Review 2' Density and percent composition

1
Review 2. Density and percent composition

• In this lesson you will review density and
  composition, and use them to solve problems

2
Definition of density

• Density is the ratio of sample mass to sample
  volume
• d m V
• Density depends only on the composition of the
  sample, not on the amount of material present

3
Density is a conversion factor

• A conversion factor is a relationship between two
  properties or quantities, expressed as a fraction
  or ratio
• Density is a conversion factor that shows the
  relationship between mass and volume.

4
Solve this density problem

• 385 mL of mercury weighs 5.23 kg. What is the
  density of mercury, in g/mL?
• 5.23 kg x 1000 g 5230 g 1 kg
• d m 5230 g 13.6 g V 385 mL
  mL

5
Solve this density problem

• Ethylene glycol has a density of 1.11 g/mL. What
  is the mass of 18.6 L of ethylene glycol?
• 18.6 L x 1000 mL x 1.11 g 20,646 g
  1 L 1 mL
• Round off answer to 3 significant digits ? 20.6
  x 103 g or 20,600 g

6
Solve this density problem

• Ethyl alcohol has a density of 0.789 g/mL. What
  is the volume of 235 g of ethyl alcohol?
• 235 g x 1 mL 297.845384 mL 298 mL
  0.789 g
• Density was inverted to make units cancel
  properly, but relationship is still valid

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Composition of a sample percent by mass

• If a sugar solution is 5 sugar by mass, it
  contains 5 g sugar 100 g solution
• Notice units are same in numerator denominator,
  but labels are different
• Composition may also be by volume (volume units
  instead of mass units)

8
Solve this composition problem

• 825 g of ore are found to contain 137 g of
  aluminum. What is the aluminum in this ore, by
  mass?
• 137 g Al x 100 16.6 Al 825 g ore

9
Solve this composition problem

• A fertilizer contains 21 nitrogen by mass. What
  mass of fertilizer (in kg) should be applied to
  supply 775 g nitrogen?
• 775 g nitrogen x 100 g fertilizer x 1 kg
  21 g nitrogen 1000 g
• 3.7 kg fertilizer needed (2 sig digits)
• Units (g) are same labels must cancel!

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Many problems require both density and
composition

• A solution that is 8.50 acetone 91.50 water,
  by mass, has a density of 0.9867 g/mL. What mass
  of acetone is present in 750. mL of the
  solution?
• 750. mL soln x 0.9867 g soln x 8.50 g
  acetone 1 mL soln 100 g soln
• 62.9 g acetone
• Labels as well as units must cancel

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Solve this problem

• A sucrosewater solution is 28.0 sucrose by mass
  and has a density of 1.118 g/mL. What mass of
  sucrose is in 2.75 L of this solution?
• 2.75 L soln x 1000 mL x 1.118 g soln x 28.0 g
  sucrose 1 L 1 mL soln 100 g soln
• 861 g sucrose

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Solve this problem

• Pool acid is 36 hydrochloric acid by mass and
  has a density of 1.179 g/mL. How many liters of
  acid solution should you use to add 5.0 x 102 g
  acid to the pool?
• 500 g acid x 100 g soln x 1mL soln x 1
  L 36 g acid 1.179 g soln 1000mL
• Add 1.2 L acid solution

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Reminder

• The most common mistake students make when
  setting up these problems is to invert a factor
  or use it in the wrong place.
• They dont catch their mistake because they
  neglect to label the units. The units cancel,
  but the labels dont.

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The bottom line

• You should be able to define density, calculate
  its value from data, and use it to convert
  between mass and volume.
• You should be able to calculate percent
  composition by mass from data and use it to solve
  problems.
• You should be able to combine density and
  composition to solve problems

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Where to get more information

• If you feel insecure on the topic of density,
  consult pages 20 and 32-33 in your text
• For more about percent composition, see page 537
  in your text or ask me for help
• If you are goofing on metric prefixes, review
  page 19 in your text