Below is a demonstration of a method that most students have found to be the key to solving the
math portion of chemistry.
Here is the golden rule, it is simple and works 100% of the time: IF YOU DO NOT HAVE THE
CORRECT UNTIS WITH YOUR ANSWER, THEN YOU DO NOT HAVE THE RIGHT ANSWER!!! If this
happens, go back through the calculation and determine what “units” you need, in order to solve the
problem.
If you do have the right units, then chances are you have the right number!
Check this out and USE IT WHEN YOU DO YOUR HOMEWORK AND EXAM. I guarantee that your scores
will increase…………
I recommend that everyone review some of your problem spots and apply this technique –
Solving Problems
When solving a problem make a list of what you know and what you need to know.
It is best to start with what you are given in the problem.
Dimensional analysis is used to cancel out units not numbers. You must always write the number and
the unit when working the problem.
When solving a problem using dimensional analysis, you need to start with a grid:
The horizontal lines are used to separate numbers from the conversion factor and represent the
mathematical operation division. The vertical lines are used to separate different conversion factors
and represent the mathematical operation multiplication.
Example 1:
1. A tile was measured to be 45.56 cm. What is the measurement in inches?
2. What do we know: tile is 45.56 cm, 1 inch = 2.54 cm.
3. Draw a grid. In the first box write 45.56 cm. The second grid to the right is used to convert
from cm to inches. Since cm is on the top in the first grid, we must put cm on the bottom in
the second grid and inches on top. We can cancel cm and we are left with inches.
4. The answer is rounded to 4 significant figures because the number from the problem has four,
while the conversion factor is not counted because it is a definition.
45.56 cm
1 in.
2.54 cm
= 17.94 in.
Example 2:
1. The mass of a lead block was measured to be 45.6 kg. What is the mass in micrograms?
2. What do we know: The mass is 45.6 kg, 1 kg = 1000 g, 1 g = 1000 mg, 1 mg = 1000 ug.
3. Start with what is given, then arrange the units so they cancel.
45.6 kg
1000 g
1 kg
1000 mg
1g
1000 ug
1 mg
= 4.56 x 1010 ug
Example 3:
1. If the speed limit is 65 miles/hr, what is the speed limit in kilometers/hr.
2. What do we know: 1 mi = 1760 yd 1.094 yd = 1 m 1000 m = 1 km
3. Start with 65 miles = 1 hour and then arrange conversion factors to change to
kilometers/hour.
65 miles
1 hour
1760 yd
1 mile
1m
1.094 yd
1 km
1000 m
= 104.6 km/hr
*** Do not let mixed units such as miles/hour or g/mL or g/cm3 cause you problems. Using this
method you can work with each unit separately, but use a clear organized approach.
Example 4:
1. A car is advertised to get 15 km/L, what is it in miles/gallon.
2. What do we know: 1 km = 1000 m 1 m = 1.094 yd 1760 yd = 1 mi 1 L = 1.06 qt 4 qt = 1 gal
3. Start with what is given in the problem (15 km/L) and then arrange conversion factors so that
the units will cancel. Convert the km to miles and then convert liters to gallons.
15 km
1L
1000 m 1.094 yd
1 km
1m
1 mi
1760 yd
1L
1.06
gt
4 qt
35 miles/gallon
1 gal
Example 5:
1. Calculate (a) the number of grams of solute in 0.250 L of 0.150 M KBr, (b) the molar
concentration of a solute containing 4.75 g of Ca(NO3)2 in 0.200L, (c) the volume of 1.50 M
Na3PO4 in milliliters that contains 5.00 g of solute.
2. What do we know: (a) 1 mol KBr = 119.0 g KBr (from the periodic table), (b) 1 mol of
Ca(NO3)2 = 164.1 g Ca(NO3)2 (from the periodic table), (c) 1 mol Na3PO4 = 163.9 g Na3PO4
and 1 L = 1000ml.
3. Start with what is given in the problem and then arrange conversion factors so that the units
will cancel.
(a)
0.150 M KBr
1L
0.250 L
1
119.0 g KBr
1 mol KBr
4.46 g KBr
(b)
4.75 g Ca(NO3)2
1
1 mol Ca(NO3)2
164.1 g Ca(NO3)2
1
0.200 L
0.145 mol Ca(NO3)2
1L
0.145 M Ca(NO3)2
(c)
5.00 g Na3PO4
1
1 mol Na3PO4
163.9 g Na3PO4
1L
1.50 mol Na3PO4
1000mL
1L
20.3 mL solution