manufacture of plastics, has the composition 71.95% C,12.08% H, and 15.97% O by mass. Determine its empirical formula.

Exercise 3.9A

Mebutamate, a diuretic (water pill) used to treat high blood pressure, has the composition51.70% C, 8.68% H, 12.06% N, and 27.55% O by mass. Determine its empirical formula.

Exercise 3.9B

Example 3.10

Diethylene glycol, used in antifreeze, as a softening agent for textile fibers and someleathers, and as a moistening agent for glues and paper, has the composition 45.27% C,9.50% H, and 45.23% O by mass. Determine its empirical formula.

StrategyFollowing our five-step procedure, we determine first the number of moles of each element and then the empirical formula.

SolutionStep 1: A 100.00-g sample of diethylene glycol contains 45.27 g C, 9.50 g H, and

45.23 g O.Step 2: We convert the masses of C, H, and O to amounts in moles:

Step 3: Now we use the numbers of moles in Step 2 as subscripts in a tentative formula:

C3.769H9.42O2.827

Step 4: Next we must divide each subscript by the smallest (2.827) in an attempt

to get integral subscripts:


Solution continuedStep 5: Finally we multiply all the subscripts from Step 4 by a common factor to

convert them all to integers. By recognizing that 1.333 = 4/3 and 3.33 = 10/3, we can see that the common factor we need is 3:

Anthracene, used in the manufacture of dyes, has the composition 94.34% C and 5.66% H by mass. Determine its empirical formula.

Exercise 3.10A

Determine the empirical formula of (a) a compound composed of 72.4% iron and 27.6%oxygen by mass; (b) a compound composed of 9.93% carbon, 58.6% chlorine, and 31.4%fluorine by mass; and (c) trinitrotoluene (TNT), an explosive that has the composition37.01% C, 2.22% H, 18.50% N, and 42.27% O by mass.

Exercise 3.10B

Example 3.11

The empirical formula of hydroquinone, a chemical used in photography, is C3H3O, and its molecular mass is 110 u. What is its molecular formula?

StrategyWe can determine the molecular formula by multiplying the subscripts of the empirical formula by the integral factor obtained by using Equation (3.7).

SolutionThe empirical formula mass is (3 x 12.0 u) + (3 x 1.0 u) + 16.0 u = 55.0 u. The multiplier we need to convert the subscripts in the empirical formula to those in the molecular formula is the integral factor from Equation (3.7):

The molecular formula is (C3H3O)2, or C6H6O2.

Ethylene (molecular mass 28.0 u), cyclohexane (84.0 u), and 1-pentene (70.0 u) all havethe empirical formula CH2. Give the molecular formula of each compound.

Exercise 3.11A

Give the empirical formula for (a) P4O6, (b) C6H9, (c) C6H8O6, (d) HOCH2COOH,(e) CH2OHCH2OH, and (f) Cu2C2O4.

Exercise 3.11B

Example 3.12

Burning a 0.1000-g sample of a carbon–hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment shows that the molecular mass of the compound is 90 u. Determine (a) the mass percent composition, (b) the empirical formula, and (c) the molecular formula of the compound.

StrategyFirst, we use conversions like those described above to calculate the mass percents in the compound. From this information, we can determine the empirical formula, integral factor, and molecular formula. Note that there is a specific order in which the calculations must be performed in order to arrive at the molecular formula.

Solution(a) First, we do the conversions outlined to calculate the mass of carbon in the CO2

produced:

This mass of carbon originated from the 0.1000-g sample. Thus, the mass percent carbon in the compound is


Solution continuedWe can use similar calculations to determine first the mass of hydrogen and then the mass percent of hydrogen in the compound:

Finally, we can calculate the mass percent oxygen by subtracting the mass percents of C and H from 100.00%:

% O = 100.00% – 53.30% – 11.19% = 35.51%

(b) Here we apply the method of Examples 3.9 and 3.10, but we need only the firstfour steps.

Step 1: A 100.00-g sample of the compound contains 53.30 g C, 11.19 g H, and 35.51 g O.

