Mole1

For on-line help with this topic, see the chemcal module "Stoichiometry" which deals with moles; balancing equations; stoichiometric calculations; molarity and solution stoichiometry. MOLE CONCEPT (General Questions 1) Your feedback on these self-help problems is appreciated. Click here to send an e-mail.		Shortcut to Questions Q: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
1	Which of the following has the greatest mass: (1.) copper (100 g), (2.) helium (6.00 mole), (3.) 12.0 x 10²³ atoms of silver?
2	Hydrogen gas (4.00 g) and chlorine gas (10.0 g) are allowed to react. (1.) Which reactant gas is in excess? (2.) What mass of hydrogen chloride is formed?
3	(1.) Name the compound of formula Na₂SO₄.10H₂O. (2.) What is its molar weight? (3.) What is the percentage by weight of sulfur in this compound? (4.) How many sodium ions are in one mole of this compound?
4	Calculate the volume of the following at 298 K and 101 kPa. (1.) Water (2.00 mol) (2.) Chlorine (1.00 mol) (3.) Mercury (3.00 mol) (4.) Neon (1.00 mol) (5.) Nitrogen (1.20 x 10²⁴ molecules)
5	Calculate the mass of the following at 298 K and 760 mmHg pressure. (1.) Helium (49.0 litre) (2.) Water (3.00 mol) (3.) Ethanol (100 cm³) (4.) Nitrogen (25.0 litre)
6	The price of chromium is 1.00 dollar gram^-1. What is the price of one mole of chromium?
7	Calculate (1.) the molar weight, and (2.) the molar volume of liquid carbon tetrachloride at 298 K given the density of carbon tetrachloride at 298 K = 1.584 g cm^-3.
8	What volume is occupied at 298 K and 101 kPa by 1.00 mole of each of the following: (1.) hydrogen(g) (2.) helium(g) (3.) aluminium(s) (4.) water(l)
9	What amount (in mole) of carbon is contained in 112 g of ethylene (C₂H₄)?
10	Which of the following has the greatest mass? (1.) lead (3.00 g) (2.) helium (22.4 litre) at 273 K and 101 kPa? (3.) 3.01 x 10²³ atoms of lithium.
11	Which of the following has the greatest mass? (1.) lead (63.0 g) (2.) fluorine (24.5 litre) at 298 K and 2.00 atmosphere. (3.) 3.01 x 10²³ atoms of caesium.
12	In the chemical reaction requiring two atoms of aluminium for every three atoms of oxygen, how many mole of oxygen atoms is required by 2.7 g of aluminium? What mass of oxygen atoms are required?
13	Hydrogen gas reacts with chlorine gas to form hydrogen chloride. If hydrogen (2.00 mol) reacts with excess chlorine, what mass of hydrogen chloride can be obtained?
14	Nickel(II) sulfate-6-water (2.63 g) is converted to nickel(II) chloride-6-water. Calculate the theoretical yield in grams.
15	(1.) What mass of sodium hydroxide reacting with excess carbon dioxide is required to prepare 53.0 g of sodium carbonate? (2.) What volume of carbon dioxide is released at 273 K and 1.00 atmosphere when this weight of sodium carbonate is treated with excess hydrochloric acid?
16	Calculate the mass of anhydrous sodium carbonate required to make 250 cm³ of 0.100 M solution.
17	A solution was prepared by dissolving nickel(II) nitrate-6-water (29.1 g) in some water and making the volume up to 1.000 litre with water. Assuming complete dissociation of the solid into ions, calculate the following for 100 cm³ of solution. (1.) The number of mole of nickel(II) ion. (2.) The number of mole of nitrate ion. (3.) The number of individual nickel(II) ions. (4.) The number of cm³ which theoretically contain one individual nickel (II) ion.

