# Chapter 7 Part II: Chemical Formulas and Equations. Mr. Chumbley Chemistry 1-2

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1 Chapter 7 Part II: Chemical Formulas and Equations Mr. Chumbley Chemistry 1-2

2 SECTION 3: USING CHEMICAL FORMULAS

3 Molecules and Formula Unit We have not yet discussed the different ways in which chemical compounds combine and form We have mentioned that there are two types of compounds: Molecular compounds Ionic compounds Since they are physically different, the identification of a simplest amount of each quantity is different Molecular compounds have a simplest amount called a molecule A molecule is neutral group of atoms that are held together by covalent bonds Ionic compounds have a simplest unit called a formula unit A formula unit is the simplest collection of atoms from which an ionic compound s formula can be written

4 Formula Mass Similar to elements, chemical compounds also have characteristic average masses This mass can be obtained by adding the average atomic mass of all of the atoms present in the simplest unit The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula average atomic mass of H: 1.01 u average atomic mass of O: u 2 H atoms 1.01 u H atom = 2.02 u 1 O atom u O atom = u average mass of H 2 O molecule = u

5 Sample Problem F (p. 226) Find the formula mass of potassium chlorate, KClO 3. The mass of the formula unit of KClO 3 is found by summing the masses of one K atom, one Cl atom, and three O atoms. 1K atom u K atom = u 1 Cl atom u Cl atom = u 3 O atoms u O atom = u formula mass of KClO 3 = u

6 Molar Mass of Compounds The molar mass of a compound can be determined by adding the molar mass of each of the moles of the elements in the compound A compound s molar mass is numerically equal to its formula mass molar mass of H: 1.01 g/mol molar mass of O: g/mol 2 mol H 1.01 g H 1 mol H 1 mol O g O 1 mol O = 2.02 g/mol = g/mol molar mass of H 2 O = g/mol

7 Sample Problem G (p. 227) Find the molar mass of barium nitrate, Ba(NO 3 ) 2. The mass of the formula unit of Ba(NO 3 ) 2 is found by summing the masses of one mole of Ba and two moles of NO 3-. The two moles of NO 3 - result in two moles N and six moles O. 1mol Ba g Ba 1 mol Ba = g Ba 2 mol N g N 1 mol N = g N 6 mol O g O 1 mol O = g O molar mass of Ba(NO 3 ) 2 = g/mol

8 Homework: Practice F (p. 226) Practice G (p. 227)

9 Whiteboarding! Get out your homework from last night! Check your playing card. Get together with the other person(s) with matching numbers and whiteboard your solution to the corresponding problem. You have 5 minute to discuss your answers and whiteboard them before we begin presentations Problem # Card # F 1a Ace F 1b 2 F 1c 3 F 1d 4 G a 5 G b 6 G c 7 Be sure that both of you are capable of explaining, because the presenter will be selected at random!!!

10 Conversions using Formula Mass and Molar Mass Pure elements could be converted back and forth from the quantities of mass, amount in moles, and number of atoms Compounds can be converted in a similar way The exception is that one mole of a compound contains Avogadro s number of molecules or formula units Mass of a compound in grams = molar mass of compound 1 mol 1 mol molar mass of compound = Amount of a compound in moles = 1 mol units units 1 mol = Number of molecules or formula units of a compound

11 Sample Problem H (p. 228) What is the mass in grams of 2.50 mol of oxygen gas? Step 1: Analyze Given: 2.50 mol of O 2 Unknown: mass of O 2 in grams Step 2: Plan amount of O 2 in moles mass of O 2 in grams To convert amount of O 2 in moles to mass od O 2 in grams, multiply by the molar mass of O 2 amount of O 2 mol molar mass of O 2 g mol = mass of O 2 (g)

