Mole 2

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 2) 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	Iron burns in air to form Fe₃O₄. (1.) Write the equation for the reaction. (2.) How many moles of oxygen gas are needed to burn Fe (1.0 mol)? (3.) What mass of oxygen gas is this? (4.) Can a piece of iron (5.6 g) burn completely to Fe₃O₄ in a vessel containing oxygen (0.050 mol)?
2	Find the empirical formula of the compound which gives the following analysis by weight - nitrogen = 26.2%; chlorine = 66.4%; hydrogen = 7.50%.
3	A compound of molar weight 62 contains C, H and O only. Analysis gives 38.7% carbon and 9.8% hydrogen by weight. Determine the empirical and molecular formulas of the compound.
4	A compound is found to contain the following weight percentages of each element: carbon = 52.5%, hydrogen = 3.7%, boron = 7.9%, nitrogen = 10.2%, chlorine = 25.6%. What is the empirical formula of the compound?
5	What is the empirical formula of a compound containing 40% sulfur and 60% oxygen by weight? What is its molecular formula if its molecular weight is 240?
6	Derive the empirical formulas of substances having the following percentage compositions by weight: (1.) Iron 63.5%; sulfur 36.5% (2.) Iron 46.5%; sulfur 53.4% (3.) Iron 53.7%; sulfur 46.3%
7	A compound is shown to be ionic as it is soluble in water, providing a solution that conducts electricity. The compound has the following percentage composition by weight: sodium = 32.4%; sulfur = 22.6%; oxygen = 45.0%. (1.)Derive the empirical formula for this compound. (2.) Why would this compound not have a molecular formula?
8	What amount (mol) of solute is there in 125 mL of a 0.864 M solution?
9	What mass of sodium chloride must be dissolved in water to give 1.50 x 10³ mL of 0.100 M sodium chloride solution?
10	Sodium hydroxide (4.62 g) is dissolved in water to give a final volume of 350 mL. What is the molarity of the solution?
11	What mass of formic acid (HCOOH) should be diluted to obtain 1.00 litre of a 0.0750 M water solution?
12	What volume of sulfuric acid (0.77 M) contains 0.50 mole of H₂SO₄?
13	What volume of silver nitrate (0.54 M) contains 0.34 g of solute?
14	A solution of 12.0 M hydrochloric acid (100 mL) is diluted to 2.000 L. What is the molarity of the solution?
15	A solution contains 0.200 mole of solute in 500 mL. What is the molarity of the solution?
16	What mass of barium chromate (BaCrO₄) can be precipitated by adding excess barium chloride (BaCl₂) to 50.0 cm³ of potassium chromate (K₂CrO₄, 0.469 M) ?
17	What volume of barium nitrate (0.280 M) is required to precipitate all the sulfate ion from 25.0 mL of aluminium sulfate (0.350 M) as barium sulfate?

MOLE CONCEPT (General Answers 2)
1	(1.) Balanced equation: 3Fe + 2O₂ → Fe₃O₄ (2.) From the equation, 3 moles Fe requires 2 moles O₂ (note that oxygen gas is diatomic), therefore 1 mole Fe requires 2/3 mole O₂. (3.) Mass O₂ = moles O₂ x molar mass O₂ = 2/3 mol x (2 x 16.00 g mol^-1) = 21 g. (4.) Moles Fe = mass iron / molar mass Fe = 5.6 g / 55.85 g mol^-1 = 0.10 mol. From part (2.), moles O₂ required = moles Fe x 2/3 = 0.10 mol x 2/3 = 0.067 mol O₂. Therefore 5.6 g iron will not burn completely in 0.050 mol O₂.
