Solution to Exercises for Study Sheet 3
⎛ 1 mol Na ⎞
1. a) ? mol Na = 23.0 g Na ⎜
= 1.00 mol Na
⎝ 23.0 g Na ⎟⎠
⎛ 5 O atoms ⎞
b) ? O atoms = 1 N 2O5 molecule ⎜
= 5 O atoms
⎝ 1 N 2O5 molecule ⎟⎠
⎛
⎞
1 mol H 2O
c) ? mol H 2O = 6.02 x 10 23 H 2O molecules ⎜
= 1.00 mol H 2O
23
⎝ 6.02 x 10 H 2O molecules ⎟⎠
⎛ 174 g K 2 SO4 ⎞
d) ? g K 2 SO4 = 1.00 mol K 2 SO4 ⎜
= 174 g K 2 SO4
⎝ 1 mol K 2 SO4 ⎟⎠
⎛ 263 amu Mg3 (PO4 )2 ⎞
e) ? amu Mg3 (PO4 )2 = 1 FU Mg3 (PO4 )2 ⎜
= 263 Mg3 (PO4 )2
⎝ 1 FU Mg3 (PO4 )2 ⎟⎠
⎛ 1 mol CO2 ⎞
2. a) ? mol CO2 = 68.0 g CO2 ⎜
= 1.55 mol CO2
⎝ 44.0 g CO2 ⎟⎠
⎛ 6.02 x 10 23 K atoms ⎞
b) ? K atoms = 0.150 mol K atoms ⎜
= 9.03 x 10 22 K atoms
⎟
1
mol
K
atoms
⎝
⎠
⎛ 1 mol Cl2 molecules ⎞
c) ? mol Cl2 = 5000 Cl2 molecules ⎜
= 8.31 x 10 −21 mol Cl2 molecules
23
⎟
⎝ 6.02 x 10 Cl2 molecules ⎠
⎛ 2 • 23.0 g Na ⎞
d) ? g Na = 0.820 mol Na2 SO4 ⎜
= 37.7 g Na
⎝ 1 mol Na2 SO4 ⎟⎠
⎛ 1 mol Ba3 (PO4 )2 ⎞ ⎛ 8 mol O atoms ⎞
e) ? mol O atoms = 125 g Ba3 (PO4 )2 ⎜
= 1.66 mol O atoms
⎝ 602 g Ba3 (PO4 )2 ⎟⎠ ⎜⎝ 1 mol Ba3 (PO4 )2 ⎟⎠
⎛ 1 mol K 2 S ⎞ ⎛ 6.02 x 10 23 K 2 S FU ⎞
f ) ? K 2 S FU = 85.0 g K 2 S ⎜
= 4.65 x 10 23 K 2 S FU
⎟
⎜
⎟
1 mol K 2 S
⎝ 110. g K 2 S ⎠ ⎝
⎠
⎛ 1 mol NH 3 molecules ⎞ ⎛
⎞
17.0 g NH 3
g) ? g NH 3 = 1 NH 3 molecule ⎜
= 2.82 x 10 −23 g NH 3
23
⎟
⎜
⎝ 6.02 x 10 NH 3 molecules ⎠ ⎝ 1 mol NH 3 molecules ⎟⎠
⎛ 58.9 amu ⎞
h) ? amu Co = 1 Co atom ⎜
= 58.9 amu
⎝ 1 Co atom ⎟⎠
÷ by smallest
number of mole
100 - 60.87 - 4.38
3. a) C60.87 g H 4.38 gO34.75 g ⇒ C
%=g
60.87 g
H
12.01 g mol
4.38 g
1.01 g mol
O
34.75 g
⇒ C5.069 mol H 4.34 molO2.172 mol
16.00 g mol
÷ by MM of each element to get the
mol ratio of each element
C7H6O3
x 3 to get
whole #s
C2.334H2.00O ⇐ C 5.069 mol H 4.34 mol O2.172 mol
2.172 mol
2.172 mol
2.172 mol
40.00 - 36.20
b) C36.20 g H 3.80 g ⇒ C
36.20 g
12.01 g mol
H
3.80 g
1.01 g mol
⇒ C3.014 mol H 3.76 mol ⇒ C 3.014 mol H 3.76 mol ⇒ CH 1.25
3.014 mol
÷ by MM of each element to get the
mol ratio of each element
C4H5
4. Inx Cly +
AgNO3 ⎯⎯
→ AgCl
3.014 mol
÷ by smallest
number of moles
x 4 to get
whole #s
(incomplete and unbalanced equation)
0.972 g
0.500 g
⎛ 35.5 g Cl ⎞
? g Cl = 0.972 g AgCl ⎜
= 0.240 g Cl
⎝ 144 g AgCl ⎟⎠
This represents the grams of Cl in AgCl, as well as,
the grams of Cl present in the 0.500 g of unknown
compound containing In and Cl.
? g In = 0.500 g Inx Cly − 0.240 g Cl = 0.260 g In
In0.260 gCl0.240 g ⇒ In 0.260 g Cl
115 g
mol
0.240 g
35.5 g mol
⇒ In0.00226 mol Cl0.00676 mol ⇒ In 0.00226 mol Cl 0.00676 mol ⇒ InCl3
÷ by MM of each element to get the
mol ratio of each element
0.00226 mol
0.00226 mol
÷ by smallest
number of moles