Data Returned: Parts List • Each resource is identified as a URL • Parts list has links to get each part’s detailed info • Key feature of REST design pattern – Client transfers from one state to next by examining and choosing from alternative URLs in the response document SOP Basics Slide 28

Molar Mass 1. Molar mass is the mass in grams of 1 mole of a substance 2. Molar mass refers to both atoms & molecules 3. Elements (atoms) Examples: 1 mole of Na has a mass of 22.99 g 1 mole of Cl has a mass of 35.45 1 mole of Cl2 has a mass of 70.90 g 4. Compounds (molecules) Examples: 1 mole of NaCl has a mass of 58.44 g • Mass of Na (22.99 g) + Mass of Cl (35.45 g) 1 mole of CO2 has a mass of 44.01 g • Mass of C (12.01 g) + 2 x Mass of O (16.00 g)

• Component Interfaces A component interface is a set of application programming interfaces (APIs) that you can use to access and modify PeopleSoft database information programmatically. Integration Tools • Integration Broker PeopleSoft Integration Broker performs asynchronous and synchronous messaging among internal systems and third-party systems, exposes PeopleSoft business logic as web services to PeopleSoft and third-party systems, and consumes and invoke web services from third-party and PeopleSoft systems. • Integration Broker Administration Integration Broker Administration performs system administration tasks in PeopleSoft Integration Broker such as: • Set up and configure integration system components, such as messaging servers, nodes, integration gateways, listening and target connectors, and so on. • Configure the integration system to handle services, including specifying namespaces, setting up UDDI repositories, and so on. • Secure the integration environment by applying security at the web server, gateway, application server, node and service operation level. • Fine tune integration system performance by employing failover, master/slave processing, load balancing, and so on. • And more. • MultiChannel Framework PeopleSoft MultiChannel Framework (MCF) provides the tools that are required to support multiple channels of communication between customers (users) and agents. Some PeopleSoft applications, such as an email response management system (ERMS) from PeopleSoft CRM, use PeopleSoft MultiChannel Framework or you

Example: Find Molar Mass A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass. Information Given: V = 0.225 L, P = 1.1658 atm, t = 328 K, m = 0.311 g Find: molar mass, (g/mol) Eq’n: PV = nRT; MM = mass/moles SM: P,V,T,R → n & mass → mol. mass • Apply the Solution Map: P V n R T PV 1.1658 atm 0.225 L n L atm RT 328 K 0.0821 mol K n 9.7 406 10 -3 mol mass in grams Molar Mass moles 0.311g Molar Mass 9.7406 10 -3 moles Molar Mass 31.9 g/mol 47

Example: Find Molar Mass A sample of a gas has a mass of 0.311 g. Its volume is 0.225 L at a temperature of 55°C and a pressure of 886 mmHg. Find its molar mass. Information Given: V = 0.225 L, P =, t = 328 K, m = 0.311 g Find: molar mass, (g/mol) Eq’n: PV = nRT; MM = mass/moles SM: P,V,T,R → n & mass → mol. mass Molar Mass = mass/moles Mass: given To find mole of gas? PV = nRT P = 886 mmHg x _____________ Mole of gas n = ____________ P = 1.1658 atm n = 0.00974 mol Molar mass = 31.9 g/mol

• Percent by Mass (%m/m): mass of solute in mass of solution multiplied by 100 • Mass % = mass solute/mass solution x 100 • Remember: mass solution=mass solute + mass solvent • Also defined as the number of grams of solute per 100g of solution Problems 1) You make a solution by dissolving 7.55 g NaCl in 52.4g H2O. What is the mass % of solute in this solution? 2) You make a solution by dissolving 2.45 g KCl in 20.6mL of water. What is the mass % of solute in this solution? 3) You have a %m/m of 10.0% sucrose. If you used 25.0 g sucrose to make the solution, how much solvent (water) did you use?

