Percent
Percent means number of parts of one type out of 100 parts of the whole. Percent can be
expressed as percent by mass, by volume, or by number. The statement that seawater
contains 3.8% sodium chloride by mass means that in 100.0 g of seawater there are 3.8 g
of sodium chloride. This statement can be used as an equivalence for use in conversion
factors. For example,
3.8 g sodium chloride
OR
100.0 g seawater
100 g seawater
3.8g sodium chloride
Different than other conversion factors though, the percent can limit the number of sig
figs in a calculation. The above sodium chloride solution has 2 sig figs determined by the
number 3.8. (The 100 parts is exact.)
Finding percent of a component in a mixture
Example - In a sucrose solution there are 75.6 grams of sucrose in 358 g of solution.
What is the percent sucrose by mass in the solution?
Then
75.6g sucrose = 0.211 g sucrose = 21.1 g sucrose = 21.1%
358 g soln
1.00 g soln
100. g soln
Using percent to calculate the quantity of a component in a mixture
Example 1 - Find the mass of sodium chloride present in 255 g seawater.
255 g seawater x 3.8 g sodium chloride
= 9.69 g sodium chloride
100.0 g seawater
Example 2 - The density of a solution that is 43.01 % sulfuric acid by mass is 1.329
g/cm3. How many grams of sulfuric acid are there in 500.0 mL of sulfuric acid solution?
500 mL sulfuric acid soln x 1 cm3 x 1.329 g = 664.5 g sulfuric acid soln
1 mL
1cm3
Then 664.5 g sulfuric acid soln x
43.01 g sulfuric acid = 285.8 g sufuric acid
100 g sulfuric acid soln
Determining the quantity of a mixture needed for a given amount of a component
Example - How many liters of seawater are needed to obtain 156 g of sodium chloride if
a 3.80 % seawater solution by mass is evaporated to dryness to obtain the sodium
chloride? The density of the seawater is 1.03 g/cm3.
156 g sodium chloride x 100.00 g seawater
= 4.11 x 103 g seawater
3.80 g sodium chloride
4.11x 103 g seawater x 1 cm3
= 3990 cm3 seawater
1.03 g
3990 cm3 seawater x 1 L
= 3.99 L seawater
3
1000cm
Borowski Sp, 97