Download Week 3 July 22, 2016 Worksheet Review III 1 mol = 6.022 × 1023 1

Week 3
July 22, 2016
Worksheet
Review III
1 mol = 6.022 × 1023
1. Which sample contains the greatest number of atoms?
a) 10.0 g Na
b) 10.0 g Fe
c) 10.0 g Ag
d) 10.0 g Cu
e) 10.0 g Ba
1 mol Na
6.022 × 1023 atoms
10.0 g Na ×
×
= 2.62 × 1023 atoms Na
22.99 g Na
1 mol
1 mol Fe 6.022 × 1023 atoms
10.0 g Fe ×
×
= 1.08 × 1023 atoms Fe
55.85 g Fe
1 mol
10.0 g Ag ×
1 mol Ag
6.022 × 1023 atoms
×
= 5.58 × 1022 atoms Ag
107.87 g Ag
1 mol
10.0 g Cu ×
10.0 g Ba ×
1 mol Cu 6.022 × 1023 atoms
×
= 9.48 × 1022 atoms Cu
63.55 g Cu
1 mol
1 mol Ba
6.022 × 1023 atoms
×
= 4.39 × 1022 atoms Ba
137.33 g Ba
1 mol
Another way to think about this problem (without actually doing any calculations) is
to think about the conversion. In all cases we start with 10.0 g of an element, and
then we divide by the mass of each element (blue numbers), and then multiply by
Avogadro’s number. So the only difference between each calculation is the mass
(blue numbers) that we use to divide 10.0 g, so since Na has the lightest mass, the
answer (conversion) will be the largest since we are dividing by the smallest number.
2. How many grams of carbon are in a 13.25 g sample of carbon suboxide (C3O2)?
a) 0.1948 g
b) 0.5843 g
c) 0.7797 g
d) 2.339 g
e) 7.017 g
First, find the molar mass of C3O2:
⎛
12.01 g ⎞ ⎛
16.00 g ⎞
1 mol C3O 2 = ⎜ 3 mol C ×
+ ⎜ 2 mol O ×
= 68.03 g C3O 2
⎟
1 mol C ⎠ ⎝
1 mol O ⎟⎠
⎝
Now we can imagine the following reaction:
3C + 2O → 1C3O2
which tells us that for every 1 mol of C3O2, we have 3 mol of C and 2 mol O. So we
can first convert the 13.25 g C3O2 into moles of C3O2, and then into moles of C using
the mole ratio, and then into grams of C.
1 mol C3O 2
3 mol C
12.01 g C
13.25 g C3O 2 ×
×
×
= 7.017 g C
68.03 g C3O 2 1 mol C3O 2 1 mol C
Week 3
July 22, 2016
Worksheet
3. A compound was analyzed to contain 79.89% carbon and 20.11% hydrogen.
Determine the empirical formula of the compound.
a) CH
b) CH2
c) CH3
d) CH4
e) C7H20
Step 1: Convert the percentages into masses (assume you have a 100 g sample):
79.89 g C
20.11 g H
Step 2: Convert these masses into moles:
79.89 g C ×
1 mol C
1 mol H
= 6.652 mol C 20.11 g H ×
= 19.911 mol H
12.01 g C
1.01 g H
Step 3: Divide both mole amounts by the smaller number (6.652):
6.652 mol C
19.911 mol H
= 1 mol C
= 2.99 mol H
6.652
6.652
Step 4: Write the empirical formula based on these numbers (2.99 ≈ 3):
C1H3
4. How many grams of aluminum sulfate are in a 0.630 mol sample?
a) 77.5 g
b) 152 g
c) 172 g
d) 185 g
e) 216 g
First, write the chemical formula for aluminum sulfate: Al2(SO4)3. Then find the
molar mass of Al2(SO4)3:
⎛
26.98 g ⎞ ⎛
32.06 g ⎞ ⎛
16.00 g ⎞
⎜⎝ 2 mol Al × 1 mol Al ⎟⎠ + ⎜⎝ 3 mol S × 1 mol S ⎟⎠ + ⎜⎝ 12 mol O × 1 mol O ⎟⎠ = 342.15 g Al2 (SO 4 )3
Now we can convert 0.630 mol Al2(SO4)3 to grams:
0.630 mol Al2 (SO 4 )3 ×
342.15 g Al2 (SO 4 )3
= 216 g Al2 (SO 4 )3
1 mol Al2 (SO 4 )3
5. If you have equal mole samples of each of the following compounds, which
compound contains the greatest number of oxygen atoms?
a)
b)
c)
d)
e)
Magnesium Nitrate
Dinitrogen Pentoxide
Iron (III) Phosphate
Barium Oxide
Potassium Acetate
Mg(NO3)2 → 6 O atoms
N2O5 → 5 O atoms
FePO4 → 4 O atoms
BaO → 1 O atoms
KC2H3O2 → 2 O atoms
If we write out the chemical formula for each compound, we can easily count the
number of oxygen atoms for each compound. For instance, H2O has 1 O atom. The
chemical formulas and O atoms are given in blue next to each compound name.
