Download More_Mole_Calculations Sep 2013 ANSWERS

Chemistry 11 (HL)
Unit 2 / IB Topic 1
Moles and Formulas 1 –More Practice Problems
Answers – Please do not print.
number
(of particles)
1.
Avogadro’s
moles
Number
molar
mass
mass
Calculating Numbers of Particles
23
a)
# NO2 molecules = 0.0045 g x (1 mol / 46.01 g) x (6.02 x 10
19
= 5.9 x 10 molecules
molecules / 1 mol)
b)
# K2SO4 formula units = 8.25 x 10 g x ( 1 mol / 174.26 g ) x ( 6.02 x 10
24
= 2.85 x 10 formula units
2
23
form units / 1 mol)
NOTE: A “formula unit” is the smallest “particle” of an ionic compound.
c)
3
# Pb atoms = 1.00 x 10 g x ( 1 mol / 207.19 g ) x ( 6.02 x 10
24
= 2.91 x 10 atoms
23
atoms / 1 mol)
d)
3
23
# molecules = 4.9 x 10 mg x ( 1 g / 1000 mg ) x (18.02 g / 1 mol) x (6.02 x 10 molecules / 1 mol)
23
= 1.6 x 10 molecules
e)
23
# O atoms = 3500 g x ( 1 mol / 44.01 g ) x (6.02 x 10 molecules / 1 mol ) x (6 O atoms / molecule)
26
= 2.9 x 10 O atoms
f)
223
2# SO4 ions = 36.85 g x ( 1 mol / 342.14 g ) x ( 6.02 x 10 molecules / 1 mol) x (3 SO4 / molecule)
23
2= 1.945 x 10 SO4 ions
2.
Calculating Masses
a)
MNH3 = 17.04 g/mol
mass = 0.0250 mol x (17.04 g / 1 mol)
= 0.426 g
b)
MH2 = 2.02 g/mol
mass = 32 mol x (2.02 g / 1 mol)
= 65 g
c)
MTNT = 227.15 g/mol
-4
mass = 5.75 x 10 mol x ( 227.15 g / 1 mol )
= 0.131 g
Chemistry 11 (HL)
d)
Unit 2 / IB Topic 1
MH2O = 18.02 g/mol
mass = 1 molecule x ( 1 mol / 6.02 x 10
-23
= 3.00 x 10
g
e)
21
atoms x ( 1 mol / 6.02 x 10
23
30
23
molecules ) x (44.01 g / 1 mol) x (1 kg / 1000 g)
MMg(OH)2 = 58.33 g/mol
18
mass = 3.25 x 10 form u x (1 mol / 6.02 x 10
–4
= 3.15 x 10 g
3.
23
form u) x (58.33 g / 1 mol)
Isopropanol (C3H8O) is the active ingredient in many hand disinfectant liquids. Brand “X” contains 70%
isopropanol by mass. If you have a 50.0 g bottle of the hand disinfectant,
a)
mass isopropanol = 50.0 g x 70/100
= 35.0 g (keep one extra sig fig because this value is used in the next Q)
b)
Misopropanol = 60.11 g/mol
# H atoms = 35.0 g x ( 1 mol / 60.11 g) x ( 6.02 x 10
24
= 2.8 x 10 H atoms (2 sig figs)
7.
atoms ) x ( 196.97 g / 1 mol)
MCO2 = 44.01 g/mol
mass = 5.0 x 10 molecules x (1 mol / 6.02 x 10
5
= 3.7 x 10 g
g)
molecules ) x (18.02 g / 1 mol )
MAu = 196.97 g/mol
mass = 8.3 x 10
= 2.7 g
f)
23
a)
23
molecules / 1 mol ) x ( 8 H atoms / 1 molecule)
Determine the molar mass of the gas.
1) mol gas = 3.01 x 10
= 50 mol
25
molecules x (1 mol / 6.02 x 10
23
molecules)
2) molar mass = (mass of sample) / (moles in sample)
= 1400 g / 50 mol
= 28 g/mol
b)
The question states the gas is diatomic, so its general formula is X2.
The atomic mass of X must be 14 g/mol.
The diatomic elements are: H2 O2 F2
Br2 I2 N2 Cl2
X must be nitrogen (N) because its atomic mass is 14.01 g/mol.