Step 2: We convert the masses of C, H, and O to numbers of moles:


Solution continued

Step 3: Next, we use the numbers of moles in Step 2 as subscripts in a tentative formula:

C4.438H11.10O2.220

Step 4: We divide all subscripts by the smallest subscript (2.220) to get integral subscripts:

(c) The empirical formula mass is (2 x 12.0 u) + (5 x 1.0 u) + 16.0 u = 45.0 u. Themultiplier we need to convert the subscripts in the empirical formula to those in the molecular formula is the integral factor in Equation (3.7):

The molecular formula is (C2H5O)2, or C4H10O2.

AssessmentAs in previous examples, we check to see that the sum of the mass percentages is equal to 100% and that the calculation of the integral factor truly yields an integer.

Complete combustion of 0.255 g of an alcohol produces 0.561 g of CO2 and 0.306 g of H2O. Calculate (a) the mass percent composition and (b) the empirical formula for thisalcohol.

Exercise 3.12A


A 0.3629-g sample of tetrahydrocannabinol (THC), the principal active component ofmarijuana, is burned to yield 1.0666 g of carbon dioxide and 0.3120 g of water. Calculate(a) the mass percent composition and (b) the empirical formula of THC.

Exercise 3.12B

Example 3.13

Balance the equation

Fe + O2 Fe2O3 (not balanced)

SolutionIt seems that the easiest place to begin is with the iron atoms. There is one on the left (Fe) and two on the right (Fe2O3), and we can balance them by placing the coefficient 2 on the left:

Fe + O2 Fe2O3 (Fe balanced, O not balanced)

To balance the oxygen atoms, we begin by noting that there are two of them on the left and three on the right. An easy way to get three O atoms on the left, thereby balancing the equation, is to use the fractional coefficient 3/2 before O2 on the left {3/22 x 2 = 3}:

2 Fe + 3/2 O2 Fe2O3 (balanced, fractional coefficient)

Fractional coefficients are not only acceptable in equations, sometimes they are desirable. If we don’t want them, however, we can easily clear an equation of them by multiplying every coefficient by the smallest integer required to clear the fractions, in this case, 2:

2 x {2 Fe + 3/2 O2 Fe2O3}becomes

4 Fe + 3 O2 2 Fe2O3 (balanced, coefficients integral)AssessmentTo verify that an equation is balanced, we multiply each subscript in a formula by the stoichiometric coefficient for that formula. This provides a count of the number of atoms of each type on both sides of the equation.


Assessment continuedFor each element, the number of atoms must be the same on the two sides of a balanced equation. In the equation above, 4 Fe and 6 O atoms are seen on each side of the equation; it is balanced.

Balance the following equations using integral coefficients:(a) SiCl4 + H2O SiO2 + HCl(b) PCl5 + H2O H3PO4 + HCl(c) CaO + P4O10 Ca3(PO4)2

Exercise 3.13A

Use integral coefficients to write a balanced equation for each of the following reactions.Use the (g), (l), (s), (aq) symbols to indicate the form of each reactant and product:(a) The formation of solid lead iodide and aqueous potassium nitrate from aqueous

solutions of lead nitrate and potassium iodide.(b) The reaction of gaseous hydrogen chloride and oxygen to form gaseous water and

chlorine.

Exercise 3.13B

Example 3.14

Balance the equationC2H6 + O2 CO2 + H2O

SolutionOxygen appears as the free element on the left, so let’s leave it for last and balance the other two elements first. To balance carbon, we place the coefficient 2 in front of CO2:

C2H6 + O2 2 CO2 + H2O (C balanced, H and O not balanced)

To balance hydrogen, we need the coefficient 3 before H2O:

C2H6 + O2 2 CO2 + 3 H2O (C and H balanced, O not balanced)

Now, if we count oxygen atoms, we find two on the left and seven on the right. We can get seven on each side by using the fractional coefficient 7/2(7/2 x 2 = 7):

C2H6 + 7/2 O2 2 CO2 + 3 H2O (balanced)

To obtain integral coefficients, we multiply each coefficient by 2:

2 x {C2H6 + 7/2 O2 2 CO2 + 3 H2O}That leads to

2 C2H6 + 7 O2 4 CO2 + 6 H2O (balanced)AssessmentWe ensure that the equation is balanced by multiplying the subscript for each element by the appropriate integral coefficient to get the number of atoms on the two sides of the equation. Note that oxygen atoms are present in two product molecules. On each side of the equation are four C atoms, twelve H atoms, and fourteen O atoms; the equation is balanced.

