Introduction to Chemical Reactions A chemical change—a chemical reaction—converts one substance into another. Chemical reactions involve: breaking bonds in the reactants (starting materials) forming new bonds in the products. CH4 and O2 CO2 and H2O

Introduction to Chemical Reactions A chemical equation is an expression that uses chemical formulas and other symbols to illustrate what reactants constitute the starting materials in a reaction and what products are formed. The reactants are written on the left. The products are written on the right. Coefficients show the number of molecules of a given element or compound that react or are formed.

Introduction to Chemical Reactions The law of conservation of mass states that atoms cannot be created or destroyed in a chemical reaction. Coefficients are used to balance an equation. A balanced equation has the same number of atoms of each element on both sides of the equation.

Introduction to Chemical Reactions

Balancing Chemical Equations HOW TO Balance a Chemical Equation Write a balanced chemical equation for the reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O). Example Step [1] Write the equation with the correct formulas. C3H8 + O2 CO2 + H2O The subscripts in a formula can never be changed to balance an equation, because changing a subscript changes the identity of a compound.

Balancing Chemical Equations HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Step [2] Balance the C’s first: Balance the H’s next:

Balancing Chemical Equations HOW TO Balance a Chemical Equation Balance the equation with coefficients one element at a time. Step [2] Finally, balance the O’s:

Balancing Chemical Equations HOW TO Balance a Chemical Equation Check to make sure that the smallest set of whole numbers is used. Step [3]

Chemistry of an Automobile Airbag A severe car crash triggers the ignition of sodium azide (NaN3) converting it to sodium (Na) and nitrogen gas (N2). 2 NaN3 2 Na + 3 N2 The nitrogen gas causes the bag to inflate fully in 40 milliseconds, protecting passengers from injury.

The Mole and Avogadro’s Number A mole is a quantity that contains 6.02 x 1023 items. 1 mole of C atoms = 6.02 x 1023 C atoms 1 mole of H2O molecules = 6.02 x 1023 H2O molecules 1 mole of vitamin C molecules = 6.02 x 1023 vitamin C molecules The number 6.02 x 1023 is Avogadro’s number. It can be used as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules: 1 mol 6.02 x 1023 atoms or 6.02 x 1023 atoms 1 mol

The Mole and Avogadro’s Number Sample Problem 5.6 How many molecules are contained in 5.0 moles of carbon dioxide (CO2)? Identify the original quantity and the desired quantity. Step [1] Author: sample problem numbering in this chapter also… 5.0 mol of CO2 original quantity ? number of molecules of CO2 desired quantity

The Mole and Avogadro’s Number Step [2] Write out the conversion factors. 1 mol 6.02 x 1023 molecules or 6.02 x 1023 molecules 1 mol Choose this one to cancel mol. Step [3] Set up and solve the problem. 6.02 x 1023 molecules 1 mol 5.0 mol x = 3.0 x 1024 molecules CO2 Unwanted unit cancels.

Mass to Mole Conversions The formula weight/ molecular mass is the sum of the atomic weights of all the atoms in a compound, reported in atomic mass units (amu). Sample Problem 5.8 What is the formula weight of nicotine (C10H14N2)? Step [1] Determine the number of atoms of each element from the subscripts in the chemical formula. C10H14N2 contains 10 C atoms, 14 H atoms, and 2 N atoms.

Mass to Mole Conversions Multiply the number of atoms of each element by the atomic weight and add the results. Step [2] 10 C atoms x 12.01 amu = 120.1 amu 14 H atoms x 1.08 amu = 14.11 amu 2 N atoms x 14.01 amu = 28.02 amu Formula weight of C10H14N2 = 162.23 amu

Mass to Mole Conversions Molar Mass The molar mass is the mass of one mole of any substance, reported in grams. The value of the molar mass of a compound in grams equals the value of its formula weight in amu.

