worked example 6.1 molar mass and avogadro’s number: number of molecules

© 2013 Pearson Education, Inc.Fundamentals of General, Organic, and Biological Chemistry, 7eJohn McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. Peterson

Worked Example 6.1 Molar Mass and Avogadro’s Number: Number of Molecules

Pseudoephedrine hydrochloride is a nasal decongestant commonly found in cold medication. (a) What is the molar mass of pseudoephedrine hydrochloride? (b) How many molecules of pseudoephedrine hydrochloride are in a tablet that contains 30.0 mg of this decongestant?

(a) The molecular weight of pseudoephedrine is found by summing the atomic weights of all atoms in the molecule:

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Remember that atomic mass in amu converts directly to molar mass in g/mol. Also, following the rules for significant figures from Sections 1.9 and 1.11, our final answer is rounded to the second decimal place.

Solution

Analysis We are given a mass and need to convert to a number of molecules. This is most easily accomplished by using the molar mass of pseudoephedrine hydrochloride calculated in part (a) as the conversion factor from mass to moles and realizing that this mass (in grams) containsAvogadro’s number of molecules .

The formula for pseudoephedrine contains 10 carbon atoms (each one of atomic weight 12.0 amu), so the molecular weight is greater than 120 amu, probably near 200 amu. Thus, the molecular weight should be near 200 g/mol. The mass of 30 mg of pseudoepinephrine HCl is less than the mass of 1 mol of this compound by a factor of roughly (0.03 g versus 200 g), which means that the number of molecules should also be smaller by a factor of (on the order of in the tablet versus in 1 mol).

Ballpark Estimate


STEP 1: Identify known information. We are given the mass of pseudoephedrine hydrochloride (in mg).

STEP 2: Identify answer and units. We are looking for the numberof molecules of pseudoephedrine hydrochloride in a 30 mg tablet.

STEP 3: Identify conversion factors. Since the molecular weight of pseudoephedrine hydrochloride is 201.70 amu, 201.70 g contains

molecules. We can use this ratio as a conversion factor to convert from mass to molecules. We will also need to convert 30 mg to g.

STEP 4: Solve. Set up an equation so that unwanted units cancel.

(b) Since this problem involves unit conversions, we can use the step-wise solution introduced in Chapter 1.

30.0 mg pseudoephedrine hydrochloride

Our estimate for the number of molecules was on the order of 1019, which is consistent with the calculated answer.

Ballpark Estimate


Worked Example 6.2 Avogadro’s Number: Atom to Mass ConversionsA tiny pencil mark just visible to the naked eye contains about atoms of carbon. What is the mass of this pencil mark in grams?

STEP 1: Identify known information. We know the number of carbon atoms in the pencil mark.

STEP 2: Identify answer and units.STEP 3: Identify conversion factors. The atomic weight of carbon is 12.01 amu, so 12.01 g of carboncontains atoms.

STEP 4: Solve. Set up an equation using the conversion

factors so that unwanted units cancel.

Solution

Analysis We are given a number of atoms and need to convert to mass. The conversion factor can be obtained by realizing that the atomic weight of carbon in grams contains Avogadro’s number

of atoms .

Since we are given a number of atoms that is six orders of magnitude less than Avogadro’s number, we should get a corresponding mass that is six orders of magnitude less than the molar mass of carbon, which means a mass for the pencil mark of about g.

Ballpark Estimate

atoms of carbon

The answer is of the same magnitude as our estimate and makes physical sense.

Ballpark Check


Worked Example 6.3 Molar Mass: Mole to Gram ConversionThe nonprescription pain relievers Advil and Nuprin contain ibuprofen , whose molecular weight is206.3 amu (Problem 6.1a). If all the tablets in a bottle of pain reliever together contain 0.082 mol of ibuprofen, what is the number of grams of ibuprofen in the bottle?

STEP 1: Identify known information.

STEP 2: Identify answer and units.STEP 3: Identify conversion factors. We use the molecular weight of ibuprofen to convert from molesto grams.

