1 / 59

Chapter 6: Measuring Matter

Chapter 6: Measuring Matter. 3.3 Conversion Factors (IMPORTANT). The same quantity can be measured or expressed in many ways 1 dollar = 4 quarters = 10 dimes = 20 nickels 1 meter = 10 decimeters = 100 centimeters Whenever 2 measurements are equal, a ratio of these measurements will equal 1

taite
Download Presentation

Chapter 6: Measuring Matter

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 6: Measuring Matter

  2. 3.3 Conversion Factors (IMPORTANT) • The same quantity can be measured or expressed in many ways • 1 dollar = 4 quarters = 10 dimes = 20 nickels • 1 meter = 10 decimeters = 100 centimeters • Whenever 2 measurements are equal, a ratio of these measurements will equal 1 • You can always multiply by 1! • Ratios of equivalent measurement is called a conversion factor

  3. 3.4-3.7 Dimensional Analysis & Conversions • One of the best methods for solving problems • Uses the units (dimensions) that are part of measurements to help solve the problem. • Let’s start with an easy problem, but SHOWING ALL WORK • To determine which unit goes on the bottom of the ratio, look at what needs canceling. • We can create infinite chains of this!

  4. Your school club has sold 600 tickets to a chili supper fund-raising event. You volunteered to make the chili (sucker). You have a very large pot, and a chili recipe that serves 10. The recipe calls for 2 teaspoons of chili powder. How much chili powder do you need for 600 servings of chili? • DON’T do this in your head. The skills to do this will be needed later!

  5. Step 1: Unknown • Number of teaspoons of chili powder • Step 2: Identify what is known • 600 servings must be made • 10 servings = 2 tsp of chili powder • Step 3: Plan a solution • Create conversion factors • 10 servings = 2 tsp chili powder • 10 servings / 2 tsp chili powder • 2 tsp chili powder / 10 servings • Change the unit “servings” into unit “teaspoons of chili powder”

  6. Step 4: Do the calculations • That’s a lot of teaspoons! • Let’s figure out how many cups that is, that would be easier. • Known: 3 tsp = 1 tbsp • 16 tbsp = 1 cup

  7. What is known? • 120 tsp chili powder • Plan a solution. • Need to convert tsp  tbsp  cups • Calculation

  8. Once you’re done, make sure all units (except for the ones you want to keep) cancel. • Check this also before doing any actual calculations. This will tell you if you’re right.

  9. Special Conversion • Volume is tricky. Show on board. Not as straight-forward as length for conversions. • 1 cm3 = 1 mL • If you mess up, and your answer is upside-down, just take the inverse of it!

  10. 6.1 Measuring Matter • One of the most common ways to measure matter is to count it • You can weigh it • Measure by volume • Some by multiple ways • You can buy nails by mass or count

  11. Some units used in measuring refer to a specific number of items • A pair • A dozen • A baker’s dozen

  12. Some items, like fruit, are commonly measured and purchased in multiple ways. • Fruit stand, 5 apples for $2.00 (count) • Grocery store, $1.00 per pound (weight) • Orchard, $9.00 per bushel (volume) • Each of these is a way to measure apples. We can further extend this to apply to a dozen apples.

  13. 1 dozen apples = 12 apples (count) • Using average sized apples, 1 dozen apples = 2 kg apples • Also using average sized apples, 1 dozen apples = 0.20 bushel apples • So what we have is a series of conversions that can be used to calculate number, mass or volume

  14. Unit Conversions! • What is the mass of 90 average-sized apples? • What is unknown? • Mass of apples • What is known? • 90 apples, 1 dozen apples = 2 kg, 1 dozen = 12 • Conversion is • Number apples  mass • Number of apples dozens of apples  mass of apples

  15. = 15 kg apples

  16. The Mole • More than a reality canceled TV show • The mole is nothing more than a certain amount of anything • Often used to count atoms, since they are so small, scales like a “dozen” were impractical • A mole represents 6.02x1023 of ANYTHING. • A mole of hydrogen molecules is 6.02x1023 hydrogen molecules • A mole of cars is 6.02x1023 cars • This is a REALLY big number. • A mole of eggs would fill the world’s oceans 30 million times over. • It would take 10 billion chickens laying 10 eggs per day more than 10 billion years to lay a mole of eggs • 1 mole of seconds would be 1.9x1016 years, which is 19,000,000,000,000,000 (compared to age of universe which is 20,000,000,000 years)

