Chapter 7 Notes: Chemical Formulas, Formula Masses, Molar Masses, and Percentage Composition

Chemical Formulas

• A chemical formula indicates:

  • The elements present in a compound.

  • The relative number of atoms or ions of each element present in a compound.

• Chemical formulas allow chemists to calculate:

  • Formula mass

  • Molar mass

  • Percentage composition

Formula Masses

• The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula.

• Example: Formula mass of water, $H_2O$

  • Average atomic mass of H: $1.01 <br>ewline amu$

  • Average atomic mass of O: $16.00 <br>ewline amu$

  • Average mass of $H_2O$ molecule: $18.02 <br>ewline amu$

• The mass of a water molecule can be referred to as a molecular mass.

• The mass of one formula unit of an ionic compound, such as NaCl, is not a molecular mass.

• The mass of any unit represented by a chemical formula ($H_2O$, NaCl) can be referred to as the formula mass.

Sample Problem F: Formula Mass of Potassium Chlorate ($KClO_3$)

• Problem: Find the formula mass of potassium chlorate, $KClO_3$.

• Solution: The mass of a formula unit of $KClO_3$ is found by adding the masses of one K atom, one Cl atom, and three O atoms.

  • Atomic masses are found in the periodic table.

  • Round each atomic mass to two decimal places.

• formula mass of $KClO_3$ = $122.55 <br>ewline amu$

Molar Masses

• The molar mass of a substance is equal to the mass in grams of one mole, or approximately $6.022 × 10^{23}$ particles, of the substance.

• Example: The molar mass of pure calcium, Ca, is $40.08 <br>ewline g/mol$ because one mole of calcium atoms has a mass of $40.08 <br>ewline g$.

• The molar mass of a compound is calculated by adding the masses of the elements present in a mole of the molecules or formula units that make up the compound.

• One mole of water molecules contains exactly two moles of H atoms and one mole of O atoms.

  • Molar mass of $H_2O$ molecule: $18.02 <br>ewline g/mol$

• A compound’s molar mass is numerically equal to its formula mass.

Calculating Molar Masses for Ionic Compounds

• $ZnCl_2$: $1 <br>ewline Zn^{2+}$ + $2 <br>ewline Cl^-$ = $(1 × 65.39 <br>ewline g/mol) + (2 × 35.45 <br>ewline g/mol) = 65.39 <br>ewline g/mol + 70.90 <br>ewline g/mol = 136.29 <br>ewline g/mol$

• $ZnSO<em>4$: $1 ewline Zn^{2+}$ + $SO</em>4^{2-}$= $(1 × 65.39 <br>ewline g/mol) + (1 × 32.07 <br>ewline g/mol) + (4 × 16.00 <br>ewline g/mol) = 65.39 <br>ewline g/mol + 32.07 <br>ewline g/mol + 64.00 <br>ewline g/mol = 161.46 <br>ewline g/mol$

• $(NH<em>4)</em>2SO<em>4$: $2 ewline NH</em>4^+$ + $SO_4^{2-}$ = $(2 × 14.01 <br>ewline g/mol) + (8 × 1.01 <br>ewline g/mol) + (1 × 32.07 <br>ewline g/mol) + (4 × 16.00 <br>ewline g/mol) = 28.02 <br>ewline g/mol + 8.08 <br>ewline g/mol + 32.07 <br>ewline g/mol + 64.00 <br>ewline g/mol = 132.17 <br>ewline g/mol$

Sample Problem G: Molar Mass of Barium Nitrate ($Ba(NO<em>3)</em>2$)

• Problem: What is the molar mass of barium nitrate, $Ba(NO<em>3)</em>2$?

• Solution: One mole of barium nitrate contains one mole of Ba, two moles of N (1 × 2), and six moles of O (3 × 2).

• Molar mass of $Ba(NO<em>3)</em>2$ = $261.35 <br>ewline g/mol$

Molar Mass as a Conversion Factor

• The molar mass of a compound can be used as a conversion factor to relate an amount in moles to a mass in grams for a given substance.

• To convert moles to grams, multiply the amount in moles by the molar mass:

  • Amount in moles × molar mass (g/mol) = mass in grams

Mole-Mass Calculations

• Grams to Moles: Mass of compound (grams) × $\frac{1}{molar <br>ewline mass <br>ewline of <br>ewline compound}$ = Amount of compound (moles)

• Moles to Grams: Amount of compound (moles) × molar mass of compound = Mass of compound (grams)

• Moles of element in Compound : Amount of compound (moles) × $\frac{molar <br>ewline mass <br>ewline of <br>ewline element}{molar <br>ewline mass <br>ewline of <br>ewline compound}$ = Amount of element in compound (moles)

• Grams of element in compound: Mass of compound (grams) × $\frac{molar <br>ewline mass <br>ewline of <br>ewline element}{molar <br>ewline mass <br>ewline of <br>ewline compound}$ = mass of element in compound (grams)

• Molecules or formula units of compound : Amount of compound (moles) × $6.022 × 10^{23}$ = Number of molecules or formula units of compound

• Number of atoms of element in compound: Amount of element in compound (moles) × $6.022 × 10^{23}$ = Number of atoms of element in compound

Sample Problem H: Grams from Moles of Oxygen Gas ($O_2$)

• Problem: What is the mass in grams of $2.50 <br>ewline mol$ of oxygen gas?