Mass to Mole Conversions Relating Grams to Moles The molar mass relates the number of moles to the number of grams of a substance. In this way, molar mass can be used as a conversion factor. Sample Problem 5.10 How many moles are present in 100. g of aspirin (C9H8O4, molar mass 180.2 g/mol)? Step [1] Identify the original quantity and the desired quantity. 100. g of aspirin original quantity ? mol of aspirin desired quantity

Mass to Mole Conversions Relating Grams to Moles Step [2] Write out the conversion factors. The conversion factor is the molar mass, and it can be written in two ways. Choose the one that places the unwanted unit, grams, in the denominator so that the units cancel: 180.2 g aspirin 1 mol 1 mol 180.2 g aspirin or Choose this one to cancel g.

Mass to Mole Conversions Relating Grams to Moles Step [3] Set up and solve the problem. 1 mol 180.2 g aspirin 100. g aspirin x = 0.555 mol aspirin Unwanted unit cancels.

Mole Calculations in Chemical Equations A balanced chemical equation also tells us: the number of moles of each reactant that combine the number of moles of each product formed. 1 N2(g) + 1 O2(g) 2 NO(g) 1 mole of N2 1 molecule N2 1 mole of O2 1 molecule O2 2 moles of NO 2 molecules NO (The coefficient “1” has been written for emphasis.)

Mole Calculations in Chemical Equations Coefficients are used to form mole ratios, which can serve as conversion factors. N2(g) + O2(g) 2 NO(g) Mole ratios: 1 mol N2 1 mol O2 1 mol N2 2 mol NO 1 mol O2 2 mol NO Use the mole ratios from the coefficients in the balanced equation to convert moles of one compound (A) into moles of another compound (B).

Mole Calculations in Chemical Equations Sample Problem 5.11 Using the balanced chemical equation, how many moles of CO are produced from 3.5 moles of C2H6? 2 C2H6(g) + 5 O2(g) 4 CO(g) + 6 H2O(g) Step [1] Identify the original and desired quantities. 3.5 mol C2H6 original quantity ? mol CO desired quantity

Mole Calculations in Chemical Equations Step [2] Write out the conversion factors. 2 mol C2H6 4 mol CO 4 mol CO 2 mol C2H6 or Choose this one to cancel mol C2H6. Step [3] Set up and solve the problem. 4 mol CO 2 mol C2H6 3.5 mol C2H6 x = 7.0 mol CO Unwanted unit cancels.

Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Using the balanced equation, how many grams of O3 are formed from 9.0 mol of O2? Example sunlight 3 O2(g) 2 O3(g) [1] [2] Moles of reactant Moles of product Grams of product mole–mole conversion factor molar mass conversion factor

Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Step [1] Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor. Convert the number of moles of product to the number of grams of product using the product’s molar mass. Step [2] 3 mol O2 2 mol O3 2 mol O3 3 mol O2 1 mol O3 48.00 g O3 48.00 g O3 1 mol O3 or or Cancel mol O2 in step [1]. Cancel mol O3 in step [2].

Mass Calculations in Chemical Equations HOW TO Convert Moles of Reactant to Grams of Product Set up and solve the conversion. mole–mole conversion factor molar mass conversion factor Moles of reactant Grams of product 2 mol O3 3 mol O2 48.00 g O3 1 mol O3 9.0 mol O2 x x = 290 g O3 Mol O2 cancel. Mol O3 cancel.

Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product Ethanol (C2H6O, molar mass 46.07 g/mol) is synthesized by reacting ethylene (C2H4, molar mass 28.05 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene? Example

Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product mole–mole conversion factor [2] Moles of reactant Moles of product molar mass conversion factor molar mass conversion factor [1] [3] Grams of reactant Grams of product

Mass Calculations in Chemical Equations HOW TO Convert Grams of Reactant to Grams of Product molar mass conversion factor mole–mole conversion factor molar mass conversion factor Grams of reactant 1 mol C2H4 28.05 g C2H4 1 mol C2H6O 1 mol C2H4 46.07 g C2H6O 1 mol C2H6O 14 g C2H4 x x x Grams C2H4 cancel. Moles C2H4 cancel. Moles C2H6O cancel. Grams of product = 23 g C2H6O