2	The empirical formula of any compound is the simplest integer ratio of the atoms of its constituent elements. It is obtained from the elemental mass percentages by using the *elements' relative atomic masses (atomic weights). The elemental mass percentages are the no. grams of each element combined in 100 g of the substance. As the atoms of each of the different elements have their own unique atomic mass, it is necessary to divide each of the experimentaly measured masses of each element present by its atomic weight to determine the relative number of each atom present. That is, the first step involves determining the number of moles of atoms of each constituent element per 100 g* of the substance: In 100 g of substance, there are 26.2 g of N atoms, 66.4 g of Cl atoms and 7.50 g of H atoms Moles N atoms = mass N atoms / molar mass of N atoms = 26.2 g / 14.01 g mol^-1 = 1.87 mol Moles Cl atoms = mass Cl atoms / molar mass Cl atoms = 66.4 g / 35.45 g mol^-1 = 1.87 mol Moles H atoms = mass H atoms / molar mass H atoms = 7.50 g/ 1.008 g mol^-1 = 7.44 mol Note again that the elements' *atomic masses* are being used here - regardless of whether the element can occur in a diatomic form. For example, the atomic mass of nitrogen is 14.01 - the fact that it exists in gaseous form as N₂, with a molecular mass 28.02, is not relevant, since we are not dealing with gaseous N₂ here. (This is a common mistake!) As moles = number of atoms / Avogadro number, the relative number of moles of atoms of each element present also represents the relative number of atoms of each of the constituent elements in the compound. As atoms must combine in integer ratios, the remaining step is to reduce these values to integers by dividing through by the lowest common denominator (in this case 1.87), to allow assignment of the empirical formula: No. atoms of N in compound formula = 1.87 / 1.87 = 1.00 No. atoms of Cl in compound formula = 1.87 / 1.87 = 1.00 No. atoms of H in compound formula = 7.44 / 1.87 = 3.98 Taking into consideration the experimental error implicit in the given mass ratios, the value of 3.98 is taken as being 4. This experimental error should not represent more than 1 % of the integer value being assigned in any formula determination problem. For example, in this question the percentage error is ((4.00 - 3.98) / 4.00) x 100 % = 0.5 % (ie less than 1 %), so taking 3.98 as 4.00 is valid. This leads to the empirical formula NH₄Cl.
3	See the previous answer for detailed discussion on the method used. Mass % O = 100 % - mass % C - mass % H = 100 % - 38.7 % - 9.8 % = 51.5 % Thus 100.0 g of compound contains 38.7 g of carbon, 9.8 g of hydrogen and 51.5 g of oxygen. Moles C atoms = mass carbon / molar mass C atoms = 38.7 g / 12.01 g mol^-1 = 3.22 mol Moles H atoms = mass hydrogen / molar mass H atoms = 9.8 g / 1.008 g mol^-1 = 9.7 mol Moles O atoms = mass oxygen / molar mass O atoms = 51.5 g / 16.00 g mol^-1 = 3.22 mol Dividing these through by 3.22: No. atoms of C in compound formula = 3.22 / 3.22 = 1.00 No. atoms of H in compound formula = 9.7 / 3.22 = 3.0 No. atoms of O in compound formula = 3.22 / 1.00 This provides the empirical formula CH₃O. The molecular formula is found from the molar weight of the substance (molar mass of the full molecule is given as 62), and the molar weight of the empirical unit: (molar mass CH₃O = 31). Therefore the ratio of the molecular to empirical formula = 62 / 31 = 2, so the molecular formula is C₂H₆O₂.
