Chapter 6 – Chemical Composition. Counting by weighing Situation: You work in a candy shop. You sell jelly beans. Somebody wants to buy 1000 jelly beans

Chapter 6 – Chemical Composition

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??


Here’s 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!!!


Problem: all jelly beans aren’t identical.Suppose we weigh 10 jelly beans and find:

Now we can find he average mass of a bean.


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?


You know:Jelly bean avg mass = 5 gMint avg mass = 15 gOne 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?


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.

Lets look at some ratios:1mint x 15g = 15g

1 bean x 5g = 5g the ratio is 3/1

10mints x 15g = 150g

10beans x 5g = 50g the ratio is 3/1

100mint x 15g = 1500g

100 bean x 5g = 500g the ratio is 3/1

How does this apply to Chemisty?

Lets say you have a pile of C and want to know how many O2 molecules you need to convert all the C into CO2.

Or you want to know how many O atoms are in a cup of H2O

Atomic Masses

How much does a C atom weight?

1.99 x 10-23g that’s really small!!!

To help with these complicated numbers we use atomic mass unit (amu).

1 amu = 1.66 x 10-24g

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.

Atomic Masses

Where can we find what the avg atomic mass of an element?

Look on the periodic table!!

Atomic Masses

So the common isotopes of C areCarbon 12Carbon 13Carbon 14

Based on the abundance found in nature C avg atom mass 12.01 amu

1 carbon atom = 12.01 amu

Back to the question

Let’s say you have a pile of C and want to know how many O2 molecules you need to convert all the C into CO2.

You weighed the pile of C and it’s 3 x 1020 amu

3x1020amu X (1carbon atom/12.01amu) = 2.5x1019atoms (see pg176)

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

Practice

Remember:

Convert 3x1020amu of C to atoms

Convert 75 Al atoms to amu

Convert 322.2 amu of N to # of atoms

HomeworkConvert:

213 Hg atoms to amu

53 P atoms to amu

172 O atoms to amu

HomeworkConvert:

12 amu of Cl to atoms

74 amu of Na to atoms

123 amu of Li to atoms

Now to a mole!

Now to a mole!

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.

The MoleThe Mole

is 6.02 X 1023 (in scientific notation)Similar to a dozen, except instead of 12, it’s 602 billion trillion 602,000,000,000,000,000,000,000This 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

What is a mole?

One mole of anything consist of 6.02x1023 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: 6.02x1023 eggs, paper clips, shoes

6.02 x 1023 particles

1 mole

or

1 mole

6.02 x 1023 particles

Note that a particle could be an atom, a molecule or

anything!!!

Avogadro’s Number as Avogadro’s Number as Conversion FactorConversion Factor

Just How Big is a Mole?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.

The MoleThis photograph shows one mole of salt(NaCl), water(H2O), and nitrogen gas(N2).

We also know…

1 mole of an elements atoms = that element’s average atomic mass (expressed in g)

We also know…

Ex. 200.6g of Hg has 6.02x1023 atoms or 200.6g of Hg has 1 mol of Hg atomsConversion factor: 200.6g Hg/1 mol Hg

Ex. 32.07g of S has 6.02x1023 atoms or 32.07g of S has 1 mol of S atomsConversion factor: 32.07g S/1 mol S

Example

We have an unknown # of H atoms.

We want to know many H atoms are present?

We weigh the sample, results = .5g

Example

So .5g H X (1mol H/1.008g H) = .496 mol of H in the sample.

Then .496 mol H atoms X (6.022x1023 H atoms/1 mol H atoms) = 2.99x1023 H atoms in the sample

(See pg 181)

PracticeConvert 10g of Al into moles of atoms

Convert 8g of S into moles of atoms

Convert 25g of Ca into moles of atoms

PracticeConvert .371 mol Al into # of atoms

Convert .249 mol S into # of atoms

Convert .624 mol Ca into # of atoms

Summary1 amu = 1.66 x 10-24g

1 ___ atom = _____ amu1 C atom = 12.01 amu1 Si atom = 28.09 amu

1 mol of ___ = 6.022x1023 ____ units1 mol F = 6.022x1023 F atom1 mol Mg = 6.022x1023 Mg atom

What are the conversion factors?

Summary1mol of ___ = ____g of _____1mol Hg = 200.6g Hg1mol Cl = 35.45g Cl

____g of __ = 6.022x1023 atoms of ____200.6g Hg = 6.022x1023 atoms Hg35.45g Cl = 6.022x1023atoms Cl

What are the conversion factors?

ClassworkHow 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?

ClassworkHow 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?

Molar Mass

Molar Mass

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.

Molar Mass

• Chemical compounds have multiple parts like the sucker.

• Ex. Methane (natural gas) CH4

1 CH4 moleculeWith 5 parts (1 C atom and 4 H atoms)

Quick Note

• In 1 mol of a molecule there are corresponding moles of its parts.

• Ex. In 1 mol of CH4 there are 1 mol of C and 4 mol of H

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

Example: SO2

• What is the molar mass of SO2?

• 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 SO2 = 64.07g

• The molar mass of SO2 is 64.07g

Example: CaCO3

• What is the molar mass of CaCO3?

• 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 CaCO3= ____

• The molar mass of CaCO3 is _____

Practice

• What is the molar mass of water, H2O?

• What is the molar mass of ammonia, NH3?

• What is the molar mass of propane, C3H8?

• What is the molar mass and name of CaSO4?

• What is the molar mass and name of Na2CO3?

• What is the molar mass and name of Ba(OH)2?

Practice

• What is the molar mass of water, H2O?

• What is the molar mass of ammonia, NH3?