MOLE CONCEPT (General answers 1)
1	The silver has the largest mass: (1.) Mass of copper = 100 g (given). (2.) Mass of helium = atomic mass He x no. moles He = 4.003 x 6.00 = 24.0 g. (3.) Moles silver = no. atoms Ag / N_A = 12.0 x 10²³ / 6.022 x 10²³ = 1.99 mol. Mass of silver = moles Ag x atomic weight Ag = 1.99 mol x 107.9 g mol^-1 = 215 g
2	(1.) MolesH₂ = mass H₂ / molar mass H₂ = 4.00 g / (2 x 1.008 g mol^-1) = 2.00 mol. Moles Cl₂ = mass Cl₂ / molar mass Cl₂ = 10.0 g / (2 x 35.45 g mol^-1) = 0.141 mol. From the balanced equation H₂ + Cl₂ → 2HCl, one mole of hydrogen will react with one mole of chlorine, so hydrogen is in excess. (2.) There are two moles HCl formed for each mole of Cl₂, hence 2 x 0.141 = 0.282 moles HCl is produced. Mass HCl = no. mol HCl x molar mass HCl = 0.282 x (1.008 + 35.45) = 10.3 g.
3	(1.) The name of Na₂SO₄.10H₂O is sodium sulfate-10-water. (2.) The molar weight of sodium sulfate-10-water is 322.21 g mol^-1. (3.) The percentage of sulfur = (atomic mass S / molar mass Na₂SO₄.10H₂O) x 100% = (32.07 g mol^-1 / 322.21 g mol^-1) x 100% = 9.95 %. (4.) The formula of Na₂SO₄.10H₂O contains two sodium ions, so one mole of Na₂SO₄.10H₂O will contain two moles of sodium ions, ie 2 x 6.022 x 10²³ sodium ions = 1.20 x 10²⁴ sodium ions.
4	The volume of the gases is found from the ideal gas approximation of 24.5 L mol^-1 (at 298K and 101 kPa), irrespective of what gas it is. There is no simple approximation for finding the density of solids and liquids, and known densities must be used for calculations. (1.) Mass of water = moles H₂O x molar mass H₂O = 2.00 mol x 18.02 g mol^-1 = 36.0 g. Volume of water = mass H₂O / density H₂O = 36.0 g / 1.00 g cm^-3 = 36.0 cm³ (Note: this is equivalent to 36.0 mL as 1cm³ is the same volume as 1 mL) (2.) Volume of chlorine = moles Cl₂ / molar volume = 1.00 mol x 24.5 L mol^-1 = 24.5 L. (3.) Mass of mercury = moles Hg x molar mass Hg = 3.00 mol x 200.6 g mol^-1 = 602 g. Volume of mercury = mass Hg / density Hg = 602 g / 13.6 g cm^-3 = 44.3 cm³. (4.) Volume of neon = moles Ne / molar volume = 1.00 mol x 24.5 L mol^-1 = 24.5 L. (5.) Moles N₂ = no. molecules N₂ / N_A = 1.20 x 10²⁴ / 6.023 x 10²³ = 1.99 mol. The volume of N₂ = moles N₂ / molar volume = 1.99 mol x 24.5 L mol^-1 = 48.8 L.
5	The mass of gases is found using the same relationships used to obtain volume in question 4. (1.) Moles He (at 298 K, 760 mmHg) = volume He / molar volume = 49.0 L / 24.5 L mol^-1 = 2.00 mol. Mass He = moles He x molar mass He = 2.00 x 4.003 = 8.00 g. (2.) Mass of H₂O = moles H₂O x molar mass H₂O = 3.00 x 18.02 = 54.1 g. (3.) Mass of CH₃CH₂OH = volume CH₃CH₂OH x density CH₃CH₂OH = 100 cm³ x 0.785 g cm^-3 = 78.5 g. (4.) Moles N₂ (at 298 K, 760 mmHg) = volume N₂ / molar volume = 25.0 L / 24.5 L mol^-1 = 1.02 mol. The mass of N₂ = moles N₂ x molar mass N₂ = 1.02 x (2 x 14.01) = 28.6 g.
6	Mass of Cr = moles Cr x molar mass Cr = 1 mol x 52.00 g mol^-1 = 52 g. The price of Cr = mass Cr x unit price = 52 g x $1.00 g^-1 = $52.
