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Chapter 6 – Chemical Composition

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Counting by weighing Situation: You work in a candy shop. You sell jelly beans. Somebody wants to buy 1000 jelly beans. How are you going to count those out? Are you going to do it one by one??

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Counting by weighing Heres an idea!! Weigh them!! Lets say that all jelly beans are identical. One jelly bean weighs 5 g. So 1000 x 5g = 5000g or 5kg Problem solved!!!

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Counting by weighing Problem: all jelly beans arent identical. Suppose we weigh 10 jelly beans and find: Now we can find he average mass of a bean.

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Counting by weighing Now lets say that there is a customer who wants a bag of jelly beans and a bag of mints. He wants the same number of jelly beans and mints. What do you do?

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Counting by weighing You know: Jelly bean avg mass = 5 g Mint avg mass = 15 g One scoop of jelly beans is 500g What mass of mints do you need, to give the same number of mints as there are jelly beans in 500g of jelly beans?

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Counting by weighing Lets compare the avg masses. Avg mass of mints/avg mass of jelly bean 15g/5g = 3 The mints mass is 3 times bigger than the jelly bean. So, 3 x 500g = 1500g of mints.

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Lets look at some ratios: 1mint x 15g = 15g 1 bean x 5g = 5gthe ratio is 3/1 10mints x 15g = 150g 10beans x 5g = 50gthe ratio is 3/1 100mint x 15g = 1500g 100 bean x 5g = 500gthe ratio is 3/1

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How does this apply to Chemisty? Lets say you have a pile of C and want to know how many O 2 molecules you need to convert all the C into CO 2. Or you want to know how many O atoms are in a cup of H 2 O

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Atomic Masses How much does a C atom weight? 1.99 x 10 -23 g thats really small!!! To help with these complicated numbers we use atomic mass unit (amu). 1 amu = 1.66 x 10 -24 g

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Atomic Masses Remember how not all jelly beans weighed the same. Not all C atoms have the same mass. Remember what an isotope is? Isotopes are atoms of element with different number of neutrons. For the atomic mass we use the average atomic mass of those isotopes.

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Atomic Masses Where can we find what the avg atomic mass of an element? Look on the periodic table!!

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Atomic Masses So the common isotopes of C are Carbon 12 Carbon 13 Carbon 14 Based on the abundance found in nature C avg atom mass 12.01 amu 1 carbon atom = 12.01 amu

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Back to the question Lets say you have a pile of C and want to know how many O 2 molecules you need to convert all the C into CO 2. You weighed the pile of C and its 3 x 10 20 amu 3x10 20 amu X (1carbon atom/12.01amu) = 2.5x10 19 atoms (see pg176)

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Practice Remember: conversion factors are all equal to one. Convert 10 ft to in Convert 5 years to days and mins Convert 289g to kg Convert 3749mm to m

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Practice Remember: Convert 3x10 20 amu of C to atoms Convert 75 Al atoms to amu Convert 322.2 amu of N to # of atoms

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Homework Convert: 213 Hg atoms to amu 53 P atoms to amu 172 O atoms to amu

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Homework Convert: 12 amu of Cl to atoms 74 amu of Na to atoms 123 amu of Li to atoms

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Now to a mole!

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Is a counting unit The mole (mol) is the SI unit used to measure the amount of a substance. It is based on the number of atoms in 12g of carbon We can convert number of particles to moles and moles to particles.

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The Mole is 6.02 X 10 23 (in scientific notation) Similar to a dozen, except instead of 12, its 602 billion trillion 602,000,000,000,000,000,000,000 This number is named in honor of Amedeo Avogadro(1776-1856) He studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present

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What is a mole? One mole of anything consist of 6.02x10 23 units of that substance. How many eggs are in a mole of eggs? How many paper clips are in a mole of paper clips? How many shoes are in a mole of shoes? Answer: 6.02x10 23 eggs, paper clips, shoes

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6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles Note that a particle could be an atom, a molecule or anything!!! Avogadros Number as Conversion Factor

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Just How Big is a Mole? Enough soft drink cans to cover the surface of the earth to a depth of over 200 miles. If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

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The Mole This photograph shows one mole of salt(NaCl), water(H 2 O), and nitrogen gas(N 2 ).