  17. 6.02x1023 is called Avagadro’s number, in honor of Amedeo Avogadro di Quarenga • When dealing with chemistry units, we often have to think about what we’re dealing with • A mole of oxygen is usually referring to 6.02x1023 oxygen molecules, not atoms • A mole of iron is 6.02x1023 atoms of iron • A mole of salt is 6.02x1023 formula units of NaCl • A mole of water is 6.02x1023 molecules of water • Table 6.1 on page 145 details more of this

  18. Some problem solving with this • Handled just like the apple example with a dozen • How many moles of magnesium are 3.01x1022 atoms of magnesium? • We now have the conversion 1 mole (mol) magnesium = 6.02x1023 atoms of magnesium

  19. Note, Avogadro's number is a conversion, so an infinite number of significant figures

  20. 6.3-6.4 The Gram Formula Mass, Gram Atomic Mass and Molar Mass • This sounds worse than it actually is • The gram atomic mass (gam) is the atomic mass of an element expressed in grams • Atomic mass found on periodic table • The gram atomic mass is the mass of one mole of atoms of any element • For single elements, the gram atomic mass = atomic mass.

  21. Mass of mole of a compound • First you must find the formula for the compound • Let’s use sulfur trioxide as an example • SO3 • One molecule of sulfur trioxide is from one atom of sulfur and three atoms of oxygen. • So one mole of sulfur trioxide contains one mole of sulfur and three moles of oxygen • To find the mass, must find mass of each part, then add together.

  22. Gram atomic mass of sulfur = • 32.1 g • Gram atomic mass of oxygen = • 16.0 g • But there’s three moles of oxygen, so must multiply by 3! • So total oxygen contribution is: 16.0x3 =48.0 g • Add them together: 32.1 + 48.0 = 80.1 g • That is the mass of one mole of sulfur trioxide. • Because we’re dealing with a molecule now, this is not the gram atomic mass, but the gram molecular mass (gmm).

  23. How about ionic compounds? • Can’t find the gram atomic mass, because not atoms. • Can’t find the gram molecular mass, because not molecules. • They use gram formula mass (gfm) • Besides the name, calculated JUST like the gram molecular mass. • So gram formula mass of NaCl is equal to the sum of the gam of sodium plus the gam of chlorine. • Generically, all of these can be called the molar mass. Not as specific, but still works, and I’ll use it.

  24. 6.5 Mole  Mass Conversions • We can use the molar mass of a substance to convert grams of a substance into moles. • We can do this since molar mass = gram/mole • We can also use a similar practice to turn moles  grams. Will still be using molar mass of the substance.

  25. 6.6 The Volume of a Mole of Gas • For many thing, the volume of a mole of a substance is different from most other substances. • For example, a volume for a mole of water is much less than the volume for a mole of sugar. • However, the volume of a mole of gases is MUCH more predictable.

  26. As we will learn later in the year, the volume of a gas varies with the temperature or pressure. • Because of this variation, the volume of a gas is usually measured at a standard temperature and pressure (STP). • This is defined as 0°C and 101.3 kPa (kilopascals) or 1 atmosphere (atm). • At STP, 1 mole of ANY gas will have a volume of 22.4 L. • Black box is 22.4 L.

  27. The volume, 22.4 L is known as the molar volume of a gas, measured at STP. • This idea of all gases containing the same number of particles in a given volume at any given pressure and temperature is what made Avogadro famous. • And is why 6.02x1023 (1 mole of anything) is named after him. • So, 22.4L of any gas at STP contains 6.02x1023 particles of that gas.

  28. Would 22.4 L of one gas (at STP) have the same mass as 22.4 L of another gas (at STP)? • No. Why? • Remember from yesterday, molar masses. • If the molar mass of the gases is different, so will the mass of 1 mole of their gas. • Only if the molar masses of two gases is the same will their mass at 22.4 L (at STP) be the same.

  29. Where does this come in handy? • Gives us another conversion • At STP, 1 mole (of a gas) = 22.4 L • For example, • Determine the number of moles in 33.6 L of He gas at STP. • Going to convert L  moles

  30. 6.7 Gas Density and the Molar Mass • Density is usually measured in units of g/L. • We can use the density of a gas at STP to determine the molar mass of that gas. • When doing so, the gas can be an element, or a compound.

  31. Example • The density of a gaseous compound of carbon and oxygen is 1.964 g/L at STP. Determine it’s molar mass, and determine if the substance is carbon dioxide or carbon monoxide (important difference here). • May seem awkward to start here, but we can do this. • Let’s tackle the steps in solving this.

