read for tuesday read for tuesday chapter 4: sections 4-6 chapter 4: sections 4-6 homework – due...

Read for Tuesday Chapter 4: Sections 4-6

HOMEWORK – DUE Thursday 9/10/15 HW-BW 3.2 (Bookwork) CH 3 #’s 41, 45, 48, 56-61 all, 68, 70, 72, 73, 78, 79,

85, 88, 95, 96, 104, 119, 124 HW-WS 4 (Worksheet) (from course website)

HOMEWORK – DUE Tuesday 9/15/15 HW-BW 4.1 (Bookwork) CH 4 #’s 3, 7, 9, 16, 17-31 odd, 42, 44, 46, 56, 62,

64, 66, 70, 72, 149 HW-WS 5 (Worksheet) (from course website)

HOMEWORK – DUE Thursday 9/17/15 HW-BW 4.2 (Bookwork) CH 4 #’s 78-81 all, 83-89 odd, 90, 93, 95, 104-112

even, 116, 156 HW-WS 6 (Worksheet) (from course website)

Lab next week – EXP 3 EXAM 1 IS THURSDAY

23

21

1 6.022 105.83

6.21 10 1

stuff

stuff stuf

tf

s uff

molec.

mole mol

g

c.

23

21

1 6.022 10 5655.83

6.21 10 1

stuff

stuff

stuff stuf

stuff stufff

molec.

molec.

mol ol

g

m g

1 463.490

244.27 1

2 22

2 2

2 22

BaCl 2H O BaCl 2

B

H O

aCl 2H O

BaCl 2H O

Hmo mo

l

m lg

g o

l 23231 4 6.022 10

63.490 6.261 10244.27 1 1

2 22

2 2

22 2

BaCl 2H O BaCl

H BaCl 2H O

BaCl 2H O

H

HH

2 O H

mol g

g

mol a

mo

tomat

olom

m l

231 4 6.022 1063.490

244.27 1 1

2

2

2

22

22

2

BaCl 2H O BaCl 2

BaCl 2H O

BaCl

H

H

H

H 2HO O

m atom mol

mo

g

ol

mollg

The mole

247.57 10 Naatoms

What is the mass of 7.57x1024 atoms of sodium?

2423

17.57 10

6.022 10

Na

Na

Na

molatoms

atoms

2423

1 22.997.57 10

6.022 10 1

Na

Na

NaNa

Naatoms

mol

mms

g

olato24

23

1 22.997.57 10 289

6.022 10 1

Na

Na

Na

Na

NaNaatoms

ato

mol

m

o s

gg

lm

How many hydrogen atoms in 63.490 g of barium chloride dihydrate?2 2BaCl 2H O = 244.27 g / mol

63.490 2 2 BaCl 2H Og 1

63.490244.27

2 2

2

2 2

2 Ba BaCl 2H O

BaCl 2H O

Cl 2H O

g

o

g

m l

What is the molar mass of a compound when 6.21x1021 molecules has a mass of 5.83g

5.83 stuffg 21

15.83

6.21 10

stuff stuff

molec. g

Empirical Formulae

The simplest whole number ratio of atoms in a molecule

Example: molecular formula = C9H18O3

empirical formula = C3H6O

Example: molecular formula = C16H32O2N4

empirical formula = C8H16ON2

Example: molecular formula = KMnO4

empirical formula = KMnO4

1 1.82934.75 2.006 2

19.00 0.9117

F FF = F F

F

mol mol g mol mol

g

1 1.82829.25 2.005 2

16.00 0.9117

O OO = O O

O

mol mol g mol mol

g

1 0.911710.95 1

12.01 0.9117

C CC = C

C

mol mol g mol

g

Calculating Empirical FormulaeWhat is the empirical formula of a compound that is made up of 10.95g C, 29.25g O, and 34.75g F?Step 1: convert each mass to MOLES

Step 2: divide EACH by the smallest!!

Calculating Empirical FormulaeWhat is the empirical formula of a compound that is made up of 10.95g C, 29.25g O, and 34.75g F?So we have 1 mol C, 2 mol O, and 2 mol F

Step 3: if each is a whole number, write the empirical formula

CO2F

2

Step 1: assume 100 g THEN convert each mass to MOLESStep 2: divide EACH by the smallest!!

1 1.29820.76 1

16.00 1.298

O O O = O

O

mol molg mol

g

1 5.1785.23 3.989 4

1.01 1.298

H H H = H H

H

mol molg mol mol

g

1 6.16274.01 4.749 4.75

12.01 1.298

C C C = C C

C

mol molg mol mol

g

What is the empirical formula of a compound made up of 74.01% C,

5.23% H, and 20.76% O?