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We also know… 1 mole of an elements atoms = that elements average atomic mass (expressed in g)

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We also know… Ex. 200.6g of Hg has 6.02x10 23 atoms or 200.6g of Hg has 1 mol of Hg atoms Conversion factor: 200.6g Hg/1 mol Hg Ex. 32.07g of S has 6.02x10 23 atoms or 32.07g of S has 1 mol of S atoms Conversion factor: 32.07g S/1 mol S

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Example We have an unknown # of H atoms. We want to know many H atoms are present? We weigh the sample, results =.5g

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Example So.5g H X (1mol H/1.008g H) =.496 mol of H in the sample. Then.496 mol H atoms X (6.022x10 23 H atoms/1 mol H atoms) = 2.99x10 23 H atoms in the sample (See pg 181)

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Practice Convert 10g of Al into moles of atoms Convert 8g of S into moles of atoms Convert 25g of Ca into moles of atoms

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Practice Convert.371 mol Al into # of atoms Convert.249 mol S into # of atoms Convert.624 mol Ca into # of atoms

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Summary 1 amu = 1.66 x 10 -24 g 1 ___ atom = _____ amu 1 C atom = 12.01 amu 1 Si atom = 28.09 amu 1 mol of ___ = 6.022x10 23 ____ units 1 mol F = 6.022x10 23 F atom 1 mol Mg = 6.022x10 23 Mg atom What are the conversion factors?

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Summary 1mol of ___ = ____g of _____ 1mol Hg = 200.6g Hg 1mol Cl = 35.45g Cl ____g of __ = 6.022x10 23 atoms of ____ 200.6g Hg = 6.022x10 23 atoms Hg 35.45g Cl = 6.022x10 23 atoms Cl What are the conversion factors?

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Classwork How many moles are in 15 grams of lithium? How many atoms? How many grams are in 2.4 moles of sulfur? How many atoms? How many moles are in 22 grams of argon? How many atoms?

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Classwork How many grams are in 88.1 moles of magnesium? How many atoms? How many moles are in 2.3 grams of phosphorus? How many atoms? How many grams are in 11.9 moles of chromium? How many atoms?

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Molar Mass

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Back to the candy shop. This time the customer wants to buy suckers. When you weigh the suckers what are you weighing? How many parts are there? Answer: 3 parts ( wrapper, candy, stick) One sucker with 3 parts.

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Molar Mass Chemical compounds have multiple parts like the sucker. Ex. Methane (natural gas) CH 4 1 CH 4 molecule With 5 parts (1 C atom and 4 H atoms)

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Quick Note In 1 mol of a molecule there are corresponding moles of its parts. Ex. In 1 mol of CH 4 there are 1 mol of C and 4 mol of H

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Molar Mass How do we figure out the mass of 1 mol of methane? We break it down in to its parts and add them up. This is called the molar mass. Molar Mass- is the mass of a molecule Total mass of all the parts of a molecule

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Example: SO 2 What is the molar mass of SO 2 ? Mass of 1 mol of S = 1 x 32.07 = 32.07g Mass of 1 mol of O = 2 x 16.00 = 32.00g Mass of 1 mol of SO 2 = 64.07g The molar mass of SO 2 is 64.07g

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Example: CaCO 3 What is the molar mass of CaCO 3 ? Mass of 1 mol of Ca = 1 x ____ = ____ Mass of 1 mol of C = 1 x _____ = ____ Mass of 1 mol of O = 3 x _____ = ____ Mass of 1 mol of CaCO 3 = ____ The molar mass of CaCO 3 is _____

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Practice What is the molar mass of water, H 2 O? What is the molar mass of ammonia, NH 3 ? What is the molar mass of propane, C 3 H 8 ? What is the molar mass and name of CaSO 4 ? What is the molar mass and name of Na 2 CO 3 ? What is the molar mass and name of Ba(OH) 2 ?

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Practice What is the molar mass of water, H 2 O? What is the molar mass of ammonia, NH 3 ? What is the molar mass of propane, C 3 H 8 ?