Parts by mass • Parts by mass – a ratio of the mass of solute to the mass of the solution, then multiplied by a multiplication factor solutemass • Percent by mass %mass 100% solutionma ss • Parts per million • Parts per billion solutemass ppm 10 6 solutionma ss solutemass ppm 109 solutionma ss

5. Using the empirical formula and molar mass that was given determine the molecular formula for the substance given the molar mass is 210.18 g/mol The empirical formula for a compound used in the past as C1H2O1 The molar mass is 210.18 g/mol . What is the molecular formula? first, calculate the “n ratio factor” “n ratio factor” = molar mass empirical mass third, fourth, second, divide multiply calculate molar empirical the mass empirical bybyempirical n mass mass 1H2Ofactor” 1] x 6 = 1C x 12.0 = 12.0 “n[Cratio 210.18 g/mol empirical formula empirical 30.0 g/mol mass 2H x 1.0 = 2.0 “n ratio factor” = 6 1O x 16.0 = 16.0 C6H12O6 empirical formula mass 30.0 g/mol molecular exists as two molecular formula 210.18 g/mol empirical formulas

5. Using the empirical formula and molar mass that was given determine the molecular formula for the substance given the molar mass is 210.18 g/mol The empirical formula for a compound used in the past as C1H2O1 The molar mass is 210.18 g/mol . What is the molecular formula? first, calculate the “n ratio factor” “n ratio factor” = molar mass empirical mass second, calculate the empirical mass 1C x 12.0 = 12.0 “n ratio factor” = 210.18 g/mol empirical mass 2H x 1.0 = 2.0 1O x 16.0 = 16.0 empirical mass 30.0 g/mol third, divide molar mass by empirical mass “n ratio factor” = 210.18 g/mol 30.0 g/mol Continued on the next slide “n ratio factor” = 6

Gram-Atomic Mass • • • • • Mass of 1 carbon-12 atom = 12 u (exactly); Mass of 1 mole of carbon-12 = 12 g; Mass of 1 oxygen atom = 16.00 u Mass of 1 mole of oxygen = 16.00 g Gram-atomic mass = mass (in grams) of 1 mole of an element – that is, the mass (in grams) that contains the Avogadro’s number of atoms of that element. • gram-atomic mass is the molar mass of an element in grams.

Empirical Formula-3 • Empirical formula from data of combustion reaction: Example: A compound is composed of carbon, hydrogen, and oxygen. When 2.32 g of this compound is burned in excess of oxygen, it produces 5.28 g of CO2 gas and 2.16 g of water. Calculate the composition (in mass percent) of the compound and determine its empirical formula. • • Solution: Find mass of C, H, and O in the sample and then calculate their mass percent: Mass of C = 5.28 g CO2 x (12.01 g C/44.01 g CO2) = 1.44 g Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1% Mass of H = 2.16 g H2O x (2 x 1.008 g/18.02 g H2O) = 0.24 g Mass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4% Mass of O = 2.32 g sample – 1.44 g C – 0.24 g H = 0.64 g Mass % of O = 100 – 62.1% C – 10.4% H = 27.5% • Derive empirical formula from these mass percent composition • (next slide)

Determination of Empirical Formula #3 Example-3: A 2.32-g sample of a compound composed of carbon, hydrogen, and oxygen is completely combusted to yield 5.28 g of CO2 gas and 2.16 g of water. Calculate the composition (in mass percent) of the compound and determine its empirical formula. •Solution: •Determine the mass of C, H, and O, respectively, in the sample: – Mass of C = 5.28 g CO2 x (12.01 g C/44.01 g CO2) = 1.44 g; – Mass of H = 2.16 g H2O x (2 x 1.008 g/18.02 g H2O) = 0.242 g; – Mass of O = 2.32 g sample – 1.44 g C – 0.24 g H = 0.64 g; •Calculate the mass percent of each element: – Mass % of C = (1.44 g C/2.32 g sample) x 100% = 62.1% ; – Mass % of H = (0.242 g H/2.32 g sample) x 100% = 10.4% ; – Mass % of O = 100 – 62.1% C – 10.4% H = 27.5% •Composition (by mass %): 62.1% C; 10.4% H, and 27.5% O; •(continue next slide for empirical formula determination)