Chapter 3 Stoichiometric

Name: Chapter 3 Stoichiometric
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Description: Chapter 3 Stoichiometric

Chapter 3 Stoichiometric
Atomic Masses, Mole concept, and Molar Mass (Average atomic mass). Number of atoms per amount of element. Percent composition and Empirical formula of molecules. Chemical equations, Balancing equations, and Stoichiometric calculations including limiting reagents. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Copyright©2000 by Houghton Mifflin Company. All rights reserved.
Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12C “weighs” 12 amu On this scale 1H = amu 16O = amu Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Measuring Atomic Mass Figure 3.1: (left) A scientist injecting a sample into a mass spectrometer. (right) Schematic diagram of a mass spectrometer. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Atomic mass of Neon Consider 100 atoms of neon
Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Average atomic mass of lithium:
Natural lithium is: 7.42% 6Li (6.015 amu) 92.58% 7Li (7.016 amu) Average atomic mass of lithium: 7.42 x x 7.016 100 = amu Copyright©2000 by Houghton Mifflin Company. All rights reserved.

The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 1 mole of anything =  1023 units of that thing Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Avogadro’s number equals 6.022  1023 units

atomic mass (amu) = molar mass (grams)
eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12C atoms = x 1023 atoms = g 1 12C atom = amu 1 mole 12C atoms = g 12C 1 mole lithium atoms = g of Li For any element atomic mass (amu) = molar mass (grams) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

EX. 3.2: Calculate the mass of 6 atoms of americium in grams
From periodic table, Am has a mass of 243 amu. 6.022x1023 atoms weigh 243 g. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Molar Mass A chemical compound is a collection of atoms. What is the mass of 1 mol of CH4? Mass of 1 mol C = g Mass of 4 mol H = 4x1.008 g Mass of 1 mol CH4 = g Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ? 1 mol C3H8O = (3 x 12) + (8 x 1) + 16 = 60 g C3H8O 1 mol C3H8O molecules = 8 mol H atoms 1 mol H = x 1023 atoms H 1 mol C3H8O 60 g C3H8O x 8 mol H atoms 1 mol C3H8O x 6.022 x 1023 H atoms 1 mol H atoms x = 72.5 g C3H8O 5.82 x 1024 atoms H Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound %C = 2 x (12.01 g) 46.07 g x 100% = 52.14% C2H6O %H = 6 x (1.008 g) 46.07 g x 100% = 13.13% %O = 1 x (16.00 g) 46.07 g x 100% = 34.73% 52.14% % % = 100.0% Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Determining Elemental Composition (Formula)
Figure 3.5: A schematic diagram of the combustion device used to analyze substances for carbon and hydrogen. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

The masses obtained (mostly CO2 and H2O and sometimes N2)) will be used to determine: % composition in compound Empirical formula Chemical or molecular formula if the Molar mass of the compound is known or given. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Collect 22.0 g CO2 and 13.5 g H2O 16 x 2 12 1mol CO2 44gCO2 1mol C 12g C 22gCO2 mc % C in CO2 collected 22gCO2 x 12gC = 6g C mc = 44gCO2 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Convert g to mole: 6g x 1molC nc = 12 gC 1mol C 12g C nmol C 6g = 0.5 mol Repeat the same for H from H2O 1mol H2O 18g H2O 2 mol H 2g H 13.5g H2O nmol H mH 2x13.5 nmol H = = 1.5 mol H 18 mol H Faster H but still need O Copyright©2000 by Houghton Mifflin Company. All rights reserved.

2x13.5 mH = = 1.5 g H 18 mO = 11.5g – mC – mH = 11.5 – 6 – 1.5 = 4g m 4 nO = = = 0.25 mol O MM 16 Empirical formula C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O Copyright©2000 by Houghton Mifflin Company. All rights reserved.