STEP 4: Solve. Set up an equation using the known information and conversion factor so that unwantedunits cancel.

Solution

Analysis We are given a number of moles and asked to find the mass. Molar mass is the conversion factor between the two.

Since 1 mol of ibuprofen has a mass of about 200 g, 0.08 mol has a mass of about 0.08 × 200 g = 16 g.

Ballpark Estimate

The calculated answer is consistent with our estimate of 16 g.

Ballpark Check


Worked Example 6.4 Molar Mass: Gram to Mole ConversionThe maximum dose of sodium hydrogen phosphate that should be taken in one day for use as a laxative is 3.8 g. How many moles of sodium hydrogen phosphate, how many moles of ions, and how many total moles of ions are in this dose?

STEP 1: Identify known information. We are giventhe mass and molecular weight of .

STEP 2: Identify answer and units. We need to find the number of moles of , and the total number of moles of ions.

STEP 3: Identify conversion factors. We can use the molecular weight of to convert from grams to moles.

Solution

Analysis Molar mass is the conversion factor between mass and number of moles. The chemical formula shows that each formula unit contains 2 ions and ion.

The maximum dose is about two orders of magnitude smaller than the molecular weight (approximately 4 g compared to 142 g). Thus, the number of moles of sodium hydrogen phosphate in 3.8 g should be about two orders of magnitude less than one mole. The number of moles of and total moles of ions, then, should be on the order of 10-2.

Ballpark Estimate


STEP 4: Solve. We use the known information and conversion factor to obtain moles of ; since 1 mol of contains 2 mol of ions and 1 mol of ions, we multiply these values by the number of moles in the sample.

The calculated answers (0.027 mol

, 0.081 mol ions) are on the order of 10-2, consistent with our estimate.

Ballpark Check


Worked Example 6.5 Balanced Chemical Equations: Mole RatiosRusting involves the reaction of iron with oxygen to form iron (III) oxide, :

(a) What are the mole ratios of the product to each reactant and of the reactants to each other?

(b) How many moles of iron (III) oxide are formed by the complete oxidation of 6.2 mol of iron?

(a) The coefficients of a balanced equation represent the mole ratios:

(b) To find how many moles of are formed, write down the known information—6.2 mol of iron—and select

the mole ratio that allows the quantities to cancel, leaving the desired quantity:

Note that mole ratios are exact numbers and therefore do not limit the number of significant figures in the result of a

calculation.

Analysis and Solution


Worked Example 6.6 Mole Ratios: Mole to Mass ConversionsIn the atmosphere, nitrogen dioxide reacts with water to produce NO and nitric acid, which contributes to pollution by acid rain:

How many grams of are produced for every 1.0 mol of NO2 that reacts? The molecular weight of is 63.0 amu.

STEP 1: Write balanced equation.

STEP 2: Identify conversion factors. We need a mole to mole conversion to find the number of moles of product, and then a mole to mass conversion to find the mass of product. For the first conversion we use the mole ratio of to as a conversion factor, and for the mole to mass calculation, we use the molar mass of (63.0 g/mol) as a conversion factor.

Analysis

Solution

Ballpark Estimate

We are given the number of moles of a reactant and are asked to find the mass of a product.Problems of this sort always require working in moles and then converting to mass, as outlined in Figure 6.2 .

The molar mass of nitric acid is approximately 60 g/mol, and the coefficientsin the balanced equation say that 2 mol of are formed for each 3 mol of that undergo reaction. Thus, 1 mol of should give about 2/3 mol , or 2/3 mol × 60 g/mol = 40 g.


STEP 3: Set up factor labels. Identify appropriate mole ratio factor labels to convert moles to moles , and moles to grams.

STEP 4: Solve.

Ballpark Check Our estimate was 40 g!


Worked Example 6.7 Mole Ratios: Mass to Mole/Mole to Mass ConversionsThe following reaction produced 0.022 g of calcium oxalate . What mass of calcium chloride was used as reactant? (The molar mass of is 128.1 g/mol, and the molar mass of is 111.0 g/mol.)

