iiiiii ii. formula calculations ch. 10 – the mole

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I II III II. Formula Calculations Ch. 10 – The Mole

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Page 1: IIIIII II. Formula Calculations Ch. 10 – The Mole

I II III

II. Formula Calculations

Ch. 10 – The Mole

Page 2: IIIIII II. Formula Calculations Ch. 10 – The Mole

Hydrates

Some compounds contain H2O in their structure. These compounds are called hydrates.

This is different from (aq) because the H2O is part of the molecule (not just surrounding it).

The H2O can usually be removed if heated.

Page 3: IIIIII II. Formula Calculations Ch. 10 – The Mole

Hydrates

A dot separates water in the formula CuSO4•5H2O

When naming a Greek prefix indicates the # of H2O groups.

copper(II) sulfate pentahydrate

Page 4: IIIIII II. Formula Calculations Ch. 10 – The Mole

Naming Hydrates

Na2SO4•10H2O

NiSO4•6H2O

CoCl2•6H2O

MgSO4•7H2O

Sodium sulfate decahydrate

Nickel (II) sulfate hexahydrate

Cobalt (II) chloride hexahydrate

Magnesium sulfate heptahydrate

Page 5: IIIIII II. Formula Calculations Ch. 10 – The Mole

Writing Formulas of Hydrates sodium carbonate

monohydrate barium chloride

dihydrate Tin (IV) chloride

pentahydrate Barium hydroxide

octahydrate

Na2CO3•H2O

BaCl2•2H2O

SnCl4•5H2O

Ba(OH)2•8H2O

Page 6: IIIIII II. Formula Calculations Ch. 10 – The Mole

What is Percent Composition?

Page 7: IIIIII II. Formula Calculations Ch. 10 – The Mole

A. Percent Composition

the percentage by mass of each element in a compound

100compound of mass total

element of mass totalelement% mass of

100compound of mass molar

compound mol 1 in element of masselement% mass of

Page 8: IIIIII II. Formula Calculations Ch. 10 – The Mole

%Fe =28 g

36 g 100 = 78% Fe

%O =8.0 g

36 g 100 = 22% O

Find the percent composition of a sample that is 28 g Fe and 8.0 g O.

A. Percent Composition

Known: Mass of Fe = 28 g Mass of O = 8.0 g Total Mass = 28 + 8.0 g = 36 g

Unknown: % Fe = ? % O = ?

Check:78% + 22%

= 100%

Page 9: IIIIII II. Formula Calculations Ch. 10 – The Mole

100 =

A. Percent Composition Find the % composition of Cu2S. Use this formula when finding % composition from a

chemical formula:

%Cu =127.10 g Cu

159.17 g Cu2S 100 =

%S =32.07 g S

159.17 g Cu2S

79.85% Cu

20.15% S

100compound of mass molar

compound mol 1 in element of masselement% mass of

Known: Mass of Cu in 1 mol Cu2S =

2(63.55g) = 127.10 g Cu Mass of S in 1 mol Cu2S = 32.07 g S Molar Mass = 127.10 g + 32.07 g = 159.17 g/mol

Unknown: % Cu = ? % S = ?

Page 10: IIIIII II. Formula Calculations Ch. 10 – The Mole

How many grams of copper are in a 38.0-gram sample of Cu2S?

Use answer from last question as a conversion factor.

Cu2S is 79.85% Cu =

A. Percent Composition

38.0 g Cu2S 79.85 g Cu

100 g Cu2S = 30.3 g Cu

79.85 g Cu

100 g Cu2S

Page 11: IIIIII II. Formula Calculations Ch. 10 – The Mole

Find the percent composition of Cu2SO4.

A. Percent Composition

Known: Mass of Cu in 1 mol Cu2SO4 = 2(63.55 g) = 127.10 g Cu Mass of S in 1 mol Cu2SO4 = 32.07 g S Mass of 0 in 1 mol Cu2SO4 = 4(16.00 g) = 64.00 g O Molar Mass = 127.10 g + 32.07 g + 64.00 g = 223.17 g/mol

Unknown: % Cu = ? % S = ? % O = ?

