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Skills Project

Skills Project. Basic Dimensional Analysis. What is dimensional analysis?. Dimensional analysis, DA, is a mathematical tool which uses the labels (or dimensions) of individual measurements to calculate solutions to problems.

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Skills Project

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  1. Skills Project Basic Dimensional Analysis

  2. What is dimensional analysis? • Dimensional analysis, DA, is a mathematical tool which uses the labels (or dimensions) of individual measurements to calculate solutions to problems. • In many ways, a DA is simply a massive fraction which is useful for making conversions between different values in chemistry.

  3. Anatomy of a DA: • Like a fraction, a DA consists of a top and bottom half. To solve a DA • Multiply the numbers in the numerator or top half of the DA, then….. • Divide by all the numbers in the denominator or bottom half of the DA.

  4. Example 1: Seconds in a Year • Convert 1 year into seconds: ___________________________________ ____ ____ ____ ____ 31,557,600 or 3.16e7 60 x 60 x 24 365.25 1 x x sec min day(s) hour(s) year(s) 31,557,600 or 3.16e7 seconds (s) x 1 1 x 1 1 year(s) x day(s) hour(s) 1 min 1 day = 24 hours 1 year = 365.25 days 1 hour = 60 minutes 1 minute = 60 seconds To solve the DA, multiply all the numbers on the top and bottom. Then, once you have each product, divide. Now, we may begin converting 1 year into seconds. To do this, we will need conversion factors. A conversion factor simply states that something is equal to something else. For example: 1 dozen = 12 Check your work. All of the dimensions (labels) should cancel out except the one we were solving for which, in this case, is seconds. Close off this “bracket” with a vertical line. The first bracket will usually only have something at the top with an “invisible” 1 underneath. All of the conversion factors that follow will have both a top and bottom, however. According to our conversion factor, we can put “1” for the years on the bottom and 365.25 days on top. To use this conversion factor, we must line it up so that the units (dimensions) cancel out. Since years are on top, they will be on the bottom of the next bracket. All DAs begin with a straight horizontal line to separate the numerator from the denominator In the problem, you were given 1 year to begin with. We’ll place this at the head of our DA as our starting point. From here on out, our job is to carry out conversions until we get to seconds. Knowing this, we can continue adding conversion factors in the right order and arranged in the correct direction.

  5. DA Mechanics: • In order to setup a proper DA, you will need to find conversion factors. • Typically, many DAs are constructed using the metric system. These relationships provide many of our conversion factors. • For example: • 100 cm = 1.00 m • 1.0 L = 1000 ml • 1.0 g = 1000 mg • 1.609 km = 1 mile • 454 g = 1.0 lb

  6. Example 2: Convert $ to cents • Convert $5.89 into cents. ___________________________________ ____ 100 cents 589 cents $ 5.89 x $ 1 1 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “cents” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: $1 = 100 cents To start this problem, place “1 dollar” at the beginning of the DA. We will use a conversion factor to find cents. 589 cents

  7. Example 3: Converting Distances • How many inches are in 1 mile? ___________________________________ ____ ____ 12 in 5280 ft 63360 in 1 mi x x 1 mi 1 x 1 ft Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with inches at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you need to know that: 1 mile = 5280 ft 1 ft = 12 in. To start this problem, place the 1 mile at the beginning. We’ll turn miles into feet and feet into inches. 63360 in

  8. Example 4: Calculating Gas Costs • You are getting ready to travel 3,500 miles across the country. Your car’s fuel economy is about 18 miles/gallon and gas will cost $2.30/gallon. How much will gas cost on this trip? ___________________________________ ____ ____ $2.30 1 gal $8050 3500 mi x x 18 mi 18 x 1 gal From the problem, we are trying to find the total cost of gas on this trip, in dollars. We also know: 18 miles = 1 gallon 1 gallon = $2.30 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with dollars at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. From the problem: 18 miles = 1 gallon 1 gallon = $2.30 To start this problem, we’ll want to use the total number of miles, 3,500 mi. All the other values are conversion factors. We can convert miles to gallons and gallons to dollars. $447.22

  9. Example 5: Travel times • A jet plane travels at 800 mph and consumes 200 gallons of fuel per hour. How far will the jet travel on 3000 gallons of fuel? ___________________________________ ____ ____ 800 mi 1 hr 2400000 mi 3000 gal x x 200 gal 200 x 1 hr Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with miles, mi, at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 200 gal = 1 hour 1 hour = 800 mi To start this problem, place the 3000 gallons of fuel at the beginning. We’ll turn gallons into hours and hours into miles. 12000 mi

