Back to Chem 101-102 Home Page
1) In (16.122 + 2.3)/ 1.461, how do we decide
significant figures?
The rule for adding and subtracting is that you use the smaller
number of decimal places. Since 16.122 has 3 decimals and 2.3 has
1 decimal, the number on top will be counted as having 1 decimal.
This is confusing since you really leave it all in your
calculator. You consider the 1 decimal at the end when rounding
the final answer. (16.122 + 2.3) = 17.422 , but really only 17.4
is significant. Again leave all (17.422) in the calculator and
divide by 1.461. The rule for division and multiplication is that
the answer will have the smaller number of sig figs (pay no
attention to where the decimal is). So 17.4 is 3 and 1.461 is 4
sig. figs so the final answer will have 3 sig figs.
1) Can you go over calculating atomic mass from isotopes again?
Element X is 98.7% X-13 and 1.3% X-12
What is the atomic mass?
Step 1: Change % to decimal (98.7% --> .987 AND 1.3% -->
.013)
Step 2: Multiply each decimal by the mass of the isotope
(.987 x 13) and (.013 X 12)
Step 3: Add the mass from each isotope to = the atomic mass
(12.831) + (0.156) = 12.987
2) How do you solve for % given the atomic mass? I know
how to go from % to atomic mass, but this is reverse.
You know that the atomic mass from the chart is 6.941 and that
this is made up of Li-6 and Li-7. The % of Li-6 + the % of Li-7 =
100. So you can let one of them = x % and the other 100-x%.
If you let Li-6 = x% and Li-7 = 100-x%, then
6.941 = (1.00 -x) 7.01600 + (x) 6.01512
Solve for x
3) How do you determine which element is largest for 2nd I.E.?
Write out the electron structure. Take away one electron. Now
which would require the most energy to remove the 2nd electron.
For example: Na is 1s2, 2s2, 2p6, 3s1 The first electron is easy
to remove, because it is in the 3s, the second electron is from a
full p level and very hard to remove. The first ionization energy
for Na is low (easy to remove), the 2nd ionization energy is high
(hard to remove).
4) How do we explain the trends?
Energy levels increase as you go down a column, and effective nuclear
charge increases as you go across a row. The trick is to apply these things to
the trends that you memorized. Let me give you an example with atomic size and
electronegativity.
You memorized that atomic size increases as you go down a column, which can be
explained by the increasing # of electron shells. You memorized that
electronegativity decreases down a column (increases as you go up), which can be
explained by the idea that with increasing # of electron shells, the tendancy to
gain an electron would decrease as the valence shell gets further from the
nucleus.
You memorized that atomic size decreases across a row (increases to the left),
which can be explained by the increasing effective nuclear charge across. With
strong effect. nuclear charge, the pull on the valence shell will make the atom
smaller. You memorized that electronegativity increases as you go across a row,
which can be explained by the increasing effective nuclear charge as you go
across a row. The more the nucleus pulls the valence shell, the more likely the
atom will have a tendancy to gain an electron.
The only variations in this is the difference a fully-filled s orbital or 1/2
filled p or d orbital makes.
1) Is there more practice on converting between molecules, moles, and grams? Here is a practice sheet that has 10 questions:
THE QUESTIONS ARE:
1. ? moles H2O = 36 g
2. ? molecules H2O = 36 g
3. ? g H2O = 0.5 mole
4. ? g H2O = 1 x 1023 molecules
5. If 88 g of X = 2 moles, what is the molar mass of X?
6. If 56 g of Y = 1.2 x 1024 molecules, what is the molar mass of Y?
7. If 3.01 x 10 24 atoms is 95 g of an element, what is the atomic mass?
8. In 66 g of CO2 there are:
how many moles?
how many molecules?
how many oxygen atoms?
9. If 1215 AMU of an element is 8.306 x 10 - 23 moles:
how many atoms are in the 1215 AMU?
what is the Atomic Mass?
how many grams are in the 1215 AMU?
