UKZN –SCHOOL OF CHEMISTRY-Howard College
CHEM 181 - Tutorial 5
QUANTITIES IN CHEMICAL REACTIONS
1.
Calculate the number of atoms in:
(a) 4.8 mol of copper
Soln:
(a)
Given:4.8 mol Cu
Find: Cu atoms
Conversion: 1 mol Cu = 6.022 x 1023 Cu atoms
4.8 mol Cu x
6.022 x 1023 Cu atoms
1 mol Cu
= 2.9 x 1024 Cu atoms
(b) 48.3 g sample of Zn
(a)
Given:
48.3 g Zn
Find:
Zn atoms
Conversion: 1 mol Zn = 6.022 x 1023 Cu atoms
No of mols Zn = mass Zn/molar mass Zn
= 48.3 g Zn/65.39 g mol-1 = 0.7386 mols
0.7386 mol Zn
x
6.022 x 1023 Zn atoms
1 mol Zn
= 4.45 x 1023 Zn atoms
2.
Calculate the number of mols of:
(a) sulfur in 57.7 g of sulfur (b) C in 0.05005 g of Ca(HCO3)2.
Soln: (a)
no of mols of S =
mass of S
molar mass of S
=
57.7 g
32.07 g
= 1.80 mols
(b) C in 0.05005 g of Ca(HCO3)2.
(b)
no of mols of C =
mass of Ca(HCO3)2
molar mass of Ca(HCO3)2
Since there are 2 mols of C in Ca(HCO3)2
=
0.05005 g
162.12 g
= 3.0872 x 10-4 mols
= 3.0872 x 10-4 mols x 2
= 6.1744 x 10-4 mols C
3.
Calculate the mass of:
(a)
1.23 x 1024 He atoms
Soln: No of mols He = No of atoms/Avogadros No
= 1.23 x 1024 He atoms/6.022 x 1023 atoms x 1 mol
= 2.04 mols
Mass = 2.04 mols x 4.003 g/mol = 8.176 g
(b)
4.78 x 1024 NO2 molecules
Soln: No of mols NO2
=
No of atoms/Avogadros No
= 4.78 x 1024 NO2 molecules /6.022 x 1023 molecules x 1 mol
= 7.94 mols
Mass = 7.94 mols x 46.01 g/mol = 365 g
4.
Calculate the mass of Al (in grams) of 6.73 mols of Al.
Mass = no of mols x molar mass
= 6.73 mols of Al x 26.98 g mol-1 = 182 g
5.
Determine the number of moles of oxygen in 7.20 mols of H2SO4.
Soln: No of mols of O = 7.20 mols x 4 (since there are 4 oxygen atoms)
= 28.8 mols of O
6.
Find the grams Fe in 79.2 g of Fe2O3.
Soln: Percentage composition of Fe =
=
2 x molar mass Fe
molar mass of Fe2O3
2 x 55.85 g/mol
159.7 g/mol
x 100
x 100 = 69.94 %
Therefore mass of Fe = 79.2 g x 0.6994 = 55.39 g
7.
If a person consumed 22 g of sodium chloride, how much of sodium
does that person consume?
NaCl is 39 % sodium by mass.
Soln: 22 g NaCl x 0.39 = 8.58 g of Na was consumed.
8.
Calculate the mass percent of:
(a)
Cl in C2Cl4F2
Soln: % Cl =
(b)
molar mass of C2Cl4F2
x 100 =
4 x 35.45 g/mol
203.82 g/mol
x 100 = 69.57 %
O in CH3COOH
Soln: % O =
9.
4 x molar mass of Cl
2 x molar mass of O
Molar mass of CH3COOH
x 100 =
2 x 16.00 g/mol
60.052 g/mol
x 100 = 53.29 %
Determine the empirical formula for the following compounds:
(a)
1.245 g Ni and 5.381 g I
Soln:
Ni
I
Mass:
1.245 g
5.381 g
molar mass:
58.69 g mol-1
126.90 g mol-1
no of mols:
0.0212
0.0424
1
2
Divide by 0.0212
Empirical formula = NiI2
(b)
2.128 g Be, 7.557 g S, 15.107 g O
Soln:
Be
S
O
Mass:
2.128 g
7.557 g
15.107 g
molar mass:
9.01 g mol-1
32.07 g mol-1
16.00 g mol-1
no of mols:
0.2362
0.2356
0.9442
Divide by 0.2356
1.003
1
4.008
Rounding off to nearest whole no. Empirical formula = BeSO4
10.
Find the molecular formula for the following compounds:
(a)
284.77 g/mol, CCl
Soln: Empricial formula mass of CCl = 12.01 g mol-1 + 35.45 g mol-1
= 47.46 g mol-1
n=
molecular mass of compound
Empirical formula mass of CCl
=
284.77 g/mol
47.46 g/mol
=6
Therefore molecular formula = CCl x 6 = C6Cl6
(b)
131.39 g/mol, C2HCl3
Soln: Empricial formula mass of C2HCl3 = 2(12.01) + 1.008 + 3(35.45)] g mol-1
= 131.38 g mol-1
n=
molecular mass of compound
Empirical formula mass of CCl
=
131.39 g/mol
131.38 g/mol
=1
Therefore molecular formula is the same as it empirical formula
= C2HCl3