  10. Example 6: Protons in Silver • How many protons are in 1 mole of silver (Ag) atoms? ___________________________________ ____ ____ 47 protons 6.022 x 1023 atoms Ag 2.830 x 1025 protons 1 mol Ag x x 1 mol Ag 1 x 1 atom Ag Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with protons at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 1 mol = 6.022 x 1023 atoms (Avogadro’s number) 1 atom Ag = 47 protons (Periodic table) 2.830 x 1025 protons (2.830e25 protons) To start this problem, place 1 mol of silver (Ag) atoms at the beginning. We’ll turn moles (mol) into atoms and atoms into protons.

  11. Example 7: Copper atoms in Cu2O • How many copper atoms are in 2.5e-10 moles of copper (I) oxide, Cu2O? ___________________________________ ____ ____ x 2.5e-10 mol Cu2O x 3.011 x 1014 atoms Cu 6.022e1023 molecules Cu2O 2 atoms Cu 1 mol Cu2O 1 x 1 molecule Cu2O Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with atoms Cu at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 1 mol = 6.022 x 1023 molecules (Avogadro’s number) 1 molecule Cu2O= 2 atoms Cu (formula) 3.011 x 1014 atoms Cu (3.011e14 atoms Cu) To start this problem, place 2.5e-10 mol of Cu2O at the beginning. We’ll turn moles (mol) into molecules and molecules into atoms.

  12. Example 8: Moles to Neutrons • How many neutrons would be found in the oxygen atoms in 7.4e34 molecules of phosphate? ___________________________________ ____ ____ x 7.4e34 molecules PO43- x 2.4e1036 neutrons 4 atoms O 8 neutrons 1 molecule PO43- 1 x 1 atom O Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 1 molecule PO43-= 4 atoms O (formula) 1 atom O = 8 neutrons 2.4 x 1036 neutrons (2.4e36 neutrons) Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with neutrons at the end. To start this problem, place 7.4e34 molecules of PO43- at the beginning. We’ll turn molecules into atoms and atoms into neutrons.

  13. Example 9: Converting Atoms to Moles • How many moles of Vanadium (V) carbonate can be made from 6.5e23 atoms of oxygen? ___________________________________ ____ ____ x 6.5e23 atoms O x 1 molecule V2(CO3)5 1 mol V2(CO3)5 6.5e23 mol V2(CO3)5 x 15 atoms O 9.033e24 6.022e23 molecules V2(CO3)5 Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 15 atoms O = 1 molecule V2(CO3)5 (Formula) 1 mol = 6.022 x 1023 molecules (Avogadro’s number) 0.072 mol V2(CO3)5 To start this problem, place 6.5e23 atoms of O (oxygen) at the beginning. We’ll turn atoms into molecules and molecules into moles. Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with moles V2(CO3)5 at the end.

  14. Example 10: Atoms from Moles • How many atoms of hydrogen will be found in 2.62e-3 mol of propane, C3H8? ___________________________________ ____ ____ x 2.62e-3 mol C3H8 x 1.26e22 atoms H 6.022e1023 molecules C3H8 8 atoms H 1 mol C3H8 1 x 1 molecule C3H8 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with atoms H at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 1 mol = 6.022 x 1023 molecules (Avogadro’s number) 1 molecule C3H8 = 8 atoms H (formula) 1.26 x 1022 atoms H (1.26e22 atoms H) To start this problem, place 2.62e-3 mol of C3H8 at the beginning. We’ll turn moles (mol) into molecules and molecules into atoms.

  15. Practice on Your Own: • Make the following conversions (you may need a conversion “cheat sheet”): • 1.45 km  miles • 181 oz  grams • 13 decades  centuries • 9.4e4 ml  gallons • 7.32 days  minutes 0.901 mi 5140 g 1.3 centuries 25 gal 10500 minutes

  16. Practice on Your Own (2): • Given 2.5 mol of H2O, determine the following: • Atoms of hydrogen: • Atoms of oxygen: • Neutrons from oxygen: • Total number of protons: • Molecules of water: 3.0e24 atoms H 1.5e24 atoms O 1.2e25 neutrons 2.6e25 protons 1.5e24 molecules H2O

  17. Congratulations! • You have just mastered one of the most powerful (and useful) tools I will be teaching you this year. • Don’t forget- you can use this method of solving problems whenever you have conversion factors. • DA is useful for a variety of subjects other than chemistry.

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