1. CHEMICAL QUANTITIES Chapter 10

2. Section Overview 10.1: The Mole: A Measurement of Matter 10.2: Mole-Mass and Mole-Volume Relationships 10.3: Percent Composition and Chemical Formulas

3. THE MOLE: A MEASUREMENT OF MATTER Section 10.1

4. Measuring Matter You often measure the amount of something by one of three different methods: count, mass, and volume. For example, you can measure the amount of apples in those three ways. By count: 1 dozen apples = 12 apples By mass: 1 dozen apples = 2.0kg apples By volume: 1 dozen apples – 0.20 bushel apples. Then you can use that information as conversion factors: 1 dozen apples/12 apples 2.0kg apples/1 dozen apples 1 dozen apples/0.20 bushel apples

5. Measuring Mass Example Problem: What is the mass of 90 average-sized apples if 1 dozen of the apples have a mass of 2.0kg?

6. Measuring Mass Example Problem: What is the mass of 90 average-sized apples if 1 dozen of the apples have a mass of 2.0kg? Solution: KnownsUnknown Number = 90 applesMass of 90 apples 12 apples = 1 dozen apples 1 dozen apples = 2.0kg Number of apples  dozens of apples  mass of apples 90 apples X 1 dozen/12 apples X 2.0kg apples/1 dozen =15kg apples

7. What is a Mole? A mole of a substance is 6.02 X 10 23 representative particles of that substance and is the SI unit for measuring the amount of a substance. The representative number is called Avogadro’s number. The term representative particle refers to the species present in the substance: usually atoms, molecules, or formula units. A mole of any substance contains Avogadro’s number of representative particles.

8. What is a Mole? Converting Number of Particles to Moles: Converting Moles to Number of Particles: Moles = representative particles X 1 mole/6.02 X 10 23 rep. particles Rep. particles = moles X 6.02 X 10 23 rep. particles/1 mole Next step = multiply the number of rep. particles by the number of atoms in each molecule or formula unit.

9. Converting Number of Atoms to Moles Example Problem: How many moles of magnesium is 1.25 X 10 23 atoms of magnesium?

10. Converting Number of Atoms to Moles Example Problem: How many moles of magnesium is 1.25 X 10 23 atoms of magnesium? Solution: KnownUnknown # of atoms = 1.25 X 10 23 moles Mg 1 mol Mg = 6.02 X 10 23 atoms Mg Conversion: atoms  moles Moles=1.25 X 10 23 atomsMg X 1molMg/6.2X10 23 atoms Mg Moles=2.08 X 10-1 mol Mg = 0.208 mol Mg

11. Converting Moles to Number of Atoms Example Problem: How many atoms are in 2.12 mol of propane (C 3 H 8 )?

12. Converting Moles to Number of Atoms Example Problem: How many atoms are in 2.12 mol of propane (C 3 H 8 )? Solution: Knowns # of moles = 2.12 mol C 3 H 8 1 mol C 3 H 8 = 6.02 X 10 23 molecules C 3 H 8 1 molecule C 3 H 8 = 11 atoms Conversion = moles  molecules  atoms 2.12 mol C 3 H 8 X 6.02 X10 23 molecules/1 mol C 3 H 8 X 11 atoms /1 molecule C 3 H 8 =1.4039 X 10 25 atoms = 1.4 X 10 25 atoms

13. The Mass of a Mole of an Element Remember that the atomic mass of an element (the mass of a single atom) is expressed in atomic mass units (amu). The atomic masses are relative values based on the mass of the most common isotope of carbon. However, 1uantities measured in grams are more convenient for use in the laboratory. The atomic mass of an element expressed in grams is the mass of a mole of the element. The mass of a mole of an element is its molar mass. The molar masses of any two elements must contain the same number of atoms.

14. The Mass of a Mole of a Compound To find the mass of a mole of a compound, you must know the formula for the compound. To calculate the molar mass of a compound, find the number of grams of each element in one mole of the compound. Then add the masses of the elements in the compound. Example: Sulfur trioxide = SO 3 = 1 S atom + 3 O atoms S =32.1 amu and O = 16.0 amu 32.1 + 3(16.0) = 80.1 amu So, 1 mole of SO 3 has a mass of 80.1g = 6.02 X 10 23 molecules of SO 3.