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Practice What is the molar mass and name of CaSO 4 ? What is the molar mass and name of Na 2 CO 3 ? What is the molar mass and name of Ba(OH) 2 ?

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Percent Composition

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Back to the candy shop A customer comes in and wants 3 different kinds of candy: jelly beans, mints and M&Ms. You give him 2 scoops of jelly beans, 6 scoops of mints, and 1 scoop of M&Ms. He now wants to know what percent of his bag has jelly beans by weight? Youre going to use Percent Composition.

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Back to the candy shop 2 jelly beans, 6 mints, 1 M&Ms. Break it down in to part of the bag. Mass of a bean = 5g x 2 = 10g Mass of a mint = 15g x 6 = 90g Mass of a M&M = 10g x 1 = 10g Total mass of 1 bag = 110g

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Back to the candy shop 2 jelly beans, 6 mints, 1 M&Ms. So: percent of beans = mass of beans x 100% mass of 1 mixed bag So: percent of beans = 10g x 100% = 9% 110g So: percent of mints = 90g x 100% = 81% 110g

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Percent Composition Percent composition consists of the mass percent of each element in a compound. Mass Percent – the percent of each element present based upon its mass. AKA: Its how much stuff is there in a percent form. Mass percent =

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For example… Ethanol(C 2 H 5 OH) – used to enhance the octane level of gasoline. Question: What is the percent mass of C? First we need to find the molar mass of the molecule.

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Ethanol(C 2 H 5 OH) First we need to find the molar mass of the molecule. Mass of 1 mol of C = 2 x 12.01 = 24.02g Mass of 1 mol of H = 6 x 1.008 = 6.048g Mass of 1 mol of O = 1 x 16.00 = 16.00g Mass of 1 mol of C 2 H 5 OH= 46.07g The molar mass of C 2 H 5 OH is 46.07g

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Ethanol(C 2 H 5 OH) Question: What is the percent mass of C? Next we divide the mass of C in 1 mole of C 2 H 5 OH by the mass of 1 mol of C 2 H 5 OH So: Mass percent of C = mass of C in 1 mol C 2 H 5 OH x 100% mass of 1 mol C 2 H 5 OH So: Mass percent of C = 24.02g x 100% = 52.14% 46.07g

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Ethanol(C 2 H 5 OH) Question: What is the percent mass of C? That means that ethanol is 52.14% by mass of carbon. That means 52.14% of ethanols mass is C. What about H and O?

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Ethanol(C 2 H 5 OH) Question: What about H and O? So: Mass percent of H = mass of H in 1 mol C 2 H 5 OH x 100% mass of 1 mol C 2 H 5 OH So: Mass percent of H = 6.048g x 100% = 13.13% 46.07g The mass percent of H is 13.13%

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Ethanol(C 2 H 5 OH) Question: What about H and O? So: Mass percent of O = mass of O in 1 mol C 2 H 5 OH x 100% mass of 1 mol C 2 H 5 OH So: Mass percent of O = 16.00g x 100% = 34.73% 46.07g The mass percent of O is 34.73%

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Ethanol(C 2 H 5 OH) Double check your answers. All mass percents should add up to 100% 52.14% - C 13.13% - H 34.73% - O 100.00%

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Practice What is the mass percent of C, H and O in the compound Carvone, C 10 H 14 O? First find the masses of the individual elements.

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Practice What is the mass percent of C, H and O in the compound Carvone, C 10 H 14 O? Now use that information to find the percent of each element. C: ______ H: ______ O: ______ Total: ______

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Practice What is the mass percent of Mg, N and O in the compound Mg(NO 3 ) 2 ? Remember to first find the molar mass then Mg: ______ N: ______ O: ______ Total: ______

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Classwork Penicillin F ( C 14 H 20 N 2 SO 4 ) – Compute the mass percent of each element in this compound. C: ______ H: ______ N: ______ S: ______ O: ______ Total: ______

