Stoichiometry and Empirical/Molecular Formulas: Quantitative Information from Balanced Equations

3.6 Quantitative Information from Balanced Equations

Stoichiometric Relationships

Balanced chemical equations provide quantitative relationships between reactants and products. The coefficients in a balanced equation indicate the relative number of molecules or moles involved in the reaction.

• Stoichiometry is the calculation of reactants and products in chemical reactions using balanced equations.

• Avogadro's number (6.022 × 1023) allows conversion between number of molecules and moles.

• The coefficients in a balanced equation represent the number of moles of each substance.

Example:

• For the reaction: 2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s), the coefficients indicate that 2 moles of iron react with 3 moles of chlorine to produce 2 moles of iron(III) chloride.

Using Mole Ratios as Conversion Factors

Mole ratios derived from balanced equations allow conversion between amounts of different substances in a reaction.

• Mole ratios act as conversion factors to relate moles of one compound to moles of another.

• This is the basis of stoichiometric calculations.

Example 3.9: How many moles of CO2 are produced when 1 mol of butane (C4H10) is burned?

• Balanced equation: 2 C4H10 + 13 O2 → 8 CO2 + 10 H2O

• Calculation: $mol_{CO_2} = mol_{C_4H_{10}} \left( \frac{8\ mol\ CO_2}{2\ mol\ C_4H_{10}} \right)$ $= 1\ mol\ C_4H_{10} \left( \frac{8\ mol\ CO_2}{2\ mol\ C_4H_{10}} \right) = 4\ mol\ CO_2$

Example 3.10: How much ammonia is produced when 1.00 g of ammonium bicarbonate is heated and decomposes?

• Reaction: NH4HCO3 → NH3 + H2O + CO2

• Calculation: $m_{NH_3} = m_{NH_4HCO_3} \left( \frac{1\ mol\ NH_4HCO_3}{79.056\ g} \right) \left( \frac{1\ mol\ NH_3}{1\ mol\ NH_4HCO_3} \right) (17.031\ g)$ $m_{NH_3} = 1.00\ g\ NH_4HCO_3 \left( \frac{1\ mol}{79.056\ g} \right) \left( \frac{1\ mol\ NH_3}{1\ mol\ NH_4HCO_3} \right) (17.031\ g)$ $= 0.215\ g\ NH_3$

Heat in Chemical Reactions

Chemical reactions may either release or absorb heat.

• Exothermic reactions produce heat: $A + B \rightarrow C + \text{heat}$

• Endothermic reactions consume heat: $A + \text{heat} \rightarrow B + C$

3.5 Empirical Formulas from Analyses

Empirical Formulas

The empirical formula of a compound gives the simplest whole-number ratio of the elements present.

• Calculated from the ratio of the number of moles of each element.

• Percent composition by mass is often used to determine empirical formulas.

Converting Percent Composition to Empirical Formula

To determine the empirical formula from percent composition:

1. Assume a 100 g sample (so percent equals grams).

2. Convert grams of each element to moles using molar mass.

3. Divide each mole value by the smallest number of moles to get the simplest ratio.

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

Stepwise Table: Percent Composition to Empirical Formula

Step	Description
1	Assume 100 g sample
2	Convert grams to moles (use molar mass)
3	Divide by smallest mole value
4	Multiply to get whole numbers (if needed)
5	Write empirical formula

Example 3.11: Empirical Formula from Percent Composition

Given: 42.88% C, 57.12% O

• Assume 100 g sample: 42.88 g C, 57.12 g O

• Convert to moles: $C: 42.88\ g \left( \frac{1\ mol}{12.011\ g} \right) = 3.5701\ mol$ $O: 57.12\ g \left( \frac{1\ mol}{15.9994\ g} \right) = 3.5701\ mol$

• Divide by smallest value: $\frac{3.5701}{3.5701} = 1$ for both

• Empirical formula: CO

Example 3.12: Empirical Formula of Vitamin C

Given: 40.92% C, 4.58% H, 54.50% O

• Assume 100 g sample: 40.92 g C, 4.58 g H, 54.50 g O

• Convert to moles: $C: 40.92\ g \left( \frac{1\ mol}{12.011\ g} \right) = 3.407\ mol$ $H: 4.58\ g \left( \frac{1\ mol}{1.00794\ g} \right) = 4.54\ mol$ $O: 54.50\ g \left( \frac{1\ mol}{15.9994\ g} \right) = 3.406\ mol$

• Divide by smallest value (3.406): $C: \frac{3.407}{3.406} = 1$ $H: \frac{4.54}{3.406} = 1.33$ $O: \frac{3.406}{3.406} = 1$

• Multiply all by 3 to get whole numbers: 3:4:3

• Empirical formula: C3H4O3

Molecular Formula

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

• To determine the molecular formula, the molar mass (molecular weight) of the compound must be known.

• Calculate the empirical formula mass.

• Find the whole number multiple:

$\text{whole number multiple} = \frac{\text{molecular weight}}{\text{empirical formula weight}}$

• Multiply the subscripts in the empirical formula by this multiple.

Example 3.13: Molecular Formula of Vitamin C

• Empirical formula: C3H4O3

• Molecular weight: 176 amu

• Empirical formula weight: $3 \times 12.0 + 4 \times 1.01 + 3 \times 16.0 = 88.0\ amu$

• Whole number multiple: $\frac{176}{88.0} = 2$

• Molecular formula: $(C_3H_4O_3) \times 2 = C_6H_8O_6$

• Ascorbic acid (Vitamin C): C6H8O6

Additional info: The empirical formula is always the simplest ratio, while the molecular formula represents the actual composition of a molecule. Stoichiometry is essential for predicting yields and understanding chemical reactions quantitatively.