15. Finding the Molar Mass of a Compound Example Problem: What is the molar mass of hydrogen peroxide (H 2 O 2 )?

16. Finding the Molar Mass of a Compound Example Problem: What is the molar mass of hydrogen peroxide (H 2 O 2 )? Solution: Knowns Molecular formula = H 2 O 2 1 molar mass H = 1 mol H = 1.0g H 1 molar mass O = 1 mol O = 16.0g O Convert moles of hydrogen and oxygen to grams and add. 2 mol H X 1.0 g H / 1 mol H = 2.0g H 2 mol O X 16.0 g O /1 mol O = 32.0g O 2.0g + 32.0g = 34.0 g H 2 O 2

17. MOLE-MASS AND MOLE-VOLUME RELATIONSHIPS Section 10.2

18. The Mole-Mass Relationship Use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance. The conversion factor is based on the relationship: molar mass = 1 mol. For example, the molar mass of NaCl is 58.5g/mol, so the mass of 3.00 mol NaCl is calculated like so: Mass of NaCl = 3.00mol X 58.5g/1mol = 176g Mass (grams) = number of moles X mass(grams) / 1 mol

19. The Mass-Mole Relationship You can also covert from mass to moles. For example, the molar mass of Na 2 SO 4 is 142.1 g/mol, so the number of moles is calculated like so: Mol of Na 2 SO 4 = 10.0 g X 1 mol/142.1g = 7.04 X 10 -2 mol Moles = mass (grams) X 1 mol / mass (grams)

20. The Mass-Volume Relationship The volumes of one mole of different solid and liquid substances are not the same (ex. The mass of one mol of water is smaller than one mol of glucose). However, the volumes of moles of gases are more similar and predictable. Avogadro's hypothesis states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles. The particles that make up different gases are not the same size, but they are so far apart that a collection of large particles does not require much more space than the same number of small particles.

21. The Mass-Volume Relationship The volume of gas does change with varieties in pressure and temperature. Due to these differences, the volume of gas is usually measured at a standard temperature and pressure (STP). STP means a temperature of 0°C and a pressure of 101.3 kPa, or 1 atmosphere (atm). At STP, 1 mole (6.02 X 1023 representative particles) of any gas occupied a volume of 22.4 L (also called molar volume of gas).

22. The Mass-Volume Relationship Calculating Volume at STP: The molar volume is used to convert a known number of moles of gas to the volume of gas at STP. For example, what volume will 0.375 mol of oxygen gas occupy at STP? Volume of O 2 = 0.375 mol X 22.4L /1 mol = 8.40 L Volume of gas = moles of gas X 22.4 L / 1 mol Moles = volume of gas X 1 mol / 22.4 L

23. The Mass-Volume Relationship Calculating the Molar Mass from Density: Different gasses have different densities. Usually, it is measured in grams per liter (g/L) and at a specific temperature. For example, what is the molar mass of a compound with a density of 1.964g/L at STP? Molar mass = 1.964g/L X 22.4L/1mol = 44.0g/mol Molar mass = density at STP X molar volume at STP grams/mole = grams/L X 22.4/1mol

24. PERCENT COMPOSITION AND CHEMICAL FORMULAS Section 10.3

25. The Percent Composition of a Compound The relative amounts of the elements in a compound are expressed as the percent composition or the percent by mass of each element in the compounds. The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%. % of mass of element = mass of element/mass of compound X 100%

26. The Percent Composition of a Compound Percent Composition from Mass Data Example Problem: When a 13.60g sample of a compound containing magnesium and oxygen is decomposed, 5.40gof oxygen is obtained. What is the percent composition of this compound?

27. The Percent Composition of a Compound Example Problem: When a 13.60g sample of a compound containing magnesium and oxygen is decomposed, 5.40gof oxygen is obtained. What is the percent composition of this compound? Solution: KnownsUnknown Mass of compound = 13.60g% O Mass of oxygen = 5.40g O% Mg Mass of magnesium = 13.60g-5.40g = 8.20g Mg

28. The Percent Composition of a Compound Solution: KnownsUnknown Mass of compound = 13.60g% O Mass of oxygen = 5.40g O% Mg Mass of magnesium = 13.60g-5.40g = 8.20g Mg % O = mass of O/mass of cmpd X 100% % O = 5.40g/13.60g X 100% = 39.7% O % Mg = mass of Mg/mass of cmpd X 100% % Mg = 8.20g/13.60gX 100% = 60.3% Mg 39.7% + 60.3% = 100%

29. The Percent Composition of a Compound Percent Composition from the Chemical Formula: The subscripts in the formula of the compound are used to calculate the mas of each element in a mole of that compound. The sum of those masses is the molar mass. % mass=mass of element in 1mol of cmpd/molar mass of cmpdX100%

30. The Percent Composition of a Compound Percent Composition from the Chemical Formula Example Problem: Calculate the percent composition of propane (C 3 H 8 ).