10. What is the mass percent composition of Ca(HSO4)2?
The answers are: 1) 2.0 mol H2O; 2) 1.2 x 1024 molecules; 3) 9 g H2O; 4) 3 g H2O; 5) 44 g/mol; 6) 28 g/mol; 7) 19 amu/atom; 8) 1.5 mol CO2; 9.0x 1023 CO2 molecules;18 x 1023 O atoms; 9)50.02 atoms; 24.3 amu/atom; 202 x 10-23 g; 10) 17.1% Ca; 0.8% H; 27.4% S; 54.7% O5 solution.
If 88 g of X = 2 moles, what is the molar mass of X?
Molar mass is the # of grams in 1 mole. You know the grams (88) in 2 moles, so
88g/2mol = 44 g/mol = molar mass7. solution
If 3.01 x 10 24 atoms is 95 g of an element, what is the atomic mass?
Again, molar mass is the # of grams in 1 mole and is the same number as the atomic mass (only the units differ).
Here you know the grams (95) and the number of atoms. Since 6.02 e23 atoms makes up 1 mole, you can find the number of moles from the atoms given. 3.01 e 24 (1 mole / 6.02 e23 ) = 5.00 moles
So use the definition of molar mass: 95 g / 5.00 moles = 19 g/mol = molar mass
Then use the idea that molar mass and atomic mass are the same # only different units:
so atomic mass = 19 amu/atom
2) In the equation to convert Celsius to Kelvin, should we
use "C+273=K" or "C+273.15=K" in our
calculations?
Often it will not matter, due to sig figs, but C + 273.15
= K is more accurate and will give the most sig figs.
3) You mentioned in class that we should use the atomic
mass rounded to the nearest tenths place. When we are adding the
atomic mass of two or more elements, should we start our
calculations by using the rounded number, or should we round
after we add the numbers together?
for PCl5
1P=30.9738 or 31.0
5Cl=177.2635 or 177.3
mass="208.2373, rounding off to 208.2" or
"208.3" if you used the masses with 1 decimal.
With this problem, you will divide 5.6 by the formula
mass; so the number of sig figs will be limited to two in the
end.
I would round with the atomic mass:
so P = 31.0
for Cl use 35.45 x 5 = 177.25 so the formula mass is 208.25 or
really 208
So mol = 6.6/208
For exams, we will usually use a table that has 1 decimal,
sometimes 2, except for the nuclear chapter..
4) How do you find molecular formulas?
The empirical formula times a whole number = the molecular
formula.
If you know the empirical and the molecular, then:
molecular formula mass/ empirical formula mass = whole number
This is the whole number that you multiply the empirical formula
by: so if the empirical formula is HX and when you are given the
molecular mass, you find the whole number is 3, the actual
molecular formula is H3X3.
5) Can you repeat the naming rules?
Click here for the
naming rules.
6) Why is H3PO4 called
phosphoric acid rather than perphosphoric acid and why is H2SO4
called sulfuric acid rather than persulfuric acid?
See the handout about acid's. when the -ate ion is mixed
with H+ to form an acid, it is called --ic acid. so SO4
(2-) is sulfate so H2SO4
is sulfuric acid. Likewise the PO4(-3)
ion is called phosphate, so the acid form is phosphoric acid. It
depends on what ion is mixed with H+,
not on the amount of O's or H's. H2SO5
, if it exists, would be persulfuric acid because it is H+
added to the persulfate ion, which has one more oxygen than the
sulfate ion that you memorized.
1)What are the answers to the practice sheet with the questions below:
1. How many grams of O2 are required to completely convert 12.0 moles of C into
CO2?
2. From: N2 (g) + 3H2 (g) --> 2NH3(g) +
22 kcal How many mol of N2will
be needed and mol of NH3 will be
produced from 7.5 moles of H2?
3. By the reaction 2SO2 + O2 --> 2SO3 how much
SO3 could be made from 12.8 g SO2 and excess O2?