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Relationships Three important 1 mol _____ = 6.022x10 23 units of _____ 1 mol H 2 0 = 6.022x10 23 molecules of H 2 0 1 mol _____ = avg mass(g) of _____ 1 mol H 2 0 = 18.016 g H 2 0 avg mass(g) of ____ = 6.022x10 23 units of ___ 18.016 g H 2 0 = 6.022x10 23 molecules of H 2 0

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Practice Mass of sample Moles of Sample Atoms in Sample 1).250 mol Al2) 25.4g Fe3)4) 5)6)2.13 x 10 24 atoms Au 7)1.28 mol Ca8) 4.28 g9)10) 11)12)3.14 x 10 23 atoms C

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Empirical Formulas

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Lets say you mix two chemicals together. From that you get a precipitate, a solid product forms. What is the chemical formula for this solid? How do you figure it out? The empirical formula is the simplest whole number ratio of the atoms present in the compound.

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Empirical Formulas Steps to find the Empirical Formula 1. Obtain the mass of each element in grams 2. Determine # of moles of each element. 3. Divide the moles by the smallest # of moles found in previous step.

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Empirical Formulas See pg 201 in your book

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Empirical Formulas To figure out the chemical formula we will first need obtain mass. This time they were given to us!!! Using methods we wont discuss you figure out that in the.2015g sample, there are:.0806g C.01353g H.1074g O.

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Step 2 0.0806g C x 1 mol C 12.01g C = 0.00671 mol C 0.01353g H x 1 mol H 1.008g H = 0.01342 mol H 0.1074g O x 1 mol O 16.00g O = 0.006713 mol O

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Step 3 0.00671 mol C 0.00671= 1 mol C 0.01342 mol H 0.00671= 2 mol H 0.006713 mol O 0.00671= 1 mol O

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Empirical Formulas Now look at the moles of each element. 1 mol C 2 mol H 1 mol O The ratio of C to H to O is 1:2:1 The empirical formula is CH 2 O!!

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Empirical Formulas The empirical formula is the simplest whole number ratio of the atoms present in the compound. So the chemical formula could not be C 3 H 6 O 3 or C 6 H 12 O 6 or C 4 H 8 C 4 or more. All of the compounds have the same empirical formula of CH 2 O. The molecular formula gives the actual number of each type of atom present.

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Lets try one… Lets say that you heat up 4.151g of Al. Because it is so hot it reacts with the O in the air. After the sample cools you find the mass is 7.843g. This means that 3.692g O reacted. What is the empirical formula? Lets do the math…

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Lets try one… 4.151g Al x 1 mol Al 26.98g Al = 0.1539 mol Al 3.692g O x 1 mol O 16.00g O = 0.2308 mol O

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Lets try one… Remember we want to use whole numbers only, so we divide our answer by the smallest # of mol. 0.1539 mol Al = 1.000 mol Al 0.1539 0.2308 mol O = 1.500 mol O 0.1539 Does that mean that our empirical formula is AlO 1.5 ? No it doesnt.

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Lets try one… Remember we want to use whole numbers only!! So we then multiply each element by 2. 1.000 Al x 2 = 2.000 = 2 Al atoms 1.500 O x 2 = 3.000 = 3 O atoms So our empirical formula is Al 2 O 3. For every 2 aluminum atoms there are 3 oxygen atoms.

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Lets try one… If we end up with a decimal we need to multiply all the sub scripts by the same # to get ride of the decimal. Some easy ones to recognize are: ½ = 0.5, ¼ =.25, ¾ =.75 2/3 =.666, 1/3 =.333

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Lets try one… To convert these decimales to whole #s just multiply by their fractions denominator. Multiply by the bottom #. Ex. 0.5 x 2 = 1 Ex. 0.666 x 3 = 2 Ex. 1.75 x 4 = 7

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Class wide practice A 1.500g sample of a compound containing only C and H is found to contain 1.198g of C. What is the empirical formula for this compound? Answer: CH 3

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Class wide practice A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813g of Pb, 0.00672g of H, 0.4995g of As, and 0.4267g of O. What is the empirical formula for lead arsenate? Answer: PbHAsO 4 (see pg 203)

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Empirical Formulas When a compound is analyzed to find the relative amounts of elements present, the results are usually given in terms of % by mass of the various elements. Now can you figure out the EF (Empirical Formula) if you are given the percent composition?

