Calculations with Chemical Formulas, Moles, and Composition

Definition and Notation
- A chemical formula represents the exact ratio of atoms for each element present in a compound.
- Subscripts located to the right of an elemental symbol denote the number of atoms of that element in the formula.
- The subscript "1" is always assumed and is not explicitly written. For example, in $H_2O$ , there are 2 hydrogen atoms and 1 oxygen atom.
Formula Mass and Molar Mass
- The formula mass is the sum of the atomic weights of every atom represented in the chemical formula.
- Molar mass is typically expressed in grams per mole ( $g/mol$ ).
- Example calculation for Ethane ( $C_2H_6$ ):
  - Sum: $2(C) + 6(H)$
  - Calculation: $2(12.011) + 6(1.0079) = 30.07\,g/mol$
Practice with Formula Masses
- Calculation of the formula mass of Sucrose ( $C_{12}H_{22}O_{11}$ ).
- Freons: These compounds contain carbon, chlorine, and fluorine. While useful, they deplete the ozone layer. In 1991, two replacement compounds were produced: ( $CH_2FCF_3$ ) and ( $CHClFCF_3$ ). Calculations require determining the molar masses for both structures.

The Mole as a Counting Unit
- The mole is a standard unit used to express the amount of a substance.
- It represents a specific quantity: exactly $6.022 \times 10^{23}$ pieces of any object. This is known as Avogadro’s number.
- Scale analogies for one mole:
  - A mole of basketballs would create a pile the size of the Earth.
  - A mole of doughnuts would cover the Earth's surface in a layer 5 miles deep.
- In chemistry, the mole facilitates the counting of atoms, molecules, ions, or formula units, serving as the bridge between the microscopic world (atoms) and the macroscopic world (grams and liters).
Historical Background and Definition
- Amedeo Avogadro (1776–1856) proposed in 1811 that equal volumes of gases at the same temperature and pressure contain an equal number of particles.
- The specific numeric value of the mole was established later through experiments involving mass, charge, and atomic theory (electrochemistry and X-ray crystallography).
- The definition of the mole was officially fixed in 2019 as precisely $6.022 \times 10^{23}$ entities per mole.

Linking Mass to Moles
- Moles cannot be measured directly; instead, mass is measured using a balance.
- Molar mass serves as the conversion factor between mass and moles.
- Numerically, the mass of one mole of an element's atoms (in grams) is equal to its atomic mass on the periodic table. For example, the atomic mass of carbon is $12.011\,g$ .
Practice Conversion Problems
- Determining the number of atoms in $3.75\,moles$ of Silver ( $Ag$ ).
- Determining the number of atoms in $158\,g$ of Phosphorus ( $P$ ).
- Calculating moles in $449\,g$ of Potassium ( $K$ ).
- Calculating moles in $2.16 \times 10^{24}$ atoms of Lead ( $Pb$ ).
- Determining the mass of $1.9 \times 10^{24}$ atoms of Zinc ( $Zn$ ).
- Determining the mass of $4.77\,moles$ of Calcium ( $Ca$ ).

General Principle
- Percent composition is determined by expressing the mass of each individual element as a percentage of the total mass of the compound.
- Formula: $\text{Percent by mass} = \left( \frac{\text{mass of element}}{\text{total mass of compound}} \right) \times 100$
Application Examples
- Nitrogen in Calcium Nitrate ( $Ca(NO_3)_2$ ).
- Copper in the superconductor discovered in 1987 ( $YBa_2Cu_3O_7$ ), which functions above the temperature of liquid nitrogen ( $77\,K$ ).
- Ranking substances by increasing mass percent of Carbon:
  - Caffeine ( $C_8H_{10}N_4O_2$ )
  - Sucrose ( $C_{12}H_{22}O_{11}$ )
  - Ethanol ( $C_2H_5OH$ )
Case Study: Hemoglobin
- Hemoglobin is the oxygen-transport protein in mammals.
- It contains $0.347\%$ Iron ( $Fe$ ) by mass.
- Each hemoglobin molecule contains exactly four iron atoms. This data is used to calculate the total molar mass of hemoglobin.