So we have 4.75 mol C, 4 mol H, and 1 mol O

Step 3: If all of the subscripts are NOT whole numbers, then multiply EACH ELEMENT by the smallest number that will make all of them whole numbers

C4.75H4O

What is the empirical formula of a compound made up of 74.01% C,

5.23% H, and 20.76% O?

multiply EACH of the subscripts by 4!

C4.75 x 4 = 19

H4 x 4 = 16

O1 x 4 = 4

C19H16O4

C4H5N2O

Caffeine is my life blood! What is the empirical formulae based on: 40.09g C, 4.172g H, 23.41g N, and 13.37g O.

Molecular Formulae: The Next Step

C4H5N2O = 97.11 g/mol

194.192

97.11

gmol

gmol

mass ratio = 2 to 1

formula ratio = 2 to 1 2(C4H5N2O) C8H10N4O2

molecular formula = C8H10N4O2

If the molar mass of caffeine is 194.19 g/mol, what is its molecular formula?

You burn 1116 g of an unknown hydrocarbon derivative (CxHyOz) and collect 2617 g of CO2(g) and 401 g of H2O(g).

a) What is the empirical formula of this compound?

b) If the molar mass of the compound is 300 g/mol, what is the molecular formula?


Where does all of the carbon in the compound go?

Where does all of the carbon in the CO2 come from?

Allows us to use mol-to-mol C/CO2

Where does all of the hydrogen in the compound go?

Where does all of the hydrogen in the H2O come from?

Allows us to use mol-to-mol H/H2O

Where does all of the oxygen in the compound go?

CANNOT use a mol-to-mol ratio for oxygen

CO2

compound

H2Ocompound

everywhere

1 1 2617 59.46

44.01 1 2

2

2 2

C CO CO

C C

O C

O

molg

g mol

l

momol

1 1 12.01 2617 59.46 714

44.01 1 1 2

2

2 2

CO CO

CO CO

C

CCC

C

molg

g mol

gmolmol

molg

1 2 401 44.60

18.02 1 2

2

2 2

H O H O

H O H O

H H

mol g

g mol

molmol

1 2 1.01 401 44.60 45.05

18.02 1 1 2

2

2 2

H H

H O H O

H O

H H

H O

H

mol g

g mol

molmol

l

gg

mo


1116 714 45.05 357.05 x y z C H OCO Hg g g g1

1116 714 45.05 357.05 22.3216.00

x y z

OO

O HH CC O O

mg

g

olm

g olgg

59.46 2.667

22.32

O

C

mol

mol 44.60 2 3

22.32

H6

O

mol

mol

22.321 3

22.32

O

3

O

mol

mol

59.46 2.667 3

22.32

C

8

Omol

mol 22.321

22.32

O

O

mol

mol

44.60 2

22.32

H

Omol

mol

C8H6O3


b) If the molar mass of the compound is 300 g/mol, what is the molecular formula?

empirical formula = C8H6O3= 150.14 g/mol

3002

150.14

gmol

gmol

2(C8H6O3) C16H12O6

molecular formula = C16H12O6

31.89 2 4 K SOg1

2 4 K SOmol

174.27 2 4 K SOg

2 2 3 2 KC H Omol

1 2 4 K SOmol

98.15

1

2 3

3

2

2 2KC H O

KC H O

g

l mo

35.92 2 3 2KC H Og

25.00 2 3 2 2 Pb(C H O )g1

2 3 2 2 Pb(C H O )mol

325.3 2 3 2 2 Pb(C H O )g

2 2 3 2 KC H Omol

1 2 3 2 2 Pb(C H O )mol

98.15

1

2 3

3

2

2 2KC H O

KC H Omol

g 15.09 2 3 2KC H Og 25.00 2 3 2 2 Pb(C H O )g

1 2 3 2 2 Pb(C H O )mol

325.3 2 3 2 2 Pb(C H O )g

2 2 3 2 KC H Omol

1 2 3 2 2 Pb(C H O )mol

98.15

1

2 3

3

2

2 2KC H O

KC H Omol

g 15.09 2 3 2KC H Og

31.89 2 4 K SOg1

2 4 K SOmol

174.27 2 4 K SOg

2 2 3 2 KC H Omol

1 2 4 K SOmol

98.15

1

2 3

3

2

2 2KC H O

KC H O

g

l mo

35.92 2 3 2KC H Og

31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?