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Empirical Formulas Remember that percent composition is how much of that part, per 100 parts, of the total Ex. 15% C means… the compound contains 15g of C per 100g of the compound. That means if we are given the % mass we can still find the EF.

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For example… Cisplatin is used to treat cancerous tumors. It has a composition of 65.02% Pt, 9.34% N, 2.02% H and 23.63% Cl. Calculate the empirical formula for Cisplatin.

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First Remember that percent composition is how much of that part, per 100 parts, of the total Recognize that: 65.02% Pt => 65.02g Pt 2.02% H => 2.02g H 9.34% N => 9.34g N 23.63% Cl => 23.63g Cl

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Second Now we are back to the same thing we were doing before. Find the # of moles of each element. 65.02g Pt x 1 mol Pt 195.1g Pt = 0.3333 mol Pt 2.02g H x 1 mol H 1.008g H = 2.00 mol H Where did the 195.1g Pf come from?

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Second Now we are back to the same thing we were doing before. Find the # of moles of each element. 9.34g N x 1 mol N 14.01g N = 0.667 mol N 23.63g Cl x 1 mol Cl 35.45g Cl = 0.6666 mol Cl Where did the 23.63g Cl come from?

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Second Now we are back to the same thing we were doing before. Find the # of moles of each element. Summary of results: 0.3333 mol Pt 2.00 mol H 0.667 mol N 0.6666 mol Cl

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Third Divide by the smallest # of moles. 0.3333 mol Pt = 1.000 mol Pt 0.3333 2.00 mol H = 6.01 mol H 0.3333 0.667 mol N = 2.00 mol N 0.3333 0.6666 mol Cl = 2.000 mol Cl 0.3333

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Finial Since all the ratio numbers came out to be whole numbers those are our numbers for our formula. Answer: PtN 2 H 6 Cl 2 Notice the only new thing we did differently was at the beginning. We changed from percentages of an element to grams of an element. Ex. 65.02% Pt => 65.02g Pt

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Class wide practice The most common form of nylon is 63.68% C, 12.38% N, 9.80% H, and 14.4% O. What is the empirical formula for nylon? Answer: C 6 NH 11 O

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Finding the molecular formula

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Since we know percent composition is how much of that part, per 100 parts, of the total. Ex. 65.02% Pt => 65.02g Pt We can now find the molecular formula if we are given the molar mass. Quick Review: What is the difference between the molecular formula and the empirical formula? What is the molar mass? How do we find the molar mass?

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Lets take a closer look… A white powder is analyzed and has and empirical formula of P 2 O 5. The compound has a molar mass of 283.88. What is the molecular formula?

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Lets take a closer look… The empirical formula mass is the total mass of the empirical formula. P 2 O 5 2 mol P: 30.97g x 2 = 61.94g 5 mol O: 16.00g x 5 = 80.00g = 141.94g of P 2 O 5 The mass of 1 mol of P 2 O 5 is 141.94g. Where did 30.97g come from? Where did the x 5 come from?

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Lets take a closer look… The empirical formula mass is the total mass of the empirical formula. The molar mass is the mass of all the atoms, not just the ones in the empirical formula. Lets compare: Molar Mass = 283.88g = 2 Empirical Formula mass 141.94g The molar mass is 2 times as big as the EF mass

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Lets take a closer look… Lets compare: Molar Mass = 283.88g = 2 Empirical Formula mass 141.94g The molar mass is 2 times as big as the EF mass That means that we multiply the compound by 2. (P 2 O 5 ) 2 => P 4 O 10 The molecular formula is P 4 O 10

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Class wide practice Caffeine is a compound containing C, H, N and O. The mass percent composition of caffeine is 49.47% C, 5.191% H, 28.86% N, and 16.48% O. The molar mass is about 194 g/mol. 1) What is the empirical formula? 2) What is the molecular formula? Answer: C 4 H 5 N 2 O Answer: C 8 H 10 N 4 O 2

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How it all relates Formula Summary for the sugar glucose:

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THE END!! WOO HOO!!

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