Balance the following equations:(a) C4H10 + O2 CO2 + H2O(b) CH3CH2CH2CH(OH)CH2OH + O2 CO2 + H2O

Exercise 3.14A


Write a balanced equation to represent the complete combustion of the compound represented by the molecular model in the margin. (Recall Figures 2.7 and 2.8.)

Exercise 3.14B

Example 3.15

Balance the equationH3PO4 + NaCN HCN + Na3PO4

SolutionNotice that the PO4 and CN groups remain unchanged in the reaction. For purposes of balancing equations, we can often treat such groups as a whole rather than breaking them down into their constituent atoms. To balance hydrogen atoms, we place a 3 before HCN:

H3PO4 + NaCN 3 HCN + Na3PO4 (not balanced)

To balance cyanide ions, we put a 3 in front of the NaCN. Doing this also balances the sodium ions:

H3PO4 + 3 NaCN 3 HCN + Na3PO4 (balanced)

Because the PO4 was balanced to begin with and we did nothing to upset that balance, we are finished; the equation is balanced.AssessmentNote that in the balanced equation, there are three Na atoms, three H atoms, one PO4 group, and three CN groups on each side of the equation.

Balance the following equations.(a) FeCl3 + NaOH Fe(OH)3 + NaCl(b) Ba(NO3)2 + Al2(SO4)3 BaSO4 + Al(NO3)3

Exercise 3.15A


Write a balanced equation for the reaction between solid calcium hydroxide and aqueousphosphoric acid to form solid calcium phosphate and water. Include symbols indicatingthe physical form of each reactant and product.

Exercise 3.15B

Example 3.16 A Conceptual Example

Write a plausible chemical equation for the reaction between water and a liquid molecular chloride of phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid. The phosphorus-chlorine compound is 77.45% Cl by mass.

Analysis and ConclusionsIn this problem, balancing a chemical equation is the last step. First, we must apply some earlier ideas:

Establishing the formula of the chloride of phosphorus.We can apply the method of Examples 3.9 and 3.10 to a compound that is 22.55% P and 77.45% Cl. A 100.00-g sample of the compound consists of 22.55 g P and 77.45 g Cl, corresponding to 0.728 mol P and 2.185 mol Cl. The formula P0.728Cl2.185 reduces to PCl3.

Establishing the formulas of hydrochloric and phosphorus acids.We described the relationship between names and formulas on pages 57–58. Hydrochloric acid, a binary acid, has the formula HCl. Table 2.5 gives H3PO4 as the formula of phosphoric acid. Phosphorus acid should have one O atom fewer per molecule than the “-ic” acid, giving the formula H3PO3.

Writing and balancing the equation.The unbalanced equation, including an indication of the physical form of each substance, is

PCl3(l) + H2O(l) HCl(aq) + H3PO3(aq)

The balanced equation requires the coefficient 3 for H2O(l) and for HCl(aq):PCl3(l) + 3 H2O(l) 3 HCl(aq) + H3PO3(aq)

Example 3.16 A Conceptual Example continued

Write a plausible chemical equation for the combustion of liquid triethylene glycol in anabundant supply of oxygen gas. Triethylene glycol is 47.99% C, 9.40% H, and 42.62% Oby mass and has a molecular mass of 150.2 u.

Exercise 3.16A

Upon heating, solid lead(II) nitrate yields solid lead(II) oxide and two gaseous products,one of which is nitrogen dioxide. Write a balanced equation for this reaction.

