# Skills Project Basic Dimensional Analysis. What is dimensional analysis? Dimensional analysis, DA, is a mathematical tool which uses the labels (or dimensions)

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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. In many ways, a DA is simply a massive fraction which is useful for making conversions between different values in chemistry.

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.

x ___________________________________ sec Example 1: Seconds in a Year Convert 1 year into seconds: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ 1 year = 365.25 days 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds 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. year(s) year(s) day(s) day(s) 1 1 365.25 24 60 60 1 1 1 hour(s) hour(s) min min 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. 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 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. According to our conversion factor, we can put “1” for the years on the bottom and 365.25 days on top. Knowing this, we can continue adding conversion factors in the right order and arranged in the correct direction. Check your work. All of the dimensions (labels) should cancel out except the one we were solving for which, in this case, is seconds. To solve the DA, multiply all the numbers on the top and bottom. Then, once you have each product, divide. x x x x x x 31,557,600 or 3.16e7 1 31,557,600 or 3.16e7 seconds (s)

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: 1100 cm = 1.00 m 11.0 L = 1000 ml 11.0 g = 1000 mg 11.609 km = 1 mile 4454 g = 1.0 lb

100 cents ___________________________________ 589 cents 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 start this problem, place “1 dollar” at the beginning of the DA. We will use a conversion factor to find cents. \$ 1 To complete this problem, you know that: \$1 = 100 cents Example 2: Convert \$ to cents Convert \$5.89 into cents. _ _ _ _ \$ 5.89 x 1 589 cents

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

Everything checks out, so multiply the top and divide by the bottom to find your answer. 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? ___________________________________ _ _ _ _ _ _ _ _ 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 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. 3500 mi Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with dollars at the end. 18 mi 1 gal 1 gal \$2.30 From the problem: 18 miles = 1 gallon 1 gallon = \$2.30 x x x \$8050 18 \$447.22

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 start this problem, place the 3000 gallons of fuel at the beginning. We’ll turn gallons into hours and hours into miles. 200 gal ___________________________________ 1 hr 2400000 mi To complete this problem, you know that: 200 gal = 1 hour 1 hour = 800 mi 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? _ _ _ _ _ _ _ _ 3000 gal 1 hr 800 mi x x x 200 12000 mi

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 10 23 atoms (Avogadro’s number) 1 atom Ag = 47 protons (Periodic table) 2.830 x 10 25 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. Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with protons at the end. 1 mol Ag ___________________________________ 6.022 x 10 23 atoms Ag 2.830 x 10 25 protons Example 6: Protons in Silver How many protons are in 1 mole of silver (Ag) atoms? _ _ _ _ _ _ _ _ 1 mol Ag 1 atom Ag 47 protons x x x 1

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. 3.011 x 10 14 atoms Cu (3.011e14 atoms Cu) To start this problem, place 2.5e-10 mol of Cu 2 O at the beginning. We’ll turn moles (mol) into molecules and molecules into atoms. ___________________________________ 6.022e10 23 molecules Cu 2 O 3.011 x 10 14 atoms Cu 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 10 23 molecules (Avogadro’s number) 1 molecule Cu 2 O = 2 atoms Cu (formula) 1 mol Cu 2 O Example 7: Copper atoms in Cu 2 O How many copper atoms are in 2.5e-10 moles of copper (I) oxide, Cu 2 O? _ _ _ _ _ _ _ _ 2.5e-10 mol Cu 2 O 1 molecule Cu 2 O 2 atoms Cu x x x 1

___________________________________ To start this problem, place 7.4e34 molecules of PO 4 3- at the beginning. We’ll turn molecules into atoms and atoms into neutrons. 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 PO 4 3- = 4 atoms O (formula) 1 atom O = 8 neutrons 2.4e10 36 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. 2.4 x 10 36 neutrons (2.4e36 neutrons) 4 atoms O 1 molecule PO 4 3- Example 8: Moles to Neutrons How many neutrons would be found in the oxygen atoms in 7.4e34 molecules of phosphate? _ _ _ _ _ _ _ _ 7.4e34 molecules PO 4 3- 1 atom O 8 neutrons x x x 1

To complete this problem, you know that: 15 atoms O = 1 molecule V 2 (CO 3 ) 5 (Formula) 1 mol = 6.022 x 10 23 molecules (Avogadro’s number) Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with moles V 2 (CO 3 ) 5 at the end. 0.072 mol V 2 (CO 3 ) 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. 15 atoms O ___________________________________ 1 molecule V 2 (CO 3 ) 5 6.5e23 mol V 2 (CO 3 ) 5 9.033e24 Everything checks out, so multiply the top and divide by the bottom to find your answer. Example 9: Converting Atoms to Moles How many moles of Vanadium (V) carbonate can be made from 6.5e23 atoms of oxygen? _ _ _ _ _ _ _ _ 6.5e23 atoms O 6.022e23 molecules V 2 (CO 3 ) 5 1 mol V 2 (CO 3 ) 5 x x x

___________________________________ 6.022e10 23 molecules C 3 H 8 1.26e22 atoms H To start this problem, place 2.62e-3 mol of C 3 H 8 at the beginning. We’ll turn moles (mol) into molecules and molecules into atoms. 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 10 23 molecules (Avogadro’s number) 1 molecule C 3 H 8 = 8 atoms H (formula) 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. 1.26 x 10 22 atoms H (1.26e22 atoms H) 1 mol C 3 H 8 Example 10: Atoms from Moles How many atoms of hydrogen will be found in 2.62e-3 mol of propane, C 3 H 8 ? _ _ _ _ _ _ _ _ 2.62e-3 mol C 3 H 8 1 molecule C 3 H 8 8 atoms H x x x 1

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

Practice on Your Own (2): Given 2.5 mol of H 2 O, 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 H 2 O

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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