CHEMISTRY 1 SK015 | Tutorial Chapter 1.0: Matter | 2022/2023

CHEMISTRY 1 SK015 | Tutorial Chapter 1.0: Matter | 2022/2023

1.1 Atoms and Molecules

1. a) Number of protons in the nucleus of an atom of an element.
i. ii. Total number of protons and neutrons in the nucleus of an atom of an element.
ii. iii. Two or more atoms of the same element having the same number of protons
iii. but different number of neutrons.
b)
13586
c) 1237 3+
1248
1326 2−

Chromium p: 24, e:24, n: 28

fluorine p: 9, e: 9, n: 10

2.
a) average atomic mass Mg

(78.9 23.98504 ) + (10.0 24.98584 ) + (11.1 25.98259 )
= (78.9 + 10.0 + 11.1)

= 24.30 amu

Relative atomic mass Mg

= 24.30
112 12.00

= 24.30

Relative atomic mass Mg is 24.30

b)

3. 65Cu : 100-X
63 Cu : X

Average atomic mass Cu is 63.5

63.5 = 63( )+65(100− )
100

X = 75%

Percentage abundance 63Cu: X: 75%

Percentage abundance 65Cu: 100-X: 25%

1.2 Mole Concept

1. Given: 10.0 g CuSO4.5H2O

a) number of moles of CuSO4.5H2O.

n CuSO4.5H2O = 10.0 g = 0.0400 mol

249.7 g mol 1

b) number of moles of copper atoms.
1 mol CuSO4.5H2O ≡ 1 mol Cu atom
0.0400 mol CuSO4.5H2O ≡ 0.0400 mol Cu atom

c) mass of anhydrous copper (II) sulphate.
1 mol CuSO4.5H2O ≡ 1 mol CuSO4
0.0400 mol CuSO4.5H2O ≡ 0.0400 mol CuSO4
∴ mass of C atoms = 0.0400 mol x 159.7 g mol-1 = 6.39 g

2. a) Definition
Empirical formula:
formula that shows the simplest ratio of atoms of different elements
present in amolecule

Molecular formula:
formula that shows the actual number of atoms of different elements
present in amolecule
b)

Element C H N O
0.198 0.025 0.661
Mass (g) 0.0165 0.025 0.116 0.0413
8.286 X10-3
Moles 2 5

(mol) =1

ratio 31
C2H3NO5
Empirical

formula

n = 121
121

Molecular formula : C2H3NO5

3. mass sample: 0.255 g

mass CO2: 0.561 g moles of CO2 = 0.561
44 /

= 0.0128 mol

Mass H2O: 0.306 g moles of H2O = 0.306
18

= 0.017 mol

Mass O = 0.255g - (0.153+ 0.034)g
= 0.068 g

Element C H O
Mass (g) 0.153 0.034 0.068
Moles (mol) 0.0128 0.034 4.25 X 10-3

ratio 3 8 1
C3H8O
Empirical
formula

Empirical formula is C3H8O
4. a) molality is number of moles of solute per kilogram of solvent.

b)

density solution: 0.954 g/mL molarity: 2.5 M

0.954 g/mL = 2.5 M =

/

Assume V solution = 1000 mL @1L

Mass solution: 954 g

Moles of solute: 2.5 mol, mass solute: 80 g

Mass solvent: 954g- 80g

= 874 g
Molality = 2.5

(874/1000)

= 2.86 mol/kg

c)

%w/w: 37.8% density solution: 1.19 g/mL

37.8 % = 100% 1.19 g/mL =
/

Assume mass solution: 1000 g

Mass solute: 378 g, moles solute: 10.36 mol

V solution: 840 mL @ 0.84L

Molarity = 10.36
0.84

= 12.33 M

d) Mass solute CaCl2: 16 g mass solvent, H2O: 64.0 g

n solute = 16 n solvent = 64.0
1101.6 18

= 0.145 mol = 3.556 mol

Mol fraction = 0.145 100%
(0.145+3.556)

= 3.92%

% w/w = 16 100%
(16+64)

= 20%

1.3 Stoichiometry 5CO2 (g) + 3H2O (g)
1.

i. C5H6O (l) + 6O2 (g)

ii. 8I-(aq) + SO42-(aq) + 10H+ (aq) 4I2(s) + H2S (aq) + 4H2O(l)

iii. 4Cl2 (aq) + 8OH- (aq) ClO4- (aq) + 7Cl- (aq) + 4H2O (l)

2.

3. a) nCO = 50 n hydrogen= 0.8 mol available
28

= 1.786 mol

1 mol CO ≡ 2 mol H2

1.786 mol CO ≡ 2 X 1.786 mol

= 3.572 mol (needed)

H2 is a Limiting reactant

b) percentage yield = 100%
ℎ

n CH3OH yield: 0.4 mol

mass CH3OH: 0.4 mol X 32 gmol-1

= 12.8g

= 11.9 100%
12.8

= 92.97%

4. Given:
Mass of copper(II) oxide,
CuO = 7.0 gV nitric acid,
HNO3 solution = 50 mL

Concentration of nitric acid, HNO3 solution = 0.20 M

a. a reactant that is completely consumed during a reaction and
limits the amountof product(s) formed.

b. CuO(s) + 2HNO3(aq) Cu(NO3)2(aq) + H2O(l)

c. Given,
mol of CuO =
0.0881 mol
mol of HNO3
= 0.01 mol

1 mol CuO ≡ 2 mol HNO3
0.0881 mol CuO ≡ 0.1762 mol HNO3

∴ mol of HNO3 given < needed. HNO3 is limiting reactant.

d. expected mass of copper (II) nitrate formed.
1 mol HNO3 ≡ 1 mol Cu(NO3)2
0.01 mol HNO3 ≡ 5 x 10-3

mol Cu(NO3)2 Mass of

Cu(NO3)2 = 0.94 g

e. Given:
actual mass of copper(II) nitrate = 0.85 g.

% yield = actual mass x100%
theoritical mass

= 0.85 g x100%
0.94 g

= 90%