Exercise 3.16B

Example 3.17

When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? The reaction is

C3H8 + 5 O2 3 CO2 + 4 H2O

StrategyWe will relate the moles of consumed oxygen to moles of propane burned, using as our conversion factor the ratio of the number of moles of each species present in the balanced chemical equation.

SolutionThe stoichiometric coefficients in the equation allow us to write the equivalence

From this equivalence, we can derive two conversion factors:

Because we are given the number of moles of propane (substance A in Figure 3.8) and are seeking the number of moles of O2 consumed (substance B in Figure 3.8), we need the factor that has the unit “mol O2” in the numerator and the unit “mol C3H8 ” in the denominator. This is the conversion factor shown in red above:


For the combustion of propane in Example 3.17:(a) How many moles of carbon dioxide are formed when 0.529 mol C3H8 is burned?(b) How many moles of water are produced when 76.2 mol C3H8 is burned?(c) How many moles of carbon dioxide are produced when 1.010 mol O2 is consumed?

Exercise 3.17A

For the combustion of 55.6 g of butane in an excess of oxygen, (a) how many moles of carbon dioxide are formed and (b) how many moles of oxygen are consumed?

Exercise 3.17B

Example 3.18

The final step in the production of nitric acid involves the reaction of nitrogen dioxide with water; nitrogen monoxide is also produced. How many grams of nitric acid are produced for every 100.0 g of nitrogen dioxide that reacts?

StrategyWorking from a balanced chemical equation and using the appropriate stoichiometric factor, we determine the moles of product produced from the given moles of reactant. The conversions from grams to moles and moles to grams are based on conversion factors associated with the respective molar masses.

SolutionBecause no equation is given, we must first write a chemical equation from the description of the reaction. Recall that we related the names and formulas of these reactants and products in Chapter 2.

To balance the equation, let’s first balance H atoms because H appears in one reactant and one product. Then we can balance N atoms because they appear on the reactant side only in NO2. At this point, O atoms will also be balanced.


Solution continued

Now, we convert the mass of the given substance, NO2, to an amount in moles, using the molar mass in the manner shown in Figure 3.9.

Next, we use coefficients from the balanced equation to establishthe stoichiometric equivalence of NO2 and HNO3 .

Because we need to convert from moles of NO2 to moles of HNO3, we should use this equivalence to write the stoichiometric factor needed for the conversion.

Finally, we can convert from moles of HNO3 to grams of HNO3 using its molar mass.

AssessmentAs you gain experience with reaction stoichiometry calculations, you will likely switch to doing all that we have done here in a combined setup that obviates the need to write intermediate answers.


How many grams of magnesium are required to convert 83.6 g TiCl4 to titanium metal? The balanced equation representing this reaction is

Exercise 3.18A

Upon being strongly heated or subjected to severe mechanical shock, ammonium nitrate decomposes into nitrogen gas, oxygen gas, and water vapor. If 75.5 g of ammonium nitrate decomposes in this way, how many grams of nitrogen and how many grams of oxygen are produced?

Exercise 3.18B

Example 3.19

Ammonium sulfate, a common fertilizer used by gardeners, is produced commercially bypassing gaseous ammonia into an aqueous solution that is 65% H2SO4 by mass and has adensity of 1.55 g/mL. How many milliliters of this sulfuric acid solution are required toconvert 1.00 kg NH3 to (NH4)2SO4?

StrategyHere we want to determine the amount of one reactant (sulfuric acid) needed to react completely with a second reactant (ammonia). As before, we will use a stoichiometric factor based on a balanced chemical equation to determine the stoichiometric equivalence between the two reactants. We will use the respective molar masses in calculating the number of moles of NH3 initially present and the mass of H2SO4 consumed. Finally, we will use the solution density and mass percentage to calculate the volume of H2SO4(aq) needed.

SolutionFirst, we must write the balanced equation for the reaction:

2 NH3(g) + H2SO4(aq) (NH4)2SO4(aq)

The required setup has 1.00 kg NH3 —the given substance—as its starting point. Because we are seeking a volume of H2SO4(aq), the setup has the general form

? mL H2SO4(aq) = 1.00 kg NH3 x conversion factors

Solution continuedWe begin by converting kilograms of NH3 to grams of NH3. Then we can apply the flow chart of Figure 3.9 to find grams of H2SO4. Finally, we use the percent composition and then the density of the H2SO4(aq) to convert from grams of H2SO4 to milliliters of H2SO4(aq):


In the combined setup, the conversions are done in the order in which they are described above:


AssessmentNote that the percent composition (65% H2SO4) and density (1.55 g/mL) are written in the inverted form. This provides for the proper cancellation of units, and it makes sense. The mass of H2SO4(aq) should be greater than that of the pure H2SO4 from which the solution is made, and the volume of H2SO4(aq) should be a smaller number than its mass because the density of the solution is greater than 1.

How many milliliters of liquid water are produced by the combustion in abundant oxygenof 775 mL of octane, C8H18(l)? Assume that the volumes of both liquids are measured at20.0 °C, where the densities are 0.7025 g/mL for octane and for 0.9982 g/mL for water.

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O (l)

Exercise 3.19A

How many milliliters of water at 20.0 °C (d = 0.9982 g/mL) are formed in the combustion of 25.0 g of a methanol–propanol mixture that is 62.5% methanol by mass? Assume there is an excess of oxygen available.

Exercise 3.19B

Example 3.20

Magnesium nitride can be formed by the reaction of magnesium metal with nitrogen gas. (a) How many grams of magnesium nitride can be made in the reaction of 35.00 g of magnesium and 15.00 g of nitrogen? (b) How many grams of the excess reactant remain after the reaction?

StrategyIn this problem, we need to determine which of the reactants is completely consumed and is therefore the limiting reactant. The quantity of this reactant, in turn, will determine the quantity of magnesium nitride produced. We will need a grams-to-moles conversion factor to convert from the given reactant masses and a moles-to-grams factor to convert to the desired product mass. The quantity of excess reactant can be calculated as the difference between the given mass of this reactant and the mass consumed in the reaction.

Solution(a) As usual, we must first write a

balanced equation, and we can do this by using ideas from this chapter and Chapter 2.

We can identify the limiting reactant by finding the number of moles of Mg3N2(s) produced by assuming first one reactant, and then the other is the limiting reactant.


Solution continuedAssuming Mg is the limiting reactant and N2 is in excess:

Assuming N2 is the limiting reactant and Mg is in excess:

Because the amount of product in the first calculation (0.4800 mol Mg3N2) is smaller than that in the second (0.5355 mol Mg3N2), we know that magnesium is the limiting reactant. When 0.4800 mol Mg3N2 has been formed, the Mg is completely consumed and the reaction stops, producing a specific mass of Mg3N2.

(b) Having found that the amount of

product is 0.4800 mol Mg3N2, we can now calculate how much must have been consumed.

From this, we calculate the mass of excess N2.


One way to produce hydrogen sulfide gas is by the reaction of iron(II) sulfide with hydrochloric acid:

FeS(s) + 2 HCl(aq) FeCl2(aq) + H2S(g)

If 10.2 g HCl is added to 13.2 g FeS, how many grams of H2S can be formed? What is the mass of the excess reactant remaining?

Exercise 3.20A

A convenient laboratory source of hydrogen gas is the reaction of an aqueous hydrochloric acid solution with aluminum metal. An aqueous solution of aluminum chloride is the other product of the reaction. How many grams of hydrogen are produced in the reaction between 12.5 g of aluminum and 250.0 mL of an aqueous hydrochloric acid solution that is 25.6% HCl by mass and has a density of 1.13 g/mL?

Exercise 3.20B

Example 3.21

Ethyl acetate is a solvent used as fingernail polish remover. What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected percent yield is 85.0%? Assume that the other reactant, ethanol, is present in excess. The equation for the reaction, carried outin the presence of H2SO4, is

StrategyThe mass we are given in this problem—252 g ethyl acetate—is an actual yield, but the stoichiometric calculation requires that we first calculate the theoretical yield of the reaction. For this we use Equation (3.8).