Week 3
July 22, 2016
Worksheet
6. How many of the following pairs of compounds have the same empirical formula?
I.
II.
III.
IV.
a) 0
C2H2 and C6H6
C2H6 and C4H10
NO2 and N2O4
C12H10O and C6H6O
b) 1
CH and CH
CH3 and C2H5
NO2 and NO2
C12H10O and C6H6O
c) 2
d) 3
e) 4
7. If you have 6.022 × 1023 water molecules present in a sample, what must be true?
I.
1 mole of water is present in the sample
True, because we know that 1 mol = 6.022 × 1023.
II.
18.016 g of water is present in the sample
True. If we have 1 mol of H2O, then:
⎛
1.008 g ⎞ ⎛
16.00 g ⎞
1 mol H 2O = ⎜ 2 mol H ×
+ ⎜ 1 mol O ×
= 18.016 g
⎟
1 mol H ⎠ ⎝
1 mol O ⎟⎠
⎝
III.
1.807 × 1024 atoms are present in the sample
True. We know that for every 1 molecule of H2O, we have 2 atoms of H
and 1 atom of O. So we every 1 molecule of H2O has 3 atoms.
3× (6.022 × 1023 molecules H 2O) = 1.807 × 1024 atoms
IV.
2.016 g of hydrogen are present in the sample
True, because 1 mol H2O has 2 mol H and 1 mol O, and 2 mol H weighs
2.016 g.
V.
2 moles of hydrogen are present in the sample
True, see IV.
a) I only
b) I, II, IV
c) I, II, V
d) II, IV
e) All
8. How many moles are in a 50.0 g sample of ammonium carbonate?
a) 0.438 mol
b) 0.520 mol
c) 0.531 mol
d) 0.641 mol
e) 3.13 mol
First, write the chemical formula for ammonium carbonate: (NH4)2CO3. Then find its
molar mass:
⎛
14.01 g ⎞ ⎛
1.01 g ⎞ ⎛
12.01 g ⎞ ⎛
16.00 g ⎞
⎜⎝ 2 mol N × 1 mol N ⎟⎠ + ⎜⎝ 8 mol H × 1 mol H ⎟⎠ + ⎜⎝ 1 mol C × 1 mol C ⎟⎠ + ⎜⎝ 3 mol O × 1 mol O ⎟⎠ = 96.09 g
Now we can convert 50.0 g (NH4)2CO3 to moles:
50.0 g (NH 4 )2CO3 ×
1 mol (NH 4 )2CO3
= 0.520 mol (NH 4 )2CO3
96.09 g (NH 4 )2CO3
Week 3
July 22, 2016
Worksheet
9. The percent by mass of nitrogen is 46.7% for a species containing only nitrogen and
oxygen. Which of the following could be this species?
a) N2O5
b) N2O
c) NO2
d) NO
e) NO3
Because 46.7% of the compound is nitrogen, we know that the remainder must be
oxygen: 100% – 46.7% = 53.3% O. Now we can follow the steps from problem 3.
Step 1: Convert the percentages into masses (assume you have a 100 g sample):
46.7 g N
53.3 g O
Step 2: Convert these masses into moles:
46.7 g N ×
1 mol N
1 mol O
= 3.333 mol N 53.3 g O ×
= 3.331 mol O
14.01 g N
16.00 g O
Step 3: Divide both mole amounts by the smaller number (3.331):
3.333 mol N
3.331 mol O
= 1 mol N
= 1 mol O
3.331
3.331
Step 4: Write the empirical formula based on these numbers:
NO
10. A compound containing only sulfur and nitrogen is 69.6% S by mass. The molar mass
is 46.07 g/mol. What is the correct name for this compound?
a)
b)
c)
d)
e)
Tetrasulfur Dinitride
Trisulfur Hexanitride
Sulfur Mononitride
Tetrasulfur Tetranitride
Disulfur Hexanitride
Because 69.6% of the compound is sulfur, we know that the remainder must be
nitrogen: 100% – 69.6% = 30.4% O. Now we can follow the steps from problem 3/9.
