Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- 48270042 biology-stpm-lower-6-chapt... by janicetiong 22458 views
- 57996218-STPM-Chemistry-Form-6 by Manniseh Yakun 5719 views
- STPM Form 6 Chemistry Solids by Sook Yen Wong 1499 views
- IB Biology Core 3.1: Chemical Eleme... by Jason de Nys 8637 views
- MUET notes by azam_hazel 454 views
- 962 Sukatan Pelajaran Kimia STPM (B... by SMK Dato' Sheikh ... 9510 views

4,532 views

Published on

STPM Form 6 Chemistry Solids

Published in:
Education

No Downloads

Total views

4,532

On SlideShare

0

From Embeds

0

Number of Embeds

5

Shares

0

Downloads

0

Comments

0

Likes

8

No embeds

No notes for slide

- 1. Mole and Avogadro Constant 1
- 2. The Mole• A counting unit• Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000• 6.02 X 1023 (in scientific notation)• This number is named in honor of Amedeo _________ (1776 – 1856), 1856) who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present
- 3. Avogadro’s Number• Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon.• Its numerical value is 6.02 × 1023.• Therefore, a 12.01 g sample of carbon contains 6.02 × 1023 carbon atoms. 3
- 4. The Mole• The mole (mol) is a unit of measure for an amount of a chemical substance.• A mole is Avogadro’s number of particles, which is 6.02 × 1023 particles. 1 mol = Avogadro’s number = 6.02 × 1023 units• We can use the mole relationship to convert between the number of particles and the mass of a substance. 4
- 5. Analogies for Avogadro’s Number• The volume occupied by one mole of softballs would be about the size of the Earth.• One mole of Olympic shotput balls has about the same mass as the Earth. 5
- 6. One Mole of Several Substances C12H22O11 H 2O lead mercury K2Cr2O7 sulfurcopper NaCl 6
- 7. Example 1• How many sodium atoms are in 0.120 mol Na? 7
- 8. Example 2• How many moles of potassium are in 1.25 × 1021 atoms K? 8
- 9. 9
- 10. Molar Mass• The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance.• The atomic mass of iron is 55.85 amu.• Therefore, the molar mass of iron is 55.85 g/mol.• Since oxygen occurs naturally as a diatomic, O2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol. 10
- 11. 11
- 12. Calculating Molar Mass• The molar mass of a substance is the sum of the molar masses of each element. 12
- 13. Example 3Calc. the molar mass: 1. Fe2O3 2. C6H8O7 3. C16H18N2O4 13
- 14. Mole Calculations• Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance. 6.02 × 1023 particles = 1 mol = molar mass• If we want to convert particles to mass, we must first convert particles to moles and then we can convert moles to mass. 14
- 15. Example 4• What is the mass of 1.33 moles of titanium, Ti? 15
- 16. Example 5• What is the mass of 2.55 × 1023 atoms of lead? 16
- 17. Example 6• How many O2 molecules are present in 0.470 g of oxygen gas? 17
- 18. Example 7• What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is 64.07 g/mol. 18
- 19. 19
- 20. Molar Volume• At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.• The volume occupied by 1 mole of gas (22.4 L) is called the molar volume.• Standard temperature and pressure are 0°C and 1 atm. 20
- 21. Molar Volume of Gases• We now have a new unit factor equation: 1 mole gas = 6.02 × 1023 molecules gas = 22.4 L gas 21
- 22. Gas Density• The density of gases is much less than that of liquids.• We can calculate the density of any gas at STP easily.• The formula for gas density at STP is: molar mass in grams/mol = density, g/L molar volume in liters/mol 22
- 23. Example 8• What is the density of ammonia gas, NH3, at STP? 23
- 24. Example 9• We can also use molar volume to calculate the molar mass of an unknown gas.• 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass?• We want g/mol; we have g/L. 24
- 25. Mole Unit Factors• We now have three interpretations for the mole: – 1 mol = 6.02 × 1023 particles – 1 mol = molar mass – 1 mol = 22.4 L at STP for a gas• This gives us 3 unit factors to use to convert between moles, particles, mass, and volume. 25