SolutionFirst, we can solve Equation (3.8) for the theoretical yield and substitute known quantities for the actual and percent yields.

Then, we can determine the mass of acetic acid required to produce 296 g ethyl acetate.


AssessmentA common source of error in a problem of this type is to misuse Equation (3.8). If we had multiplied 252 g ethyl acetate by 85% rather than dividing by it, we would have obtained 214 g ethyl acetate as the theoretical yield. Clearly, this cannot be so—the actual yield can never be greater than the theoretical yield, that is, never greater than 100%.

Isopentyl acetate is the main component of banana flavoring. Calculate the theoretical yield of isopentyl acetate that can be made from 20.0 g isopentyl alcohol and 25.0 g acetic acid.

CH3COOH + HOCH2CH2CH(CH3)2 CH3COOCH2CH2CH(CH3)2 + H2O Acetic acid Isopentyl alcohol Isopentyl acetate

If the percent yield of the reaction is 90.0%, what is the actual yield of isopentyl acetate?

Exercise 3.21A

How many grams of isopentyl alcohol are needed to make 433 g isopentyl acetate in the reaction described in Exercise 3.21A if the expected yield is 78.5%? Assume the acetic acid is in excess.

Exercise 3.21B

Example 3.22 A Conceptual Example

What is the maximum yield of CO(g) obtainable from 725 g of C6H14(l) regardless of thereaction(s) used, assuming no other carbon-containing reactant or product?

Analysis and ConclusionsIf the maximum yield is independent of the reaction(s) used, then we ought to be able to determine this quantity without writing any chemical equations and without using stoichiometric factors based on chemical equations. While at first this may not seem feasible, recall that in Section 3.6 we dealt with a similar idea. There we needed to relate the mass of CO2 produced in combustion analysis to the mass of a carbon-containing compound from which it formed. According to the law of conservation of mass, all the C atoms in the C6H14 have to be accounted for, and the maximum yield results if all the C atoms end up in CO. Let’s rephrase the original question. What mass of CO contains the same number of C atoms as 725 g C6H14? We can start the calculation by converting grams of C6H14 to moles of C6H14, and we can end it by converting moles of CO to grams of CO. The critical link between these two “ends” of the calculation are the factors that relate moles of C to moles of C6H14 on the one hand and moles of CO to moles of C on the other. These are the factors shown in red in the following equation.

What is the maximum yield of the important commercial fertilizer ammonium hydrogenphosphate that can be obtained per kilogram of phosphoric acid?

Exercise 3.22A

Example 3.22 A Conceptual Example continued

Without doing detailed calculations, determine whether calcium, magnesium, or aluminum yields the most hydrogen per gram of metal when the metal reacts with an excess of hydrochloric acid. An aqueous solution of the metal chloride is the other product of the reaction.

Exercise 3.22B

Example 3.23

What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution?

StrategyAccording to Equation (3.9), the molarity of a solution is calculated from the number of moles of solute and the volume of solution in liters. Because we are given the volume of solution (10.0 L), the only requirement prior to substitution into Equation (3.9) is to convert the quantity of solute from mass in grams to number of moles.

SolutionFirst, let’s prepare the setup that converts mass to number of moles of KHCO3.

Now, without solving this expression, let’s use it as the numerator in the defining equation for molarity. The solution volume, 10.0 L, is the denominator.

Calculate the molarity of the solute in each of the following solutions.(a) 3.00 mol KI in 2.39 L of solution (b) 0.522 g HCl in 0.592 L of solution(c) 2.69 g C12H22O11 in 225 mL of solution

Exercise 3.23A


Calculate the molarity of (a) glucose, C6H12O6, in 100.0 mL of solution containing 126 mg of glucose; (b) ethanol, CH3CH2OH, in a solution containing 10.5 mL of ethanol (d = 0.789 g/mL) in 25.0 mL of solution; and (c) urea, CO(NH2)2 in a solution of urea whose concentration is expressed as 9.5 mg N/mL solution.