4. By the reaction 2SO2 + O2 --> 2SO3 how much
SO3 could be made from 6.4 g O2 and excess SO2?
5. From the reaction of Hg and S to produce Hg2S, how much Hg would be needed to produce 217.0 g Hg2S?
6. If 3.00 mole of C6H6 and 10.0 mole of O2
are combined, how many moles of H2O is produced? The balanced equation is :
2C6H6 + 15O2 --> 12 CO2 + 6 H2O
7. If 159 g Na2CO3 are reacted with 146 g HCl via the unbalanced reaction:
Na2CO3 + HCl --> NaCl + H2O + CO2, how many moles
of CO2 are produced?
8. If 30.0 g NH3 are obtained by
reaction of 1.0 mole of N2 with
3.0 moles of H2 via the reaction
:
N2 (g) + 3H2 (g) --> 2NH3(g)
What is the % yield?
ANSWERS:
The answers are: 1) 384g Oxygen ; 2)5.0mol NH3,
2.5mol of N2 ; 3)
16.0g of SO3 ; 4) 32g of SO3;
5) 201.0g of Hg; 6) 4.00mol Water; 7) 1.50mol of CO2;
8) 88.0%
CLICK HERE FOR THE SOLUTIONS TO THE 8
QUESTIONS
2) Can you give us some practice with limiting reagents?
PRACTICE
Balance the following equation and answer questions 1-4 based on the balanced
equation.
NaOH + H2SO4 ----> Na2SO4 + H2O
1. 120 g NaOH + excess H2SO4 yields how many grams of Na2SO4 ?
2. 120 g NaOH + 49 H2SO4 yields how many grams of Na2SO4 ?
3. 120 g NaOH requires how many grams of H2SO4 for neutralization?
4. If 80 g NaOH and 98 g H2SO4 reacted, producing 30 g H2O, what was the % yield of water?
The answers are: 1) 210 g Na2SO4; 2)
71 g Na2SO4; 3) 150 g
H2SO4; 4) 83%.
CLICK HERE FOR THE SOLUTIONS TO THE 4
QUESTIONS.
3) I was just wondering which solubility guidelines we need to memorize in order to determine if a salt is soluble or not.
Good Question! I usually ask students to know the basics (almost all sodium, potassium, ammonium, chloride, and nitrate salts are soluble.) If you need more than that, the problem will say "soluble potassium hydroxide" or KOH (aq) or will give you a solubility table.
4) Describe what is meant by oxidation/reduction?
When an atom or ion loses electrons, it
is oxidized, it is the reducing agent, and it increases in
oxidation number (the number becomes more positive). When an atom
or ion gains electrons, it is reduced, it is the oxidizing agent,
and it decreases in oxidation number (the number becomes more
negative).
For example if Cu+
loses another electron to become Cu2+,
it is oxidized, it is the reducing agent, and its oxidation
number goes from +1 to +2 or becomes more positive. IF Cu+
gains an electron to become Cu, it is reduced, it is the
oxidizing agent, and its oxidation number goes from +1 to 0 or
becomes more negative.
5) What about some practice on net-ionic equations ? Click Here for a problem set with equations (a) through (h) AND the solutions.
1) What is core notation? When writing
out the electron conf. of K (or anything thereafter) You have a
choice:
The complete electron config for K is:
1s2, 2s2, 2p6, 3s2, 3p6, 4s1
A condensed way to show K is to show the nearest noble
gas in brackets. [Ar] 4s1
This works because [Ar] = the electron conf of Ar = 1s2,
2s2, 2p6, 3s2, 3p6,
It can be done will all noble gases, so a condensed or
core notation way to show Sr is: [Kr] 5s2
instead of the full electron conf. 1s2, 2s2, 2p6, 3s2,
3p6, 4s2, 3d10, 4p6, 5s2
2) Why does Ni end in a 4s2 instead of a full
3d10?
