Free Response Question


Free Response Question #1 a)

The empirical formula is CH₂O and the molecular formula is C₆H₁₂O₆. Assume we have 100 g of the compound. Then we have 40.00 g of C, 6.72 g of H, and 53.28 g of O. Moles of C = 40.00 g C ×

1mol C12.01g C

= 3.331 mol C Moles of H = 6.72 g H ×

1 mol H1.008g H

= 6.67 mol H Moles of O = 53.28 g O ×

1 mol O16.00g O

= 3.330 mol H Moles of C:Moles of H:Moles of O = 3.331:6.67:3.330=1.000:2.00:1 ≈ 1:2:1 The empirical formula is CH₂O. The empirical formula is the simplest formula of a compound. The actual formula is an integral multiple of the empirical formula. If the empirical formula is CH₂O, the actual formula is



n or






n , where


= 1, 2, 3, … Our job is to determine the value of


. The empirical formula mass of CH₂O is 30.03 u. The molecular mass of 180.18 u must be some multiple of this number.

n=180.18 u30.03 u

= 6.000 ≈ 6 ∴ The molecular formula =






n = C₆H₁₂O₆.


molar mass = 180.15 C = 40% of 180.15 = 72 g or 6 moles of carbon H = 6.7 % of 180.15 = 12.07 g = 12 moles O = 53.3% of 180.15 = 96 g or 6 moles of O your molecular formula is C6H12O6 1

Free Response Question #2


Au + HNO3 + 4 HCl → HAuCl4 + NO + 2 H2O (2 medals) x (175 g/medal) x (23/24) / (196.966569 g Au/mol) x (1 mol HAuCl4 / 1 mol Au) x (339.7865 g HAuCl4/mol) = 579 g HAuCl4


(2 medals) x (175 g/medal) x (23/24) / (196.966569 g Au/mol) x (1 mol HNO3 / 1 mol Au) = 1.7029 mol HNO3 reacted (2 medals) x (175 g/medal) x (23/24) / (196.966569 g Au/mol) x (4 mol HCl / 1 mol Au) = 6.8116 mol HCl reacted (0.200 L) x (16 mol/L HNO3) - 1.7029 mol HNO3 = 1.4971 mol excess HNO3 (0.800 L) x (12 mol/L HCl) - 6.8116 mol HCl = 2.7884 mol excess HCl Both HCl and HNO3 are monoprotic. H{+} + NaHCO3 → Na{+} + H2O + CO2 (1.4971 mol + 2.7884 mol) x (1 mol NaHCO3 / 1 mol H{+}) x (84.0066 g NaHCO3/mol) = 360 g NaHCO3 2

Free Response Question #4 – Part 1

Alkali Metals: As you go down group, there are extra electron shells (for example Lithium has 2 shells, Francium has 7) which cause electron shielding. Because the force of attraction must extend further and through these shells to the outermost shell, the force becomes less so there is less strong bonding occurring. Thus, it is easier to break these less strong bonds, and because the melting point is a measure of the point at which all the bonds holding a metallic structure together are broken, it will decrease because this point is reached a lot sooner as the bonds are not as strong. 1.

the melting point of a substance is determined by the force of attraction between the nuclei of the atoms in a substance. The stronger of force of attraction between the nuclei of atoms in a substance, the higher the melting point of the substance. 2.


Both of these metals are in the solid state so the atoms in both metals are packed tightly. Then according to the size of atoms for sodium is much larger than for lithium. 4.

So, the force of attraction between nuclei of two atoms for sodium is weaker than for lithium. 5.

Thus, the melting point of sodium is lower than that of lithium. Halogens: The melting and boiling points increase down the group because of the van der Waals forces. The size of the molecules increases down the group. This increase in size means an increase in the strength of the van der Waals forces. F < Cl < Br < I < At All the halogens exist as diatomic molecules temporary dipoles that create these forces. —F2, Cl2, and so on. van der Waals dispersion forces are the primary intermolecular attractions between one molecule and its neighbors. Larger molecules farther down the group have more electrons which can move around and form the The stronger intermolecular attractions down the group require more heat energy for melting or vaporizing, increasing their melting or boiling points.

Free Response Question #4 – Part 2



Free Response Question #5

a) N 2 H 4 +3 O 2 b) 2 NO 2 +2H 2 O 5


Free Response Question #6


Free Response Question #7



Free Response Question #12

a) solution 2. HCl dissolves completely in solution. b) solution 3. Since it has 3 hydogens as present in NH3. c) solution 5. One is hydrogen ion and another is HSO4- ion. d) solution 6. It has the same number of hydrogen ions attached to larger sized sphere. e) solution 6 has the lowest pH. Since, it has no free hydrogen ion in solution. Thank you 10