Chapter 10 Chemical Quantities Yes you will need
- Slides: 65
Chapter 10 “Chemical Quantities” Yes, you will need a calculator for this chapter! NOT TODAY, but every other day of Chapter 10!!! TODAY YOU NEED: -PERIODIC TABLE 1 -PAPER FOR NOTES
Chapter 10’s essential question: n. BR : How do scientists measure quantities of chemicals and compounds? (Answer this question as well as you can, with as many examples as you can think of) 2
Agenda Bellringer (essential question) n 10. 1 objectives n Lecture n Exit slip 10. 1 n Homework: – Tonight n » Section 10. 1 WS » 10. 1 notes 3
Before we start… if a slide’s TITLE starts with yellow words then write the title down on the left, write what’s underneath on the right of your notes n If a slide starts with WHITE WORDS then just read what’s on the slide n (If something is in parentheses then you don’t need to write it, but you should read and consider it) n 4
Section 10. 1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe methods of measuring the amount of something. 5
Section 10. 1 The Mole: A Measurement of Matter n OBJECTIVES: –Define Avogadro’s number. 6
Section 10. 1 The Mole: A Measurement of Matter n OBJECTIVES: –Distinguish between the atomic mass of an element and its molar mass. 7
Section 10. 1 The Mole: A Measurement of Matter n OBJECTIVES: –Describe how the mass of a mole of a compound is calculated. 8
How do we measure items? You can measure mass § In grams § or volume § In liters § or you can count pieces. § In moles § 9
What is the mole? NOT this kind of mole! 10
Moles (is abbreviated: mol) 23 § 1 mole = 6. 02 x 10 of the representative particles. § It is based on the number of carbon atoms in exactly 12 grams of carbon-12. § Like a much larger dozen § 6. 02 x 1023 is called: 11 Avogadro’s number.
6. 02 x 1023 n 602, 000, 000, 000 n You DON’T want to write this EVERYTIME!!!!! 12
Similar Words for an amount Pair: 1 pair of shoelaces = 2 shoelaces § Dozen: 1 dozen oranges = 12 oranges § Gross: 1 gross of pencils = 144 pencils § Ream: 1 ream of paper 13 = 500 sheets of paper §
What are Representative Particles? § 14 The smallest pieces of a substance: 1) Molecular compound: it is the molecule. 2) Ionic compound: it is the formula unit (made of ions). 3) Element: it is the atom. » Remember the 7 diatomic elements? (made of molecules)
Counting Representative Particles n How many oxygen atoms in the following? 3 atoms of oxygen Ca. CO 3 12 (3 x 4) atoms of oxygen Al 2(SO 4)3 n How many ions in the following? 2+ ion and 2 Cl 1 - ions) 3 total ions (1 Ca Ca. Cl 2 2 total ions (1 Na 1+ ion and 1 OH 1 - ion) Na. OH Al 2(SO 4)3 5 total ions (2 Al 3+ + 3 SO 42 - ions) 15
Counting Representative Particles n Counting Representative Particles is useful in MOLE CONVERSION because: – 1 mole = 6. 02 x 1023 molecules – 1 mole = 6. 02 x 1023 atoms 16
Practice problems (coverting moles to molecules and/or atoms) § How many molecules of CO 2 are in 4. 56 moles of CO 2? 2. 75 x 10 molecules How many moles of water is 5. 87 x 1022 molecules? 0. 0975 mol (or 9. 75 x 10 ) How many atoms of carbon are in 1. 23 moles of C 6 H 12 O 6? 4. 44 x 10 atoms C How many moles is 7. 78 x 1024 formula units of Mg. Cl 2? 12. 9 moles 24 § -2 § 24 § 17
Wednesday Bellringer § How many molecules of O 2 are in 5. 56 moles of O 2? § How many moles of water is 3. 87 x 1022 molecules? 18
What to use? A)moles molecules B)molecules moles (×) (÷) ___6. 022 x 1023 ____ 1 MOLE 19 __1 MOLE___ 6. 022 x 1023
Measuring Moles § § 20 Remember relative atomic mass? - The relative atomic mass is found on the periodic table. The decimal number on the periodic table is also the mass of 1 mole of those atoms in grams. For example 1 mole of Cl = 35. 453 g of Cl (35. 453 g is found on the periodic table) Knowing that 1 mole equals the mass of the periodic table is helpful in calculating problems
Examples (converting mole to mass or mass to mole) § § § How much would 2. 34 moles of carbon weigh? 28. 1 grams C How many moles of magnesium is 24. 31 g of Mg? 1 mol Mg How many atoms of lithium is 1. 00 g of Li? 8. 72 x 10 atoms Li How much would 3. 45 x 1022 atoms of U weigh? 13. 6 grams U 22 § 21
What about compounds? § § 22 in 1 mole of H 2 O molecules there are two moles of H atoms and 1 mole of O atoms (think of a compound as a molar ratio) To find the mass of one mole of a compound –determine the number of moles of the elements present –Multiply the number times their mass (from the periodic table) –add them up for the total mass
Calculating Formula Mass Calculate the formula mass of magnesium carbonate, Mg. CO 3. 24. 3 g 23 + 12 g + 3 x (16. 00 g) = 84. 3 g Thus, 84. 3 grams is the formula mass for Mg. CO 3.