• Given: $2.50 <br>ewline mol <br>ewline O_2$

• Unknown: Mass of $O_2$ in grams

• Solution: moles $O<em>2$ → grams $O</em>2$

  • Amount of $O<em>2$ (mol) × molar mass of $O</em>2$ (g/mol) = mass of $O_2$ (g)

• Calculate the molar mass of $O_2$.

• Use the molar mass of $O_2$ to convert moles to mass.

Converting Between Amount in Moles and Number of Particles

• Steps:

  1. Decide which quantity you are given: amount (in moles) or number of particles (in atoms, molecules, formula units, or ions).

  2. If you are converting from amount to number of particles (going left to right), use the top conversion factor.

  3. If you are converting from number of particles to amount (going right to left), use the bottom conversion factor.
     *Conversion factors:
     *amount → number of particles: amount × $\frac{6.022 x 10^{23} <br>ewline particles}{1 <br>ewline mol}$

  • number of particles → amount: number of particles × $\frac{1 <br>ewline mol}{6.022 x 10^{23} <br>ewline particles}$

Sample Problem I: Ibuprofen ($C<em>{13}H</em>{18}O_2$)

• Problem: Ibuprofen, $C<em>{13}H</em>{18}O_2$, is the active ingredient in many nonprescription pain relievers. Its molar mass is $206.31 ewline g/mol$.

  • a. If the tablets in a bottle contain a total of $33 <br>ewline g$ of ibuprofen, how many moles of ibuprofen are in the bottle?

  • b. How many molecules of ibuprofen are in the bottle?

  • c. What is the total mass in grams of carbon in $33 <br>ewline g$ of ibuprofen?

• Given: $33 <br>ewline g$ of $C<em>{13}H</em>{18}O_2$, molar mass $206.31 <br>ewline g/mol$

• Unknown:

  • a. moles $C<em>{13}H</em>{18}O_2$

  • b. molecules $C<em>{13}H</em>{18}O_2$

  • c. total mass of C

• Solution:

  • a. grams → moles

  • b. moles → molecules

  • c. moles $C<em>{13}H</em>{18}O_2$ → moles C → grams C

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Percentage Composition

• It is often useful to know the percentage by mass of a particular element in a chemical compound.

• To find the mass percentage of an element in a compound, the following equation can be used.

• The mass percentage of an element in a compound is the same regardless of the sample’s size.

• The percentage of an element in a compound can be calculated by determining how many grams of the element are present in one mole of the compound.

• The percentage by mass of each element in a compound is known as the percentage composition of the compound.

Percentage Composition of Iron Oxides

• Iron(III) oxide, $Fe<em>2O</em>3$:

  • Iron: $69.9 <br>ewline$

  • Oxygen: $30.1 <br>ewline$

• Iron(II) oxide, FeO:

  • Iron: $77.7 <br>ewline$

  • Oxygen: $22.3 <br>ewline$

Percentage Composition Calculations

• Percentage by mass = $\frac{mass <br>ewline of <br>ewline each <br>ewline element (units: g)}{total <br>ewline mass} × 100$

• Amount of each element (moles) × molar mass (g) = mass of each element (g)

• Empirical formula subscripts: obtained from percentage composition without conversion necessary

• Percentage composition: obtained from empirical formula using $\frac{x <br>ewline molar <br>ewline mass (g)}{smallest <br>ewline amount} × 1 <br>ewline mole$

Sample Problem J: Percentage Composition of Copper(I) Sulfide ($Cu_2S$)

• Problem: Find the percentage composition of copper(I) sulfide, $Cu_2S$.

• Given: Formula, $Cu_2S$

• Unknown: Percentage composition of $Cu_2S$

• Solution: Formula → molar mass → mass percentage of each element

• Molar mass of $Cu_2S = 159.2 <br>ewline g$

• Percentage of Cu $\frac{127.1 <br>ewline g <br>ewline Cu}{159.2 <br>ewline g <br>ewline Cu_2S} × 100 = 79.85$

• Percentage of S $\frac{32.07 <br>ewline g <br>ewline S}{159.2 <br>ewline g <br>ewline Cu_2S} × 100 = 20.15$