Chemistry 111 for Engineering students Dr. Ayman H. Kamel Office: 33

Chapter 3 Stoichiometry: Chemical Calculations

3 Objectives. Define Atomic masses. Molecular Masses and Formula Masses. Define the mole. Calculate the number of moles. Calculate the molar mass of any compound. Determine the percent composition of compounds. Determine the formula of compounds. Balance chemical equations. Solutions and solution stoichiometry

4 3.1.Molecular masses [ Molecular Weight] Molecular mass is the sum of the masses of the atoms represented in a molecular formula. Ex.1. M. Wt. of O 2 = 2x atomic mass of O = 2 x u = u. Ex. 2. Calculate the molecular mass of glycerol CH 2 (OH)CH(OH)CH 2 OH 3 x atomic mass of C = 3 x u = u. 3 x atomic mass of O = 3 x u = u. 8 x atomic mass of H = 8 x u = u. Molecular Mass of C 3 H 8 O 3 = u.

5 3.2.The Mole A mole is an amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of C 12. Avogadro’s Number: is the number of carbon atoms in exactly 12 grams of pure C 12. This No. was found to be x SO, ONE MOLE OF SOMETHING CONSISTS OF AVOGADRO’S NUMBER OF THAT SUBSTANCE i.e x10 23 units of that substance. The mass of 1 mole of an element is equal to its atomic mass in grams.

6 Example 1 Aluminum (Al) is a metal with a high strength to mass ratio and high resistance to corrosion. Compute both the number of moles of atoms and the number of atoms in a 10 g sample of aluminum. Solution The mass of 1 mole (6.022x10 23 atoms) of aluminum is g. 1 mole ? mole g 10.0 x 1/ = mol Al atoms 1 mole x10 23 atoms 0.37 mole Al ? atoms The no. of atoms in 10.0 g (0.37 mol) of aluminum is x 6.022x10 23 /1 =2.23x10 23

7 Example 2 Cobalt Co is metal that is added to steel to improve its resistance to corrosion. Calculate both the number of moles in a sample of cobalt containing 5.00x10 20 atoms and the mass of the sample. Solution 1 mole Co x10 23 atoms Co ? mole x10 20 atoms Co No of mole of Co is = 5.00x10 20 x1/6.022x10 23 = 8.30x10 -4 Since the mass of 1 mole of cobalt is mole Co g 8.30x10 -4 mole Co ?g The mass of Co is = x 8.30x10 -4 /1 = 4.89x10 -2 g Co.

8 Example 3 A silicon chip used in an integrated circuit of a microcomputer has a mass of 5.68 mg. How many silicon (Si) atoms are present in the chip? Solution First we have to convert weights to g, so the weight of silicon chip is 5.68x10 -3 g. 1 mol Si g Si ? Mol x10 -3 g Si. No of moles of Si is 5.68x10 -3 x1/28.09 = 2.02x10 -4 mol Si. 1mol Si x10 23 atoms 2.02x10 -4 mol Si ? Atoms The no of silicon atoms is 2.02x10 -4 x6.022x10 23 /1 = 1.22x10 23 atoms.

9 3.3.Molar Mass A chemical compound is a collection of atoms, for example, methane consists of molecules that each contain one carbon and four hydrogen atoms (CH 4 ). How to calculate the mass of 1 mole of methane? Mass of 1 mol C = g Mass of 4 mol H = 4x g  Mass of 1 mol CH 4 = g Thus the molar mass of a substance is the mass in grams of one mole of the compound

10 Example 1 For the natural dye, Jug lone, C 10 H 6 O 3,Calculate the molar mass of jug lone. How many moles of jug lone are found in a 1.56x10 -2 g sample of pure dye? Solution The molar mass can be obtained by summing the masses of the component atoms, so, we have 10 C: 10 x g = 120.1g 6 H: 6 x g = g 3 O: 3 x g = g Mass of I mol C 10 H 6 O 3 = g. 1 mol jug lone g ? mol x10 -2 g juglone  No. of moles of jug lone is 1.56x10 -2 x 1/174.1 = 8.96x10 -5 mol juglone.

11 Example 2 Calculate the molar mass of calcium carbonate CaCO 3 (calcite). For a certain sample of calcite containing 4.86 moles, what is the mass in grams of this sample? And what is the mass of carbonate ions present. Solution The molar mass Ca(40) + C (12) + O (3x16 =48) = g 1 mol g 4.86 mol ? g  The no. of moles of Calcium carbonate is 4.86 x1/ = 486 gm CaCO 3 To find the mass of present in this sample, we must realize that 4.86 moles of CaCO 3 contains 4.86 moles of Ca 2+ ions and 4.86 moles of carbonate ions. The mass of 1 mole carbonate ion is C (1x 12.01) + O (3x16.00 =48.00) = g 1 mol carbonate ion g 4.86 mol ? The mass of 4.86 moles is 4.86 x 60.01/1 = 292 g

Percent composition of compounds There are two common ways of describing the composition of a compound. In terms of the number of its constituent atoms and in terms of the percentage (by mass ) of its elements. The mass percents of the elements can be obtained by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole of the compound.

