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1 Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas.

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1 1 Chemical Quantities The Mole, % Composition, Empirical and Molecular Formulas

2 2 How you measure how much? How you measure how much? n You can measure mass, or volume, or you can count pieces. n We measure mass in grams. n We measure volume in liters. n We count pieces in MOLES.

3 3 Moles n Defined as the number of carbon atoms in exactly 12 grams of carbon grams of carbon-12. n 1 mole is 6.02 x particles. n Treat it like a very large dozen n 6.02 x is called Avagadro’s number.

4 4 Representative particles n The smallest pieces of a substance. n For a molecular compound it is a molecule. n For an ionic compound it is a formula unit. n For an element it is an atom.

5 5 Types of questions n How many oxygen atoms in the following? –CaCO 3 –Al 2 (SO 4 ) 3 n How many ions in the following? –CaCl 2 –NaOH –Al 2 (SO 4 )

6 6 Types of questions using the equality; 1 mole = 6.02 x n How many molecules of CO 2 are the in 4.56 moles of CO 2 ? n 4.56 mole x 6.02x10 23 mc = 1 1 mole 1 1 mole n How many moles of water is 5.87 x molecules? n 5.87 x mc x 1 mole = x10 23 mc x10 23 mc 2.75x10 24 mc mole

7 7 Types of questions using the equality; 1 mole = 6.02 x n How many atoms of carbon are there in 1.23 moles of C 6 H 12 O 6 ? n 1.23 moles x 6.02x10 23 mc x 6 atoms = 1 1 mole 1 mc 1 1 mole 1 mc n How many moles is 7.78 x formula units of MgCl 2 ? 7.78x10 24 FU x 1 mole = 7.78x10 24 FU x 1 mole = 12.9 mole x10 24 FU x10 24 FU 4.44x10 24 atoms

8 8 Measuring Moles n Remember relative atomic mass? n The amu was one twelfth the mass of a carbon 12 atom. n Since the mole is the number of atoms in 12 grams of carbon-12, n the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.

9 9 Gram Atomic Mass n The mass of 1 mole of an element in grams. n grams of carbon has the same number of pieces as grams of hydrogen and grams of iron. n We can right this as g C = 1 mole n We can count things by weighing them.

10 10 Examples n How much would 2.34 moles of carbon weigh? n 2.34 moles C x 12 g C 1 1mole 1 1mole n How many moles of magnesium in g of Mg? g Mg x 1 mole g Mg x 1 mole = 1 24g Mg 1 24g Mg = g mole

11 11 n How many atoms of lithium in 1.00 g of Li? n 1.00 g Li x 1 mole x 6.02x1023 atoms 1 7 g Li 1 mole 1 7 g Li 1 mole n How much would 3.45 x atoms of U weigh? n 3.45x1022 atoms U x 1 mole x 238 g U x1023atoms 1 mole x1023atoms 1 mole 8.60x10 22 atoms 13.6 g

12 12 What about compounds? n in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms n To find the mass of one mole of a compound n determine the moles of the elements they have n Find out how much they would weigh n add them up

13 13 What about compounds? n What is the mass of one mole of CH 4 ? n 1 mole of C = 12 g n 4 mole of H x 1 g = 4g n 1 mole CH 4 = = 16g n The Gram Molecular mass of CH 4 is 16.05g n The mass of one mole of a molecular compound.

14 14 Gram Formula Mass n The mass of one mole of an ionic compound. n Calculated the same way. n What is the GFM of Fe 2 O 3 ? n 2 moles of Fe x 56 g = 112 g n 3 moles of O x 16 g = 48 g n The GFM = 112 g + 48 g = 160g

15 15 Molar Mass n The generic term for the mass of one mole. n The same as gram molecular mass, gram formula mass, and gram atomic mass.

16 16 Examples n Calculate the molar mass of the following and indicate what type it is. n Na 2 S n 2 (23) + 32 = nN2O4nN2O4nN2O4nN2O4 n 2(14) + 4(16) = nCnCnCnC = 78g Gram Formula Mass = 92g Gram Molecular Mass 12g 1mole Na 2 S = 78g 1 mole N 2 O 4 = 92g 1 mole C = 12 g Gram Atomic Mass

17 17 Molar Mass Cont. n Ca(NO 3 ) 2 n (14) + 6(16) = = 164g n 1 mole Ca(NO 3 ) 2 = 164g n C 6 H 12 O 6 n 6(12) + 12(1) + 6(16) = = 180g 1 mole C6H12)6 = 180g 1 mole C6H12)6 = 180g Gram Molecular Mass n (NH 4 ) 3 PO 4 n 3(14) + 12(1) (16) = = 149g n 1 mole (NH 4 ) 3 PO 4 = 149g Gram Formula Mass Gram Formula Mass

18 18 Using Molar Mass Finding moles of compounds Counting pieces by weighing

19 19 Molar Mass n The number of grams of 1 mole of atoms, ions, or molecules. n We can make conversion factors from these. To change grams of a compound to moles of a compound.

