1. ( ) Error and % Error Page 6 Pages 1-5 are for a Table of Contents
A science student is trying to measure the speed of light. She measures the speed of light as 3.20X108 m/s. The accepted value for the speed of light is known to be 3.00X108 m/s. Find the Error and % Error for her work.
Click to see solution
Error = Experimental Value – Accepted Value
Error = 3.20X108 m/s – 3.00X108 m/s
= 2.0X107 m/s
% Error = ( ) *
100
Experimental Value−Accepted Value Accepted Value
3.20X108 m/s – 3.00X108 m/s
3.00X108 m/s
* 100
% Error =
% Error = %
% Error = 6.7%

2. Law of Conservation of Mass Page 7
When solid calcium carbonate is heated to very high temperatures in a crucible, it decomposes into gaseous carbon dioxide and solid calcium oxide according to the equation shown below.
CaCO3 (s) → CO2 (g) CaO (s)
Use the information in the laboratory table below to find the mass of carbon dioxide produced when a student attempts to decompose some calcium carbonate.
Mass of empty crucible g
Mass of crucible and CaCO3 before heating g
Mass of crucible and CaO after heating g
Click to see solution
g g = g (mass of CaCO3 used in experiment)
g g = g (mass of CaO produced in experiment)
CaCO3 (s) → CO2 (g) CaO (s)
2.298 g = x g
x = g g = g CO2 produced

3. x x 100 100 m%Na = mNa mNa2SO4 m%Na = m%Na = mNa mNa2SO4 mNa = m%Na
Mass Percent Problems
Page 8
When a g sample of sodium sulfate is analyzed, it is found to contain g of sodium. Find the mass percent of sodium in the sodium sulfate sample.
m%Na =
mNa
mNa2SO4 x
100
Click to see solution
m%Na = g g x
100 = % Na
If a chemist analyzed g of sodium sulfate, how many grams of Na would the sample be found to contain?
Click to see solution
m%Na =
mNa
mNa2SO4 x
100
mNa =
m%Na
mNa2SO4
100
mNa =
(32.38%)(14.62 g)
100
mNa = g Na

4. Moles to Mass
Page 9
What is the mass in grams of moles of silver?
Click to see Solution
First we will solve this using the equivalence statement method:
Start with what was given and then use the equivalence statement from the periodic table: 1 mole Ag = g Ag
moles Ag
( )
1 mol Ag
107.9 g Ag
( )
= g Ag
This is a conversion factor based on the equivalence statement from the periodic table.
Now we can solve the same problem using the equation method:
Given: n = mol
MW = g/mol (form periodic table)
MW = m n rearrange to get equation for “m”
m = ( mol Ag)(107.9 g Ag/mol Ag) = 7.50 g Ag

5. Mass to Moles
Page 10
How many moles of copper (Cu) are present in a sample of copper that
has a mass of 127.3g?
Click to see Solution
First we will solve this using the equivalence statement method:
Start with what was given and then use the equivalence statement from the periodic table: 1 mole Cu = g Cu
127.3 g Cu
( )
63.55 g Cu
1 mol Cu
( )
= mol Cu
Now we can solve the same problem using the equation method:
Given: m = g
MW = g/mol (form periodic table)
MW = m n then rearrange to get “n”
n = m MW = g g/mol
= mol Cu

6. ( ) ( ) ( ) ( ) Page 11 Particles to Moles and Moles to Particles
How many atoms of Zn are contained in moles of Zn?
Click to see Solution
Given: moles Zn
1 mole Zn = 6.022X1023 atoms Zn
moles Zn
( )
1 mol Zn
6.022X1023 atoms Zn
( )
= 5.2X1021 atoms Zn
How many moles of copper are present if you have a sample
containing 3.8X1021 atoms of copper?
Click to see Solution
Given: X1021 atoms Cu
1 mole Cu = 6.022X1023 atoms Cu
3.8X1021 atoms Cu
( )
1 mol Cu
6.022X1023 atoms Cu
( )
= mol Cu

7. Particles to Mass and Mass to Particles
Page 12
How many atoms of Zn are contained in 14.3 g of Zn?
Click to see Solution
Given: g Zn mole Zn = 6.022X1023 atoms Zn mol Zn = g Zn
14.3 g Zn
( )
( )
1 mol Zn
65.38 g Zn
1 mol Zn
6.022X1023 atoms Zn
( )
= 1.32X1023 atoms Zn
How many grams of copper are present if you have a sample
containing 7.430X1022 atoms of copper?
Given: 7.430X1022 atoms Cu mole Cu = 6.022X1023 atoms Cu
1 mol Cu = g Cu
Click to see Solution
7.430X1022 atoms Cu
( )
1 mol Cu
6.022X1023 atoms Cu
( )
( )
1 mol Cu
63.55 g Cu
= g Cu

8. What is the frequency of light that has a wavelength of 7.32X10-6 m?
Wavelength and Frequency
Page 13
What is the frequency of light that has a wavelength of 7.32X10-6 m?
Click to see Solution
Given: l = 7.32X10-6 m n = ? c = 3.00X108 m/s ln = c
n = c l
n = (3.00X108 m/s) (7.32X10−6m)
ln = c
n = 4.10X1013 1/s
What is the wavelength of light that has a frequency of 4.83X105 1/s?
Click to see Solution
Given: n = 4.83X105 1/s l = ? c = 3.00X108 m/s ln = c
l = c n
l = (3.00X108 m/s) (4.83X105 1/s)
ln = c
l = 621 m

9. What is the energy of light that has a wavelength of 3.92X10-6 m?
Energy of light (or electron transitions)
Page 14
What is the energy of light that has a wavelength of 3.92X10-6 m?
Given: l = 3.92X10-6 m E = ?
Click to see Solution
E = hc l
E = (6.626X10−34 Js)(3.00X108 m/s) (3.92X10−6m)
E = 5.07X10-20 J
What is the energy of light that has a frequency of 7.34X107 1/s?
Click to see Solution
Given: n = 7.34X107 1/s E = ?
E = hn
E = (6.626X10−34 Js)(7.34X107 1/s)
E = 4.86X10-26 J
What is the wavelength of light having an energy of 8.93X10-20 J?
Click to see Solution
Given: E = 8.93X10-20 J l = ?
E = hc l
l = hc E
l = (6.626X10−34 Js)(3.00X108 m/s) (8.93X10−20J)
l = 2.23X10-6 m