Step 1: Convert the percentages into masses (assume you have a 100 g sample):
69.6 g S
30.4 g N
Step 2: Convert these masses into moles:
69.6 g S ×
1 mol S
1 mol N
= 2.170 mol S 30.4 g N ×
= 2.170 mol N
32.07 g S
14.01 g N
Step 3: Divide both mole amounts by the smaller number (2.170):
2.170 mol S
2.170 mol N
= 1 mol S
= 1 mol N
2.170
2.170
Step 4: Write the empirical formula based on these numbers:
SN
Step 5: Name the compound, which is covalent: sulfur mononitride
Week 3
July 22, 2016
Worksheet
11. Using solubility rules, predict which of the following pairs will produce a precipitate.
a)
b)
c)
d)
e)
NH4Cl and Na2S
HNO3 and CuSO4
FeCl2 and KOH
NiCl2 and (NH4)2SO4
All of the above
(NH4)2S (aq) and NaCl (aq)
H2SO4 (aq) and CuNO3 (aq)
Fe(OH)2 (s) and KCl (aq)
NiSO4 (aq) and NH4Cl (aq)
False.
You can determine the product of each reaction by either using a chart with the
cations/anions, or swapping the two ions in each reaction. Then you want to use
solubility rules to determine which products are soluble or insoluble (solid or
precipitate). The products and their solubility are written next to each pair of
reactants. The only insoluble (solid) compound is Fe(OH)2.
12. The correct net ionic equation for the reaction between nickel (II) chloride and
sodium sulfide is:
a)
b)
c)
d)
e)
Ni (aq) + S (aq) → NiS (s)
NiCl2 (aq) + Na2S (aq) → NiS (s) + 2NaCl (aq)
Ni2+ (aq) + S2– (aq) → NiS (s)
Na+ (aq) Cl– (aq) → NaCl (aq)
Ni2+ (aq) + Cl– (aq) + Na+ (aq) + S2– (aq) → NiS (s) + NaCl (aq)
First determine the chemical formulas for nickel (II) chloride and sodium sulfide,
which are the reactants:
NiCl2 (aq) + Na2S (aq) →
These are both soluble so they will exist as ions in solution:
Ni2+ (aq) + Cl– (aq) + Na+ (aq) + S2– (aq) →
Now you want to determine the products of this reaction. You can either set up a
chart with the cations/anions or swap the two ions. Use the solubility rules to
determine the solubility of each of the products as aqueous (soluble) or solid
(insoluble):
Ni2+ (aq) + Cl– (aq) + Na+ (aq) + S2– (aq) → NaCl (aq) + NiS (s)
In a net ionic equation we only care about the precipitate (solid) and the ions that
form it. The others (Na+ and Cl–) are spectactor ions and do not react. We can now
write:
Ni2+ (aq) + S2– (aq) → NiS (s)
13. Which of the following compounds is soluble?
a)
b)
c)
d)
e)
Lead (II) Sulfate
Magnesium Sulfate
Iron (III) Phosphate
Nickel (II) Hydroxide
Magnesium Hydroxide
Week 3
July 22, 2016
Worksheet
14. According to the following unbalanced chemical reaction
1C12H22O11 (s) + 12O2 (g) → 12CO2 (g) + 11H2O (g)
How many moles of oxygen gas are required to react completely with 2.0 moles of
C12H22O11 (s)?
a) 12 mol
b) 24 mol
c) 30 mol
d) 35 mol
e) Other
After we have balanced the chemical equation (hint: start by trying to balance the
number of oxygens), we see that for every 1 mole of C12H22O11, 12 moles of O2 are
needed for the reaction. We can use mole ratios to convert now:
2.0 mol C12 H 22O11 ×
12 mol O 2
= 24 mol O 2
1 mol C12 H 22O11
15. According to the following unbalanced chemical reaction
1CH4 + 2O2 → 1CO2 + 2H2O
If 5.00 g of CH4 is burned, what mass of water can be produced?
a) 5.62 g
b) 10.0 g
c) 11.2 g
d) 18.0 g
e) 36.0 g
After we have balanced the chemical equation (hint: start by trying to balance the
number of hydrogens), we see that for every 1 mole of CH4, 2 moles of H2O are
produced. We can now convert 5.00 g CH4 into moles of CH4 using its molar mass
(16.05 g), then use mole ratios to convert to moles of water, and then convert moles
of H2O into grams using its molar mass (18.02 g):
5.00 g CH 4 ×
1 mol CH 4 2 mol H 2O 18.02 g H 2O
×
×
= 11.2 g H 2O
16.05 g CH 4 1 mol CH 4
1 mol H 2O