Exercise 3.23B

Example 3.24

We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH).(a) How many moles of NaOH are required to prepare 0.500 L of 6.68 M NaOH?(b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH?

Solution(a) This calculation requires only a one-step conversion from liters of solution to

moles of NaOH, with the molarity of the solution as the conversion factor:

(b) The central conversion factors in this calculation are the inverse of the molar mass of NaOH to convert from grams of NaOH to moles of NaOH and the inverse of the molarity—1 L soln/6.68 NaOH—to convert from moles of NaOH to liters of solution.We must also convert from kilograms of NaOH to grams of NaOH. In all, the required conversions are

kg NaOH g NaOH mol NaOH L soln

which are set up as follows:


How many grams of potassium hydroxide are required to prepare each of the followingsolutions?(a) 2.00 L of 6.00 M KOH(b) 10.0 mL of 0.100 M KOH(c) 35.0 mL of 2.50 M KOH

Exercise 3.24A

How many milliliters of 1-butanol, CH3CH2CH2CH2OH, (d = 0.810 g/mL) are requiredto prepare 725 mL of a 0.350 M aqueous solution of this solute?

Exercise 3.24B

Example 3.25

The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH3 by mass and has a density of 0.898 g/mL. Calculate the molarity of the solution.

StrategyWe will find it most convenient to base the calculation on a 1.00-L volume of solution. When we have found the number of moles of NH3 in this 1.00 L of solution, we will have found the molarity. The way in which these factors enter into the calculation is outlined below.

Solution


A stock bottle of aqueous formic acid indicates that the solution is 90.0% HCOOH bymass and has a density of 1.20 g/mL. Calculate the molarity of the solution.

Exercise 3.25A

A concentrated solution of perchloric acid, HClO4, is 11.7 M and has a density of 1.67 g/mL. What is the mass percent perchloric acid in this solution?

Exercise 3.25B

Example 3.26

How many milliliters of a 2.00 M CuSO4 stock solution are needed to prepare 0.250 L of 0.400 M CuSO4?

StrategyWe have considered two ways of viewing situations in which a dilute solution is prepared from a more concentrated one. One way is based on the principle of dilution illustrated in Figure 3.13, and the other employs Equation (3.10). If we work the same problem both ways, we expect to obtain the same result, thus providing us with an answer and a check of the answer at the same time.

SolutionApplying the principle of dilution:The key is to note that all the solute in the unknown volume of the stock solution appears in the 0.250 L of 0.400 M CuSO4. First, let’s calculate that amount of solute:

Now we need to answer the question, “What volume of 2.00 M CuSO4 contains 0.100 mol CuSO4?” In doing so, we will have answered the original question.

Of course, we could have done all of this in a single setup:


Solution continuedUsing the dilution equation:First, we can identify the terms we need for Equation (3.10):

Then, we can substitute these terms into the equation:


AssessmentAs expected, the two methods yield the same result: 50.0 mL 2.00 M CuSO4(aq). To prepare the dilute solution, we should measure out 50.0 mL of 2.00 M CuSO4 and add it to enough water to make 0.250 L of solution, as illustrated in Figure 3.14.


How many milliliters of a 10.15 M NaOH stock solution are needed to prepare 15.0 L of0.315 M NaOH?

Exercise 3.26A

How many milliliters of a 5.15 M CH3OH stock solution are needed to prepare 375 mL of a solution having 7.50 mg of methanol per mL of solution?

Exercise 3.26B

Example 3.27

A chemical reaction familiar to geologists is that used to identify limestone. The reaction of hydrochloric acid with limestone, which is largely calcium carbonate, is seen through an effervescence—a bubbling due to the liberation of gaseous carbon dioxide:

CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)

How many grams of CaCO3(s) are consumed in a reaction with 225 mL of 3.25 M HCl?