20 20 For example n How many moles is 5.69 g of NaOH?

21 21 For example n How many moles is 5.69 g of NaOH?

22 22 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles

23 23 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH

24 24 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 23g 1 mol O = g 1 mole of H = 1 g

25 25 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g l 1 mole NaOH = 40 g

26 26 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g l 1 mole NaOH = 40 g

27 27 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH l 1mole Na = 23g 1 mol O = 16 g 1 mole of H = 1 g l 1 mole NaOH = 40 g

28 28 Examples n How many moles is 4.56 g of CO 2 ? n 4.56g CO 2 x 1 mole 1 44gCO gCO 2 n How many grams is 9.87 moles of H 2 O? n 9.87 moles x 18g H 2 O 1 1 mole 1 1 mole = moles = 178g

29 29 Examples n How many molecules in 6.8 g of CH 4 ? n 6.8g CH 4 x 1 mole x 6.02x1023 mc 1 16g CH 4 1mole 1 16g CH 4 1mole n 49 molecules of C 6 H 12 O 6 weighs how much? n 49 mc x 1 mole x 180g C 6 H 12 O 6 = x10 23 mc 1 mole x10 23 mc 1 mole n 8820 g____= 1.5x g 6.02x x10 23 = 2.56x10 23 mc

30 30 Gases and the Mole

31 31 Gases n Many of the chemicals we deal with are gases. n They are difficult to weigh. n Need to know how many moles of gas we have. n Two things effect the volume of a gas n Temperature and pressure n Scientists compare gases at Standard Temperature and Pressure

32 32 Standard Temperature and Pressure n 0ºC and 1 atm pressure n abbreviated STP n At STP 1 mole of gas occupies 22.4 L n Called the molar volume n Avogadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

33 33 Examples n What is the volume of 4.59 mole of CO 2 gas at STP? n 4.59mole x 22.4L 1 1mole 1 1mole n How many moles is 5.67 L of O 2 at STP? n 5.67L x 1 mole L L n What is the volume of 8.80g of CH 4 gas at STP? n 8.80g CH 4 x 1mole x 22.4L 1 16g CH 4 1mole 1 16g CH 4 1mole = = 103L =.523moles = = 12.3L

34 34 We have learned how to n change moles to grams n moles to atoms n moles to formula units n moles to molecules n moles to liters n molecules to atoms n formula units to atoms n formula units to ions

35 35 Moles Mass Periodic Table

36 36 Moles Mass Volume Periodic Table

37 37 Moles Mass Volume 22.4 L Periodic Table

38 38 Moles Mass Volume Representative Particles 22.4 L Periodic Table

39 x Moles Mass Volume Representative Particles 22.4 L Periodic Table

40 40 Moles Mass Volume Representative Particles 6.02 x Atoms 22.4 L Periodic Table

41 41 Moles Mass Volume Representative Particles 6.02 x Atoms Ions 22.4 L Periodic Table

42 42 Percent Composition n Like all percents n Part x 100 % whole n Find the mass of each component, n divide by the total mass.

43 43 Example n Calculate the percent composition of each element in a compound that is 29.0 g of Ag with 4.30 g of S. n Ag 29.0g n S g n 33.3g /33.3=.8709 x 100 = 87.09% /33.3=.1291 x 100 = 12.91%

44 44 Getting % from the formula n If we know the formula, assume you have 1 mole. n Then you know the pieces and the whole.

