Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107	Fundamentals of Chemistry	Fall 2009
Lecture Notes:: Dimen Anal	© R. Paselk 2000
	Worked Examples
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How long is a one foot ruler?

Known: 1 ft = 12 inches (defined, therefore exactly); 2.54 cm = 1 inch (defined).

Set up: (1 ft)(12 inches/ft)(2.54 cm/inch)

note that ft cancel ft and inches cancel inches to give cm!

Solve: (1 ft)(12 inches/ft)(2.54 cm/inch) = 30.48 cm.

How about sig figs? In this problem there are no significant figures the way its set up, because there are no measurements! That is all of the numbers are part of definitions, so they are exact, and that means the answer is exact as well.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A student found that 20.0 mL of a liquid weighed 35.987 g. What is its density?

Known: Density = mass/volume, generally expressed as g/mL = g/cm³

Solve: (35.987 g) / (20.0 mL) = 1.79935 g/mL

note that the units are those of density so we are confident we set it up correctly.

How about sig figs? Use multiplication/division rules, so count: 3 for 20.0 and 5 for 35.987, therefore should have three sig figs:

1.79935 g/mL = 1.80 g/mL

 

Using a jewelers balance a student found that a coin weighed 2.34 carats in air. By weighing it again submerged in water she found it had a volume of 0.034 mL. What is its density?

Known: 1 carat = 200 mg (defined), density is g/mL

Solve: (2.34 carats)(200 mg/carat)(1 g/1,000 mg) / 0.034 mL = 13.764706 g/mL

How about sig figs? Both conversion factors are defined, so exact. Two measurements: 2.34 and 0.034 = 3.4 x 10^-2. Thus the answer will have only two sig figs since using counting rule - least number of sig figs.

13.764706 g/mL = 14 g/mL

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

750 calories of heat is transferred to 100.0 g of water at 20.00 °C. What will the new temperature of the water be assuming no heat is lost to the container of the surroundings?

Known: heat capacity of water = 1 cal / (g°C) [assume exact]; q = mC_spT

Rearranging equations gives: T = q/ (mC_sp)

Substituting values into the equations get: T = 750 cal / {(100.0 g)(1 cal / (g°C)} = 7.50 °C

Adding the difference to the original temperature gives: 20.00 °C + 7.50 °C = 27.50 °C

How about sig figs? Heat capacity of water is assumed to be exact. Three measurements: 750, 20.00°C, 100.0g. Thus the calculation will have only two sig figs since using multiplication: 7.5°C

Next, using the addition rule, we look at decimal: 20.00 °C + 7.5°C = 27.50 °C = 27.5 °C

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Last modified 3 September 2009