STEP 1: Write balanced equation.

STEP 2: Identify conversion factors. Convert the mass of into moles, use a mole ratio to find moles of , and convert the number of moles of to mass. We willneed three conversion factors.

Analysis

Solution

Ballpark Estimate

Both the known information and that to be found are masses, so this is a mass to mass conversion problem. The mass of is first converted into moles, a mole ratio is used to find moles of , and the number of moles of is converted into mass.

The balanced equation says that 1 mol of is formed for each mole of that reacts. Because the formula weights of the two substances are similar, it should take about 0.02 g of to form 0.02 g of .


STEP 3: Set up factor labels. We will need to perform gram to mole and mole to mole conversions to get from grams to grams .

Ballpark Check The calculated answer (0.019 g) isconsistent with our estimate (0.02 g).

STEP 4: Solve.


Worked Example 6.8 Percent Yield

The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.

Analysis

The combustion of acetylene gas produces carbon dioxide and water, as indicated in the following reaction:

When 26.0 g of acetylene is burned in sufficient oxygen for complete reaction, the theoretical yield of is 88.0 g. Calculate the percent yield for this reaction if the actual yield is only 72.4 g .

Solution

The theoretical yield (88.0 g) is close to 100 g. The actual yield (72.4 g) is about 15 g less than the theoretical yield. The actual yield is thus about 15% less than the theoretical yield, so the percent yield is about 85%.

Ballpark Estimate

The calculated percent yield agrees very well with our estimate of 85%.Ballpark Estimate


Worked Example 6.9 Mass to Mole Conversions: Limiting Reagent and Theoretical Yield

The element boron is produced commercially by the reaction of boric oxide with magnesium at high temperature:

What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? The molar masses of boric oxide and magnesium are 69.6 g/mol and 24.3 g/mol, respectively.

STEP 1: Identify known information. We have the masses and molar masses of the reagents.

STEP 2: Identify answer and units. We are solving for the theoretical yield of boron.

STEP 3: Identify conversion factors. We can use the molar masses to convert from masses to moles of reactants ( , Mg). From moles of reactants, we can use mole ratios from the balanced chemical equation to find the number of moles of Bproduced, assuming complete conversion of a given reactant. is the limiting reagent, since complete conversion of thisreagent yields less product (67.6 mol B formed) than doescomplete conversion of Mg (98.0 mol B formed).

Solution

Analysis To calculate theoretical yield, we first have to identify the limiting reagent. The theoretical yield in grams is then calculated from the amount of limiting reagent used in the reaction. The calculation involves the mass to mole and mole to mass conversions discussed in the preceding section.


STEP 4: Solve. Once the limiting reagent has been Identified ( ), the theoretical amount of B that should be formed can be calculated using a mole to mass conversion.


Worked Example 6.10 Mass to Mole Conversion: Percent Yield

Treat this as a typical mass relationship problem to find the amount of ethyl alcohol that can theoretically be formed from 25.0 g of ethylene, and then multiply the answer by 0.785 (the fraction of the theoretical yield actually obtained) to find the amount actually formed.

Analysis

The reaction of ethylene with water to give ethyl alcohol occurs with 78.5% actual yield. How many grams of ethyl alcohol are formed by reaction of 25.0 g of ethylene? (For ethylene, MW = 28.0 amu; for ethyl alcohol, MW = 46.0 amu. )

SolutionThe theoretical yield of ethyl alcohol is:

and so the actual yield is:

41.1 g ethyl alc. × 0.785 = 32.3 g ethyl alcohol

The 25.0 g of ethylene is a bit less than 1 mol; since the percent yield is about 78%, a bit less than 0.78 mol of ethyl alcohol will form—perhaps about 3/4 mol, or 3/4 × 46 g = 34 g.

Ballpark Estimate

The calculated result (32.3 g) is close to our estimate (34 g).Ballpark Check

worked example 6.1 molar mass and avogadro’s number: number of molecules

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