Review of 10. 1 Measurement is by – Count (moles) – Mass (grams) – Volume (liters) n Avogadro’s number – 6. 022 x 1023 of anything n Atomic mass Mass of one atom (very small) n Molar mass Mass of one mole of an atom, molecule, or formula unit n 24 What you see on periodic table is either one) n
Review of 10. 1 n Calculating molar mass of a formula – Add molar mass of each atom in a formula – Example: H 2 O has: » 2 hydrogen atoms (1. 0079 x 2) = 2. 0158 amu » 1 oxygen atom (15. 999) = 15. 999 amu H 2 O molar mass = 18. 0148 amu (rounded to 4 sig figs = 18. 01 amu) 25
What to use? A)moles molecules B)molecules moles (×) (÷) ___6. 022 x 1023 ____ 1 MOLE __1 MOLE___ 6. 022 x 1023 C)moles mass (×) D)Mass moles (÷) ___Molar mass___ 1 MOLE ___ Molar mass 26
Exit slip Calculate the average molar mass of: – Oxygen gas: O 2 – Hydrochloric acid: HCl – Sugar: C 6 H 12 O 6 – Aluminum Oxide Al 2 O 3 n Calculate how many GRAMS of each of the above would be in 3. 45 moles of each n 27
Thursday BR 11. 24. 11 n 28 Find the molar mass of the following. –C –O –H – CO – H 2 O 2 – C 6 H 12 O 6
What to use? A)moles atoms/molecules (×) B) atoms/molecules moles (÷) ___6. 02 x 1023 ____ 1 MOLE __1 MOLE___ 6. 02 x 1023 C)moles mass(grams) (×) D) mass(grams) moles (÷) ___Molar mass___ 1 MOLE ___ Molar mass 29
Section 10. 2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: 30 –Describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass.
Section 10. 2 Mole-Mass and Mole-Volume Relationships n OBJECTIVES: –Identify the volume of a quantity of gas at STP. 31
Molar Mass § Molar mass is the generic term for the mass of one mole of any substance (expressed in grams/mol) § The same as: 1) Gram Molecular Mass (for molecules) 2) Gram Formula Mass (ionic compounds) 3) Gram Atomic Mass (for elements) 32 – molar mass is just a much broader term than these other specific masses
Examples Calculate the molar mass of the following and tell what type it is: = 78 g/mol gram formula mass Na 2 S = 92 g/mol gram molecular mass N 2 O 4 C = 12 g/mol gram atomic mass Ca(NO 3)2 = 164 g/mol gram formula mass C 6 H 12 O 6 = 180 g/mol gram molecular mass (NH 4)3 PO 4 = 149 g/mol gram formula mass § 33
Since Molar Mass is… § The number of grams in 1 mole of atoms, ions, or molecules, § We can make conversion factors from these. - To change between grams of a compound and moles of a compound. 34
For example § How many moles is 5. 69 g of Na. OH? (Solution on next slides) 35
For example n 36 How many moles is 5. 69 g of Na. OH?
For example 37 n How many moles is 5. 69 g of Na. OH? l We need to change 5. 69 grams Na. OH to moles
For example n How many moles is 5. 69 g of Na. OH? We need to change 5. 69 grams Na. OH to moles l 1 mole Na = 23 g 1 mol O = 16 g 1 mole of H = 1 g l 38
For example n How many moles is 5. 69 g of Na. OH? We need to change 5. 69 grams Na. OH to moles l 1 mole Na = 23 g 1 mol O = 16 g 1 mole of H = 1 g l 1 mole Na. OH = 40 g l 39
For example n How many moles is 5. 69 g of Na. OH? We need to change 5. 69 grams Na. OH to moles l 1 mole Na = 23 g 1 mol O = 16 g 1 mole of H = 1 g l 1 mole Na. OH = 40 g l 40
For example n How many moles is 5. 69 g of Na. OH? We need to change 5. 69 grams Na. OH to moles l 1 mole Na = 23 g 1 mol O = 16 g 1 mole of H = 1 g l 1 mole Na. OH = 40 g l 41
The Mole-Volume Relationship § Many of the chemicals we deal with are in the physical state as: gases. - They are difficult to weigh (or mass). § But, we may still need to know how many moles of gas we have. § Two things effect the volume of a gas: a) Temperature and b) Pressure § We need to compare all gases at the 42 same temperature and pressure.