7	(1) The molecular formula of carbon tetrachloride is CCl₄. Molar mass CCl₄= molar mass C + (4 x molar mass Cl) = 12.01 + (35.45 x 4) = 153.81. (2) Molar volume = volume of one mole = molar mass / density = 153.81 /1.584 = 97.10 g cm^-3.
8	From the ideal gas approximation the volume of the gases can be obtained: (1.) Volume H₂ (at 298 K, 101 kPa) = moles H₂ x molar volume = 1.00 mol x 24.5 L mol^-1 = 24.5 L. (2.) Volume He (at 298 K, 101 kPa) = moles He x molar volume = 1.00 mol x 24.5 L mol^-1 = 24.5 L. For solids and liquids, volume calculations involve the particular density of the substance: (3.) Mass Al = moles Al x molar mass Al = 1.00 mol x 26.98 g mol^-1 = 26.98 g. Volume Al = mass Al / density Al = 26.98 g / 2.7 g cm^-3 = 10 cm³. (4.) Mass H₂O = moles H₂O x molar mass H₂O = 1.00 mol x 18.02 g mol^-1 = 18.0 g. Volume H₂O = mass H₂O / density H₂O = 18.0 g / 1.00 g cm^-3 = 18.0 cm³.
9	Moles C₂H₄ = mass C₂H₄ / molar mass C₂H₄ = 112 g / 28.05 g mol^-1 = 3.99 mol. Each molecule of C₂H₄ contains 2 C atoms, therefore 3.99 moles of C₂H₄ will contain 2 x 3.99 moles of C = 7.98 moles C.
10	The helium has the largest mass: Mass of lead = 3.00 g (given). Using the ideal gas approximation, 1 mole of gas occupies 22.4L at 273K, 101 kPa. Moles helium = volume He / molar volume = 22.4 L / 22.4 L mol^-1 = 1.00 mol. The mass of He = moles He x molar mass He = 1.00 x 4.003 = 4.00 g. Moles lithium = no. atoms Li / N_A = 3.01 x 10²³ / 6.022 x 10²³ = 0.500 mol. The mass of Li = moles Li x molar mass Li = 0.500 x 6.941 = 3.47 g.
11	The fluorine has the largest mass: (1.) Mass of lead = 63.0 g (given). (2.) Moles fluorine, F₂ , in 24.5 L at 298 K and *1.00 atm* is found from volume F₂ / molar volume = 24.5 L / 24.5 L mol^-1 = 1.00 mol. At a pressure of *2.00 atm, there will be twice as many molecules (assuming the volume and temperature are the same), ie 2.00 mol F₂. Mass F₂ = moles F₂ x molar mass F₂ = 2.00 mol x (2 x 19.00 g mol^-1) = 76 g. (An equation that can be used to calculate this directly is covered later in the course.) (3.) Moles caesium, Cs, = no. atoms Cs / N_A* = 3.01 x 10²³ / 6.022 x 10²³ = 0.500 mol. Mass Cs = moles Cs x molar mass Cs = 0.500 x 132.9 = 66.5 g.
12	Moles Al = mass Al / molar mass Al = 2.7 / 26.98 = 0.10 mol. The stoichiometric ratio of Al to O is 1 to 1.5, therefore for each mole of Al, 1.5 moles O is required. No. mol O required = no. mol Al (0.10) x 1.5 = 0.15 mol O. Mass of O atoms is 0.15 mol x 16.0 g mol^-1 = 2.4 g. Note that the question specifies oxygen atoms, not molecules, so the calculations are based on monatomic O, rather than the gaseous diatomic O₂.
13	Note that the question refers to the reaction of gaseous H₂ and Cl₂, so the diatomic nature of these gases must be taken into account in subsequent calculations. Balanced equation: H₂ + Cl₂ → 2HCl Therefore, in the presence of an excess of Cl₂, 2.00 mol H₂ will react completely to give 4.00 mol HCl. Mass HCl = moles HCl x molar mass HCl = 4.00 mol x 36.46 g mol^-1 = 146 g.