45 45 Examples n Calculate the percent composition of C 2 H 4 ? n C 2(12g)=24 n H 4(1g) = +4 n 28g /28 =.8571 x 100 = 85.71% =.1429 x 100 = 14.29%

46 46 Example n Calculate the percent composition of Aluminum carbonate. Aluminum carbonate. n Al 2 (CO 3 ) 3 n Al 2(27g)= 54 n C 3(12g)= 36 n O 9(16)= 144 n 234g /234 =.2308 x 100 = 23.08% = x 100 = 15.38% =.6154 x 100 = 61.54%

47 47 n You can also calculate the mass of an element in a given amount of a compound using % composition. nStep 1: calculate the % comp. only of the element you want to find the mass of. nStep 2: Multiply the elements %, by the mass of the compound given. n Example: Calculate the mass of sulfur in 3.54g of H 2 S. n MM of H 2 S = H 2 (1) = 2 n S 1(32) = +32 n 34g H 2 S n % S = 32/34 x 100 = 94.1% S n 94.1% x 3.54g = 3.33g S

48 48 Calculate the mass of nitrogen in 25g of (NH 4 ) 2 CO 3. n Calculate the mass of nitrogen in 25g of (NH 4 ) 2 CO 3. n N 2(14g) = 28 n H 8(1g) = 8 n C 1(12g) = 12 n O 8(16g) = +128 n 176g (NH 4 ) 2 CO 3 n %N = 28/176 x 100 = 15.91% n 15.91% x 25g = 4.0g N

49 49 Calculate the mass of magnesium in 97.4g of Mg(OH) 2. n Mg 1 (24g) = 24 n O 2(16g) = 32 n H 2 ( 1g) = +2 n 58g Mg(OH) 2 n %Mg = 24/58 x 100 = 41.38% n 41.38% x 97.4g = 40.3g Mg

50 50 Empirical Formula From percentage to formula

51 51 The Empirical Formula n The lowest whole number ratio of elements in a compound. n The molecular formula the actual ration of elements in a compound. n The two can be the same. n CH 2 empirical formula n C 2 H 4 molecular formula n C 3 H 6 molecular formula n H 2 O both

52 52 Calculating Empirical n Just find the lowest whole number ratio n C 6 H 12 O 6 n CH 4 N n It is not just the ratio of atoms, it is also the ratio of moles of atoms. n In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. n In one molecule of CO 2 there is 1 atom of C and 2 atoms of O.

53 53 Calculating Empirical n Means we can get ratio from percent composition. n Assume you have a 100 g. n The percentages become grams. n Can turn grams to moles. n Find lowest whole number ratio by dividing by the smallest moles.

54 54 Example n Calculate the empirical formula of a compound composed of % C, % H, and %N. n Assume 100 g so n g C x 1mol C = mole C 12 g C n g H x 1mol H = mole H 1 g H n g N x 1mol N = mole N 14 g N

55 55 Example n The ratio is mol C = 1 mol C mol N 1 mol N n The ratio is mol H = 5 mol H mol N 1 mol N nC1H5N1nC1H5N1nC1H5N1nC1H5N1

56 56 n A compound is % P and % O. What is the empirical formula? n g P x 1mol P = mole P 31 g P n g O x 1mol O = mole O 16 g O n The ratio is mol O = 2.5 mol O mol P 1 mol P n Can not have 2.5 atoms! Double nP2O5nP2O5nP2O5nP2O5

57 57 n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula? n g C x 1mol C = mole C 12 g C n 5.15 g H x 1mol H = 5.15 mole H 1 g H n g N x 1mol N = mole N 14 g N n g O x 1mol O = mole O 16 g O

58 58 n The ratio is mol C = 4 mol C mol O 1 mol O n The ratio is 5.15 mol H = 5 mol H mol O 1 mol O n The ratio is mol N = 2 mol N mol O 1 mol O nC4H5N2O1nC4H5N2O1nC4H5N2O1nC4H5N2O1

59 59 Empirical to molecular n Since the empirical formula is the lowest ratio the actual molecule would weigh more. n By a whole number multiple. n Divide the actual molar mass by the the mass of one mole of the empirical formula. n Caffeine has a molar mass of 194 g. what is its molecular formula? n C 4 H 5 N 2 O 1 = 97g n 194/97 = 2 n Molecular Formula = C 8 H 10 N 4 O 2

60 60 Example n A compound is known to be composed of % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be g. What is its molecular formula? n g Cl x 1mol Cl = mole Cl 35 g Cl n g C x 1mol C = mole C 12 g C n 4.07 g H x 1mol H = 4.07 mole H 1 g H

61 61 n The ratio is mol Cl = 1 mol Cl mol C 1 mol C n The ratio is 4.07 mol H = 4 mol H mol C 1 mol C n Empirical Formula = CClH 4 n = 51g n 98.96/51 = 2 n Molecular Formula = C 2 Cl 2 H 8


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