Common lab calculations involving percents
Finding Percent Change
Percentages are often used to compare two quantities. You often hear statements such as
the following:
The price of steel rose 3% over last year's price.
The weights of two cars differed by 20%.
Production dropped 5% from last year.
When the two numbers being compared involve a change from one to the other, the
original value is usually taken as the base.
new value - original value
Percent Change =
×100
original value
Be sure to show the direction of change with a plus or a minus sign, or with words such
as increase or decrease.
Find the percent change when a quantity changes from 29.3 to 57.6
A student earns $680.00 per week working in a bank, and receives a raise of $95.00.
What is their percentage increase in pay?
The temperature in a building rose from 19.0ºC to 21.0ºC during the day. Find the
percent change in temperature.
Practise: Calter, pg 46 #50-58
Common lab calculations involving percents
Finding Percent Efficiency
The power output of any machine or device is always less than the power input, because
of inevitable power losses within the device. The efficiency of the device is a measure of
those losses.
output
Percent Efficiency =
×100
input
1. A man expends 100 joules of work to move a box up an inclined plane. The amount of
work produced is 80 joules.
2. A solar panel absorbs 110 kilowatt hours of electricity from the sun. It is 37% efficient.
How much electricity does it produce?
Finding Percent Error
The accuracy of measurements is often specified by the percent error. The percent error
is the difference between the measured value and the known or “true” value, expressed as
a percent of the known value.
measured value - known value
Percent Error =
×100
known value
1. Working in the laboratory, a student find the density of a piece of pure aluminum to
be 2.85 g/cm3. The accepted value for the density of aluminum is 2.699 g/cm3. What
is the student's percent error?
2. A student takes an object with an accepted mass of 200.00 grams and masses it on his
own balance. He records the mass of the object as 196.5 g. What is his percent
error?
Practise: Calter, pg 46 # 59 – 66
Common lab calculations involving percents
Finding Percent Concentration
The following equation applies to a mixture of two or more ingredients:
amount of ingredient
Percent Concentration =
×100
total amount of mixture
As with percent change, be sure to specify the direction of the error.
1. Suppose that 20.0 g of sugar are dissolved in 165 g of water. Calculate the
% concentration.
2. Calculate the mass of slat that must be dissolved to prepare 5.00 x 10 2 g of a
7.00% solution.
3. Calculate the percentage of K2CO3 in a solution that is made by dissolving 15
g of K2CO3 in 60 g of water.
4. Suppose that a mixture of aluminum and iron weighs 14.5 g. If there are 0.14
moles of aluminum in the mixture then determine the mass percent of
aluminum in the mixture. (Aluminum = 26.98 g/mol; Iron = 55.85 g/mol)
Practise: Calter, pg 46 # 67 – 70
Common lab calculations involving percents
Solutions
A solution is a mixture in which one substance, the solute, is uniformly distributed in
another substance, the solvent.
Percentages are usually given in terms of mass, volume and mass-volume.
Types of concentration expression:
1) Percent by weight
% (w/w) =
g solute
100 g (solute + solvent)
Ex. A student is asked to prepare 100 g of a 15% (w/w) KCl solution.
The student would weigh 15 g of KCl, and would weigh 85 g of water (100-15g), and
mix to dissolve.
2) Percent by weight per volume
Percent by % (w/v) =
g solute
100 mL total solution
Ex. A student is asked to prepare 100 mL of a 15%(w/v) KCl solution.
The student would weigh 15 g of KCl, and using a 100 mL volumetric flask, would
fill to the mark, and mix to dissolve. Thus, the amount of water added is not 85 g
3) Percent by volume
% (v/v) =
mL solute
100 mL total solution
Ex. A student is asked to prepare 100 mL of a 15% (v/v) ethanol solution.
The student would measure 15 mL (pipette) ethanol, and using a 100 mL volumetric
flask, would fill to the mark, and mix to dissolve. Thus, the amount of water is not
85 g.
Parts per Million:
For trace components of a mixture, concentration is sometimes expressed in parts per
million (ppm). Thus, we might describe the level of mercury in a contaminated sample
of tunafish by saying the concentration is 1.05 ppm. This description implies that if we
analyzed a million (106) g of tunafish, we would find 1.05 g of mercury.
Common lab calculations involving percents
Molarity (M) is the number of moles of a substance per litre of solution.
Concentration =
amount of solute (moles)
x 100
volume of solution (L)
Ex. A solution contains 5.85 g of NaCl dissolved in 5.00 x 103 mL of solution.
Determine its molarity. (Given Na = 22.99 g/mol; Cl = 35.45 g/mol)
Ex. What mass of KOH is required to prepare 6.00 x 102 mL of a solution with
concentration of 0.225 mol/L? (Given K = 39.10 g/mol; 0 = 16.00 g/mol; H = 1.01
g/mol)
Challenge question:
A 12.0% (w/w) aqueous solution of acetone has a density of 0.9849 g/mL at room temp.
The molar mass of acetone is 58.08 g/mol and the density is 0.7899 g/mL.
For the solute acetone, calculate the following:
(i) % (w/v) (ii) %(v/v) (iii) molarity(mol/L)
(iv) ppm (mg/L)