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

Worked Example 6.1 Molar Mass and Avogadro’s Number: Number of Molecules. Pseudoephedrine hydrochloride is a nasal decongestant commonly found in cold medication.

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

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  1. 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? 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) contains Avogadro’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 Solution (a) The molecular weight of pseudoephedrine is found by summing the atomic weights of all atoms in the molecule: _____________________________________________ 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.

  2. (b) Since this problem involves unit conversions, we can use the step-wise solution introduced in Chapter 1. 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 number of 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. 30.0 mg pseudoephedrine hydrochloride Ballpark Estimate Our estimate for the number of molecules was on the order of 1019, which is consistent with the calculated answer.

  3. Worked Example 6.2 Avogadro’s Number: Atom to Mass Conversions A tiny pencil mark just visible to the naked eye contains about atoms of carbon. What is the mass of this pencil mark in grams? 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 Solution 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 carbon contains atoms. STEP 4:Solve. Set up an equation using the conversion factors so that unwanted units cancel. atoms of carbon Ballpark Check The answer is of the same magnitude as our estimate and makes physical sense.

  4. Worked Example 6.3 Molar Mass: Mole to Gram Conversion The nonprescription pain relievers Advil and Nuprin contain ibuprofen , whose molecular weight is 206.3 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? We are given a number of moles and asked to find the mass. Molar mass is the conversion factor between the two. Analysis 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 Solution 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 moles to grams. STEP 4:Solve. Set up an equation using the known information and conversion factor so that unwanted units cancel. The calculated answer is consistent with our estimate of 16 g. Ballpark Check

  5. Worked Example 6.4 Molar Mass: Gram to Mole Conversion The 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? 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. Analysis 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 Solution STEP 1:Identify known information. We are given the 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.

  6. 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

  7. Worked Example 6.5 Balanced Chemical Equations: Mole Ratios Rusting 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? Analysis and Solution (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.

  8. Worked Example 6.6 Mole Ratios: Mole to Mass Conversions In 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. Analysis 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 . Ballpark Estimate The molar mass of nitric acid is approximately 60 g/mol, and the coefficients in 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. Solution 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.

  9. 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!

  10. Worked Example 6.7 Mole Ratios: Mass to Mole/Mole to Mass Conversions The 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.) Analysis 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. Ballpark Estimate 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 . Solution 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 will need three conversion factors.

  11. 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 . STEP 4:Solve. Ballpark Check The calculated answer (0.019 g) is consistent with our estimate (0.02 g).

  12. Worked Example 6.8 Percent Yield 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 . Analysis The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. Ballpark Estimate 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%. Solution Ballpark Estimate The calculated percent yield agrees very well with our estimate of 85%.

  13. 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. 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. Solution 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 B produced, assuming complete conversion of a given reactant. is the limiting reagent, since complete conversion of this reagent yields less product (67.6 mol B formed) than does complete conversion of Mg (98.0 mol B formed).

  14. 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.

  15. Worked Example 6.10 Mass to Mole Conversion: Percent Yield 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. ) Analysis 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. Ballpark Estimate 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. Solution The theoretical yield of ethyl alcohol is: and so the actual yield is: 41.1 g ethyl alc. × 0.785 = 32.3 g ethyl alcohol Ballpark Check The calculated result (32.3 g) is close to our estimate (34 g).

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