It depends on whether you write it in the order the
levels are from the nucleus OR in the order of increasing energy
(they way electrons fill).
Ni = [Ar] 4s2, 3d8 is the same as [Ar] 3d8, 4s2 The first is in
energy order, the second is in order from the nucleus.
3)Can you explain the electron configuration and orbital configuration
for Sc & Cd
?
Sc = [Ar] 4s2, 3d7 is the same as [Ar] 3d7, 4s2 so you
would show all orbitals filled with an up arrow and a down arrow,
except 3d. For 3d you have 5 orbitals, but 7 electrons. Since
electrons fill with 1 in each orbital, you will have 3 of the d orbitals with both arrows, and
3 of the d orbitals with a single
arrow.
Cd =[Kr] 5s2, 4d10 which means all filled with 2 arrows for the
[Kr] part, then 2 arrows in the single 5s orbital and 2 arrows in the five d orbitals.
4) Why do the electron configurations for Tl and Bi contain 4f14? I got the [Xe] and the last sublevels,
but not the f sublevel in my answer.
Ti is element # 81, so the nearest noble gas is [Xe],
then 6s2, then notice than that element number 57 is followed by
element 58, which is in the lower f series. So Ti = [Xe] 6S2,
4f14, 5d10, and 6p1. Also note that Xe is element # 54 so
54 electrons (Xe) +2 (6s) + 14 (4f) + 10 (5d) + 1 (6p) = 81
If you leave out any of these you won't have 81 electrons.
5) Can you explain how to determine the orbital spin for
a specific sublevel?
Each orbital can contain 2 electrons, one will spin one
direction, one the opposite direction. If there is 1 electron in
the orbital, we usually put it in an orbital diagram as an up
arrow, but I would not expect you to say if it's spin quantum
number is +1/2 or -1/2
6) How many orbitals in any atom can have the
given quantum number or designtion D) n=5 ? I came up with 23
orbitals since energy level 5 has s, p, d, and f 1+3+5+7=16..how
did solutions manual get 25
The 5th level has s, p, d, f so 1, 3, 5, and 7 = 16. The
5th level is predicted to have s, p, d, f, and g (although no
known elements have the g). So 1, 3, 5, 7, and 9 =25.
7) What are the answers to the practice sheet that has 3
questions about electron configurations?
Click here for the
quantum numbers worksheet!
The answers are:
1) The Cl- ion puts its last electron in the 3p level so:
n = 3;
l=1;
ml= 1, 0, or -1;
ms = +1/2 or -1/2n = main shell;
l = shape or sublevel;
ml = orientation;
ms = spin2) Cu = [Ar] 4s2, 3d9 from the chart, but is an exception, so really Cu = [Ar] 4s1, 3d10
The Cu 1+ = [Ar] 3d10 and is diamagnetic;3) size = Rb; trend is larger down & to left
size = Br-; Rb+, Sr2+; and Br- have same # of electrons and Br has fewer protons.
1st IE = O; trend is larger up and to right
2nd IE = K; both K and Rb have a noble gas config. after 1 electron has been removed, so MUCH energy to remove 2nd electron, but K is smaller and nucleus can pull harder
EA = O; trend is larger up and to left
1) I'm emailing you because I have a question
about the polarity of molecules. Is it set that, for example, all linear shaped molecules are nonpolar, all bent/angular shaped molecules are polar, all triangular plane
shaped molecules are nonpolar, etc?
This is basically true, if the atoms that attach to the
central one are all the same. For example: CCl4 and CHCl3 are
both tetrahedral, but CCl4 is nonpolar & CHCl3 is polar.
IF the atoms attached to the central on are all the same:
tetrahedral molecules are nonpolar
triangular bipyramidal are non polar
octahedral are nonpolar
Otherwise ( & for other shapes), consider each bond.
Find the negative end of each bond. Use the "horse"
analogy, assume that the 'horses' are all running toward the
negative end. If the center will move, the molecule is polar. If
the center will not move, the molecule is nonpolar.
