General Chemistry I: Composition of Compounds and Chemical Reactions

Composition of Compounds

Mass Percent Composition

Understanding the composition of compounds is fundamental in chemistry, as it allows us to quantify the amount of each element present in a substance. Mass percent composition (or mass percent) expresses the percentage by mass of each element in a compound.

• Definition: The mass percent of an element in a compound is the fraction of the element's mass relative to the total mass of the compound, multiplied by 100%.

• Formula:

$\text{Mass percent of element X} = \frac{\text{mass of element X in 1 mol of compound}}{\text{mass of 1 mol of the compound}} \times 100\%$

$\text{Mass \% element} = \frac{(\text{number of atoms})(\text{atomic weight})}{\text{formula weight (FW) of the compound}} \times 100\%$

• Example: For the compound CCl2F2, the mass percent of Cl is calculated as:

$\text{Mass percent Cl} = \frac{2 \times \text{molar mass Cl}}{\text{molar mass CCl}_2\text{F}_2} \times 100\%$

Determining Empirical Formulas from Mass Percent

Empirical formulas represent the simplest whole-number ratio of atoms in a compound. They can be determined from mass percent data obtained through laboratory analysis.

• Steps to Determine Empirical Formula:

  1. Convert the mass of each element to moles using their atomic masses.

  2. Write a pseudoformula using the mole amounts as subscripts.

  3. Divide all subscripts by the smallest number of moles to get the simplest ratio.

  4. If necessary, multiply all subscripts by a small whole number to obtain whole numbers.

• Example: A sample of water produces 2.02 g of hydrogen and 16.00 g of oxygen. Convert masses to moles, write the pseudoformula, and simplify to obtain H2O.

Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecule and is a whole-number multiple of the empirical formula.

• Formula:

$\text{Molecular formula} = (\text{Empirical formula})_n$

$\text{Molar mass} = n \times \text{Empirical formula molar mass}$

• Example: Fructose has an empirical formula CH2O and a molar mass of 180.2 g/mol. The empirical formula mass is 30.03 g/mol, so:

$n = \frac{180.2}{30.03} = 6$

$\text{Molecular formula} = (\text{CH}_2\text{O})_6 = \text{C}_6\text{H}_{12}\text{O}_6$

Combustion Analysis

Combustion analysis is a laboratory technique used to determine the empirical formula of compounds containing carbon, hydrogen, and oxygen. The compound is burned in oxygen, and the masses of CO2 and H2O produced are measured.

• Steps:

  1. Convert the masses of CO2 and H2O to moles.

  2. Determine the moles of C and H from the moles of CO2 and H2O.

  3. If the compound contains other elements (e.g., O), subtract the masses of C and H from the original sample mass to find the mass of the other element.

  4. Convert all masses to moles and write the empirical formula.

• Example: Combustion of a compound yields 1.83 g CO2 and 0.901 g H2O. Calculate moles of C and H, then determine the empirical formula.

$\text{Moles of CO}_2 = \frac{1.83}{44.01} = 0.0416$ $\text{Moles of H}_2\text{O} = \frac{0.901}{18.02} = 0.0500$ $\text{Moles of C} = 0.0416$ $\text{Moles of H} = 0.100$

Divide by the smallest value to get the ratio, and multiply to obtain whole numbers if necessary.

Chemical Reactions

Introduction to Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of chemical bonds. The substances that react are called reactants, and the substances formed are called products.

• Chemical Equation: A symbolic representation of a chemical reaction.

• Example: $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$

State Symbols in Chemical Equations

The physical state of each reactant and product is indicated by symbols in parentheses:

• (s): solid

• (l): liquid

• (g): gas

• (aq): aqueous solution

Example: $\text{Fe}(s) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{FeOH}(aq) + \text{H}_2(g)$

Balancing Chemical Equations

Balancing chemical equations ensures the law of conservation of mass is obeyed: the number of atoms of each element must be the same on both sides of the equation.

• Steps:

  1. Write the skeletal equation.

  2. Balance the atoms in the most complex substance first.

  3. Balance the remaining atoms, usually starting with elements that appear in only one reactant and one product.

  4. Adjust coefficients as needed; never change subscripts.

  5. If fractional coefficients are used, multiply all coefficients by the denominator to obtain whole numbers.

• Example: Combustion of butane:

$\text{C}_4\text{H}_{10}(g) + \frac{13}{2}\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 5\text{H}_2\text{O}(g)$ Multiply all coefficients by 2: $2\text{C}_4\text{H}_{10}(g) + 13\text{O}_2(g) \rightarrow 8\text{CO}_2(g) + 10\text{H}_2\text{O}(g)$

Subscripts vs. Coefficients

In chemical equations:

• Subscripts indicate the number of atoms of each element in a molecule (e.g., H2O has 2 H and 1 O).

• Coefficients indicate the number of molecules (e.g., 2H2O means 2 molecules of water).

Sample Balanced Equations

• $\text{Fe} + \text{O}_2 \rightarrow \text{FeO}_2$

• $\text{Fe}_2\text{O}_3 + 6\text{HCl} \rightarrow 2\text{FeCl}_3 + 3\text{H}_2\text{O}$

• $\text{FeCl}_3 + 3\text{NaOH} \rightarrow \text{Fe(OH)}_3 + 3\text{NaCl}$

• $2\text{Fe(OH)}_3 + 3\text{H}_2\text{SO}_4 \rightarrow \text{Fe}_2(\text{SO}_4)_3 + 6\text{H}_2\text{O}$

• $\text{Fe}_2(\text{SO}_4)_3 + 6\text{Ba(NO}_3)_2 \rightarrow 2\text{Fe(NO}_3)_3 + 3\text{BaSO}_4$

Summary Table: Key Concepts

Concept	Definition	Formula/Example
Mass Percent Composition	Percentage by mass of each element in a compound	$\frac{\text{mass of element}}{\text{mass of compound}} \times 100\%$
Empirical Formula	Simplest whole-number ratio of atoms	H2O, CH2O
Molecular Formula	Actual number of atoms in a molecule	C6H12O6 (fructose)
Combustion Analysis	Lab technique to determine empirical formula	Measure CO2 and H2O produced
Chemical Equation	Symbolic representation of a chemical reaction	$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$
Balancing Equations	Ensuring equal number of atoms on both sides	$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$

Additional info: Some context and examples have been expanded for clarity and completeness, including stepwise procedures and sample calculations.