12 Sample Problem H (p. 228) What is the mass in grams of 2.50 mol of oxygen gas? Step 3: Solve First, find the molar mass of O 2 2 mol O g O 1 mol O = g mass of one mole of O 2 The molar mass of O 2 is therefore g/mol. Now calculate the mass of 2.50 mol of O mol O g O 2 1 mol O 2 = 80.0 g O mol of oxygen gas has a mass of 80.0 g

13 Sample Problem I (p ) Ibuprofen, C 13 H 18 O 2, is the active ingredient in many nonprescription pain relievers. Its molar mass is g/mol. a. If the tablets in a bottle contain a total of 33g of ibuprofen, how many moles of ibuprofen are in the bottle? b. How many molecules of ibuprofen are in the bottle? c. What is the total mass is grams of carbon in 33g of ibuprofen?

14 Sample Problem I (p ) a. If the tablets in a bottle contain a total of 33g of ibuprofen, how many moles of ibuprofen are in the bottle? Given: 33 g of C 13 H 18 O 2 molar mass = g/mol The mass in grams can be converted to amount in moles using the molar mass. mass moles g C 13 H 18 O 2 1 mol C 13 H 18 O 2 mol C 13 H 18 O 2 Unknown: g C 13 H 18 O 2 moles C 13 H 18 O 2 33 g C 13 H 18 O 2 1 mol C 13 H 18 O mol C 13 H 18 O g C 13 H 18 O 2

15 Sample Problem I (p ) b. How many molecules of ibuprofen are in the bottle? Given: 33 g of C 13 H 18 O 2 molar mass = g/mol 0.16 mol C 13 H 18 O 2 The number of molecules can be found using Avogadro s number. moles molecules mol C 13 H 18 O molecules molecules C 13 H 18 O 2 Unknown: molecules C 13 H 18 O 2 1 mol 0.16 mol C 13 H 18 O molecules 1 mol molecules C 13 H 18 O 2

16 Sample Problem I (p ) c. What is the total mass is grams of carbon in 33g of ibuprofen? Given: 33 g of C 13 H 18 O 2 molar mass = g/mol 0.16 mol C 13 H 18 O molecules C 13 H 18 O 2 Unknown: grams C The mass in grams of carbon can be found by determining the amount in moles of carbon in ibuprofen, and using the molar mass of carbon. moles C 13 H 18 O 2 moles C grams C mol C 13 H 18 O 2 13 mol C g C g C 1 mol C 13 H 18 O 2 1 mol C 0.16 mol C 13 H 18 O 2 13 mol C g C 25 g C 1 mol C 13 H 18 O 2 1 mol C

17 Homework: Practice I (p. 230) Answers: 1a mol 1b. 61 mol 2a molecules 2b molecules g

18 Practice! Begin working on the worksheet on your desk: Molar Mass of Compounds Practice. Feel free to use the book, any example problems, notes, etc. You have until 1:00 to work individually to finish all 10 problems. Pair up with the person either in front of or behind you, depending on your seat. Compare your answers.

19 Answers! 1. Fe 2 O g/mol 2. HClO g/mol 3. MnO g/mol 4. (NH 4 ) 2 SO g/mol g NH g C 3 H g Ca(NO 3 ) mol SO mol CaCl mol CuSO 4

20 On a half sheet of paper (share with a neighbor if you are so inclined), solve the following problem as I pass out papers. How many moles of calcium phosphate are in a g sample? Formula: Ca 3 (PO 4 ) 2 Molar mass: g/mol g Ca 3 (PO 4 ) 2 1 mol Ca 3 (PO 4 ) mol Ca 3 (PO 4 ) g Ca 3 (PO 4 ) 2

21 Percent Composition A useful way to evaluate the amount of an element is its percentage by mass The percent composition of a compound is the percentage by mass of each element in a compound To find the percent composition, the mass of the element in a sample of a compound is divided by the mass of the whole sample and multiplied by 100 Since the mass percentage is independent of the size of the sample, it is often calculated using a sample size of one mole mass of element in 1 mol of a compound molar mass of compound 100 = % element in compound

