Chapter 6 Notes: Chemical Reactions - Mole and Mass Relationships

Chapter 6: Chemical Reactions - Mole and Mass Relationships

6.1 The Mole and Avogadro’s Number

Atomic Weight: The average mass of an element’s atoms.
Molecular Weight (MW): The average mass of a substance’s molecules.
Molecular weight (or formula weight for ionic compounds) is the sum of the atomic weights of all atoms in the molecule or formula unit.
- Example: Ethylene (C2H4) has a molecular weight of 28.0 amu.
Samples of different substances contain the same number of molecules or formula units when their mass ratio equals their molecular or formula weight ratio.
Mole: The amount of a substance whose mass in grams is numerically equal to its molecular or formula weight.
Molar Mass: The mass in grams of 1 mole of a substance, numerically equal to its molecular weight.
One mole of any substance contains $6.022 × 10^{23}$ formula units (Avogadro’s number $N_A$ ).

Example 6.1: Pseudoephedrine Hydrochloride

Pseudoephedrine hydrochloride (C10H16ClNO) is a nasal decongestant.
- (a) What is its molar mass?
- (b) How many molecules are in a 30.0 mg tablet?
Analysis: Convert mass to number of molecules using molar mass and Avogadro’s number.
Solution:
- (a) Molar mass calculation:
  - C: 10 × 12.01 amu = 120.1 amu
  - H: 16 × 1.01 amu = 16.16 amu
  - Cl: 1 × 35.45 amu = 35.45 amu
  - N: 1 × 14.01 amu = 14.01 amu
  - O: 1 × 16.00 amu = 16.00 amu
  - Total: 201.72 amu
- (b) Unit conversions:
  - Step 1: Known information: 30.0 mg pseudoephedrine hydrochloride
  - Step 2: Answer and units: molecules of pseudoephedrine hydrochloride
  - Step 3: Conversion factors: 201.70 g contains $6.022 × 10^{23}$ molecules; convert 30 mg to grams.
  - Step 4: Solve. Set up an equation so that unwanted units cancel.
    $30.0 \,\text{mg} \times \frac{1 \,\text{g}}{1000 \,\text{mg}} \times \frac{1 \,\text{mol}}{201.70 \,\text{g}} \times \frac{6.022 \times 10^{23} \,\text{molecules}}{1 \,\text{mol}} = 8.96 \times 10^{19} \,\text{molecules}$

Example 6.2: Pencil Mark

A tiny pencil mark contains about $3 × 10^{17}$ atoms of carbon. What is the mass in grams?
Analysis: Convert number of atoms to mass using Avogadro’s number and atomic weight.
Solution:
- Step 1: Known information: $3 × 10^{17}$ atoms of carbon
- Step 2: Answer and units: Mass of carbon = ?? g
- Step 3: Conversion factors: 12.01 g of carbon contains $6.022 × 10^{23}$ atoms.
- Step 4: Solve. Set up an equation using the conversion factors so that unwanted units cancel.
  $3 \times 10^{17} \,\text{atoms} \times \frac{12.01 \,\text{g}}{6.022 \times 10^{23} \,\text{atoms}} = 6.0 \times 10^{-6} \,\text{g}$

6.2 Gram-Mole Conversions

Molar mass = Mass of 1 mole of substance = Mass of $6.022 × 10^{23}$ molecules = Molecular weight in grams
Molar mass serves as a conversion factor between numbers of moles and mass.

Example 6.3: Ibuprofen in Pain Relievers

Advil and Nuprin contain ibuprofen (C13H18O2, MW = 206.3 amu). If a bottle contains 0.082 mol of ibuprofen, what is the mass in grams?
Analysis: Convert moles to mass using molar mass.
Solution:
- Step 1: Known information: 0.082 mol ibuprofen
- Step 2: Answer and units: mass ibuprofen = ?? g
- Step 3: Conversion factor: 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.
  $0.082 \,\text{mol} \times \frac{206.3 \,\text{g}}{1 \,\text{mol}} = 17 \,\text{g}$

