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CHM_Sept. 9

by: charlotteee

CHM_Sept. 9 25954

GPA 4.0
CHM Concepts, Systems, Practice 1

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CHM Concepts, Systems, Practice 1
Class Notes
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This 3 page Class Notes was uploaded by charlotteee on Wednesday September 9, 2015. The Class Notes belongs to 25954 at University of Rochester taught by Hafensteiner in Summer 2015. Since its upload, it has received 32 views. For similar materials see CHM Concepts, Systems, Practice 1 in Chemistry at University of Rochester.


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Date Created: 09/09/15
Jasmine Han Chemistry 131 Course Number 25954 Sept 9th 2015 31 Atomic Masses Mass 13C divided by Mass 12C 10836129 which is the ratio of the masses Atomic Mass Unit is de ned such that the mass of 12C is exactly 12 atomic mass units Mass of 13C 1 083612912 amu 13003355 amu Calculating atomic mass for natural carbon 9889 of 12 amu 111 of 130034 amu 0988912 amu 00111 130034 amu 1201 amu 32 The Mole Aluminum has 602210quot23 atoms and 2698 grams lron has 602210quot23 atoms and 5585 grams Calculate the mass in grams of a sample of americium containing six atoms 6 atoms times 243 amuatom 14610quot3 amu 14610quot3 amu times 1 gram602210quot23 amu 24210quot 21 grams Molar Mass is gramsmoles 34 Finding the Percent Composition of Compounds Formula C2H5OH Mass of C 2 moles times 12011 gmol 24022 grams Mass of H 6 moles times 1008 gmol 6048 grams Mass of O 1 mole times 159999 gmol 159999 grams Mass of 1 mole of C2H50H 46069 grams Mass percent of carbon in ethanol will be found by comparing the mass of carbon in 1 mole of ethanol with the total mass of 1 mole of ethanol and multiplying the result by 100 Mass percent of C mass of C in 1 mol C2H50Hmass of 1 mol C2H5OH times 100 24022 g46069 g times 100 52144 The mass percents of hydrogen and oxygen in ethanol are obtained in the same manner and noticecheck that the percentages add up to 100 35 Determining the Formula of a Compound To determine the mass of carbon in 01638 grams of C02 01638 grams C02 times 12011 grams of C44009 grams of C02 004470 grams of C The mass percent of carbon in this compound will then be 004470 g C01156 g compound times 100 3867 To determine the mass percent of hydrogen in an unknown compound We will assume that hydrogen present in the original 01156 gram of compound was converted to H20 Molar mass of H20 is 18015 grams and the fraction of hydrogen by mass in H20 is 2016 grams of H per 18015 grams H20 so the mass of hydrogen in 01676 grams of H20 is 01676 g H20 times 2016 g H18015 g H20 001876 g H Therefore the mass percent of hydrogen in the compound is 001876 g H01156 g compound times 100 1623 H The formula of a compound indicates the number of atoms in the compound so you need to convert the masses of the elements to numbers of atoms Example 3867 g C times 1 mol C12011 g C 3220 mol C Let s say that in this 1000 grams compound it contains 3220 moles of carbon atoms 1610 moles of hydrogen atoms and 3220 moles of nitrogen atoms We nd the smallest WholeNumber Ratio of atoms in the compound by dividing each of the mole values by the smallest of the three C 32203220 1 H 16103220 5 N 32203220 1 Therefore we can write the formula of his compound as CH5N Empirical Formula represents the simplest wholenumber ratio of the various types of atoms in a compound Must only contain whole numbers You cannot have a fraction of an atom You can calculate this formula from mass percent because we know the molar mass gmo You can also use the mole ratio Molecular Formula to be able to specify the exact formula of the molecule involved need to know the molar mass Use ratio of molar massempirica formula mass Mass percent gives the number of grams of a particular element per 100 grams of compound 36 Chemical Equations In a chemical reaction atoms are neither created nor destroyed The symbol aq stands for quotdissolved in water in an aqueous solution Chemical equations give two important types of information the nature of the reactants and products and the relative numbers of each 38 Stoichiometric Calculations Coefficients in chemical equations represent the numbers of molecules and not the masses of molecules Balance the equation for the reaction rst Then convert the known masses of the substances to moles Use the balanced equation to set up the appropriate mole ratios Use the appropriate mole ratios to calculate the number of moles of the desired reactant or product Finally convert from moles back to gram if required by problem 39 Calculations involving a Limiting Reactant Stoichiometric Quantities chemicals are often mixed in these quantities in correct amounts so that all the reactants run out and are used up at the same time Limiting ReactantReagent reactant that runs out and limits the amount of products that can be formed First write and balance the equation for the reaction Then convert the known masses of substances to moles and by comparing the mole ratio of reactants required by the balanced equation with the mole ratio of reactants actually present determine which reactant is limiting Using the amount of the limiting reactant and the appropriate mole ratios compute the number of moles of the product Convert from moles to grams using the molar mass WalkIn tutoring Sundays 48 pm Carlson 102 rochmywconinecom Homework Ch 3 Practice Problems on Sapling Look over problems inside the textbook Oxygen has an atomic mass of 160048 amu What is the chance that you will pick up an atom of O that has a mass of 160048 out of a random sample of O atoms There is 0 that you will pick up an atom of O that has a mass of 160048 out of a random sample of O atoms Change Proton different element Change Neutron Isotope Change Electron Ions Empirical Formula XCH20 gt C3H603 Empirical formula gtsum of the mass of all the elements Molecular formula gtmoar mass Stoichiometry Example 7239 H 9145 N 2089 0 What is the empirical formula You can nd the percent of carbon by adding them all up to 100 Mass percent of carbon is 62737 Use 100 gram sample Must convert grams to moles because not all of the elements have the same molar mass and we need to compare them in ratios So there is 7239 g of H 9145 g of N 6273 g of C and 2089 g of 0 Let s say that we do it for hydrogen Convert grams into moles so 7239 g H times 1 mol H1008 g H 7182 mol H Repeat with the other elements to get 5238 mol C 06527 mol N and 1306 mol 0 Divide the lowest number which is nitrogen Carbon 523806527 8025 Hydrogen 11 Nitrogen 1 Oxygen 2 Therefore C8H11N02 which is an empirical formula Balancing Reactions conservation of mass Number of type of atoms in reactants must be same as number of type of atoms on the right Theoretical yield vs actual yield Theoretical yield is the maximum amount we can made Percent yield is the actual yield divided by the theoretical yield multiplied by 100 Theoretical yield is based on the limiting reagent For every 1 mole of Carbon there is 12011 grams of carbon And 1 mole is 602210 to the 23rd things


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