4	As per the previous questions, 100.0 g of compound contains 52.5 g carbon, 3.7 g hydrogen, 7.9 g boron, 10.2 g nitrogen and 25.6 g chlorine. Moles C atoms = mass carbon / molar mass C atoms = 52.5 g / 12.01 g mol^-1 = 4.37 mol Moles H atoms = mass hydrogen / molar mass H atoms = 3.7 g / 1.008 g mol^-1 = 3.6 mol Moles B atoms = mass boron / molar mass B atoms = 7.9 g / 10.81 g mol^-1 = 0.73 mol Moles N atoms = mass nitrogen / molar mass N atoms = 10.2 g/ 14.01 g mol^-1 = 0.728 mol Moles Cl atoms = mass chlorine / molar mass Cl atoms = 25.6 g/ 35.45 g mol^-1 = 0.722 mol Dividing these through by 0.722: Molar ratio C in compound formula = 6.05 Molar ratio H in compound formula = 5.0 Molar ratio B in compound formula = 1.0 Molar ratio N in compound formula = 1.00 Molar ratio Cl in compound formula = 1.00 which is the same as the ratio of the number of constituent atoms. Allowing for experimental errors, the empirical formula is C₆H₅BNCl
5	100 g of compound contains 40 g of sulfur and 60 g of oxygen Moles S atoms = mass sulfur / molar mass S atoms = 40 g / 32.07 g mol^-1 = 1.2 mol Moles O atoms = mass oxygen / molar mass O atoms = 60 g / 16.00 g mol^-1 = 3.8 mol Dividing these through by 1.2: No. atoms of S in compound formula = 1.0 No. atoms of O in compound formula = 3.0 This gives the empirical formula SO₃. The molar mass of SO₃ is 80.07, and the molar mass of the full molecule is given as 240. Therefore the ratio of molecular to empirical weight = 240 / 80.07 = 3, so the molecular formula is S₃O₉.
6	(1.) Moles Fe atoms per 100 g = mass iron / molar mass Fe atoms = 63.5 g / 55.85 g mol^-1 = 1.14 mol Moles S atoms per 100 g = mass sulfur / molar mass S atoms = 36.5 g / 32.07 g mol^-1 = 1.14 mol Dividing these through by 1.14: No. atoms of Fe in compound formula = 1.00 No. atoms of S in compound formula = 1.00 This gives the empirical formula FeS. (2.) Moles Fe atoms per 100 g = mass iron / molar mass Fe atoms = 46.5 g / 55.85 g mol^-1 = 0.833 mol Moles S atoms per 100 g = mass sulfur / molar mass S atoms = 53.4 g / 32.07 g mol^-1 = 1.67 mol Dividing these through by 0.838: No. atoms of Fe in compound formula = 1.00 No. atoms of S in compound formula = 2.00 This gives the empirical formula FeS₂. (3.) Moles Fe atoms per 100 g = mass iron / molar mass Fe atoms = 53.7 g / 55.85 g mol^-1 = 0.962 mol Moles S atoms per 100 g = mass sulfur / molar mass S atoms = 46.3 g / 32.07 g mol^-1 = 1.44 mol Dividing these through by 0.962: No. atoms of Fe in compound formula = 1.00 No. atoms of S in compound formula = 1.50 In some cases this division process does not provide integer values suitable for assigning the empirical formula. However, it is usually easy to manipulate these values into integers. In this case it can be seen that multiplying the values obtained above by 2 will provide suitable integer values for the assignment of the empirical formula Fe₂S₃.
7	(1.) Moles Na atoms per 100 g = mass sodium / molar mass Na atoms = 32.4 g / 22.99 g mol^-1 = 1.409 mol Moles S atoms per 100 g = mass sulfur / molar mass S atoms = 22.6 g / 32.07 g mol^-1 = 0.705 mol Moles O atoms per 100 g = mass oxygen / molar mass O atoms = 45.0 g / 16.00 g mol^-1= 2.812 mol Dividing these through by 0.705: No. atoms of Na in compound formula = 2.00 No. atoms of S in compound formula = 1.00 No. atoms of O in compound formula = 4.00 This gives the empirical formula Na₂SO₄. (2.) As this compound is ionic, there are no "molecules" present - ionic compounds consist of 3-dimensional arrays of ions packed into a crystal lattice with each ion having a number of nearest neighbours. In covalent compounds, each atom is bonded to one or more other atoms to form the molecule of the compound. Consequently ionic compounds cannot have a molecular formula, only an empirical formula is possible, but covalent compounds have both an emprical and a molecular formula.