K2SO4(aq) + Pb(C2H3O2)2(aq) 2 KC2H3O2(aq) + PbSO4(s)

174.27 g/mol 325.3 g/mol 98.15 g/mol 303.3 g/mol

limiting reactant theoretical yield of KC2H3O2

theoretical yield = 15.09g KC2H3O2

Pb(C2H3O2)2 is the L.R. (0 left)


(Hint: start with the given amount of the limiting reactant.)

Now calculate how much of the other product(s) will be formed.

25.00 2 3 2 2 Pb(C H O )g1

2 3 2 2 Pb(C H O )mol

325.3 2 3 2 2 Pb(C H O )g

1 4 PbSOmol

1 2 3 2 2 Pb(C H O )mol

303.3

1

4

4PbSO

PbSO

g

m

ol

23.31 4PbSOg

31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?



23.31 g 23.31 g

15.09 g 15.09 g

31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?

reactants products

K2SO4 = 18.49 g

Pb(C2H3O2)2 = 0 g (L.R.)

KC2H3O2 =

PbSO4 =

Mass must be the same before and after the reaction!!mass before reaction 31.89 + 25.00 = 56.89 g

mass after reaction 15.09 + 23.31 = 38.40 g

difference will be the mass of excess reactant left over = 18.49 g

25.00 2 3 2 2 Pb(C H O )g1

2 3 2 2 Pb(C H O )mol

325.3 2 3 2 2 Pb(C H O )g

1 2 4 K SOmol

1 2 3 2 2 Pb(C H O )mol

174.27

1 4

2 4

2K S

K

O

SO

g

mol

13.39 2 4K SOg


31.89 g K2SO4 (STARTED) – 13.39 g K2SO4 (USED) = 18.50 g K2SO4 (LEFT)

14.71 3 CrClg1

3 CrClmol

158.35 3 CrClg

3 2 PbClmol

2 3 CrClmol

278.1

1

2

2PbCl

PbCl

g

mol

38.75 2PbClg

23.41 3 2 Pb(NO )g1

3 2 Pb(NO )mol

331.2 3 2 Pb(NO )g

3 2 PbClmol

3 3 2 Pb(NO )mol

278.1

1

2

2PbCl

PbCl

g

m

ol

19.66 2PbClg

14.71 3 CrClg1

3 CrClmol

158.35 3 CrClg

3 2 PbClmol

2 3 CrClmol

278.1

1

2

2PbCl

PbCl

g

mol

38.75 2PbClg

23.41 3 2 Pb(NO )g1

3 2 Pb(NO )mol

331.2 3 2 Pb(NO )g

3 2 PbClmol

3 3 2 Pb(NO )mol

278.1

1

2

2PbCl

PbCl

g

m

ol

19.66 2PbClg

2 CrCl3(aq) + 3 Pb(NO3)2(aq) 2 Cr(NO3)3(aq) + 3 PbCl2(s)

theoretical yield = 19.66g PbCl2(s)Pb(NO3)2 is the L.R. (0 left)

14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction?



23.41 3 2 Pb(NO )g1

3 2 Pb(NO )mol

331.2 3 2 Pb(NO )g

2 3 3 Cr(NO )mol

3 3 2 Pb(NO )mol

238.03

1 3 3

3 3 Cr(NO )

Cr(NO )

mol

g11.22 3 3Cr(NO )g


(Hint: start with the given amount of the limiting reactant.)

Now calculate how much of the other product(s) will be formed.





19.66 g

11.22 g

reactants products

CrCl3 = 7.24 g

Pb(NO3)2 = 0 g (L.R.)

Cr(NO3)3 = 11.22 g

PbCl2 = 19.66 g

Mass must be the same before and after the reaction!!mass before reaction 14.71 + 23.41 = 38.12 g

mass after reaction 11.22 + 19.66 = 30.88 g

difference will be the mass of excess reactant left over = 7.24 g


14.71 g CrCl3 (STARTED) – 7.462 g CrCl3 (USED) = 7.24 g CrCl3 (LEFT)

23.41 3 2 Pb(NO )g1

3 2 Pb(NO )mol

331.2 3 2 Pb(NO )g

2 3 CrClmol

3 3 2 Pb(NO )mol

158.35

1

3

3CrC

l

CrCl

g

m

ol

7.462 3CrClg

read for tuesday read for tuesday chapter 4: sections 4-6 chapter 4: sections 4-6 homework – due...

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