Page 12: IIIIII II. Formula Calculations Ch. 10 – The Mole

Find the percent composition of Cu2SO4.

A. Percent Composition

%Cu =127.10 g

223.17 g 100 = 56.95% Cu

%S =32.07 g

223.17 g 100 = 14.37% S

Check:56.95% + 14.37% + 28.68% = 100%

%O =64.00 g

223.17 g 100 = 28.68% O

Page 13: IIIIII II. Formula Calculations Ch. 10 – The Mole

100 =%H2O = 36.04 g/mol

147.02 g/mol 24.51%

H2O

Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O.

A. Percent Composition

Known: Mass of H2O in 1 mol compound =

2(2(1.01g) + 16.00 g) = 36.04 g H2O Molar Mass = 40.08 g + 2(35.45 g) +

36.04 g = 147.02 g

Unknown: % H2O = ?

Page 14: IIIIII II. Formula Calculations Ch. 10 – The Mole

B. Empirical Formula

C2H6

CH3

reduce subscripts

Smallest whole number ratio of atoms in a compound

Page 15: IIIIII II. Formula Calculations Ch. 10 – The Mole

B. Empirical Formula1. Find mass (or %) of each element.

2. Find moles of each element.

3. Divide moles by the smallest # to find subscripts.

4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

Page 16: IIIIII II. Formula Calculations Ch. 10 – The Mole

B. Empirical Formula

Find the empirical formula for a sample of 25.9% N and 74.1% O.

25.9 g N 1 mol N

14.01 g N = 1.85 mol N

74.1 g O 1 mol O

16.00 g O = 4.63 mol O

1.85 mol

1.85 mol

= 1 N

= 2.5 O

N1.85O4.63

Page 17: IIIIII II. Formula Calculations Ch. 10 – The Mole

B. Empirical Formula

N1O2.5Need to make the subscripts whole

numbers multiply by 2

N2O5

Page 18: IIIIII II. Formula Calculations Ch. 10 – The Mole

B. Empirical Formula

Find the empirical formula for a sample of 94.1% O and 5.9% H.

94.1 g O 1 mol O

16.00 g O = 5.88 mol O

5.9 g H 1 mol H

1.01 g H = 5.84 mol H

5.84 mol

5.84 mol

= 1 O

= 1 H

EF = OH

Page 19: IIIIII II. Formula Calculations Ch. 10 – The Mole

C. Molecular Formula “True Formula” - the actual number of atoms

in a compound, either the same as or a whole-number multiple of the empirical formula

CH3

C2H6

empiricalformula

molecularformula

?

Page 20: IIIIII II. Formula Calculations Ch. 10 – The Mole

C. Molecular Formula1. Find the empirical formula.2. Find the empirical formula mass.3. Divide the molar mass by the empirical formula mass.4. Multiply each subscript in your EF by the answer from step 3.

nmass EF

mass molar nEF

Page 21: IIIIII II. Formula Calculations Ch. 10 – The Mole

C. Molecular Formula The empirical formula for ethylene is CH2. Find the

molecular formula if the molar mass is 28.1 g/mol?

28.1 g/mol

14.03 g/mol = 2.00

empirical formula mass = 14.03 g/mol

(EF)n = (CH2)2 C2H4

n =

Page 22: IIIIII II. Formula Calculations Ch. 10 – The Mole

D. Put it all together!! 1,6-diaminohexane is 62.1% C, 13.8% H,

and 24.1% N. What is the empirical formula? If the molar mass is 116.21 g/mol, what is the molecular formula?

62.1 g C 1 mol C

12.01 g C

= 5.17 mol C

13.8 g H 1 mol H

1.01 g H

= 13.7 mol H

= 3 C

= 8 H

24.1 g N= 1 N

1 mol N

14.01 g N

= 1.72 mol N

1.72 mol

1.72 mol

1.72 mol EF = C3H8N

Page 23: IIIIII II. Formula Calculations Ch. 10 – The Mole

C3H8N

116.21 g/mol

58.12 g/mol = 2.00

empirical formula mass = 58.12 g/mol

(C3H8N)2

n =

C6H16N2

D. Put it all together!!