• What is the molar mass of propane, C3H8?

Practice

• What is the molar mass and name of CaSO4?

• What is the molar mass and name of Na2CO3?

• What is the molar mass and name of Ba(OH)2?

Percent Composition

Back to the candy shop

• A customer comes in and wants 3 different kinds of candy: jelly beans, mints and M&M’s.

• You give him 2 scoops of jelly beans, 6 scoops of mints, and 1 scoop of M&M’s.

• He now wants to know what percent of his bag has jelly beans by weight?

• You’re going to use Percent Composition.


• 2 jelly beans, 6 mints, 1 M&M’s. • 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


• 2 jelly beans, 6 mints, 1 M&M’s. • 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

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: It’s how much stuff is there in a percent form.

mass of a given element in 1 mol of compound100%

mass of 1 mol of compound

Mass percent =

For example…

• Ethanol(C2H5OH) – 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.

Ethanol(C2H5OH)

• 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 C2H5OH= 46.07g

• The molar mass of C2H5OH is 46.07g

Ethanol(C2H5OH)

• Question: What is the percent mass of C?• Next we divide the mass of C in 1 mole of

C2H5OH by the mass of 1 mol of C2H5OH

• So: Mass percent of C = mass of C in 1 mol C2H5OH x 100%

mass of 1 mol C2H5OH

• So: Mass percent of C = 24.02g x 100% = 52.14%

46.07g

Ethanol(C2H5OH)

• Question: What is the percent mass of C?• That means that ethanol is 52.14% by mass of

carbon.• That means 52.14% of ethanol’s mass is C.• What about H and O?

Ethanol(C2H5OH)

• Question: What about H and O?• So: Mass percent of H =

mass of H in 1 mol C2H5OH x 100%


• So: Mass percent of H = 6.048g x 100% = 13.13%

46.07g

The mass percent of H is 13.13%

Ethanol(C2H5OH)

• Question: What about H and O?• So: Mass percent of O =

mass of O in 1 mol C2H5OH x 100%


• So: Mass percent of O = 16.00g x 100% = 34.73%

46.07g

The mass percent of O is 34.73%

Ethanol(C2H5OH)

• Double check your answers. All mass percents should add up to 100%

• 52.14% - C• 13.13% - H• 34.73% - O• 100.00%

Practice

• What is the mass percent of C, H and O in the compound Carvone, C10H14O?

• First find the masses of the individual elements.

Practice

• What is the mass percent of C, H and O in the compound Carvone, C10H14O?

• Now use that information to find the percent of each element.

• C: ______• H: ______• O: ______• Total: ______

Practice

• What is the mass percent of Mg, N and O in the compound Mg(NO3)2 ?

• Remember to first find the molar mass then • Mg: ______• N: ______• O: ______• Total: ______

Classwork

• Penicillin F ( C14H20N2SO4) – Compute the mass percent of each element in this compound.

• C: ______• H: ______• N: ______• S: ______• O: ______• Total: ______

Relationships

• Three important• 1 mol _____ = 6.022x1023 units of _____

1 mol H20 = 6.022x1023 molecules of H20

• 1 mol _____ = avg mass(g) of _____1 mol H20 = 18.016 g H20

avg mass(g) of ____ = 6.022x1023 units of ___18.016 g H20 = 6.022x1023 molecules of H20

Practice

Mass of sample

Moles of Sample

Atoms in Sample

1) .250 mol Al 2)

25.4g Fe 3) 4)

5) 6) 2.13 x 1024 atoms Au

7) 1.28 mol Ca 8)

4.28 g 9) 10)

11) 12) 3.14 x 1023 atoms C

Empirical Formulas

Empirical Formulas

• 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.

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.

Empirical Formulas

• See pg 201 in your book

Empirical Formulas

• To figure out the chemical formula we will first need obtain mass.

• This time they were given to us!!! • Using methods we won’t discuss you figure

out that in the .2015g sample, there are: .0806g C .01353g H .1074g O.

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

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

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 CH2O!!

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 C3H6O3 or C6H12O6 or C4H8C4 or more.

• All of the compounds have the same empirical formula of CH2O.

• The molecular formula gives the actual number of each type of atom present.

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…

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

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

AlO1.5? No it doesn’t.

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 Al2O3. For every 2 aluminum atoms there are 3 oxygen atoms.

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

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

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: CH3

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: PbHAsO4(see pg 203)

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?

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.

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.

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 N23.63% Cl => 23.63g Cl

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?

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?

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 Pt2.00 mol H0.667 mol N0.6666 mol Cl

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

Finial

• Since all the ratio numbers came out to be whole numbers those are our numbers for our formula.

• Answer: PtN2H6Cl2• 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

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: C6NH11O

Finding the molecular formula

Finding the molecular formula

• 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?

Lets take a closer look…

• A white powder is analyzed and has and empirical formula of P2O5. The compound has a molar mass of 283.88. What is the molecular formula?


• The empirical formula mass is the total mass of the empirical formula. P2O5

• 2 mol P: 30.97g x 2 = 61.94g• 5 mol O: 16.00g x 5 = 80.00g

= 141.94g of P2O5

The mass of 1 mol of P2O5 is 141.94g.

Where did 30.97g come from?

Where did the “x 5” come from?


• 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


• 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.

(P2O5)2 => P4O10

• The molecular formula is P4O10

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: C4H5N2O

• Answer: C8H10N4O2

How it all relates

Formula Summary for the sugar glucose:

THE END!! WOO HOO!!

Documents

Chapter 6 – Chemical Composition. Counting by weighing Situation: You work in a candy shop. You sell jelly beans. Somebody wants to buy 1000 jelly beans