- 26. Example 10• A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present? 26
- 27. Example 11• What is the mass of 3.36 L of ozone gas, O3, at STP? 27
- 28. Example 12• How many molecules of hydrogen gas, H2, occupy 0.500 L at STP? 28
- 29. ConcentrationWhich beaker oftea is stronger?Which beaker oftea is moreconcentrated?
- 30. ConcentrationConcentration = amount of solute in aspecific volumeMore solute = higher concentration
- 31. ConcentrationConcentration can be measured in three ways: 1. Percent by Volume 2. Percent by Mass 3. Molarity
- 32. Percent Composition• The percent composition of a compound lists the mass percent of each element.• For example, the percent composition of water, H2O is: – 11% hydrogen and 89% oxygen• All water contains 11%hydrogen and 89% oxygenby mass. 32
- 33. Calculating Percent Composition• There are a few steps to calculating the percent composition of a compound. Let’s practice using H2O. – Assume you have 1 mole of the compound. – One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. – 2(1.01 g H) + 1(16.00 g O) = molar mass H2O – 2.02 g H + 16.00 g O = 18.02 g H2O 33
- 34. Calculating Percent Composition• Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water: 2.02 g H %H= × 100% = 11.2% H 18.02 g H2O 16.00 g O % O = 18.02 g H O × 100% = 88.79% O 2 34
- 35. Example 13• TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3. 35
- 36. Percent Composition of TNT 36
- 37. Empirical Formulas• The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule.• The molecular formula of benzene is C6H6. – The empirical formula of benzene is CH.• The molecular formula of octane is C8H18. – The empirical formula of octane is C4H9. 37
- 38. Calculating Empirical Formulas• We can calculate the empirical formula of a compound from its composition data.• We can determine the mole ratio of each element from the mass to determine the formula of radium oxide, Ra? O?.• A 1.640 g sample of radium metal was heated to produce 1.755 g of radium oxide. What is the empirical formula?• We have 1.640 g Ra and 1.755-1.640 = 0.115 g O. 38
- 39. Calculating Empirical Formulas• The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol. 1 mol Ra 1.640 g Ra × = 0.00726 mol Ra 226.03 g Ra 1 mol O 0.115 g O × = 0.00719 mol O 16.00 g O• We get Ra0.00726O0.00719. Simplify the mole ratio by dividing by the smallest number.• We get Ra1.01O1.00 = RaO is the empirical formula. 39
- 40. Example 14• Determine the empirical formula of sodium sulfate given the following dataA 3.00 g sample of NaxSyOz is comprised of 0.9722 g of sodium, 0.678 g of sulfur and 1.35 g of oxygen 40
- 41. Empirical Formulas from Percent Composition• We can also use percent composition data to calculate empirical formulas.• Assume that you have 100 grams of sample.• Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula?• If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen. 41
- 42. Empirical Formula for Acetylene• Calculate the moles of each element: 1 mol C 92.2 g C × = 7.68 mol C 12.01 g C 1 mol H 7.83 g H × = 7.75 mol H 1.01 g H • The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the formula: 7.68 7.75 C 7.68 H 7.68 = C1.00H1.01 = CH 42
- 43. Example 15• Find the empirical formula of Iron(III) chloride from the following data• In Iron(III) chloride – Fe 34.43 % by mass – Cl 65.57 % by mass 43
- 44. Molecular Formulas• The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene.• The actual molecular formula is some multiple of the empirical formula, (CH)n.• Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula: (CH)n 26 g/mol n = 2 and the molecular = CH 13 g/mol formula is C2H2. 44
- 45. Example 16• Find the empirical and molecular formula for hydrazine, NxHy• The molar mass of hydrazine is 32.046 g/mol• In hydrazine – N is 87.42% by mass – H is 12.58 % by mass 45
- 46. Chapter Summary• We can use the following flow chart for mole calculations: 46

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment