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ATOMIC MASS – The mass of an individual atom ATOMIC MASS UNIT (amu or u) – One twelfth the mass of a carbon-12 atom, equal to 1.66 x 10 -24 g Atomic masses.

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Presentation on theme: "ATOMIC MASS – The mass of an individual atom ATOMIC MASS UNIT (amu or u) – One twelfth the mass of a carbon-12 atom, equal to 1.66 x 10 -24 g Atomic masses."— Presentation transcript:

1 ATOMIC MASS – The mass of an individual atom ATOMIC MASS UNIT (amu or u) – One twelfth the mass of a carbon-12 atom, equal to 1.66 x 10 -24 g Atomic masses are measured today with mass spectrometers CHAPTER 8 – CHEMICAL COMPOSITION 4B-1 (of 30)

2 20 Ne19.99 u90.92 % 21 Ne 20.99 u0.26 % 22 Ne 21.99 u8.82 % 4B-2

3 Any sample of neon contains the same percentage of each isotope ELEMENT’S ATOMIC MASS - Found on the Periodic Table, the average atomic mass of all of the element’s naturally occurring isotopes The atomic mass of neon is 20.18 u The atomic mass of calcium is 40.08 u 4B-3

4 Gummy WormsJordan Almonds 8 g each 4 g each 200 g These 2 samples contain the same number of pieces of candy 100 g Atomic masses allowed chemists to count atoms by weighing them 4B-4

5 1 Ne atom 20.18 u 1 Ca atom 40.08 amu 20.18 g Ne40.08 g Ca These two samples contain the same number of atoms 4B-5

6 MOLAR MASS – The mass of one mole of atoms of an element For neon: 20.18 g Ne = 1 mol Ne atoms For calcium: 40.08 g Ca = 1 mol Ca atoms The number of atoms contained in 20.18 g neon or 40.08 g calcium (numerically the element’s atomic mass, but measured in grams) is called a MOLE 4B-6

7 1 pair = 2 shoes 1 dozen = 12 eggs 1 mole = 6.022  10 23 atoms = The number of atoms in 20.18 g Ne = The number of atoms in 40.08 g Ca For shoes: For eggs: For atoms: MOLE – The number of atoms contained in 12 g of carbon-12 Experimentally it has been shown that a mole is 6.022  10 23 atoms AVOGADRO’S NUMBER – 6.022  10 23 4B-7

8 The Element’s Atomic Mass The average atomic mass of the element’s isotopes, measured in amu’s 1 average O atom = 15.9994 u The Element’s Molar Mass The mass necessary to have 1 mole of atoms of the element, measured in grams 1 mole O atoms = 15.9994 g A Molar mass may be rounded to the hundredths place as long as it doesn’t limit the number of significant figures in a calculation 4B-8

9 Calculate the mass of a sample containing 2.50 moles of phosphorus atoms For phosphorus: 30.97 g P = 1 mol P x 30.97 g P _____________ 1 mol P = 77.4 g P Calculate the moles of atoms contained in a 10.0 g sample of potassium 2.50 mol P 4B-9

10 Calculate the number of atoms contained in 4.50 moles of sodium For any matter: 6.022 x 10 23 atoms = 1 mol 2.50 mol Na x 6.022 x 10 23 atoms Na _____________________________ 1 mol Na = 2.71 x 10 24 atoms Na Calculate the moles of atoms in a sample of 1.50  10 23 tungsten atoms 4B-10

11 Calculate the number of atoms contained in a 1.00 g sample of iron For iron: 55.85 g Fe = 1 mol Fe For any matter: 6.022 x 10 23 atoms = 1 mol 1.00 g Fe x 6.022 x 10 23 atoms Fe _____________________________ 1 mol Fe = 1.08  10 22 atoms Fex 1 mol Fe _______________ 55.85 g Fe 4B-11

12 MOLAR MASSES OF COMPOUNDS MOLAR MASS – The mass of one mole of molecules of a molecular substance, or one mole of formula units of an ionic substance Sodium chloride, NaCl 1 mole of NaCl contains 1 mole of Na + and 1 mole of Cl - 1 mol Na + (22.99 g/mol)= 22.99 g 1 mol Cl - (35.45 g/mol)=35.45 g 58.44 g The mass necessary to have 1 mole of NaCl formula units (1 mole of Na + ions and 1 mole of Cl - ions) 4B-12

13 Sucrose, C 12 H 22 O 11 1 mole of C 12 H 22 O 11 contains 12 moles C, 22 moles H, and 11 moles of O 12 mol C (12.01 g/mol)= 144.12 g 22 mol H (1.01 g/mol)=22.22 g 11 mol O (16.00 g/mol)=176.00 g 342.34 g The mass necessary to have 1 mole of C 12 H 22 O 11 molecules (12 moles C atoms, 22 moles H atoms, 11 moles O atoms) 4B-13

14 Calcium hydroxide, Ca(OH) 2 1 mol Ca (40.08 g/mol)= 40.08 g 2 mol O (16.00 g/mol)=32.00 g 2 mol H (1.01 g/mol)=2.02 g 74.10 g 4B-14

15 Calculate the mass of a sample containing 3.50 moles of magnesium fluoride. MgF 2 1 mol Mg (24.31 g/mol)= 24.31 g 2 mol F (19.00 g/mol)=38.00 g 62.31 g  62.31 g MgF 2 = 1 mol MgF 2 x 62.31 g MgF 2 __________________ 1 mol MgF 2 = 218 g MgF 2 3.50 mol MgF 2 4B-15