Example 6.4: Sodium Hydrogen Phosphate Dose

The maximum daily dose of sodium hydrogen phosphate (Na2HPO4, MW = 142.0 amu) as a laxative is 3.8 g. How many moles of sodium hydrogen phosphate, how many moles of Na+ ions, and how many total moles of ions are in this dose?
Analysis: Convert mass to moles using molar mass. Each formula unit contains 2 Na+ ions and 1 $HPO_4^{2-}$ ion.
Solution:
- Step 1: Known information: 3.8 g Na2HPO4; MW = 142.0 amu
- Step 2: Answer and units:
 - Moles of Na2HPO4 = ?? mol
 - Moles of $Na^+$ ions = ?? mol
 - Total moles of ions = ?? mol
- Step 3: Conversion factor: Use the molecular weight of Na2HPO4 to convert from grams to moles.
- Step 4: Solve.
 $3.8 \,\text{g} \times \frac{1 \,\text{mol}}{142.0 \,\text{amu}} = 0.027 \,\text{mol Na}2\text{HPO}4$
 $0.027 \,\text{mol Na}2\text{HPO}4 \times \frac{2 \,\text{mol Na}^+}{1 \,\text{mol Na}2\text{HPO}4} = 0.054 \,\text{mol Na}^+$
 $0.027 \,\text{mol Na}2\text{HPO}4 \times \frac{1 \,\text{mol HPO}4^{2-}}{1 \,\text{mol Na}2\text{HPO}4} = 0.027 \,\text{mol HPO}4^{2-}$
 $0.054 + 0.027 = 0.081 \,\text{mol ions}$

6.3 Mole Relationships and Chemical Equations

The mole specifies the relationship between reactants and products in chemical reactions.
Coefficients in a chemical equation indicate how many molecules/moles of each reactant are needed and how many of each product are formed.
Mole ratios from coefficients act as conversion factors.
- For the synthesis of ammonia: 3 H2 + N2 → 2 NH3, the mole ratios are:
 $\frac{2 \,\text{mol NH}3}{3 \,\text{mol H}2}, \frac{2 \,\text{mol NH}3}{1 \,\text{mol N}2} , \frac{3 \,\text{mol H}2}{1 \,\text{mol N}2}$

Example 6.5: Rusting of Iron

Rusting: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
- (a) What are the mole ratios of product to each reactant and reactants to each other?
- (b) How many moles of Fe2O3 are formed from 6.2 mol of iron?
Analysis: Coefficients represent mole ratios.
Solution:
- (a) Mole ratios:
 $\frac{2 \,\text{mol Fe}2\text{O}3}{4 \,\text{mol Fe}}, \frac{2 \,\text{mol Fe}2\text{O}3}{3 \,\text{mol O}2}, \frac{4 \,\text{mol Fe}}{3 \,\text{mol O}2}$
- (b) Moles of Fe2O3 formed:
 $6.2 \,\text{mol Fe} \times \frac{2 \,\text{mol Fe}2\text{O}3}{4 \,\text{mol Fe}}= 3.1 \,\text{mol Fe}2\text{O}3$

6.4 Mass Relationships and Chemical Equations

Coefficients represent molecule-to-molecule or mole-to-mole relationships.
Actual amounts are weighed in grams.
Three types of conversions are needed:
- Mole-to-mole: Use mole ratios.
- Mole-to-mass and mass-to-mole: Use molar mass.
- Mass-to-mass: Cannot be carried out directly; must convert to moles first
Four steps for determining mass relationships:
- Step 1: Write the balanced chemical equation.
- Step 2: Choose molar masses and mole ratios to convert known to needed information.
- Step 3: Set up the factor-label expression.
- Step 4: Calculate the answer and check against ballpark estimate.

Example 6.6: Nitrogen Dioxide and Acid Rain

3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
How many grams of HNO3 are produced for every 1.0 mole of NO2 that reacts? (MW of HNO3 = 63.0 amu)
Analysis: Convert moles of reactant to mass of product through moles.
Solution:
- Step 1: Balanced equation: 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g)
- Step 2: Conversion factors: mole ratio of HNO3 to NO2, molar mass of HNO3 (63.0 g/mol)
- Step 3: Set up factor labels. Identify appropriate mole ratio factor labels to convert moles NO2 to moles HNO3 and moles HNO3 to grams.
- Step 4: Solve:
 $1.0 \,\text{mol NO}2 \times \frac{2 \,\text{mol HNO}3}{3 \,\text{mol NO}2} \times \frac{63.0 \,\text{g HNO}3}{1 \,\text{mol HNO}3}= 42 \,\text{g HNO}3$