31. The Percent Composition of a Compound Percent Composition from the Chemical Formula Example Problem: Calculate the percent composition of propane (C 3 H 8 ). Solution: Knowns:Unknowns: Mass of C in 1 Mol = 36.0g (12x3) % of C Mass of H in 1 Mol = 8.0g (2x4) % of H Molar mass of propane = 44.0g/mol (36.0g+8.0g)

32. The Percent Composition of a Compound Solution: Knowns:Unknowns: Mass of C in 1 Mol = 36.0g (12x3) % of C Mass of H in 1 Mol = 8.0g (2x4) % of H Molar mass of propane = 44.0g/mol (36.0g+8.0g) % C = mass of C/mass of propane X 100% % C = 36.0g/44.0g X 100% = 81.8% C % H = mass of H/mass of propane X 100% % H = 8.0g/44.0g X 100% = 18% H

33. The Percent Composition of a Compound Percent Composition as a Conversion Factor: You can use percent composition to calculate the number of gram of any element in a specific mass of a compound. To do this, multiply the mass of the compound by a conversion factor based on the percent composition of the element in the compound. For example, suppose you want to know how much carbon and hydrogen are contained in 82.0g of propane. You already know propane is 81.8% C and 18% hydrogen. 82.0g propane X 81.8g C/100g propane = 67.1g C 82.0g propane X 18g H/100g propane = 15g H

34. Empirical Formula The percent composition is the data you need to calculate the ratio of the elements contained in a compound. The empirical formula gives the lowest whole-number ratio of the atoms of the elements in a compound. The empirical formula may or may not be the same as the molecular formula. The molecular formula tells the actual number of each kind of atoms present in a compound and is not simplified, however, the ratio is the same.

35. Determining the Empirical Formula of a Compound Example Problem: A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?

36. Determining the Empirical Formula of a Compound Example Problem: A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound? Solution: KnownsUnknown Percent of N= 25.9% Empirical Formula = N ? O ? Percent of O = 74.1% (% means per 100, you can assume 100.g of the compound contains 25.9g of N and 74.1g of O and use as conversion factors).

37. Determining the Empirical Formula of a Compound Solution: KnownsUnknown Percent of N= 25.9% Empirical Formula = N ? O ? Percent of O = 74.1% 25.9g N X 1mol N/14.0g N = 1.85 mol N 74.1 O X 1mol O/16.0g O = 4.63 mol O Must be whole numbers, so divide each by lowest. 1.85 mol N/1.85mol N = 1 mol N and 4.63mol O/1.85 mol N = 2.50mol O O still not a whole number. To obtain lowest ratio, multiply each by 2. Empirical Formula = N 2 O 5.

38. Molecular Formulas The molecular formula of a compound is either the same as its experimentally determined empirical formula, or it is a simple whole-number multiple of its empirical formula. Once you know the empirical formula for a compound, you can determine its molecular formula, but you need to know the compound’s molar mass. From the empirical formula, you can calculate the empirical formula mass (efm) and then divide the experimentally determined molar mass by the efm.

39. Finding the Molecular Formula for a Compound Example Problem: Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH 4 N.

40. Finding the Molecular Formula for a Compound Example Problem: Calculate the molecular formula of a compound whose molar mass is 60.0g/mol and empirical formula is CH 4 N. Solution: KnownsUnknown Empirical formula = CH 4 NMolecular formula = ? Molar mass = 60.0g/mol First, find the efm then divide by the molar mass to get a whole number. To get the molecular formula, multiply the subscripts by the value. Efm = 12 + 1(4) + 14 = 30 Molar mass/efm = 30/60 = 2 CH4N X 2 = C 2 N 8 N 2