Tutorial 1 Answer

1. Express the answers to the following calculations in scientific notation:

a) 0.0095 + (8.5 x 10-3)

9.5 x 10-3 + 8.5 x 10-3 = 18.0 x 10-3 = 1.8 x 10-2

b) 653 ÷ (5.75 x 10-8)

653 = 6.53 x 102

(6.53 ÷ 5.75) x 102- (-8) = 1.14 x 1010

c) 850000 – (9.0 x 105)

850000 = 8.5 x 105

8.5 x 105 - 9.0 x 105 = -0.5 x 105 = -5 x 104

d) (3.6 x 10-4) x (3.6 x 106)

(3.6 x 3.6) x 10-4 + (+6) = 13 x 102 = 1.3 x 103

2. Carry out the following conversions:

a)185 nm to m

185 nm x 1 x 10-9 m = 1.85 x 10-7 m 1 nm

b) 71.2 cm3 to m3

71.2 cm3 x 0.01 m 3 = 7.12 x 10-5 m3

1 cm

c) 88.9 m3 to liters

88.9 m3 x 1 cm 3 x 1 L = 8.89 x 104 L 1 x 10-2 m 1000 cm3

d) 7.3 x 102 K to 0C and oF

(7.3 x 102 - 273) = 4.6 x 102 0C (4.6 x 102) x 9 + 32 oF = 8.6 x 102 oF 5 d) 0.625 g/L to g/cm3

0.625 g x 1 L x 1 mL = 6.25 x 10-4 g/cm3

1 L 1000mL 1 cm3

Symbol

Proton 8 26 53 17 12 6

Electron 8 26 54 17 10 6

Neutron 8 30 74 18 12 7

Charge 0 0 -1 0 +2 0

Atomic Number 8 26 53 17 12 6

Mass Number 16 56 127 35 24 13

3. Complete the following table:

Atomic number (Z) = number of protons in nucleus

Mass number (A) = number of protons + number of neutrons

= atomic number (Z) + number of neutrons

4. Urea (NH2)2CO is used for fertilizer and many other things. Calculate the number of molecules and atoms in 1.68 x 104 g of urea. Also, calculate the number of N, C, O and H atoms.

# of molecules = mole of molecule x NA

# of mole (NH2)2CO = 1.68 x 104 g = 280 mole [2(14) + 4 (1.0) + 12.0 + 16.0]g/mole

# of molecules = 280 mole (NH2)2CO x 6.022 x 1023 molecule (NH2)2CO mole (NH2)2CO = 1.69 x 1026 molecule

# of atoms = 280 mole (NH2)2CO x 6.022 x 1023 molecule (NH2)2CO x 8 atoms mole (NH2)2CO molecule (NH2)2CO = 1.35 x 1027 atoms

# of N atoms = 280 mole (NH2)2CO x 6.022 x 1023 molecule (NH2)2CO x 2 N atoms mole (NH2)2CO molecule (NH2)2CO

= 3.37 x 1026 N atoms

# of C atoms = 280 mole (NH2)2CO x 6.022 x 1023 molecule (NH2)2CO x 1 C atoms mole (NH2)2CO molecule (NH2)2CO

= 1.69 x 1026 N atoms

# of O atoms = 280 mole (NH2)2CO x 6.022 x 1023 molecule (NH2)2CO x 1 O atoms mole (NH2)2CO molecule (NH2)2CO

= 1.69 x 1026 N atoms

# of H atoms = 280 mole (NH2)2CO x 6.022 x 1023 molecule (NH2)2CO x 4 H atoms mole (NH2)2CO molecule (NH2)2CO

= 6.74 x 1026 H atoms

5. In 18.05 g of magnesium phosphate Mg3(PO4)2, calculate:

a) # of Mg2+ ions = 18.05 g x 6.022 x 1023 Mg3(PO4)2 x 3 Mg2+ ions [3(24.3) + 2(31.0) + 8(16.0)]g/mol 1 mole Mg3(PO4)2 Mg3(PO4)2