3) What are the regions of electron density, the
shape, the angles and the polarity for PCl3?
there are 4 regions of electron density,
the structure is pyramidal (triangular pyramidal),
the bond angles are less than 109.5 degrees (the lone pair pushes the bonds
closer),
the molecule is polar.
1) When determining the pressure of a certain gas over water,
will the P of water always be given?
You will either
have a chart or be given the vapor pressure of water at the
temperature. Vapor pressure is related to temperature so the 24 torr is for the specific temperature in the problem.
2) Do the gas particles at one temperature have the same speed?
No, at the same temp all have the same KE; but not the
same speed (or velocity).
KE = 1/2 mV2
3)
In my notes from lecture I wrote that gas particles that take up
the most space (have a greater MW of # of electrons) will deviate
the most from ideal. Which will be nearest ideal
H2 , F2 or HF? I
think H2 b/c it is smaller and will move faster
and therefore have fewer attractions. So when asked which
would deviate most from ideal I answered F2 b/c it
was heaviest and slowest. But this is not the correct answer. HF is the answer. Why is this? Is the MW/# electrons
method not always the best way to determine the answer?
For now you can think of this as HF being a polar molecule, and having more
attraction between molecules. Always use the type of IM force to separate first. H2
and F2 have only London forces, while HF has hydrogen bonding. ...so
HF will deviate the most (more attraction between molecules) because of the
stronger IM force. Then between the 2 with London forces, use the MW to
determine that H2 will behave most like ideal because it is the
lightest.
4) At 333K, which of the pairs of gases below would have the most nearly identical rates of effusion??
a) N_2_O and NO_2_
b) CO and CO_2_
c) CO and N_2_
d) N_2_ and O_2_
e) NO_2_ and N_2_O_4_
this question is from fall 2000, the correct answer is C..
rates of effusion depend on molar mass. (same KE, but KE = 1/2 mv2 ; heavy particles move more slowly, etc.)
a) 44 g/mol compared with 46
b) 30 g/mol to 44 g/mol
c) 28 g/mol compared to 28 -----has the same Molar Mass.
d) 28 g/mol to 32 g/mol
e) 46 g/mol to 92 g/mol
5)What is boiling? Boiling is the point
where the vapor pressure of a liquid equals the atmospheric
pressure. A liquid with high vapor pressure at room temperature
will need just a little added heat to increase the vapor pressure
to the atmospheric, so boiling pt will be low. (higher IMF =
higher bp).
6) What is viscosity, surface tension,
capilary action, and vapor pressure?
Viscosity is the resistance to flow of a liquid (higher IMF =
more viscosity); surface tension is the 'beading' up of water, it
is caused by the attraction of surface molecules to interior
molecules (higher IMF = higher surface tension); capillary action
is when a liquid climbs up a wick or rises in a glass tube, this
depends on the IMF of the liquid molecules compared to those
between the liquid and the glass. (higher IMF = lower capillary
action); vapor pressure is the pressure created by gas molecules
above the surface of the liquid from which they have escaped
(higher IMF = lower vp or slower evaporation).
7) HEATING GRAPH SHOWING PHASE CHANGES: On the graph from page
498 comparing the temperature of water at different states, why
was the temperature leveled out for the solid and water phase?.
Is that saying that ice water is a constant temperature of 0
degrees Celsius?
This is a good question. During the level parts of the graph, heat is being used to break up the IM forces to change the state rather than being used to increase the temperature.
8) How do the IM Forces compare?
Ionic > hydrogen b.> dipole-dipole > London f.
Differentiate between IONIC by:
Charge on ions
Size
Requires ions
Differentiate between DIPOLE-DIPOLE by:
Charge on Dipole Moment
EN Difference
Requires polar molecules
Requires H bonded to F,O,or N for HYDROGEN BONDING
Differentiate between LONDON FORCES by:
# of electrons (Molar Mass)
Requires nonpolar molecules