CH 115 Fall 2014
Worksheet 1
1. Fill in the table below:
Subatomic Particle
Mass
Proton
1.6 x 10-27 kg; 1
amu
Electron
9.1 x 10-31 kg
Neutron
Same as a proton
Charge
+1
Location
In nucleus
-1
In electron clouds
surrounding the
nucleus
In nucleus
0
2. Convert 486 meters into nanometers and micrometers.
1 m = 109 nm  486 m = 486 x 109 nm or 4.86 x 1011 nm
1 m = 106 micrometers  486 m = 486 x 106 micrometers or 4.86 x 108
micrometers
3. List the four components of Dalton’s atomic theory in your own words (so, what do
they really mean and how can you best explain them?)
When forming compounds, elements have to present in constant ratios 
change the ratios, and you change the chemical nature of the compound
All compounds must be made of 2 or more elements otherwise they are just
considered to be in their elemental form.
You can’t change, destroy, or transform atoms into other elements  can’t
take lead and turn it into gold. Exception – nuclear reactions
Atoms of a particular element have SIMILAR properties (not identical because
each has various isotopes).
4. Determine the number of protons, neutrons, and electrons in each of the following
chemical species.
a). 40Ar – 18 protons, 22 neutrons, 18 electrons
b). 25Mg – 12 protons, 13 neutrons, 12 electrons
c). 67Cu –29 protons, 38 neutrons, 29 electrons
d). 35Cl1- - 17 protons, 18 neutrons, 18 electrons
e). 112Ba2+ - 56 protons, 56 neutrons, 54 electrons
5. In each of the following, indicate which element is defined by the given information.
a). 11 protons, 12 neutrons, 11 electrons – sodium (Na)
b). 20 protons, 20 neutrons, 20 electrons – calcium (Ca)
c). 17 protons, 20 neutrons, 17 electrons – chlorine (Cl)
d). 35 protons, mass number = 80 – bromine (Br)
e). 92 protons, mass number = 235 – uranium (U)
CH 115 Fall 2014
Worksheet 1
6. What is an isotope? Give an example.
Two elements that have the same number of protons but differing numbers of
neutrons (or a different mass number) are considered to be isotopes of each
other. Ex. Carbon-14
7. Calculate the atomic mass of silver if silver has 2 naturally occurring isotopes with
the following masses and natural abundances:
Ag-107
106.90509 amu
51.84%
Ag-109
108.90476 amu
48.46%
(.5184) * (106.90509 amu) + (.4846) * (108.90476 amu) = 108.2 amu
8. List an example of an alkaline earth metal, a metalloid, an alkali metal, a halogen, a
transition metal, and a noble gas in this order.
Magnesium, boron, lithium, chlorine, copper, argon (or any other combination
– many possible answers)
9. The Aurora Borealis is a night display in the Northern latitudes caused by ionizing
radiation interacting with the Earth's magnetic field and the upper atmosphere. The
distinctive green color is caused by the interaction of the radiation with oxygen and
has a wavelength of 5577 angstroms. What is the frequency of this light? How much
energy is released?
Given wavelength, we are asked to find frequency. The equation that relates
the two of these is c (speed of light) = wavelength (lambda) * frequency (nu).
The speed of light is a constant that we approximate as 3.0 x 108 m/s. Given
the wavelength, we can now solve for the frequency.
1 m = 1010 angstroms  5577 A = 5577 x 10-10 m  5.577 x 10-7 m
frequency = 3.0 x 108 / 5.577 x 10-7 = 5.379236148 x 1014 1/s (Hertz, Hz) 
remember NOT to round until the end of a problem as it will mess up your
significant figures (we’ll talk more about this next week)
Now, we are asked to find the amount of energy released. The equation we
want to use here is E (energy) = h (Planck’s constant) * frequency (nu).
Planck’s constant is always 6.626 * 10-34 Js. We have the frequency we figured
out above, so we can multiply Planck’s constant times that to get the amount of
energy in Joules that was released.
E = (6.626 x 10-34) x (5.379236147 x 1014) = 3.564 x 1019 J (now you can
round)
NOTE: Make sure ALL of your units match up and cancel out before you do ANY
math. The most common mistake in doing these types of problems is just using
the numbers as given, rather than converting them to the units you need!