22 Sample Problem J (p. 231) Find the percentage composition of copper (I) sulfide, Cu 2 S. Given: Cu 2 S Unknown: % comp. Cu 2 S formula molar mass mass percentage of each element 2 mol Cu g Cu 1 mol Cu = g Cu 1 mol S g S 1 mol S = g S % Cu = molar mass Cu 2 S = g/mol g Cu 100 = 79.85% Cu g Cu 2 S % S = g S 100 = 20.15% S g Cu 2 S

23 You try it! Find the mass percentage of water in sodium carbonate decahydrate, Na 2 CO 3 10H 2 O. Given: Na 2 CO 3 10H 2 O Unknown: molar mass of hydrate % water by mass 2 mol Na Molar mass of Hydrate g Na 1 mol Na = g Cu 1 mol C g C 1 mol C = g C 13 mol O g O 1 mol O = g O 20 mol H 1.01 g H 1 mol H = g H molar mass Na 2 CO 3 10H 2 O = g/mol formula mass water per mole % water 10 mol H 2 O g H 2O 1 mol H 2 O = g H 2O In each mole of Na 2 CO 3 10H 2 O, there is g of H 2 O. % H 2 O = g H 2 O g Na 2 CO 3 10H 2 O 100 = % H2O

24 SECTION 4: DETERMINING CHEMICAL FORMULAS

25 Empirical Formulas Substances can be analyzed quantitatively to determine the chemical formula of the quantity An empirical formula consists of the symbols for the elements combined in a compound with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound This is often times done with percentage composition, but can also be done using known masses of an element within the compound

26 Sample Problem L (p. 234) Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound. Given: 32.38% Na 22.65% S 44.99% O Unknown: Na x S x O x percent comp. mass comp. molar comp. empirical formula g Na g S 1 mol S g S g O 1 mol O g O 1 mol Na = mol Na g Na = mol S = mol O divide each molar amount by the smallest number mol N mol S mol O : : mol Na 1 mol S mol O The empirical formula of the compound is Na 2 SO 4

27 Sample Problem M (p. 235) Analysis of a g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of g. What is the empirical formula of the compound? Given: g P x O x 4.433f P Unknown: P x O x

28 Molecular Formula While the empirical formula frequently provides a compound s chemical formula, it is not always the case Some molecular compounds share an empirical formula, but have a different chemical formula that makes up a molecule of that compound A molecular formula consists of the symbols for the elements combined in a compound with subscripts showing the actual number of the different atoms in the compound

29 Empirical versus Molecular Empirical Formula Compound Molecular Formula methylene CH 2 ethene C 2 H 4 CH 2 (85.6% C, 14.4% H) cyclopropane C 3 H 6 butene C 4 H 8 cyclohexane C 6 H 12 formaldehyde CH 2 O CH 2 O (40.0% C, 6.7% H, 53.3% O) acetic acid C 2 H 4 O glyceraldehyde C 3 H 6 O 3 glucose C 6 H 12 O 6

30 Molecular Formula Molecular formulas will always be whole number multiples of empirical formulas Since the formulas are whole number multiples, the molecular formula mass is the same whole number multiple of the empirical formula mass x empirical formula mass = molecular formula mass This shows that multiple compounds can have the same empirical formula but can be distinguished based on the mass of the different molecular formulas

31 Sample Problem N (p ) The empirical formula of a compound of phosphorus and oxygen was found to be P 2 O 5. Experimentation shows that the molar mass of this compound is g/mol. What is the compound s molecular formula? Given: Empirical formula P 2 O 5 Molar mass = g/mol Unknown: Molecular formula x empirical formula mass = molecular formula mass molecular molar mass = g mol molecular formula mass = u determine the empirical formula mass 2 P atoms u 1 P atom = u 5 O atom u 1 O atom = g u formula mass = u u u = The compound s molecular formula is 2 times its empirical formula, P 4 O 10

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