1. Copyright © Houghton Mifflin Company. All rights reserved.4–14–1 5.6 Chemical Equations A chemical equation uses symbols and for- mulas to describe the changes that take place in a chemical reaction. 2 H 2 + O 2  2 H 2 O Reactants Products There must be the same number of each type of atom on both sides of the equation.

2. Copyright © Houghton Mifflin Company. All rights reserved.4–24–2 5.6 Chemical Equations A skeleton equation has the same type, but not the same number, of each type of atom on each side of the equation. It must be balanced before it is useful in chemical calculations. Fe 2 O 3 + CO  Fe + CO 2

3. Copyright © Houghton Mifflin Company. All rights reserved.4–34–3 5.6 Chemical Equations More equations to balance: C 3 H 8 + O 2  CO 2 + H 2 O C 4 H 10 + O 2  CO 2 + H 2 O Al 2 O 3 + HCl  AlCl 3 + H 2 O II –1

4. Copyright © Houghton Mifflin Company. All rights reserved.4–44–4 5.6 Chemical Equations We can read equations as referring to individual atoms and molecules, but in the real world, we work with very large groups of atoms and molecules.

5. Copyright © Houghton Mifflin Company. All rights reserved.4–54–5 5.2 The Mole: A Counting Unit for Chemists We often use words that refer to a specific number of items: a pair of socks = 2 socks a dozen eggs = 12 eggs fourscore years = 4 x 20 years It would be convenient to work with a specific number of atoms in chemical calculations.

6. Copyright © Houghton Mifflin Company. All rights reserved.4–64–6 5.2 The Mole In 12.0 g of C-12, there are 6.022 x 10 23 atoms. 6.022 x 10 23 objects is one mole of objects 6.022 x 10 23 is “Avogadro’s number” A mole is a huge number, but we can use it like we use dozens and scores.

7. Copyright © Houghton Mifflin Company. All rights reserved.4–74–7 5.3 Molar Masses 6.022 x 10 23 atoms of C-12 have a mass of 12.0 grams. The mass of a mole of atoms of all other elements is defined relative to the mass of 1 mole of C-12. Atomic mass of Mg = 24.305 amu Molar mass of Mg = 24.305 g/mol

8. Copyright © Houghton Mifflin Company. All rights reserved.4–84–8 5.3 Molar Masses We can obtain a mole of an element by weigh- ing it. If we need more or less than a mole, we can calculate the desired amount. We can also determine how many atoms are present. Atoms MolesMolecules Avogadro’s # Molar mass 6.022 x 10 23 grams mole mole

9. Copyright © Houghton Mifflin Company. All rights reserved.4–94–9 5.3 Molar Masses Use the molar mass of magnesium and Avogadro’s number as needed to make these calculations: How many moles of Mg in 12.8 g of Mg? How many grams of Mg in 2.50 mol of Mg? How many atoms in 2.50 mol of Mg? How many atoms in 42.8 g of Mg?

10. Copyright © Houghton Mifflin Company. All rights reserved.4–10 5.4 Moles and Chemical Formulas 1 Molecule of H 2 O2 H atoms 1 O atom 1 Mole of H 2 O2 moles H atoms 1 mole O atom We work in moles, rather than molecules, because we can’t see individual molecules!

11. Copyright © Houghton Mifflin Company. All rights reserved.4–11 5.4 Moles and Chemical Formulas In a reaction equation: 2 H 2 + O 2  2 H 2 O 2 H 2 molecules and 1 O 2 molecule produce 2 H 2 O molecules. 2 moles of H 2 molecules and 1 mole of O 2 molecule produce 2 moles of H 2 O molecules.

12. Copyright © Houghton Mifflin Company. All rights reserved.4–12 5.4 Moles and Chemical Formulas What is the mass of 1.00 mole of H 2 O? How many H 2 O molecules are present in one mole of water?

13. Copyright © Houghton Mifflin Company. All rights reserved.4–13 5.5 Calculations with Molar Mass A can of soda contains 340 g of water. How many moles of water is this? How many grams of hydrogen are present in this amount of water? How many hydrogen atoms are pres- ent in this amount of water?

14. Copyright © Houghton Mifflin Company. All rights reserved.4–14 5.7 Chemical Equations and Moles N 2 + 3 H 2  2 NH 3 3 moles of H 2 produce 2 moles of NH 3 3 mol H 2 2 mol NH 3 2 mol NH 3 3 mol H 2

15. Copyright © Houghton Mifflin Company. All rights reserved.4–15 5.8 Mass Calculations for Chemical Equations A person eats a candy bar that contains 14.2 g of sucrose (C 12 H 22 O 11 ). How much water will be produced by the sucrose? It is metabolized according to the equation shown below: C 12 H 22 O 11 + 12 O 2  12 CO2 + 11 H 2 O II -2

16. Copyright © Houghton Mifflin Company. All rights reserved.4–16 5.8 Mass Calculations How much Fe 2 O 3 must I use if I want 500 g of Fe metal? Fe 2 O 3 + 3 CO  2 Fe + 3 CO 2 Na I and CH 4 O react accordingly: Na I + CH 4 O  CH 3 I + NaOH If one starts with 750 g each of Na I and CH 4 O, how much CH 3 I will be produced?