StrategyTo relate the quantity of CaCO3 to that of HCl, we need to express the amount of HCl in moles and multiply by the stoichiometric factor 1 mol CaCO3/2 mol HCl. First, though, we have to relate the number of moles of HCl to the volume of HCl(aq) and its molarity. Thus, we use molarity as a conversion factor before introducing the stoichiometric factor. The setup below is outlined in the stoichiometry diagram in Figure 3.15.

Solution


How many milliliters of 0.100 M AgNO3(aq) are required to react completely with 750.0 mL of 0.0250 M Na2CrO4(aq)?

2 AgNO3(aq) + Na2CrO4(aq) Ag2CrO4(s) + 2 NaNO3(aq)

Exercise 3.27A

In a reaction similar to that in which baking soda (NaHCO3) neutralizes stomach acid,175 mL of 1.55 M NaHCO3 is added to 235 mL of 1.22 M HCl:

NaHCO3(aq) + HCl(aq) NaCl(aq) + H2O(l) + CO2(g)

(a) How many grams of CO2 are liberated?(b) What is the molarity of the NaCl(aq) produced? Assume that the solution volume is

175 mL + 235 mL = 410 mL.

Exercise 3.27B

Cumulative Example

The combustion in oxygen of 1.5250 g of an alkane-derived compound composed of carbon, hydrogen, and oxygen yields 3.047 g CO2 and 1.247 g H2O. The molecular mass of this compound is 88.1 u. Draw a plausible structural formula for this compound. Is there more than one possibility? Explain.

StrategyExample 3.12 provides much of the initial guidance for this problem. We can use the masses of CO2 and H2O from the combustion analysis to determine first the number of moles of carbon and hydrogen in the sample and then the masses of carbon and hydrogen. Next, we can find the mass of oxygen by subtraction and convert that mass to moles of oxygen. Now, we can use the numbers of moles of the elements to establish the empirical formula of the sample. Once we have the empirical formula, we can use the molecular mass to lead us to the molecular formula. Finally, using our knowledge of alkanes and functional groups (Section 2.9), we can draw structural formulas and determine whether more than one structure is possible.

SolutionWe begin by finding the moles of carbon in 3.047 g of CO2 and the moles of hydrogen in 1.247 g of H2O.

Cumulative Example continued

Solution continuedWe will need these numbers of moles for the empirical formula determination later. Next, we convert moles of each element to mass.We find the mass of oxygen by subtracting the masses of carbon and hydrogen from the sample mass.

Then we convert the mass of oxygen to moles.

Now we use the moles of carbon, hydrogen, and oxygen to construct a tentative formula.

Then we divide each subscript by the subscript for oxygen (because this is the smallest of the three subscripts).

The empirical formula mass is 44.053 u.

Next, we use Equation (3.7) to calculate the ratio of the molecular mass to the empirical formula mass.

We multiply each subscript by the factor 2 to obtain the molecular formula.

Cumulative Example continued

Solution continuedRecall from Section 2.9 that a carboxylic acid has two oxygen atoms. Butanoic acid fits the molecular formula (I). However, there are many other possibilities. The three-carbon chain of butanoicacid could be replaced by a branched group, as shown in structure (II).The compounds (III) and (IV) also fit the molecular formula.

These are but a few of the possible structures.

AssessmentBecause the masses of CO2 and H2O formed in the combustion are considerably less than their molar masses, we expect the moles of carbon and of hydrogen to be less than 1, and they are. Also, although it is not compelling evidence, the fact that we obtained small integral values for the subscripts in the empirical formula suggests that it is a reasonable formula, as is the molecular formula. Another interesting observation is that alkane-based alcohols (ROH) and ethers [(ROR´),where R and R´ represent alkyl groups] are not possible structures for the formula C4H8O2. There are not enough hydrogen atoms in the molecule for this to be the case.

Example 3.1 Calculate the molecular mass of glycerol (1,2,3-propanetriol). Strategy To determine a...

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Transcript of Example 3.1 Calculate the molecular mass of glycerol (1,2,3-propanetriol). Strategy To determine a...