OR for C mC 1. Fraction of C in CO2 = = = MMCO 2. Mass of C in compound = mass of CO2 x Fraction of C mC 3. % C in compound = x 100 msample 4. If N then % N = % C - % H Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Then Empirical Formula
Using the previously calculated % in compound: % in gram a. Number of mole of C = Atomic mass of C b. Number of mole of H = Atomic mass of H a b c Then divide by the smallest number: smallest smallest smallest : : Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Note If results are : 0.99 : 2.01 : 1.00 Then you have to convert to whole numbers: 1 : 2 : 1 CH2N If results are : 1.49 : 3.01 : 0.99 Then you have to multiply by 2: 3 : 6 : 2 C3H6N2 Hence, empirical formula is the simplest formula of a compound Copyright©2000 by Houghton Mifflin Company. All rights reserved.

71.65% Cl; % C ; % H 2.021 mol Cl mol C mol H : : Empirical formula is ClCH2 Molar mass = g/mol Molar mass/ empirical formula mass = 98.96/49.48 = 3 Molecular formula – (ClCH2)2 = Cl2C2H4 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Formulas molecular formula = (empirical formula)n [n = integer] molecular formula = C6H6 = (CH)6 empirical formula = CH Then Molecular Mass = n Empirical Mass Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 3.6: Examples of substances whose empirical and molecular formulas differ. Notice that molecular formula = (empirical formula)n, where n is a integer. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Exercise 3.12 A compound analyzed and found to contain only P and O. Mass % of P = 43.64% Molar mass = g/mol. What are the empirical and molecular formulas? Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 3.8 The Structure of P4O10.

Chemical Equation C2H5OH + 3O2  2CO H2O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Methane Reacts with Oxygen to Produce Flame

S.Ex. 3.14 Balance the equation (NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g) + H2O(g) N and Cr are balanced; 4 for H2O balances H It also balances O. (NH4)2Cr2O7 (s) → Cr2O3(s) + N2(g)+4H2O(g) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Decomposition of Ammonium Dichromate

S. Ex.3.15 Balance the equation NH3(g) + O2(g)→ NO (g) + H2O (g) All molecules are of equal complexity. Balance H first; 2 for NH3 and 3 for H2O N can be balanced with a coeff. of 2 for NO O can be balanced by 5/2 for O2 2NH3(g) + 5/2 O2(g)→ 2NO (g) + 3H2O (g) x2 4NH3(g) + 5O2(g)→ 4NO (g) + 6H2O (g) Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Calculating Masses of Reactants and Products
1. Balance the equation. 2. Convert mass to moles. 3. Set up mole ratios. 4. Use mole ratios to calculate moles of desired substituent. 5. Convert moles to grams, if necessary. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Methanol burns in air according to the equation 2CH3OH + 3O CO2 + 4H2O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH3OH moles CH3OH moles H2O grams H2O molar mass CH3OH coefficients chemical equation molar mass H2O 1 mol CH3OH 32.0 g CH3OH x 4 mol H2O 2 mol CH3OH x 18.0 g H2O 1 mol H2O x = 209 g CH3OH 235 g H2O Copyright©2000 by Houghton Mifflin Company. All rights reserved.

S. Ex 3.16 What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? Unbalanced eq. LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l) Balanced equation 2 LiOH(s) + CO2 (g) → Li2CO3 (s) + H2O(l) 23.95 g/mol g/mol Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Figure 3.12 Hydrogen and Nitrogen React to Form Ammonia

Solving a Stoichiometry Problem
1. Balance the equation. 2. Convert masses to moles. 3. Determine which reactant is limiting. 4. Use moles of limiting reactant and mole ratios to find moles of desired product. 5. Convert from moles to grams. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Limiting Reactant N2 (g) H2 (g) → 2 NH3 (g) 25.0kg 5.00 kg ?? 8.93x x103 8.93x102 mol N2 x 3 mol H2 = 2.68x103 mol H2 1 mol N2 H2 is limiting; x103 mol H2 x 2 mol NH3 = 1.65x103 3 mol H2 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Limiting Reactant Calculations
What mss of molten iron is produced by 1 kg each of the reactants? Fe2O3(s) Al(s)  Al2O3(s) + 2 Fe(l) 1 mol 2 mol 6.26mol ?? If all Fe2O3(s) reacts x6.26 = 12.5 mol Fe -Limiting If all Al reacts mol Fe -- Excess The 6.26 mol Fe2O3 will Disappear first Copyright©2000 by Houghton Mifflin Company. All rights reserved.