= 1.24 x 1023 Mg2+ ions

b) # of PO43- ions = 0.069 mol x 6.022 x 1023 Mg3(PO4)2 x 2 PO4

3- ions 1 mole Mg3(PO4)2 1 Mg3(PO4)2

= 8.27 x 1022 PO43- ions

c) The total # of ions = 0.069 mol x 6.022 x 1023 Mg3(PO4)2 x 5 ions 1 mole Mg3(PO4)2 1 Mg3(PO4)2

= 2.08 x 1023 ions

6. The percentage composition by mass of a compound is 40.01% C, 6.67% H and 53.32% O. 0.25 mole of this compound weights 44.5 g. Determine the empirical and the molecular formula of this compound.

Element C H O

Mass (g) 40.01 g 6.67 g 53.32 g

# of moles 40.01 g = 3.33112.01 g/mol

6.67 g = 6.6041.01 g/mol

53.32 g = 3.33316 g/mol

÷ smallest # 3.331 = 13.331

6.604 = 1.9833.331

3.333 = 13.331

1 2 1

Empirical formula = CH2O

Molar mass of a compound = 44.5 g = 178 g/mol 0.25 mol

n = molar mass = 178 = 5.9 = 6 empirical formula (12.01) + 2(1.01) + (16)

Molecular formula = (CH2O)6 = C6H12O6

7. An organic compound that contains only carbon, hydrogen and nitrogen has 73.59% C, 7.966% H and 8.584% N by mass. What is the empirical formula of this compound if the mass of 5.03 x 1022 molecules of this compound is 27.26 g, determine its molecular formula.

Element C H N

Mass (g) 73.59g 7.966 g 8.584 g

# of moles 73.59 g = 6.12712.01 g/mol

7.966 g = 7.8871.01 g/mol

8.584 g = 0.61314.01 g/mol

÷ smallest # 6.127 = 100.613

7.887 = 12.90.613

0.613 = 10.613

10 13 1

Empirical formula = C10H13N# of molecules = moles x NA

5.03 x 1022 = moles x 6.022 x 1023

Moles = 8.35 x 10-2

Molar mass of a compound = 27.26 g = 326.5 g/mol 8.35 x 10-2 mol

n = molar mass = 326.5 = 2.2 = 2 empirical formula 10(12.01) + 13(1.01) + (14.01)

Molecular formula = (C10H13N)2= C20H26N2

8. A compound contains 60.0% C, 4.48% H and 35.5% O. If the molecular weight of this compound is 180.2, what is its molecular formula?

Element C H O

Mass (g) 60.0 g 4.48 g 35.5 g

# of moles 60.0 g = 4.99612.01 g/mol

4.48 g = 4.4361.01 g/mol

35.5 g = 2.21916 g/mol

÷ smallest # 4.996 = 2.252.219

4.436 = 22.219

2.219 = 12.219

2.25 x 4 = 9 2 x 4 = 8 1 x 4 =4

Empirical formula = C9H8O4

n = molar mass = 180.2 = 1 empirical formula 9(12.01) + 8(1.01) + 4(16)

Molecular formula = (C9H8O4)1 = C9H8O4

9. A sample of phosphorus oxide has a mass of 6.28g. The empirical formula for the phosphorus is P2O5, and its molar mass is 284 g/mol. Answer the following questions on the phosphorus oxide. Relative atomic mass: P = 31, O = 16.

a) Determine the molecular formula for the phosphorus oxide.

b) Calculate the percentage by mass of the element phosphorus in the molecule.

mol P4O10 = 6.28 g = 0.0221 mol [4(31) + 10(16)]g/mol

In 0.00221 mol P4O10 there have 0.0221 x 4 mol P = 0.0884 mol P

mass of P = 0.0884 mol x 31 g/mol = 2.7404 g

% by mass of P = 2.7404 g x 100 = 43.6% 6.28 g

c) What is the mass of oxygen in the sample?