  32. What are we looking for? • Molar mass • What do we know? • Density is 1.964 g/L • 1 mole of gas = 22.4 L • Outline path • Density  Molar mass • Which is the same as Density  g/mol

  33. So the molar mass of this gas is 44.0 g/mol • Almost done, just need to determine if this is CO or CO2 • Calculate molar mass of CO: • 28.0 g/mol • Calculate molar mass of CO2 • 44 g/mol • This unknown substance is CO2

  34. 6.8 Converting Between Units with Moles • We have now examined a mole in terms of particles, mass, and for a gas, volume at STP. • 22.4 L = 1 mole • Molar mass = 1 mole • 6.02x1023 particles = 1 mole • All of these have 1 mole in common, so we can always use the mole to convert from one to the other!

  35. 6.9 Percent Composition • When we make a new compound in the laboratory, we need to determine its formula. • One of the first steps is to find the relative amounts of the elements in the compound. • These relative amounts are expressed as the percent composition. • This looks at the mass of each element in the compound • There are as many percent values in a compound as there are different elements in the compound. • For example, K2Cr2O7 is K = 26.5%, Cr = 35.4% and 0 = 38.1 %. • These percents must add up to 100%.

  36. The percent by mass of an element in a compound is the number of grams of the element divided by the grams of the compound, multiplied by 100% • Or,

  37. Example • An 8.20-g piece of Mg combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? • Since we know the mass of each element, we can add them to find the mass of the compound: • 8.20g + 5.40g = 13.60g • Now we divide the mass of each element by the mass of the compound, then multiply by 100%

  38. Check your percentages to make sure they add to up 100%... • And we’re done! 60.3% magnesium and 39.7% oxygen by mass.

  39. % Composition of a Known Compound • To do this, we will use the chemical formula of the compound to calculate the molar mass. • This gives me the mass of one mole of the compound. • Then for each element, calculate the percent by mass in ONE MOLE of the compound. • The subscripts in the formula are used to calculate the grams of each element in a mole of that compound.

  40. Example • Calculate the percent composition of propane, C3H8 • First, calculate the molar mass of propane • 3 mol C = • 36.0g C • 8 mol H = • 8.0 g H • Molar mass of propane = 44.0 g • Now percent by mass for each element can be calculated, just like before!

  41. % Composition with a specific amount of a substance • So far, these amounts have been generic amounts for the compound. • Propane in general, etc. • You can use the % composition to calculate the number of grams of an element in a specific amount of a compound. • To do so, you multiply the mass of the compound by the conversion factor based on the % composition.

  42. Example • Using propane from previous example. • Calculate mass of carbon in 82.0 g of propane, C3H8 • From the previous example, I know that % composition of propane is 81.8% C and 18.2% H. • This means that for every 100 g of propane, you will have 81.8 g of carbon • And for every 100 g of propane, you will have 18.2 g H.

  43. So we convert…. • So 82g of propane, there will be 67.1 g of carbon.

  44. 6.10 Calculating Empirical Formulas • Once a new compound has been made in the lab, you can usually determine its % composition experimentally. • Then, from % composition data, you can calculate its empirical formula. • The empirical formula gives the lowest whole-number ratio of the atoms of the elements in the compound

  45. Example • A compound may have an empirical formula of CO. • This tells me that C and O have a 1:1 ratio • However, empirical formula may or may not be the same as the actual molecular formula. • If not the same as the molecular formula, the molecular formula will be a simple multiple of the empirical formula • Dinitrogen tetrahydride has formula of • N2H4 • But has empirical formula of • NH2

  46. How to Calculate Empirical Formulas • This is a little hard. Just giving you a head’s up. • First, you must know the % composition. • Let’s give this a shot... We’ll go through step-by-step. • Example: • What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

  47. Because 25.9% nitrogen, and 74.1% oxygen, we can say that in 100 g of the compound • There will be 25.9 g nitrogen • There will be 74.1 g oxygen • Need to convert these masses into moles • To do this, I use molar mass of these elements

  48. We now have a mole ratio of nitrogen to oxygen of: N1.85O4.63 • This doesn’t work. We want whole numbers. • To start out, divide both numbers by the smaller number of moles. • This this case, divide both by 1.85 • 1.85 / 1.85 = 1 mol N • 4.63 / 1.85 = 2.5 mol O • Now have a mole ratio of NO2.5 • Still doesn’t work, need whole numbers.

More Related