16 Calculate the number of moles of zinc bromide present in a 10.0 g sample of the compound. 4B-16

17 PERCENT COMPOSITION OF COMPOUNDS BY MASS Fe 2 O 3 2 mol Fe (55.85 g/mol)= 111.70 g 3 mol O (16.00 g/mol)=48.00 g 159.70 g Calculate the percent composition by mass of iron (III) oxide. 111.70 g Fe  100 ________________________________ 159.70 g Fe 2 O 3 % Fe == 69.944 % Fe 48.00 g O  100 ________________________________ 159.70 g Fe 2 O 3 % O == 30.06 % O 4B-17

18 Al 2 (SO 4 ) 3 2 mol Al (26.98 g/mol)= 53.96 g 3 mol S (32.07 g/mol)=96.21 g 12 mol O (16.00 g/mol)= 192.00 g 342.17 g Calculate the percent composition by mass of aluminum sulfate. 4B-18

19 Mass of Compound Mass of Mn Mass of O A sample of a compound containing Mn and O had a mass of 2.59 g. The compound was heated to drive off the O, and the final residue had a mass of 1.63g. Find the percent composition by mass of the compound. 1.63 g Mn  100 ______________________________________ 2.59 g compound % Mn == 62.9 % Mn 0.96 g O  100 ______________________________________ 2.59 g compound % O == 37 % O 2.59 g 1.63 g 2.59 g - 1.63 g = 0.96 g 4B-19

20 EMPIRICAL FORMULA CALCULATIONS EMPIRICAL FORMULA – The simplest whole-number ratio of the atoms of different elements in a compound C6H6C6H6 C1H1C1H1 C 6 H 12 O 6 C1H2O1C1H2O1 H2OH2O H2OH2O Molecular Formula: Empirical Formula: 4B-20

21 1)Assume you have 100 g of the compound Find the empirical formula of a compound that is 75.0% carbon and 25.0% hydrogen by mass.  75.0 g C and 25.0 g H 2)Calculate the moles of atoms of each element x 1 mol C _____________________ 12.01 g C = 6.24 mol C 75.0 g C x 1 mol H ____________________ 1.01 g H = 24.8 mol H 25.0 g H 3)Divide each number of moles by the smallest number of moles 6.24 mol C _______________________ 6.24 24.8 mol H _______________________ 6.24 = 1.00 mol C= 3.97 mol H 4)The integer mole ratio must be the atom ratio:CH 4 4B-21

22 Find the empirical formula of a compound that is 30.4% nitrogen and 69.6% oxygen by mass. 4B-22

23 Find the empirical formula of a compound that is 81.7% carbon and 18.3% hydrogen by mass. x 1 mol C _____________________ 12.01 g C = 6.80 mol C 81.7 g C x 1 mol H ___________________ 1.01 g H = 18.1 mol H 18.3 g H 6.80 mol C _______________________ 6.80 18.1 mol H _______________________ 6.80 = 1.00 mol C= 2.66 mol H If the moles of all elements are not within ± 0.1 moles of an integer, they must all be multiplied by a constant number until they are integers  2 = 2.00 mol C= 5.32 mol H 4B-23

24 Find the empirical formula of a compound that is 81.7% carbon and 18.3% hydrogen by mass. x 1 mol C _____________________ 12.01 g C = 6.80 mol C 81.7 g C x 1 mol H ___________________ 1.01 g H = 18.1 mol H 18.3 g H 6.80 mol C _______________________ 6.80 18.1 mol H _______________________ 6.80 = 1.00 mol C= 2.66 mol H If the moles of all elements are not within ± 0.1 moles of an integer, they must all be multiplied by a constant number until they are integers  3 = 3.00 mol C= 7.98 mol H Empirical formula: C 3 H 8 4B-24

25 MOLECULAR FORMULA CALCULATIONS MOLECULAR FORMULA – The actual number of the atoms of different elements in a molecule C1H1C1H1 or C 2 H 2 Empirical Formula: Molecular Formula: C1H1C1H1 or C 3 H 3 or C 4 H 4 etc. 4B-25

26 Find the molecular formula of a compound that is 5.9% hydrogen and 94.1% oxygen by mass, and has a molar mass of 34.0 g/mol. x 1 mol H _____________________ 1.01 g H = 5.8 mol H 5.9 g H x 1 mol O _____________________ 16.00 g O = 5.88 mol O 94.1 g O 5.8 mol H _______________________ 5.8 5.88 mol O _______________________ 5.8 = 1.0 mol H= 1.0 mol O Empirical formula: HO Find the molar mass of the empirical formula 4B-26

27 HO 1 mol H (1.01 g/mol)= 1.01 g 1 mol O (16.00 g/mol)=16.00 g 17.01 g Divide the compound’s actual molar mass by the empirical formula’s molar mass – it should be very close to an integer 34.0 g/mol _________________________ 17.01 g/mol ≈ 2  The molecular formula is 2 times the empirical formula Molecular formula: H 2 O 2 4B-27

28 Find the molecular formula of a compound that is 26.7% P, 12.1% N, and 62.1% Cl by mass, and has a molar mass of about 700 g/mol. 4B-28

29 4B-29

30 REVIEW FOR TEST Polyatomic Ions Names and Formulas of Ionic Compounds, Covalent Compounds, and Acids Atomic Mass Mole, Avogadro’s Number Molar Mass of Elements Conversions from Mass to Moles to Atoms Molar Mass of Compounds Conversions from Mass to Moles to Molecules Percentage Composition by Mass Empirical Formula Calculations Molecular Formula Calculations 4B-30


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