13 Example 1 Carvone is a substance with the molecular formula (C 10 H 14 O), compute the mass percent of each element in carvone. Solution Mass of C = 10 x = g Mass of H = 14 x = g Mass of O = 1 x = g  Mass of 1 mol (C 10 H 14 O) = = 150.2g So the mass percent of each component will be: Mass percent of C= 120.1/150.2 x 100% = 79.96% Mass percent of H = 14.11/150.2 x 100% = 9.394% Mass percent of O = 16.00/150.2 x 100% = 10.65%

14 Example 2 Penicillin, has the formula C 14 H 20 N 2 SO 4. Compute the mass percent of each element. Solution The molar mass of penicillin is computed as follows, C: 14 x = g H: 20 x = g N: 2 x = g S: 1 x = g O: 4 x = g  Mass of 1 mol C 14 H 20 N 2 SO 4 = g, then, Mass percent of C: 168.1/312.4 x 100% = 53.81% Mass percent of H: 20.16/312.4 x 100% = 6.543% Mass percent of N: 28.02/312.4 x 100% = 8.969% Mass percent of S: 32.07/312.4 x 100% = 10.27% Mass percent of O: 64.00/312.4 x 100% = 20.49%

Determining the formula of a compound The formula of any substance can either be represented by, Empirical formula, which is the simplest whole number ratio of atoms in a compound (C n H n O n..) or Molecular formula, which is the exact formula of a molecule, giving the types of atoms and the exact number of each type. (C n H n O n ….) n

16 Example 1 Determine the empirical and molecular formulas for a compound that gives the following analysis in mass percents: 71.65% Cl, 24.27%C, 4.07%H, The molar mass is known to be g/mol. Solution From the above given values we can say that in a 100 g sample of the compound we have g Cl, g C and 4.07 g H, so to calculate the no of moles of each component we should divide each by its atomic mass So, No of moles of Cl is = 71.65/35.45 = mol Cl No of moles of C is = 24.27/ = mol C No of moles of H is = 4.07/1.008 = 4.04 mol H To get the empirical formula we should divide each of the above values by the smallest number of moles that is (2.021)

17 Example 1 (cont.) 2.021/2.021 CL :2.021/2.021 C : 4.04/2.021H 1 Cl: 1C: 2H (notice that you should approximate the values to the nearest whole numbers, i.e. no decimal values should be written the formula) So the empirical formula can be represented as ClCH 2. To determine the molecular formula, we must compare the empirical formula mass to the molar mass. The empirical formula mass is ( (1.008))., Molar mass/empirical formula mass = 98.96/49.48 = 2  Molecular formula is (ClCH 2 ) 2 = Cl 2 C 2 H 4

18 Example 2 Caffeine, is a stimulant found in coffee, tea and chocolate, contains 49.48% carbon, 5.51% hydrogen, 28.87% nitrogen and 16.49% oxygen by mass and has a molar mass of g/mol. Determine the molecular formula of caffeine. Another way of solving molecular formula problems: We will first determine the mass of each element in 1 mole (194.2 g) of caffeine, Mass of C = 49.48/100 x = g/mol C Mass of H = 5.15/100 x = 10.0 g/mol H Mass of N = 28.87/100 x = g/mol N Mass of O = 16.49/100 x = g/mol O Now we will convert to moles, No of moles of C = 96.09/12.01 = mol C/mol caffeine. No of moles of H = 10.0/1.008 = 9.92 mol H/ mol caffeine. No of moles N = 56.07/14.01 = mol N / mol caffeine. No of moles of O = 32.02/16.00 = mol O/ mol Caffeine.  the molecular formula of caffeine can be given as: C 8 H 10 N 4 O 2

Writing and balancing Chemical equations Chemical reactions are represented by chemical equations, which identify reactants and products. Formulas of reactants appear on the left side of the equation and those of products are written on the right. In a balanced chemical equation, there are the same number of atoms of a given element on both sides. You can’t write an equation unless you know what happens in the reaction that it represents. So all the reactants and products should be identified.