In 0.0221 mol P4O10, there have 0.0221 x 10 mol O = 0.221 mol O

mass of O = 0.221 mol x 16 g/mol = 3.54 g

n = molar mass = 284 = 2 empirical formula 2(31) + 5(16)

Molecular formula = (P2O5)2 = P4O10

10. 74.8 g MnO2 reacts with the 48.2 g HCl to produce MnCl2, Cl2 and H2O according to this chemical equation: MnO2 + 4HCl MnCl2 + Cl2 + 2H2O a) Determine the limiting reagent

mol MnO2 = 74.8 g = 0.86 mol [55 + 2(16)]g/mol

mol HCl = 48.2 g = 1.32 mol [1 + 35.5]g/mol

From the balanced equation

4 mol HCl 1 mol MnO2

1.32 mol HCl 1.32 mol HCl x 1 mol MnO2 = 0.33 mol MnO2

4 mol HCl

MnO2 is the excess reagent since 0.86 mol present, more than needed. Therefore HCl is the limiting reagent.

b) How much is the mass of MnCl2 formed? From the balanced equation

4 mol HCl 1 mol MnCl2

1.32 mol HCl 1.32 mol HCl x 1 mol MnCl2 = 0.33 mol MnCl2

4 mol HCl

Mass of MnCl2 = 0.33 mol x [55 + 2(35.5)]g/mol = 41.58 g

b) How much of the excess reagent (in gram) is left at the end of the reaction?

mol of excess reagent = 0.86 mol – 0.33 = 0.53 mol

mass of the excess reagent (MnO2) = 0.53 mol x 87 g/mol = 46.11 g

11. a) Calculate the mass of K2Cr2O7 required to prepare exactly 200.0 cm3 of a 0.750 M K2Cr2O7 solution in water.

mol K2Cr2O7 = (0.750 mol/L) x (200.0 x 10-3 L) = 0.15 mol

mass of K2Cr2O7 = 0.15 mol x [2(39.1) + 2(52) + 7(16)] g/mol = 44.13 g

b) 1.325 g dried sodium carbonate, Na2CO3 is dissolved in water until it reached volume of 250.0 mL. 25.0 mL of this solution is titrated with H2SO4. Volume of acid needed is 23.4 mL. Calculate the concentration (molarity) of sulphuric acid.

H2SO4 + Na2CO3 Na2SO4 + CO2 + H2O

mol Na2CO3 = 1.325 g = 0.0125 mol Na2CO3

[2(23) + 12 + 3(16)]g/mol

Molarity Na2CO3 = 0.0125 mol ÷ 250 x 10-3 L = 0.05 M

In titration

MaVa = MbVb

Ma(23.4 mL) = (0.05 M)(25.0)

Ma = 0.053 M

12. Balance the redox equation below:

a) Fe2+ + Cr2O72- Fe3+ + Cr3+ (acidic medium)

Fe2+ Fe3+

Fe2+ Fe3+ + e (oxidation)

x6 6Fe2+ 6 Fe3+ + 6e (1)

Cr2O72- Cr3+

Cr2O72- + 14 H+ + 6e 2Cr3+ + 7H2O (reduction) (2)

Combine (1) and (2)

6Fe2+ + Cr2O72- + 14 H+ + 6e 6 Fe3+ + 6e + 2Cr3+ + 7H2O

6Fe2+ + Cr2O72- + 14 H+ 6 Fe3+ + 2Cr3+ + 7H2O

b) I- + MnO4- I2 + MnO2 (basic medium)

2I- I2 + 2e

x 3 6I- 3I2 + 6e (1)

MnO4- + 3e + 4H+ MnO2 + 2 H2O

x 2 2MnO4- + 6e + 8H+ 2MnO2 + 4H2O (2)

In basic medium (+ 8OH-) in both side of (2)

2MnO4- + 6e + 8H2O 2MnO2 + 4H2O + 8OH-

2MnO4

- + 6e + 4H2O 2MnO2 + 8OH- (3)

Combine (1) and (3)

6I- + 2MnO4- + 6e + 4H2O 3I2 + 6e + 2MnO2 + 8OH-

6I- + 2MnO4- + 4H2O 3I2 + 2MnO2 + 8OH-