## CHEMISTRY THE CENTRAL SCIENCE

**3 STOICHIOMETRY: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS**

**CHAPTER SUMMARY AND KEY TERMS**

**INTRODUCTION AND ****SECTION 3.1** The study of the quantitative relationships between chemical formulas and chemical equations is known as **stoichiometry**. One of the important concepts of stoichiometry is the law of conservation of mass, which states that the total mass of the products of a chemical reaction is the same as the total mass of the reactants. The same numbers of atoms of each type are present before and after a chemical reaction. A balanced **chemical equation** shows equal numbers of atoms of each element on each side of the equation. Equations are balanced by placing coefficients in front of the chemical formulas for the **reactants** and **products** of a reaction, *not* by changing the subscripts in chemical formulas.

**SECTION 3.2** Among the reaction types described in this chapter are (1) **combination reactions**, in which two reactants combine to form one product; (2) **decomposition reactions**, in which a single reactant forms two or more products; and (3) **combustion reactions** in oxygen, in which a hydrocarbon or related compound reacts with O_{2} to form CO_{2} and H_{2}O.

**SECTION 3.3** Much quantitative information can be determined from chemical formulas and balanced chemical equations by using atomic weights. The **formula weight** of a compound equals the sum of the atomic weights of the atoms in its formula. If the formula is a molecular formula, the formula weight is also called the **molecular weight**. Atomic weights and formula weights can be used to determine the elemental composition of a compound.

**SECTION 3.4** A mole of any substance is **Avogadro's number** (6.02 × 10^{23}) of formula units of that substance. The mass of a **mole** of atoms, molecules, or ions (the **molar mass**) equals the formula weight of that material expressed in grams. The mass of one molecule of H_{2}O, for example, is 18 amu, so the mass of 1 mol of H_{2}O is 18 g. That is, the molar mass of H_{2}O is 18 g/mol.

**SECTION 3.5** The empirical formula of any substance can be determined from its percent composition by calculating the relative number of moles of each atom in 100 g of the substance. If the substance is molecular in nature, its molecular formula can be determined from the empirical formula if the molecular weight is also known.

**SECTIONS 3.6**** AND ****3.7** The mole concept can be used to calculate the relative quantities of reactants and products in chemical reactions. The coefficients in a balanced equation give the relative numbers of moles of the reactants and products. To calculate the number of grams of a product from the number of grams of a reactant, first convert grams of reactant to moles of reactant. Then use the coefficients in the balanced equation to convert the number of moles of reactant to moles of product. Finally, convert moles of product to grams of product.

A **limiting reactant** is completely consumed in a reaction. When it is used up, the reaction stops, thus limiting the quantities of products formed. The **theoretical yield** of a reaction is the quantity of product calculated to form when all of the limiting reactant reacts. The actual yield of a reaction is always less than the theoretical yield. The **percent yield** compares the actual and theoretical yields.

**KEY SKILLS**

• Balance chemical equations. (Section 3.1)

• Predict the products of simple combination, decomposition, and combustion reactions. (Section 3.2)

• Calculate formula weights. (Section 3.3)

• Convert grams to moles and moles to grams using molar masses. (Section 3.4)

• Convert number of molecules to moles and moles to number of molecules using Avogadro's number. (Section 3.4)

• Calculate the empirical and molecular formulas of a compound from percentage composition and molecular weight. (Section 3.5)

• Calculate amounts, in grams or moles, of reactants and products for a reaction. (Section 3.6)

• Calculate the percent yield of a reaction. (Section 3.7)

**KEY EQUATIONS**

This is the formula to calculate the mass percentage of each element in a compound. The sum of all the percentages of all the elements in a compound should add up to 100%.

This is the formula to calculate the percent yield of a reaction. The percent yield can never be more than 100%.