S. Ex 3.18 2NH3(g) +3CuO(s) → N2(g) +3Cu(s) +3H2O(g) 1.06 mol mol ??g If CuO is limiting; 1.14 mol CuO x 1 mol N2 3 mol CuO = mol N2 If NH3 is limiting; 1.06 mol NH3 x 1 mol N2 2 mol NH3 = mol N2 Least amount of product obtained with CuO; Cuo is the limiting reactant Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Its amount is Calculated using the balanced equation. Actual Yield is the amount of product actually obtained from a reaction. It is usually given. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Percent Yield Example excess 14.4 g 0.104 mol Actual yield = 6.26 g 1 mol S.A. → 1 mol aspirin 0.104 mol S.A. → mol aspirin = 18.7 g aspirin Percent yield = g x 100 = % 18.7 g Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Sample exercise 3.19 - % yield calculation
Methanol can be manufactured by combination of gaseous carbon monoxide and hydrogen. Suppose 68.5Kg CO(g) is reacted with 8.60Kg H 2(g). Calculate the theoretical yield of methanol. If 3.57x104g CH3OH is actually produced, what is the percent yield of methanol? Copyright©2000 by Houghton Mifflin Company. All rights reserved.

2 H2 (g) CO(g) → CH3OH (l) 8.60 kg 68.5 kg 35.7 kg(actual yield) 4.27x103mol x103 mol If all H2 are converted to methanol, 4.27x103mol H2 x 1 mol methanol = 2.14x103 mol methanol 2 mol H2 If all CO are converted to methanol, 2.44x103 mol CO x 1 mol methanol = 2.44x103 mol methanol 1 mol CO H2 is the limiting reactant; theoretical yield = 2.14x103 mol methanol = 68.6 kg methanol % yield = actual yield x = 35.7 kg x100 = 52.0% theoretical yield kg Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Sample Exercise Titanium tetrachloride, TiCl4, can be made by combining titanium-containing ore (which is often impure TiO2) with carbon and chlorine - TiO2(s) Cl2(g) + C(s) TiCl4(l) + CO2(g) If one begins with 125 g each of Cl2 and C, but plenty of titanium-containing ore, which is the limiting reagent in the reaction? What quantity of TiCl4 can be produced? Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Practice Example 1 A compound contains C, H, N. Combustion of 35.0mg of the compound produces 33.5mg CO2 and 41.1mg H2O. What is the empirical formula of the compound? Solution: 1. Determine C and H, the rest from 33.5mg is N. 2. Determine moles from masses. 3. Divide by smallest number of moles. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Practice Example 2 Caffeine contains 49.48% C, 5.15% H, 28.87% N and 16.49% O by mass and has a molar mass of g/mol. Determine the molecular formula. Solution: Convert mass to moles. Determine empirical formula. Determine actual formula. C8H10N4O2 Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Practice Example 3 Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1g of NH3 is reacted with 90.4g of CuO, which is the limiting reactant? How many grams of N2 will be formed. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Practice Example 5 SnO2(s) + 2 H2(g) ® Sn(s) + 2 H2O(l)
a) the mass of tin produced from moles of hydrogen gas. b) the number of moles of H2O produced from 339 grams of SnO2. c) the mass of SnO2 required to produce 39.4 grams of tin. d) the number of atoms of tin produced in the reaction of 3.00 grams of H2. e) the mass of SnO2 required to produce 1.20 x 1021 molecules of water. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

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