Example 6.7: Calcium Oxalate Reaction

CaCl2(aq) + Na2C2O4(aq) → CaC2O4(s) + 2 NaCl(aq)
0.022 g of calcium oxalate (CaC2O4) is produced. What mass of calcium chloride was used as reactant? (Molar mass of CaC2O4 = 128.1 g/mol, CaCl2 = 111.0 g/mol)
Analysis: Mass-to-mass conversion: convert mass of CaC2O4 to moles, use mole ratio to find moles of CaCl2, then convert to mass.
Solution:
- Step 1: Balanced equation: CaCl2(aq) + Na2C2O4(aq) → CaC2O4(s) + 2 NaCl(aq)
- Step 2: Conversion factors: Convert the mass of CaC2O4 into moles, use a mole ratio to find moles of CaCl2, and convert the number of moles of CaCl2 to mass. We will need three conversion factors.
- Step 3: Set up factor-labels. We will need to perform gram to mole and mole to mole conversions to get from grams CaC2O4 to grams CaCl2.
- Step 4: Solve:
 $0.022 \,\text{g CaC}2\text{O}4 \times \frac{1 \,\text{mol CaC}2\text{O}4}{128.1 \,\text{g CaC}2\text{O}4} \times \frac{1 \,\text{mol CaCl}2}{1 \,\text{mol CaC}2\text{O}4} \times \frac{111.0 \,\text{g CaCl}2}{1 \,\text{mol CaCl}2} = 0.019 \,\text{g CaCl}2$

6.5 Limiting Reagent and Percent Yield

Rarely are all reactants converted to products.
Limiting reagent: The reactant that runs out first.
Theoretical yield: The amount of product formed assuming complete reaction of the limiting reagent.
Actual yield: The amount of product actually formed.
Percent yield: $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

Example 6.8: Combustion of Acetylene

2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(g)
26.0 g of acetylene is burned with sufficient oxygen. The theoretical yield of CO2 is 88.0 g. If the actual yield is 72.4 g CO2, calculate the percent yield.
Analysis: Percent yield = (Actual yield / Theoretical yield) × 100
Solution:
$\frac{72.4 \,\text{g CO}2}{88.0 \,\text{g CO}2} \times 100 = 82.3 \%$

Example 6.9: Production of Boron

B2O3(l) + 3 Mg(s) → 2 B(s) + 3 MgO(s)
What is the theoretical yield of boron when 2350 g of boric oxide is reacted with 3580 g of magnesium? (Molar masses: B2O3 = 69.6 g/mol, Mg = 24.3 g/mol)
Analysis: Identify the limiting reagent and calculate the theoretical yield from it.
Solution:
- Step 1: Known information: 2350 g B2O3 (69.6 g/mol), 3580 g Mg (24.3 g/mol)
- Step 2: Answer and units: Theoretical mass of B = ?? g
- Step 3: Conversion factors:
 - Convert masses to moles. Then use mole ratios to find moles of B produced from each reactant to identify the limiting reagent.
 - Moles of $B2O3$ :
 $2350 \,\text{g B}2\text{O}3 \times \frac{1 \,\text{mol B}2\text{O}3}{69.9 \,\text{g}} = 33.8 \,\text{mol B}2\text{O}3$
 - Moles of $B$ from $B2O3$ :
 $33.8 \,\text{mol B}2\text{O}3 \times \frac{2 \,\text{mol B}}{1 \,\text{mol B}2\text{O}3} = 67.6 \,\text{mol B}$
 - Moles of $Mg$ :
 $3580 \,\text{g Mg} \times \frac{1 \,\text{mol Mg}}{24.3 \,\text{g}} = 147.3 \,\text{mol Mg}$
 - Moles of $B$ from $Mg$ :
 $147.3 \,\text{mol Mg} \times \frac{2 \,\text{mol B}}{3 \,\text{mol Mg}} = 98.2 \,\text{mol B}$
 - $B2O3$ is the limiting reagent
- Step 4: Solve. Once the limiting reagent has been identified ( $B2O3$ ), the theoretical amount of B that should be formed can be calculated using a mole to mass conversion.
 $67.6 \,\text{mol B} \times \frac{10.8 \,\text{g B}}{1 \,\text{mol B}} = 730 \,\text{g B}$

Example 6.10: Reaction of Ethylene with Water

H2C=CH2 + H2O → CH3CH2OH (78.5% actual yield)
How many grams of ethyl alcohol are formed by reaction of 25.0 g of ethylene? (MW: ethylene = 28.0 amu, ethyl alcohol = 46.0 amu)
Analysis: Find the theoretical yield, then multiply by 78.5%.
Solution: The theoretical yield of ethyl alcohol is
$25.0 \,\text{g C}2\text{H}4 \times \frac{1 \,\text{mol C}2\text{H}4}{28.0 \,\text{g C}2\text{H}4} \times \frac{1 \,\text{mol C}2\text{H}5\text{OH}}{1 \,\text{mol C}2\text{H}4} \times \frac{46.0 \,\text{g C}2\text{H}5\text{OH}}{1 \,\text{mol C}2\text{H}5\text{OH}} = 41.1 \,\text{g C}2\text{H}5\text{OH}$
So, the actual yield is as follows:
$41.1 \,\text{g C}2\text{H}5\text{OH} \times 0.785 = 32.2 \,\text{g C}2\text{H}5\text{OH}$