Definition
- The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.
- Different compounds can share the same empirical formula. For example, Benzene ( $C_6H_6$ ), Ethyne ( $C_2H_2$ ), and 1,3,5,7-cyclooctatetraene ( $C_8H_8$ ) all have different molecular formulas but share the same empirical formula: $CH$ (a 1:1 ratio).
Standard Calculation Procedure
- 1. If given percentages, assume a $100\,g$ sample to convert percentages directly to grams.
- 2. Convert the mass of each element into moles using atomic weights.
- 3. Identify the smallest molar value and divide all results by that number to find the mole ratio.
- 4. If the resulting ratio includes a decimal representing a fraction (e.g., $.500$ , $.333$ , $.250$ ), multiply all values by the appropriate integer (2, 3, or 4) to achieve a whole-number ratio.
Practice Problems
- Determining the empirical formula for a compound with $40.1\%$ Carbon, $6.60\%$ Hydrogen, and $53.3\%$ Oxygen.
- Calculating empirical formulas for Iron Oxides:
  - (a) $77.78\%\,Fe$ , $22.22\%\,O$
  - (b) $70.00\%\,Fe$ , $30.00\%\,O$
  - (c) $72.40\%\,Fe$ , $27.60\%\,O$
- Identifying valid empirical formulas from a list: $CH$ , $CH_2$ , $C_4H_4$ , $CH_3O$ , $C_3H_6O_2$ .
- Nylon-6 analysis: $63.68\%$ Carbon, $12.38\%$ Nitrogen, $9.80\%$ Hydrogen, and $14.14\%$ Oxygen.
- Gold and Oxygen compound: $89.14\%\,Au$ and $10.80\%\,O$ .

Technique for Organic Compounds
- This method is primarily used for compounds containing Carbon and Hydrogen.
- The sample is burned in an excess of Oxygen gas ( $O_2$ ).
- Combustion products, Carbon Dioxide ( $CO_2$ ) and Water ( $H_2O$ ), are trapped and weighed separately.
Deduction of Formula
- All Carbon in the resulting $CO_2$ is assumed to have originated from the original sample.
- All Hydrogen in the resulting $H_2O$ is assumed to have originated from the original sample.
- Knowing the masses of $CO_2$ and $H_2O$ allows for the calculation of the moles of Carbon and Hydrogen, leading to the empirical formula.
Practice Problems
- A $1.500\,g$ hydrocarbon sample produces $4.400\,g$ of $CO_2$ and $2.700\,g$ of $H_2O$ .
- A $35.0\,mg$ sample containing C, H, and N produces $33.5\,mg$ of $CO_2$ and $41.1\,mg$ of $H_2O$ .

Relation to Empirical Formula
- The molecular formula indicates the exact number of atoms of each element in a molecule.
- It is always a whole-number multiple of the empirical formula.
- For an empirical formula of $CH_2O$ (mass near $30\,g/mol$ ), possible molecular formulas include $C_2H_4O_2$ ( $2 \times$ ) or $C_3H_6O_3$ ( $3 \times$ ).
Calculation Method
- To find the molecular formula, divide the experimental molar mass of the compound by the molar mass of the empirical formula. Multiply the subscripts of the empirical formula by the resulting integer.
Practice Problems
- A hydrocarbon is $7.690\%\,H$ and $92.31\%\,C$ . Its molar mass is $78.00\,g/mol$ . Determine the empirical and molecular formulas.
- Given an empirical formula of $CH_2O$ and a molar mass of $180$ , determine the molecular formula.
- A $3.585\,g$ sample contains $1.388\,g\,C$ , $0.345\,g\,H$ , and $1.850\,g\,O$ with a molar mass of $62\,g$ . Determine the molecular formula.
- A compound contains $26.7\%\,P$ , $12.1\%\,N$ , and $61.2\%\,Cl$ with a molar mass of $580\,g/mol$ . Determine the molecular formula.