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1 Chapter 8 Chemical Quantities. 2 How you measure how much? How you measure how much? n You can measure mass, n or volume, n or you can count pieces.

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Presentation on theme: "1 Chapter 8 Chemical Quantities. 2 How you measure how much? How you measure how much? n You can measure mass, n or volume, n or you can count pieces."— Presentation transcript:

1 1 Chapter 8 Chemical Quantities

2 2 How you measure how much? How you measure how much? n You can measure mass, n or volume, n 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- 12. n 1 mole is 6.02 x 10 23 particles. n Treat it like a very large dozen n 6.02 x 10 23 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 ) 3

6 6 Types of questions n How many molecules of CO 2 are the in 4.56 moles of CO 2 ? n How many moles of water is 5.87 x 10 22 molecules? n How many atoms of carbon are there in 1.23 moles of C 6 H 12 O 6 ? n How many moles is 7.78 x 10 24 formula units of MgCl 2 ?

7 7 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.

8 8 Gram Atomic Mass n The mass of 1 mole of an element in grams. n 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron. n We can right this as 12.01 g C = 1 mole n We can count things by weighing them.

9 9 Examples n How much would 2.34 moles of carbon weigh? n How many moles of magnesium in 24.31 g of Mg? n How many atoms of lithium in 1.00 g of Li? n How much would 3.45 x 10 22 atoms of U weigh?

10 10 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 –determine the moles of the elements they have –Find out how much they would weigh –add them up

11 11 What about compounds? n What is the mass of one mole of CH 4 ? n 1 mole of C = 12.01 g n 4 mole of H x 1.01 g = 4.04g n 1 mole CH 4 = 12.01 + 4.04 = 16.05g n The Gram Molecular mass of CH 4 is 16.05g n The mass of one mole of a molecular compound.

12 12 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 55.85 g = 111.70 g n 3 moles of O x 16.00 g = 48.00 g n The GFM = 111.70 g + 48.00 g = 159.70g

13 13 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.

14 14 Examples n Calculate the molar mass of the following and tell me what type it is. n Na 2 S nN2O4nN2O4nN2O4nN2O4 nCnCnCnC n Ca(NO 3 ) 2 n C 6 H 12 O 6 n (NH 4 ) 3 PO 4

15 15 Using Molar Mass Finding moles of compounds Counting pieces by weighing

16 16 Molar Mass n The number of grams of 1 mole of atoms, ions, or molecules. n We can make conversion factors from these. n To change grams of a compound to moles of a compound.

17 17 For example n How many moles is 5.69 g of NaOH?

18 18 For example n How many moles is 5.69 g of NaOH?

19 19 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles

20 20 For example n How many moles is 5.69 g of NaOH? l need to change grams to moles l for NaOH

21 21 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 = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g

22 22 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 = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

23 23 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 = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

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 = 22.99g 1 mol O = 16.00 g 1 mole of H = 1.01 g l 1 mole NaOH = 40.00 g

25 25 Examples n How many moles is 4.56 g of CO 2 ? n How many grams is 9.87 moles of H 2 O? n How many molecules in 6.8 g of CH 4 ? n 49 molecules of C 6 H 12 O 6 weighs how much?

26 26 Gases and the Mole

27 27 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 Compare at the same temp. and pressure.

28 28 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 Avagadro’s Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.

29 29 Examples n What is the volume of 4.59 mole of CO 2 gas at STP? n How many moles is 5.67 L of O 2 at STP? n What is the volume of 8.8g of CH 4 gas at STP?

30 30 Density of a gas n D = m /V n for a gas the units will be g / L n We can determine the density of any gas at STP if we know its formula. n To find the density we need the mass and the volume. n If you assume you have 1 mole than the mass is the molar mass (PT) n At STP the volume is 22.4 L.

31 31 Examples n Find the density of CO 2 at STP. n Find the density of CH 4 at STP.

32 32 The other way n Given the density, we can find the molar mass of the gas. n Again, pretend you have a mole at STP, so V = 22.4 L. n m = D x V n m is the mass of 1 mole, since you have 22.4 L of the stuff. n What is the molar mass of a gas with a density of 1.964 g/L? n 2.86 g/L?

33 33 All the things we can change

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

36 36 Moles Mass PT

37 37 Moles Mass Volume PT

38 38 Moles Mass Volume PT 22.4 L

39 39 Moles Mass Volume Representative Particles PT 22.4 L

40 40 6.02 x 10 23 Moles Mass Volume Representative Particles PT 22.4 L

41 41 Moles Mass Volume Representative Particles 6.02 x 10 23 PT Atoms 22.4 L

42 42 Moles Mass Volume Representative Particles 6.02 x 10 23 PT Atoms Ions 22.4 L

43 43 Percent Composition n Like all percents n Part x 100 % whole n Find the mass of each component, n divide by the total mass.

44 44 Example n Calculate the percent composition of a compound that is 29.0 g of Ag with 4.30 g of S.

45 45 Getting it from the formula n If we know the formula, assume you have 1 mole. n Then you know the pieces and the whole.

46 46 Examples n Calculate the percent composittion of C 2 H 4 ? n Aluminum carbonate.

47 47 Empirical Formula From percentage to formula

48 48 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

49 49 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.

50 50 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.

51 51 Example n Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. n Assume 100 g so n 38.67 g C x 1mol C = 3.220 mole C 12.01 gC n 16.22 g H x 1mol H = 16.09 mole H 1.01 gH n 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

52 52 Example n The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N n The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N nC1H5N1nC1H5N1nC1H5N1nC1H5N1 n A compound is 43.64 % P and 56.36 % O. What is the empirical formula? n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?

53 53 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 mass?

54 54 Example n A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mas is known (from gas density) is known to be 98.96 g. What is its molecular formula?


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