20 Example 1 At 1000 °C, ammonia gas, NH 3, reacts with oxygen gas to form gaseous nitric oxide, NO and water vapour. Balance the equation for this reaction. Solution The unbalanced equation for the reaction is NH 3 + O 2 NO + H 2 O To start balancing, we can first chose hydrogen, as follows 2NH 3 + O 2 NO + 3H 2 0 The nitrogen can be balanced by a coefficient of 2 for NO 2NH 3 + O 2 2NO + 3H 2 O Finally note that there are two atoms of oxygen on the left and 5 on the right so oxygen can be balanced with a coefficient 5/2 for O2 2NH 3 + 5/2O 2 2NO + 3H 2 O However, the usual custom is to have whole number coefficients, so we multiply the entire equation by 2 4NH 3 + 5O 2 4NO + 6H 2 O

Reaction Stoichiometry The principal reason for writing balanced equations is to make it possible to relate the masses of reactants and products, taking into consideration that: “the coefficient of a balanced equation represent numbers of moles of reactants and products”. For the reaction, 2N 2 H 4 + N 2 O 4 3N 2 + 4H 2 O 2 molecules N 2 H molecule N 2 O 4 gives 3 molecules N molecules H 2 O. A balanced equation remains valid even if each coefficient is multiplied by the same number including Avogadro’s number: 2AN molecules N 2 H 4 + 1AN molecule N 2 O 4 gives 3AN molecules N 2 + 4AN molecules H 2 O.

22 Example 1 Ammonia can be prepared by the following reaction N 2 + 3H 2 2NH 3 Determine The number of moles of ammonia formed when 1.34 mol of N 2 reacts. The mass in grams of N 2 required to form 1.00x10 3 g NH 3. The number of molecules of ammonia formed when 1.34 g of H 2 reacts. Solution The coefficients of the balanced equation shows that each mol of N 2 produces 2 moles of ammonia thus 1.34 mol N 2 should produce 2.68 mol NH 3. I mole N moles NH 3 (2 x 14.01) = 28.02g ( x1.008) = 34.06g ? g x10 3 g NH 3 Mass of N 2 = 1.00x10 3 x 28.02/ = 823 g N 2

23 Example 1(Cont.) 3 mol H mol NH 3 3( 2 x 1.008) = g ( x 1.008) =34.02 g 1.34 g H ? g NH 3 Mass of NH 3 = x 1.34/6.048 = g To get the number of molecules, 1 mol [17 gm] x molecule gm ? The number of molecules is = x 6.022x10 23 = 2.67 x molecule NH 3 17

24 Example 2 Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide by forming lithium carbonate and water. What mass of carbon dioxide can be absorbed by 1kg of lithium hydroxide? Solution LiOH + CO 2 Li 2 CO 3 + H 2 O We should first balance the equation 2LiOH + CO 2 Li 2 CO 3 + H 2 O 1 mol LiOH g LiOH ? mol g Mass of LiOH is 1000 x 1/23.95= 41.8 mol LiOH From the equation 1 mol of CO 2 can be produced from 2 moles LiOH 1 mol CO mol LiOH (44.0) (2 x 23.95) ?g LiOH The mass of CO 2 = 44.0 x 41.8/2x23.95 = 920 g CO 2

3.11 Solutions and solution Stoichiometry Molarity, M: i. The Molecular Weight, MW: It is the summation of the atomic weights of all the atoms composing the molecule ● Example: MW of NaOH = = 40 Where: at wt of Na = 23, O = 16, and H = 1

iv. Molarity: It is the Number of Moles of solute per one liter of solution  M x V L = wt of solute / its MW No of moles of solute sol.n Volume in Liters = wt / MW VLVL  M = M =

We want to prepare a 6.68 molar solution of NaOH ( 6.68 M NaOH). A- How many moles of NaOH are required to prepare L of 6.68 M NaOH? B- How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH? Solution A. moles of NaOH = L x 6.68 mol NaOH/ 1 L solution = 3.34 mol NaOH B. gm NaOH/ Molar mass of NaOH = M x Volume (L) 2.35 x 1000 gm / 40 = 6.68 x V L V L = 8.79 L Exercise : How many grams of 1- butanol, CH3CH2CH2OH, are required to prepare 725 mL of a M aqueous solution of this solute? 27

 Dilution: By diluting a solution, the number of moles doesn’t change as we don’t increase or decrease the amount of solute  n o of moles of solute before dilution = no of moles of solute after dilution, ∵ no of moles = wt(g)/MW = M x V L  (M x V) Before dilution = (M’ x V’) After dilution

● Example: Calculate the molarity of a solution prepared by diluting 10 ml of 0.1 M with 90 ml of water - Answer: ∵ (M x V) Before dilution = (M’ x V’) After dilution  (0.1 x 10) = (M’ x (90+10))  M’ = 1/100 = 0.01 moles/L