Standard Temperature and Pressure 0ºC and 1 atm pressure - is abbreviated “STP” § At STP, 1 mole of any gas occupies a volume of 22. 4 L - Called the molar volume § This is our fourth equality: § 43 1 mole of any gas at STP = 22. 4 L
Practice Examples § What is the volume of 4. 59 mole of CO 2 gas at STP? = 103 L § How many moles is 5. 67 L of O 2 at STP? = 0. 253 mol § What is the volume of 8. 8 g of CH 4 gas at STP? = 12. 3 L 44
Density of a gas D = m / V (density = mass/volume) - for a gas the units will be: g / L § We can determine the density of any gas at STP if we know its formula. § To find the density we need: 1) mass and 2) volume. § If you assume you have 1 mole, then the mass is the molar mass (from periodic table) 45§ And, at STP the volume is 22. 4 L. §
Practice Examples (D=m/V) § Find the density of CO 2 at STP. D = 44 g/22. 4 L = 1. 96 g/L § Find the density of CH 4 at STP. D = 16 g/22. 4 L = 0. 714 g/L 46
Another way: § § 47 § If given the density, we can find the molar mass of the gas. Again, pretend you have 1 mole at STP, so V = 22. 4 L. m=Dx. V modify: D = m/V to show: “m” will be the mass of 1 mole, since you have 22. 4 L of the stuff. What is the molar mass of a gas with a density of 1. 964 g/L? = 44. 0 g/mol How about a density of 2. 86 g/L? = 64. 0 g/mol
Summary n 48 These four items are all equal: a) 1 mole b) molar mass (in grams/mol) c) 6. 02 x 1023 representative particles (atoms, molecules, or formula units) d) 22. 4 L of gas at STP Thus, we can make conversion factors from these 4 values!
Section 10. 3 Percent Composition and Chemical Formulas n OBJECTIVES: –Describe how to calculate the percent by mass of an element in a compound. 49
Section 10. 3 Percent Composition and Chemical Formulas n OBJECTIVES: –Interpret an empirical formula. 50
Section 10. 3 Percent Composition and Chemical Formulas n OBJECTIVES: –Distinguish between empirical and molecular formulas. 51
Calculating Percent Composition of a Compound § 1) 2) 52 Like all percent problems: part x 100 % = percent whole Find the mass of each of the components (the elements), Next, divide by the total mass of the compound; then x 100
Example § Calculate the percent composition of a compound that is made of 29. 0 grams of Ag with 4. 30 grams of S. 53 29. 0 g Ag X 100 = 87. 1 % Ag 33. 3 g total 4. 30 g S X 100 = 12. 9 % S 33. 3 g total Total = 100 %
Getting it from the formula If we know the formula, assume you have 1 mole, § then you know the mass of the elements and the whole compound (these values come from the periodic table!). § 54
Examples § Calculate the percent composition of C 2 H 4? 85. 7% C, 14. 3 % H § How about Aluminum carbonate? 23. 1% Al, 15. 4% C, and 61. 5 % O § Sample Problem 10. 10, p. 307 § We can also use the percent as a conversion factor § Sample Problem page 308 55
Formulas Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound. • Example: molecular formula for benzene is C 6 H 6 (note that everything is divisible by 6) • Therefore, the empirical formula = (the lowest whole number ratio) 56 CH
Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced). Examples: Na. Cl 57 Mg. Cl 2 Al 2(SO 4)3 K 2 CO 3
Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: (Correct formula) Empirical: (Lowest whole number ratio) 58 H 2 O C 6 H 12 O 6 C 12 H 22 O 11 H 2 O CH 2 O C 12 H 22 O 11
Calculating Empirical Just find the lowest whole number ratio C 6 H 12 O 6 = CH 2 O CH 4 N = this is already the lowest ratio. § A formula is not just the ratio of atoms, it is also the ratio of moles. § In 1 mole of CO 2 there is 1 mole of carbon and 2 moles of oxygen. § In one molecule of CO 2 there is 1 atom 59 of C and 2 atoms of O. §
Calculating Empirical § 1) We can get a ratio from the percent composition. Assume you have a 100 g sample - the percentage become grams (75. 1% = 75. 1 grams) 2) 3) 60 Convert grams to moles. Find lowest whole number ratio by dividing each number of moles by the smallest value.
Example § Calculate the empirical formula of a compound composed of 38. 67 % C, 16. 22 % H, and 45. 11 %N. § Assume 100 g sample, so 38. 67 g C x 1 mol C = 3. 22 mole C 12. 0 g C 16. 22 g H x 1 mol H = 16. 22 mole H 1. 0 g H 45. 11 g N x 1 mol N = 3. 22 mole N 14. 0 g N § § § 61 Now divide each value by the smallest value
Example § § The ratio is 3. 22 mol C = 1 mol C 3. 22 mol N 1 mol N The ratio is 16. 22 mol H = 5 mol H 3. 22 mol N 1 mol N = C 1 H 5 N 1 § § 62 which is = CH 5 N A compound is 43. 64 % P and 56. 36 % O. What is the empirical formula? = P 2 O 5 Caffeine is 49. 48% C, 5. 15% H, 28. 87% N and 16. 49% O. What is its empirical formula? = C 4 H 5 N 2 O
Empirical to molecular Since the empirical formula is the lowest ratio, the actual molecule would weigh more. §By a whole number multiple. § Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula § Caffeine has a molar mass of 194 g. 63 what is its molecular formula? = C 8 H 10 N 4 O 2 §
Note page 313 – Gas Chromatography used for chemical analysis 64
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