14	Empirical formula of nickel(II) sulfate-6-water: NiSO₄.6H₂O (molar mass 262.86) Empirical formula of nickel(II) chloride-6-water: NiCl₂.6H₂O (molar mass 237.69) Since both formulae contain one nickel ion, each mole of NiSO₄.6H₂O will produce one mole of NiCl₂.6H₂O. Moles NiSO₄.6H₂O = mass NiSO₄.6H₂O / molar mass NiSO₄.6H₂O = 2.63 g / 262.86 g mol^-1 = 0.0100 mol, and so 0.0100 mol NiCl₂.6H₂O will be produced (theoretically, ie ignoring losses due to experimental technique). Mass NiCl₂.6H₂O = moles NiCl₂.6H₂O x molar mass NiCl₂.6H₂O = 0.0100 mol x 237.69 g mol^-1 = 2.38 g.
15	(1.) Balanced equation: 2NaOH + CO₂ → Na₂CO₃ + H₂O From the equation, each mole of Na₂CO₃ produced requires two moles of NaOH. Moles Na₂CO₃ = mass Na₂CO₃ / molar mass Na₂CO₃ = 53.0 g / 105.98 g mol^-1 = 0.500 mol. Therefore moles NaOH required = 2 x 0.500 mol = 1.00 mol. Mass NaOH = moles NaOH x molar mass NaOH = 1.00 mol x 40.00 g mol^-1 = 40.0 g. (2.) Balanced equation: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ Moles Na₂CO₃ = mass Na₂CO₃ / molar mass Na₂CO₃ = 53.0 g / 105.98 g mol^-1 = 0.500 mol. With excess hydrochloric acid present, all the sodium carbonate will react, each mole of Na₂CO₃ producing one mole of CO₂, therefore 0.500 mol CO₂ is produced. At 273 K and 1.00 atm, 1 mol of gas occupies 22.4 L (ideal gas approximation), volume CO₂ = moles CO₂ / molar volume = 0.500 mol x 22.4 L mol^-1 = 11.2 L.
16	0.100 M implies 0.100 mol of substance per L of solution. Therefore 250 mL of solution will require 0.0250 mol Na₂CO₃. Mass Na₂CO₃ = moles Na₂CO₃ x molar mass Na₂CO₃ = 0.0250 mol x 105.98 g mol^-1 = 2.65 g.
17	Empirical formula of nickel(II) nitrate-6-water: Ni(NO₃)₂.6H₂O. Ni(NO₃)₂.6H₂O will dissolve in water as shown below: Ni(NO₃)₂.6H₂O(s) → Ni²⁺(aq) + 2NO₃^-(aq) + 6H₂O (Note that the water of crystallisation is now part of the water solution). Therefore each mole of Ni(NO₃)₂.6H₂O dissolved will produce one mole of Ni²⁺, and two moles of NO₃^-. Moles Ni(NO₃)₂.6H₂O = mass Ni(NO₃)₂.6H₂O / molar mass Ni(NO₃)₂.6H₂O = 29.1 g / 290.8 g mol^-1 = 0.100 mol. This amount of Ni(NO₃)₂.6H₂O is dissolved in 1.00 L of water, so in 100 mL there will be 0.100 x 0.100 = 0.0100 moles of Ni(NO₃)₂.6H₂O dissolved. (1) Moles Ni²⁺in 100 mL = moles Ni(NO₃)₂.6H₂O dissolved as the formula contains only one Ni²⁺ ion = 0.0100 mol, (2) Moles NO₃^-in 100 mL = 2 x moles Ni(NO₃)₂.6H₂O dissolved = 0.0100 x 2 = 0.0200 mol. (3) No. of Ni²⁺ions = moles of Ni²⁺ions x N_A = 0.0100 x 6.022 x 10²³ = 6.02 x 10²¹ ions. (4) If there are 6.02 x 10²¹ ions present in 100 mL, then on average, 100 / 6.02 x 10²¹ mL would contain a single Ni²⁺ion.