8	The molarity of a solution is the number of moles of the solute present in exactly one litre of solution. i.e. molarity = moles / litres or M = n / V where V must be in litres. Rearranging, n = M x V Moles in 125 mL is 0.125 L x 0.864 mol L^-1 = 0.108 mol.
9	Moles in 1.50 x 10³ mL is 1.50 L x 0.100 mol L^-1 = 0.150 mol. Mass sodium chloride = moles NaCl x molar mass NaCl = 0.15 mol x 58.44 g mol^-1 = 8.77 g.
10	Moles NaOH = mass sodium hydroxide / molar mass NaOH = 4.62 g / 40.00 g mol^-1 = 0.116 mol. Molarity = moles NaOH / volume of solution (L) = 0.116 mol / 0.350 L = 0.331 M.
11	As the volume is one litre, moles HCOOH required = 0.0750 mol. Mass formic acid = moles HCOOH x molar mass HCOOH = 0.0750 mol x 46.03 g mol^-1 = 3.45 g.
12	molarity = moles / volume(litres) therefore volume = moles / molarity Volume = 0.50 mol / 0.77 mol L^-1 = 0.65 L.
13	Moles AgNO₃ in 0.34 g = mass silver nitrate / molar mass AgNO₃ = 0.34 g / 169.91 g mol^-1 = 0.0020 mol. M = n / V Volume containing 0.0020 mol (0.34 g) = 0.0020 mol / 0.54 mol L^-1 = 0.0037 L = 3.7 mL.
14	Moles HCl = molarity hydrochloric acid x volume hydrochloric acid = 12.0 mol L^-1 x 0.100 L = 1.20 mol. Molarity of new solution = moles HCl / volume of solution = 1.20 mol / 2.000 L = 0.60 M. An alternative method for calculating the molarity of a solution that has been diluted is to use the expression M1V1 = M2V2 where M1 and V1 are the volume and molarity of the original solution while M2 and V2 are the new molarity and volume. Note that the units of V do not need to be in litres - they will necessarily be the same on both sides of the expression so other units such as mL can be used. This expression can only be used for dilution calculations and is not applicable to other types of calculation.
15	Molarity = moles / volume(L) = 0.200 mol / 0.500 L = 0.400 M.
16	Balanced equation: K₂CrO₄ + BaCl₂ → BaCrO₄ + 2KCl Hence one mole of K₂CrO₄ results in the formation of one mole of BaCrO₄ and since BaCl₂ is in excess, the number of moles K₂CrO₄ will determine the amount of BaCrO₄ produced. Moles K₂CrO₄ = molarity of K₂CrO₄ x volume of K₂CrO₄ solution = 0.469 x 0.0500 mol = 0.02345 mol Mass of barium chromate = moles BaCrO₄ x molar mass BaCrO₄ = 0.02345 mol x 253.3 g mol^-1 = 5.94 g
17	Balanced equation: 3Ba(NO₃)₂ + Al₂(SO₄)₃ → 3BaSO₄ + 2Al(NO₃)₃ Hence one mole of Al₂(SO₄)₃ requires three moles of Ba(NO₃)₂ for complete reaction. First the moles of aluminium sulfate in the solution is calculated as it is the reactant for which both the molarity and volume are given: Moles Al₂(SO₄)₃ dissolved = molarity of Al₂(SO₄)₃ x volume of Al₂(SO₄)₃ solution = 0.350 mol L^-1 x 0.0250 L = 0.00875 mol. As pointed out above, 1 mole Al₂(SO₄)₃ requires 3 mole Ba(NO₃)₂ or molesl Ba(NO₃)₂ required = 3 x moles Al₂(SO₄)₃ = 3 x 0.00875 mol = 0.0263 mol. M = n / V or volume of 0.280 M Ba(NO₃)₂ solution required = moles Ba(NO₃)₂ / molarity of Ba(NO₃)₂ solution = 0.0263 mol / 0.280 mol L^-1 = 0.0939 L = 93.9 mL.